UPGRADE YOUR PHYSICS

Transcription

1 Correctons March 7 UPGRADE YOUR PHYSICS NOTES FOR BRITISH SIXTH FORM STUDENTS WHO ARE PREPARING FOR THE INTERNATIONAL PHYSICS OLYMPIAD, OR WISH TO TAKE THEIR KNOWLEDGE OF PHYSICS BEYOND THE A-LEVEL SYLLABI. A. C. MACHACEK Page

2 Correctons March 7 Introducton The Internatonal Physcs Olympad s an annual nternatonal physcs competton for pre-unversty students. Teams of fve from each partcpatng naton attend, and recently over 6 countres have taken part. Each naton has ts own methods for selectng ts team members. In Brtan, ths s by means of a seres of wrtten and practcal exams. The queston paper for the frst round s crculated to all secondary schools. Once the team has been chosen, t s necessary for ts members to broaden ther horzons. The syllabus for the Internatonal Physcs Olympad s larger than that of the Brtsh A-level, and ndeed forms a convenent steppng-stone to frst year undergraduate work. For ths reason, tranng s provded to help the Brtsh team brdge the gap. The Brtsh Olympad Commttee recognzes the need for teachng materal to help canddates prepare for the nternatonal competton. Furthermore, ths materal ought to have greater potental n the hands of students who wsh to develop ther physcs, even f they have no desre to take part n the examnatons. It s my hope that these notes make a start n provdng for ths need. A.C. Machacek, About the author: Anton Machacek has served on the Brtsh Physcs Olympad Commttee snce 997, and has been nvolved regularly n tranng the Brtsh team and n wrtng Physcs Challenge examnatons. He served on the academc commttee for the Internatonal Physcs Olympad n Lecester n. Anton s Head of Physcs at the Royal Grammar School, Hgh Wycombe, and s an Academc Vstor n the subdepartment of Atomc and Laser Physcs, Unversty of Oxford. Page

3 Correctons March 7 Contents LINEAR MECHANICS...5. MOTION IN A LINE...5. GOING ORBITAL FLUIDS WHEN THINGS GET STICKY QUESTIONS... FAST PHYSICS...4. THE PRINCIPLE OF RELATIVITY...5. HIGH SPEED OBSERVATIONS RELATIVISTIC QUANTITIES THE LORENTZ TRANSFORMS QUESTIONS ROTATION ANGLE ANGULAR VELOCITY ANGULAR ACCELERATION TORQUE ANGULAR FORCE MOMENT OF INERTIA ANGULAR MASS ANGULAR MOMENTUM ANGULAR MOMENTUM OF A SINGLE MASS MOVING IN A STRAIGHT LINE ROTATIONAL KINETIC ENERGY SUMMARY OF QUANTITIES ROTATIONAL MECHANICS WITH VECTORS MOTION IN POLAR CO-ORDINATES MOTION OF A RIGID BODY QUESTIONS VIBES, WIGGLES & LIGHT OSCILLATION WAVES & INTERFERENCE RAYS FERMAT S PRINCIPLE...69 Page 3

4 Correctons March QUESTIONS HOT PHYSICS THE CONSERVATION OF ENERGY THE SECOND LAW HEAT ENGINES AND FRIDGES ENTROPY IRREVERSIBLE PROCESSES AND THE SECOND LAW RE-STATEMENT OF FIRST LAW THE BOLTZMANN LAW PERFECT GASES RADIATION OF HEAT QUESTIONS SPARKS & GENERATION ELECTROSTATICS WHEN THINGS ARE STILL MAGNETISM WHEN THINGS MOVE CIRCUITS PUTTING IT TOGETHER QUESTIONS SMALL PHYSICS WAVES AND PARTICLES UNCERTAINTY ATOMS LITTLE NUTS QUESTIONS PRACTICAL PHYSICS ERRORS, AND HOW TO MAKE THEM ERRORS, AND HOW TO MAKE THEM WORSE SYSTEMATIC ERRORS WHICH GRAPH? QUESTIONS APPENDIX MULTIPLYING VECTORS DIMENSIONAL ANALYSIS...35 Page 4

5 Correctons March 7 Lnear Mechancs. Moton n a Lne.. The Fundamentals... Knematcs Mechancs s all about moton. We start wth the smplest knd of moton the moton of small dots or partcles. Such a partcle s descrbed completely by ts mass (the amount of stuff t contans) and ts poston. There s no nternal structure to worry about, and as for rotaton, even f t tred t, no-one would see. The most convenent way of labellng the poston s wth a vector r showng ts poston wth respect to some convenent agreed statonary pont. If the partcle s movng, ts poston wll change. If ts speed and drecton are steady, then we can wrte ts poston after tme t as r = s + ut, where s s the startng pont (the poston of the partcle at t=) and u as the change n poston each second otherwse known as the velocty. If the velocty s not constant, then we can t measure t by seeng how far the object goes n one second, snce the velocty wll have changed by then. Rather, we say that u how far the object would go n one second f the speed or drecton remaned unchanged that long. In practce, f the moton remans constant for some small tme (called t), and durng ths small tme, the partcle s poston changes r, then the change n poston f ths were mantaned for a whole second (otherwse known as the velocty) s u = r number of t perods n one second = r t. Smlarly, f the velocty s changng, we defne the acceleraton as the change n velocty each second (f the rate of change of acceleraton were constant. Accordngly, our equaton for acceleraton becomes a = u t. Hopefully, t s apparent that as the moton becomes more complex, and the t perods need to be made shorter and shorter, we end up wth the dfferental equatons lnkng poston, velocty and acceleraton: d u r dt d a u dt r u u dt. a dt Page 5

6 Correctons March 7... Dynamcs Now we have a way of descrbng moton, we need a way of predctng or explanng the moton whch occurs changng our queston from what s happenng? to why? and our explanaton s gong to nvolve the actvty of forces. What do forces do to an object? The frst essental pont s that forces are only needed to change (not mantan) moton. In other words unless there s a change of velocty, no force s needed. But how much force s needed? Newton made the assumpton (whch we fnd to be helpful and true) that the force causes a change n what he called the moton we now call t momentum. Suppose an object has mass m and velocty u (we shall clarfy what we mean by mass later) then ts momentum s equal to mu, and s frequently referred to by physcsts by the letter p. Newton s second law states that f a constant force F s appled to an object for a short tme t, then the change n the momentum s gven by F t. In dfferental notaton d(mu)/dt = F. In the case of a sngle object of constant mass t follows that d mu du F m ma. dt dt Hs next assumpton tells us more about forces and allows us to defne mass properly. Imagne two brcks are beng pulled together by a strong sprng. The brck on the left s beng pulled to the rght, the brck on the rght s beng pulled to the left. Newton assumed that the force pullng the left brck rghtwards s equal and opposte to the force pullng the rght brck leftwards. To use more mathematcal notaton, f the force on block no. caused by block no. s called f, then f =f. If ths were not the case, then f we looked at the brcks together as a whole object, the two nternal forces would not cancel out, and there would be some left over force whch could accelerate the whole object. It makes sense that f the brcks are dentcal then they wll accelerate together at the same rate. But what f they are not? Ths s where Newton s second law s helpful. If the resultant force on an object of If you want to prove that ths s rdculous, try lftng a large bucket whle standng n t. Page 6

7 Correctons March 7 constant mass equals ts mass tmes ts acceleraton, and f the two forces are equal and opposte, we say f m a a a f m m m a, and so the more massve block accelerates less. Ths s the defnton of mass. Usng ths equaton, the mass of any object can be measured wth respect to a standard klogram. If a mystery mass experences an acceleraton of m/s whle pushng a standard klogram n the absence of other forces, and at the same tme the klogram experences an acceleraton of 4m/s the other way, then the mystery mass must be kg. When we have a group of objects, we have the opton of applyng Newton s law to the objects ndvdually or together. If we take a large group of objects, we fnd that the total force F total d F dt m u changes the total momentum (just lke the ndvdual forces change the ndvdual momenta). Note the smplfcaton, though there are no f j n the equaton. Ths s because f j + f j =, so when we add up the forces, the nternal forces sum to zero, and the total momentum s only affected by the external forces F....3 Energy and Power Work s done (or energy s transferred) when a force moves somethng. The amount of work done (or amount of energy transformed) s gven by the dot product of the force and the dstance moved. W = F r = F r cos () where s the angle between the force vector F and the dstance vector r. Ths means that f the force s perpendcular to the dstance, there s no work done, no energy s transferred, and no fuel supply s needed. If the force s constant n tme, then equaton () s all very well and good, however f the force s changng, we need to break the moton up nto lttle parts, so that the force s more or less constant for each part. We may then wrte, more generally, W = F r = F r cos (a) Two useful dfferental equatons can be formed from here. Page 7

8 Correctons March Vrtual Work From equaton (a) t s clear that f the moton s n the drecton of the force appled to the object (.e. =), then W F, r where W s the work done on the object. Accordngly, we can calculate the force on an object f we know the energy change nvolved n movng t. Let s gve an example. An electron (wth charge q) s forced through a resstor (of length L) by a battery of voltage V. As t goes through, t must lose energy qv, snce V s the energy loss per coulomb of charge passng through the resstor. Therefore, assumng that the force on the electron s constant (whch we assume by the symmetry of the stuaton), then the force must be gven by W / d = qv / L. If we defne the electrc feld strength to be the force per coulomb of charge (F/q), then t follows that the electrc feld strength E = V/L. So far, we have gnored the sgn of F. It can not have escaped your attenton that thngs generally fall downwards n the drecton of decreasng [gravtatonal] energy. In equatons () and (a), we used the vector F to represent the externally appled force we use to drag the object along. In the case of lftng a hodful of brcks to the top of a wall, ths force wll be drected upwards. If we are nterested n the force of gravty G actng on the object (whether we drag t or not), ths wll be n the opposte drecton. Therefore F = G, and W = G r, (b) W G. r In other words, f an object can lose potental energy by movng from one place to another, there wll always be a force tryng to push t n ths drecton....5 Power Another useful equaton can be derved f we dfferentate (a) wth respect to tme. The rate of workng s the power P, and so W P t F δr t r F. t As we let the tme perod tend to zero, r/t becomes the velocty, and so we have: Page 8

9 Correctons March 7 P = F u = F u cos () where s now best thought of as the angle between force and drecton of moton. Agan we see that f the force s perpendcular to the drecton of moton, no power s needed. Ths makes sense: thnk of a bke gong round a corner at constant speed. A force s needed to turn the corner - that s why you lean nto the bend, so that a component of your weght does the job. However no work s done you don t need to pedal any harder, and your speed (and hence knetc energy) does not change. Equaton () s also useful for workng out the amount of fuel needed f a workng force s to be mantaned. Suppose a car engne s combatng a frcton force of N, and the car s travellng at a steady 3m/s. The engne power wll be N 3m/s = 6 kw. Our equaton can also be used to derve the knetc energy. Thnk of startng the object from rest, and calculatng the work needed to get t gong at speed U. The force, causng the acceleraton, wll be F=ma. The work done s gven by W P dt F mv dv v dt m v dt U mv mu dv dt (3) although care needs to be taken justfyng the ntegraton stage n the mult-dmensonal case... Changng Masses The applcaton of Newton s Laws to mechancs problems should pose you no trouble at all. However there are a couple of extra consderatons whch are worth thnkng about, and whch don t often get much attenton at school. The frst stuaton we ll consder s when the mass of a movng object changes. In practce the mass of any self-propellng object wll change as t uses up ts fuel, and for accurate calculatons we need to take ths nto account. There are two cases when ths must be consdered to get the answer even roughly rght jet aeroplanes and rockets. In the case of rockets, the fuel probably makes up 9% of the mass, so t must not be gnored. The proof s nterestng. It turns out that v dv v dv cos v dv snce the change n speed dv s equal to dv cos where dv (note the bold type) s the vector gvng the change n velocty. Page 9

10 Correctons March 7 Changng mass makes the physcs nterestng, because you need to thnk more carefully about Newton s second law. There are two ways of statng t ether () () Force on an object s equal to the rate of change of ts momentum Force on an object s equal to mass acceleraton The frst says F d( mu) dt mu mu ma m u, whereas the second smply states F=ma. Clearly they can t both be correct, snce they are dfferent. Whch s rght? The frst: whch was actually the way Newton stated t n the frst place! The good old F=ma wll stll work but you have to break the rocket nto parts (say grams of fuel) so that the rocket loses parts, but each part does not lose mass and then apply F=ma to each ndvdual part. However f you want to apply a law of moton to the rocket as a whole, you have to use the more complcated form of equaton. Ths may be the frst tme that you encounter the fact that momentum s a more frendly and fundamental quantty to work wth mathematcally than force. We shall see ths n a more extreme form when lookng at specal relatvty. Let us now try and calculate how a rocket works. We ll gnore gravty and resstve forces to start wth, and see how fast a rocket wll go after t has burnt some fuel. To work ths out we need to know two thngs the exhaust speed of the combuston gas (w), whch s always measured relatve to the rocket; and the rate at whch the motor burns fuel (n kg/s), whch we shall call. We ll thnk about one part of the moton, when the rocket starts wth mass (M+m), burns mass m of fuel, where m s very small, and n dong so ncreases ts speed from U to U+u. Ths s shown below n the dagram. Before M+m m After M U U-w U+u Notce that the velocty of the burnt fuel s U-w, snce w s the speed at whch the combuston gas leaves the rocket (backwards), and we need to take the rocket speed U nto account to fnd out how fast t s gong relatve to the ground. Conservaton of momentum tells us that Page

11 Correctons March 7 (M+m) U = m (U-w) + M (U+u) so m w = M u. (4) We can ntegrate ths expresson for u to evaluate the total change n speed after burnng a large amount of fuel. We treat the u (change n U) as an nfntesmal calculus du, and the m as a calculus dm. Notce the mnus sgn clearly the rocket must lose mass as fuel s burnt. Equaton (4) now tells us Ths can be ntegrated to gve dm w du (5) M U w M w fnal dm ln M U U ntal du M wln M ntal fnal (6) Ths formula (6) s nterestng because t tells us that n the absence of other forces, the gan n rocket speed depends only on the fracton of rocket mass that s fuel, and the exhaust speed. In ths calculaton, we have gnored other forces. Ths s not a good dea f we want to work out the moton at blast off, snce the Earth s gravty plays a major role! In order to take ths, or other forces, nto account, we need to calculate the thrust force of the rocket engne a task we have avoded so far. The thrust can be calculated by applyng F=ma to the (fxed mass) rocket M n our orgnal calculaton (4). The acceleraton s gven by du/t = u/t, where t s the tme taken to burn mass m of fuel. The thrust s T u mw mw m M M w w (7) t Mt t t gven by the product of the exhaust speed and the rate of burnng fuel. For a rocket of total mass M to take off vertcally, T must be greater than the rocket s weght Mg. Therefore for lft off to occur at all we must have w Mg. (8) Ths explans why heavy hydrocarbon fuels are nearly always used for the frst stage of lqud fuel rockets. In the later stages, where absolute Page

12 Correctons March 7 thrust s less mportant, hydrogen s used as t has a better kck per klogram because of ts hgher exhaust speed...3 Fcttous Forces Fcttous forces do not exst. So why do we need to gve them a moment s thought? Well, sometmes they make our lfe easer. Let s have a couple of examples...3. Centrfugal Force You may have travelled n one of those farground rdes n whch everyone stands aganst the nsde of the curved wall of a cylnder, whch then rotates about ts axs. After a whle, the floor drops out and yet you don t fall, because you re stuck to the sde. How does ths work? There are two ways of thnkng about ths. The frst s to look at the stuaton from the statonary perspectve of a frend on the ground. She sees you rotatng, and knows that a centrpetal force s needed to keep you gong round a force pontng towards the centre of the cylnder. Ths force s provded by the wall, and pushes you nwards. You feel ths strongly f you re the rder! And by Newton s thrd law t s equally true that you are pushng outwards on the wall, and ths s why you feel lke you are beng thrown out. Whle ths approach s correct, sometmes t makes the maths easer f you analyse the stuaton from the perspectve of the rder. Then you don t need to worry about the rotaton! However n order to get the workng rght you have to nclude an outwards force to balance the nward push of the wall. If ths were not done, the force from the walls would throw you nto the central space. The outward force s called the centrfugal force, and s our frst example of a fcttous force. It doesn t really exst, unless you are workng n a rotatng reference frame, and nsst that you are at rest. The dfference between the two vewponts s that n one case the nward push of the wall provdes the centrpetal acceleraton. In the other t opposes the centrfugal force - gvng zero resultant, and keepng the rder stll. Therefore the formulae used to calculate centrpetal force also gve the correct magntude for centrfugal force. The two dfferences are: () Centrfugal force acts outwards, centrpetal force acts nwards () Centrfugal force s only consdered f you are assumng that the cylnder s at rest (n the cylnder s reference frame). On the other hand, you only have centrpetal acceleratons f you do treat the cylnder as a movng object and work n the reference frame of a statonary observer. Ths example also shows that fcttous forces generally act n the opposte drecton to the acceleraton that s beng gnored. Here the Page

13 Correctons March 7 acceleraton s an nward centrpetal acceleraton, and the fcttous centrfugal force ponts outward...3. Inertal Force The second example we wll look at s the moton of a lft (elevator) passenger. You know that you feel heaver when the lft accelerates upwards, and feel lghter when t accelerates downwards. Therefore f you want to smplfy your maths by treatng the lft car as a statonary box, you must nclude an extra downward force when the lft s actually acceleratng upwards, and vce-versa. Ths fcttous force s called the nertal force. We see agan that t acts n the opposte drecton to the acceleraton we are tryng to gnore. We shall look more closely at ths stuaton, as t s much clearer mathematcally. Suppose we want to analyse the moton of a ball, say, thrown n the ar n a lft car whle t s acceleratng upwards wth acceleraton A. We use the vector a to represent the acceleraton of the ball as a statonary observer would measure t, and a to represent the acceleraton as measured by someone n the lft. Therefore, a = A + a. Now ths ball won t smply travel n a straght lne, because forces act on t. Suppose the force on the ball s F. We want to know what force F s needed to get the rght moton f we assume the lft to be at rest. Newton s second law tells us that F=ma, f m s the mass of the ball. Therefore F=m(A+a ), and so F-mA = ma. Now the force F must be the force needed to gve the ball acceleraton a (the moton relatve to the lft car), and therefore F =ma. Combnng these equatons gves F = F ma. (9) In other words, f workng n the reference frame of the lft, you need to nclude not only the forces whch are really actng on the ball (lke gravty), but also an extra force ma. Ths extra force s the nertal force. Let us contnue ths lne of thought a lttle further. Suppose the only force on the ball was gravty. Therefore F=mg. Notce that F = F ma = m (g-a) () and therefore f g=a (that s, the lft s fallng lke a stone, because some nasty person has cut the cable), F =. In other words, the ball behaves as f no force (not even gravty) were actng on t, at least from the pont of vew of the unfortunate lft passengers. Ths s why weghtlessness s experenced n free fall. Page 3

14 Correctons March 7 A smlar argument can be used to explan the weghtlessness of astronauts n orbtng spacecraft. As statonary observer (or a physcs teacher) would say that there s only one force on the astronauts gravty, and that ths s just the rght sze to provde the centrpetal force. The astronaut s perspectve s a lttle dfferent. He (or she) experences two forces gravty, and the fcttous centrfugal force. These two are equal and opposte, and as a result they add to zero, and so the astronaut feels just as weghtless as the doomed lft passengers n the last paragraph.. Gong Orbtal.. We have the potental We shall now spend a bt of tme revewng gravty. Ths s a frequent topc of Olympad questons, and s another area n whch you should be able to do well wth your A-level knowledge. Gravtaton causes all objects to attract all other objects. To smplfy matters, we start wth two small compact masses. The sze of the force of attracton s best descrbed by the equaton F r GMm () R Here G s the Gravtatonal constant ( Nm /kg ), M s the mass of one object (at the orgn of coordnates), and m s the mass of the other. The equaton gves the force experenced by the mass m. Notce the r subscrpt and the mnus sgn the force s radal, and drected nwards toward the orgn (where the mass M s). It s possble to work out how much work s needed to get the mass m as far away from M as possble. We use ntegraton R R R GMm GMm GMm F dx Fr dr dr r r. R Notce the use of F r n the second stage. In order to separate the masses we use a force F whch acts n opposton to the gravtatonal attracton F r. The equaton gves the amount of work done by ths force as t pulls the masses apart. We usually defne the zero of potental energy to be when the masses have nothng to do wth each other (because they are so far away). Accordngly, the potental energy of the masses m and M s gven by R GMm E( R). () R Page 4

15 Correctons March 7 That s, GMm R joules below zero energy. Notce that de( R) F r ( R). (3) dr Ths s a consequence of the defnton of work as W F dx, and s generally true. It s useful because t tells us that a forces always pont n the drecton of decreasng energy. The potental energy depends on the mass of both objects as well as the poston. The gravtatonal potental V(R) s defned as the energy per unt mass of the second object, and s gven by V R E( R, m) GM lm. (4) m R m Accordngly, the potental s a functon only of poston. The zero lmt on the mass m s needed (n theory) to prevent the small mass dsturbng the feld. In practce ths wll not happen f the masses are fxed n poston. To see the consequences of breakng ths rule, thnk about measurng the Earth s gravtatonal feld close to the Moon. If we do ths by measurng the force experenced by a kg mass, we wll be fne. If we do t by measurng the force experenced by a 8 kg planet put n place for the job, we wll radcally change the moton of Earth and Moon, and thus affect the measurement. In a smlar way, we evaluate the gravtatonal feld strength as the force per klogram of mass. Wrtng the feld strength as g gves MG g (5) R and equaton (3) may be rewrtten n terms of feld and potental as dv g( R). (6) dr.. Orbtal trcks There s a useful shortcut when dong problems about orbts. Suppose that an object of mass m s orbtng the centre of co-ordnates, and n experences an attractve force Fr Ar, where A s some constant. Therefore n=- for gravty, and we would have n=+ for moton of a partcle attached to a sprng (the other end fxed at the orgn). Page 5

16 Correctons March 7 If the object s performng crcular orbts, the centrpetal acceleraton wll be u R where R s the radus of the orbt. Ths s provded by the attractve force mentoned, and so: mu R mu F r AR n AR n (7) Now the potental energy E(R) s such that de dr F r AR n, so E( R) n AR n (8) f we take the usual conventon that E(R) s zero when the force s zero. Combnng equatons (7) and (8) gves mu n E( R) (9) so that Knetc Energy = Potental Energy (n+). () Ths tells us that for crcular gravtatonal orbts (where n=), the potental energy s twce as large as the knetc energy, and s negatve. For ellptcal orbts, the equaton stll holds: but now n terms of the average 3 knetc and potental energes. Equaton () wll not hold nstantaneously at all tmes for non-crcular orbts...3 Kepler s Laws The moton of the planets n the Solar system was observed extensvely and accurately durng the Renassance, and Kepler formulated three laws to descrbe what the astronomers saw. For the Olympad, you won t need to be able to derve these laws from the equatons of gravty, but you wll need to know them, and use them (wthout proof).. All planets orbt the Sun n ellptcal orbts, wth the Sun at one focus. 3 By average, we refer to the mean energy n tme. In other words, f T s the orbtal perod, the average of A s gven by T T A t ) ( dt. Page 6

17 Correctons March 7. The area traced out by the radus of an orbt n one second s the same for a planet, whatever stage of ts orbt t s n. Ths s another way of sayng that ts angular momentum s constant, and we shall be lookng at ths n Chapter The tme perod of the orbt s related to the [tme mean] average radus of the orbt: T R 3. It s not too dffcult to show that ths s true for crcular orbts, but t s also true for ellptc ones...4 Large Masses In our work so far, we have assumed that all masses are very small n comparson to the dstances between them. However, ths s not always the case, as you wll often be workng wth planets, and they are large! However there are two useful facts about large spheres and sphercal shells. A sphercal shell s a shape, lke the skn of a balloon, whch s bounded by two concentrc spheres of dfferent radus.. The gravtatonal feld experenced at a pont outsde a sphere or sphercal shell s the same as f all the mass of the shape were concentrated at ts centre.. A sphercal shell has no gravtatonal effect on an object nsde t. These rules only hold f the sphere or shell s of unform densty (strctly f the densty has sphercal symmetry). Therefore let us work out the gravtatonal force experenced by a mner down a very very very deep hole, who s half way to the centre of the Earth. We can gnore the mass above hm, and therefore only count the bt below hm. Ths s half the radus of the Earth, and therefore has one eghth of ts mass (assumng the Earth has unform densty whch t doesn t). Therefore the M n equaton () has been reduced by a factor of eght. Also the mner s twce as close to the centre (R has halved), and therefore by the nverse-square law, we would expect each klogram of Earth to attract hm four tmes as strongly. Combnng the factors of /8 and 4, we arrve at the concluson that he experences a gravtatonal feld ½ that at the Earth s surface, that s 4.9 N/kg..3 Fluds when thngs get stcky Questons about fluds are really classcal mechancs questons. You can tackle them wthout any detaled knowledge of flud mechancs. There are a few ponts you need to remember or learn, and that s what ths secton contans. Perfect gases are also fluds, but we wll deal wth them n chapter 5 Hot Physcs..3. Floatng and... the opposte The most mportant thng to remember s Archmedes Prncple, whch states that: Page 7

18 Correctons March 7 When an object s mmersed n a flud (lqud or gas), t wll experence an upwards force equal to the weght of flud dsplaced. By weght of flud dsplaced we mean the weght of the flud that would have been there f the object was not n poston. Ths upward force (sometmes called the buoyant upthrust) wll be equal to Force = Weght of flud dsplaced = g Mass of flud dsplaced = g Densty of flud Volume of flud dsplaced () For an object that s completely submerged, the volume of flud dsplaced s the volume of the object. For an object that s only partly submerged (lke an ceberg or shp), the volume of flud dsplaced s the volume of the object below the waterlne. Ths allows us to fnd out what wll float, and what wll snk. If an object s completely submerged, t wll have two forces actng on t. Its weght, whch pulls downwards, and the buoyant upthrust, whch pulls upwards. Volume V Mass M Upthrust = V g Object floats f: V > M > M/V Flud Densty Weght = M g Therefore, thngs float f ther overall densty (total mass / total volume) s less than the densty of the flud. Notce that the overall densty may not be equal to the actual densty of the materal. To gve an example a shp s made of metal, but contans ar, and s therefore able to float because ts overall densty s reduced by the ar, and s therefore lower than the densty of water. Puncture the hull, and the ar s no longer held n place. Therefore the densty of the shp = the densty of the steel, and the shp snks. For an object that s floatng on the surface of a flud (lke a shp on the ocean), the upthrust and weght must be equal otherwse t would rse Page 8

19 Correctons March 7 or fall. From Archmedes prncple, the weght of water dsplaced must equal the total weght of the object. There s a bran-teaser queston lke ths: A boat s floatng n the mddle of a lake, and the amount of water n the lake s fxed. The boat s carryng a large rock. The rock s lfted out of the boat, and dropped nto the lake. Wll the level of water n the lake go up or down? Answer: Level goes down whle the rock was n the boat (and therefore floatng) ts weght of water was beng dsplaced. When t was dropped nto the depths, ts volume of water was dsplaced. Now the densty of rock s hgher than that of water, so the water level n the lake was hgher n the frst case..3. Under Pressure What s the pressure n a flud? Ths must depend on how deep you are, because the deeper you are, the greater weght of flud you are supportng. We can thnk of the pressure (=Force/Area) as the weght of a square prsm of flud above a horzontal square metre marked out n the depths. Pressure = Weght of flud over m square = g Densty Volume of flud over m = g Densty Depth Cross sectonal area of flud (m ) Pressure = g Densty Depth () Of course, ths equaton assumes that there s nothng pushng down on the surface of the lqud! If there s, then ths must be added n too. Therefore pressure m under the surface of the sea = atmospherc pressure + weght of a m hgh column of water. It s wse to take a bt of cauton, though, snce pressures are often gven relatve to atmospherc pressure (.e. MPa more than atmospherc) and you need to keep your wts about you to spot whether pressures are relatve (vacuum = - kpa) or absolute (vacuum = Pa)..3.3 Contnuty Contnuty means conservatsm! Some thngs just don t change lke energy, momentum, and amount of stuff. Ths gves us a useful tool. Thnk about the dagram below, whch shows water n a cm [dameter] ppe beng forced nto a 5cm ppe. Page 9

20 Correctons March 7 cm 5 cm Water, lke most lquds, doesn t compress much so t can t form bottlenecks. The rate of water flow (cubc metres per second) n the bg ppe must therefore be equal to the rate of water flow n the lttle ppe. You mght lke to draw an analogy wth the current n a seres crcut. The lght bulb has greater resstance than the wre but the current n both s the same, because the one feeds the other. How can we express ths mathematcally? Let us assume that the ppe has a cross sectonal area A, and the water s gong at speed u m/s. How much water passes a pont n second? Let us put a marker n the water, whch moves along wth t. In one second t moves u metres. Therefore volume of water passng a pont = volume of cylnder of length u and cross sectonal area A = u A. Therefore Flow rate (m 3 /s) = Speed (m/s) Cross sectonal area (m ). (3) Now we can go back to our orgnal problem. The flow rate n both wde and narrow ppes must be the same. So f the larger one has twce the dameter, t has four tmes the cross sectonal area; and so ts water must be travellng four tmes more slowly..3.4 Bernoull s Equaton Somethng odd s gong on n that ppe. As the water squeezes nto the smaller radus, t speeds up. That means that ts knetc energy s ncreasng. Where s t gettng the energy from? The answer s that t can only do so f the pressure n the narrower ppe s lower than n the wder ppe. That way there s an unbalanced force on the flud n the cone-shaped part speedng t up. Let s follow a cubc metre of water through the system to work out how far the pressure drops. The flud n the larger ppe pushes the flud n the cone to the rght. The force = pressure area = P L A L. A cubc metre of flud occupes length /A L n the ppe, where A L s the cross sectonal area of the ppe to the left of the constrcton. Accordngly, the work done by the flud n the wder ppe on the flud n the cone n pushng the cubc metre through s Force Dstance = P L A L /A L = P L. However ths cubc metre does work P R A R /A R = P R n gettng out the other sde. Thus the net energy gan of the cubc metre s P L P R, and ths must equal the change n the cubc metre s knetc energy u R / u L /. Page

21 Correctons March The Flow Equaton Equaton (3) s also useful n the context of electrc currents, and can be adapted nto the so-called flow equaton. Let us suppose that the flud contans charged partcles. Suppose that there are N of these partcles per cubc metre of flud, and each partcle has a charge of q coulombs, then: Current = Flow rate of charge (charge / second) = Charge per cubc metre (C/m 3 ) flow rate (m 3 /s) = N q Area Speed. (4) Among other thngs, ths equaton shows why the free electrons n a semconductng materal travel faster than those n a metal. If the semconductor s n seres wth the metal, the current n both must be the same. However, the free charge densty N s much smaller n the semconductor, so the speed must be greater to compensate..4 Questons. Calculate the work done n pedallng a bcycle 3m up a road nclned at 5 to the horzontal.. Calculate the power of engne when a locomotve pulls a tran of kg up a nclne at a speed of 3m/s. Ignore the frcton n the bearngs A trolley can move up and down a track. It s potental energy s gven by V = k x 4, where x s the dstance of the trolley from the centre of the track. Derve an expresson for the force exerted on the trolley at any pont A ball bearng rests on a ramp fxed to the top of a car whch s acceleratng horzontally. The poston of the ball bearng relatve to the ramp s used as a measure of the acceleraton of the car. Show that f the acceleraton s to be proportonal to the horzontal dstance moved by the ball (measured relatve to the ramp), then the ramp must be curved upwards n the shape of a parabola Use arguments smlar to equaton (3) to prove that the knetc energy s stll gven by mu even when the force whch has caused the acceleraton from rest has not been appled unformly n a constant drecton Calculate the fnal velocty of a rocket 6% of whose launch mass s propellant, where the exhaust velocty s m/s. Repeat the calculaton for a rocket where the propellant makes up 9% of the launch mass. In both cases neglect gravty. Page

22 Correctons March 7 7. Repeat queston 6, now assumng that rockets need to move vertcally n a unform gravtatonal feld of 9.8N/kg. Calculate the velocty at MECO (man engne cut-off) and the greatest heght reached. Assume that both rockets have a mass of kg on the launch pad, and that the propellant s consumed evenly over one mnute A 7kg woman stands on a set of bathroom scales n an elevator. Calculate the readng on the scales when the elevator starts acceleratng upwards at m/s, when the elevator s gong up at a steady speed, and when the elevator decelerates at m/s before comng to a halt at the top floor of the buldng. 9. The woman n q8 s a juggler. Descrbe how she mght have to adjust her throwng technques n the elevator as t accelerates and decelerates.. Archtectural models can not be properly tested for strength because they appear to be stronger than the real thng. To see why, consder a half-scale model of a buldng made out of the same materals. The weght s /8 of the real buldng, but the columns are ¼ the cross sectonal area. Accordngly the stress on the columns s half of that n the full sze buldng, and accordngly the model can wthstand much more severe load before collapsng. To correct for ths, a :3 archtectural model s put on the end of a centrfuge arm of radus m whch s spun around. The spnnng smulates an ncreased gravtatonal force whch allows the model to be accurately tested. How many tmes wll the centrfuge go round each mnute?. Consder an ncompressble flud flowng from a 5cm dameter ppe nto a 5cm dameter ppe. If the velocty and pressure before the constrcton are m/s and N/m, calculate the velocty and pressure n the constrcted ppe. Neglect the effects of vscosty and turbulent flow. To work out the new pressure, remember that the ncrease n speed nvolves an ncrease of knetc energy, and ths energy must come from somewhere so there wll be a drop n pressure.. Calculate the orbtal tme perod T of a satellte skmmng the surface of a planet wth radus R and made of a materal wth densty. Calculate the orbtal speed for an astronaut skmmng the surface of a comet wth a km radus. 3. The alcohol percentage n wne can be determned from ts densty. A very lght glass test tube (of cross sectonal area.5cm ) has 5g of lead pellets fxed to the bottom. You place the tube n the wne, lead frst, and t floats wth the open end of the tube above the surface of the wne. You can read the % alcohol from markngs on the sde of the tube. Calculate how far above the lead the % and % marks should be placed. The densty of water s.g/cm 3, whle that of ethanol s Page

23 Correctons March 7.98g/cm 3. Where should the 5% lne go? Remember that alcohol percentages are always volume percentages. + Page 3

24 Correctons March 7 Fast Physcs Imagne a summer s day. You are sunbathng by the sde of a busy motorway whle you wat for a pckup truck to rescue your car, whch has broken down. All of a sudden, an rresponsble person throws a used drnks can out of ther car wndow, and t heads n your drecton. To make thngs worse, they were speedng at the tme. Ouch. The faster the car was gong, the more t wll hurt when the can hts you. Ths s because the can automatcally takes up the speed the car was travellng at. Suppose the rresponsble person could throw the can at mph, and ther car s gong at 8mph. The speed of the can, as you see t, s 9mph f t was thrown forwards, and 7mph f t was thrown backwards. To sum ths up, Velocty as measured by you = Velocty of car + Velocty of throwng where we use veloctes rather than speeds so that the drectonalty can be taken nto account. So far, ths probably seems very obvous. However, let s extend the logc a bt further. Rather than a car, let us have a star, and n place of the drnks can, a beam of lght. Many stars travel towards us at hgh speeds, and emt lght as they do so. We can measure the speed of ths lght n a laboratory on Earth, and compare t wth the speed of ordnary lght made n a statonary lght bulb. And the worryng thng s that the two speeds are the same. No matter how hard we try to change t, lght always goes at the same speed. 4 Ths tells us that although our deas of addng veloctes are nce and straghtforward, they are also wrong. In short, there s a problem wth the Newtonan pcture of moton. Ths problem s most obvous n the case of lght, but t also occurs when anythng else starts travellng very quckly. Whle ths s not the way Ensten approached the problem, t s our way nto one of hs early theores the Specal Theory of Relatvty and t s part of the Olympad syllabus. Before we go further and talk about what does happen when thngs go fast, please be aware of one thng. These observatons wll seem very 4 Lght does travel dfferent speeds n dfferent materals. However f the measurement s made n the same materal (say, ar or vacuum) the speed regstered wll always be the same, no matter what we do wth the source. Page 4

25 Correctons March 7 werd f you haven t read them before. But don t dsmss relatvty as nonsense just because t seems werd t s a better descrpton of Nature than classcal mechancs and as such t demands our respect and attenton.. The Prncple of Relatvty The theory of specal relatvty, lke all theores, s founded on a premse or axom. Ths axom cannot be derved t s a guessed statement, whch s the startng pont for the maths and the phlosophy. In the case of specal relatvty, the axom must be helpful because ts logcal consequences agree well wth experments. Ths prncple, or axom, can be stated n several ways, but they are effectvely the same.. There s no method for measurng absolute (non-relatve) velocty. The absolute speed of a car cannot be measured by any method at all. On the other hand, the speed of the car relatve to a speed gun, the Earth, or the Sun can all be determned.. Snce t can t be measured there s no such thng as absolute velocty. 3. The laws of physcs hold n all non-acceleratng laboratores 5, however fast they may be gong. Ths follows from statement, snce f experments only worked for one partcular laboratory speed, that would somehow be a specal speed, and absolute veloctes could be determned relatve to t. 4. Maxwell s theory of electromagnetsm, whch predcts the speed of lght, counts as a law of physcs. Therefore all laboratores wll agree on the speed of lght. It doesn t matter where or how the lght was made, nor how fast the laboratory s movng.. Hgh Speed Observatons In ths secton we are gong to state what relatvty predcts, as far as t affects smple observatons. Please note that we are not dervng these statements from the prncples n the last secton, although ths can be done. For the moment just try and understand what the statements mean. That s a hard enough job. Once you can use them, we shall then worry about where they come from. 5 We say non-acceleratng for a good reason. If the laboratory were acceleratng, you would feel the nertal force, and thus you would be able to measure ths acceleraton, and ndeed adjust the laboratory s moton untl t were zero. However there s no equvalent way of measurng absolute speed. Page 5

26 Correctons March 7.. Speedng objects look shortened n the drecton of moton. A metre stck comes hurtlng towards you at hgh speed. Wth a clever arrangement of cameras and tmers, you are able to measure ts length as t passes you. If the stck s length s perpendcular to the drecton of travel, you stll measure the length as metre. However, f the stck s parallel to ts moton, t wll seem shorter to you. If we call the stck s actual length (as the stck sees t) as L, and the apparent length (as you measure t) L a, we fnd L a u L, () c where u s the speed of the metre stck relatve to the observer. The object n the square root appears frequently n relatvstc work, and to make our equatons more concse, we wrte () u c so that equaton () appears n shorter form as L L a. (3).. Speedng clocks tck slowly A second observaton s that f a clock whzzes past you, and you use another clever arrangement of tmers and cameras to watch t, t wll appear (to you) to be gong slowly. We may state ths mathematcally. Let T be a tme nterval as measured by our (statonary) clock, and let T a be the tme nterval as we see t measured by the whzzng clock. T T a (4)..3 Slowng and shrnkng go together Equatons (3) and (4) are consstent you can t have one wthout the other. To see why ths s the case, let us suppose that Andrew and Betty both have excellent clocks and metre stcks, and they wsh to measure ther relatve speed as they pass each other. They must agree on the relatve speed. Andrew tmes how long t takes Betty to travel along hs metre stck, and Betty does the same. Page 6

27 Correctons March 7 The queston s: how does Andrew settle hs mnd about Betty s calculaton? As far as he s concerned, she has a short metre stck, and a slow clock how can she possbly get the answer rght! Very easly provdng that her clock runs slow by the same amount that her metre stck s short 6. An expermental example may help clarfy ths. Muons are charged partcles that are not stable, and decay wth a half-lfe of s. Because they are charged, you can accelerate them to hgh speeds usng a large electrc feld n a partcle accelerator. You can then measure how far they travel down a tube before decayng. Gven that the laws of physcs are the same n all reference frames, ths must mean that muon and expermenter agree on the poston n the tube at whch the muon passes away. The muon gets much further down the tube than a classcal calculaton would predct, however the reason for ths can be explaned n two ways: Accordng to the expermenter, the muon s travellng fast, so t has a slow clock, and therefore lves longer so t can get further. Accordng to the muon, t stll has a woefully short lfe, but the tube (whch s whzzng past) s shorter so t appears to get further along n the s. For the two calculatons to agree, the clock slowng must be at the same rate as the tube shrnkage...4 Speedng adds weght to the argument The most useful observaton of them all, as far as the Olympad syllabus s concerned s ths: f someone throws a kg bag of sugar at you at hgh speed, and you (somehow) manage to measure ts mass as t passes, you wll regster more than kg. If the actual mass of the object s M, and the apparent mass s M a, we fnd that M a. (5) M The actual mass s usually called the rest mass n other words the mass as measured by an observer who s at rest wth respect to the object. 6 Note that slow and short are placed n quotaton marks. Betty s clock and metre stck are not defectve however to Andrew they appear to be. Page 7

28 Correctons March The Unversal Speed Lmt Ths formula has mportant consequences. Frst of all, ths s the orgn of the unversal speed lmt, whch s a well-known consequence of specal relatvty. Ths states that you wll never measure the speed of an object (relatve to you) as beng greater than the speed of lght. Let us pause for a moment to see why. Suppose the object concerned s an electron n a partcle accelerator (electrons currently hold the speed record on Earth for the fastest humanly accelerated objects). It starts at rest wth a mass of about -3 kg. We turn on a large, constant electrc feld, and the electron starts to move relatve to the accelerator. However, as t gets close to the speed of lght, t starts to appear more massve. Therefore snce our electrc feld (hence acceleratng force) s constant, the electron s acceleraton decreases. In fact, the acceleraton tends to zero as tme passes, although t never reaches zero exactly after a fnte tme. We are never able to persuade the electron to break the lght-barrer, snce when u c,, and the apparent mass becomes very large (so the object becomes mpossble to accelerate any further). Please note that ths does not mean that faster-than-lght speeds can never be obtaned. If we accelerate one electron to.6c Eastwards, and another to.6c Westwards, the approach speed of the two electrons s clearly superlumc (.c) as we measure t wth Earth-bound speedometers. However, even n ths case we fnd that the velocty of one of the electrons as measured by the other s stll less than the speed of lght. Ths s a consequence of our frst observaton namely that relatve veloctes do not add n a smple way when the objects are movng quckly. In fact the approach speed, as the electrons see t, s.88c. If you want to perform these calculatons, the formula turns out to be u AC u AB u BC, (6) u u c AB BC where u AB means the velocty of B as measured by A. Equaton (6) only apples when all three relatve veloctes are parallel (or antparallel)...4. Newton s Law of moton Our second consequence s that we need to take great care when usng Newton s laws. We need to remember that the correct form of the second law s d F momentum (7) dt Page 8

29 Correctons March 7 Why s care needed? Look closely for the trap f the object speeds up, ts mass wll ncrease. Therefore the tme dervatve of the mass needs to be ncluded as well as the tme dervatve of the velocty. We shall postpone further dscusson untl we have had a better look at momentum..3 Relatvstc Quanttes Now that we have mentoned the busness of relatvstc mass ncrease, t s tme to address the relatvstc forms of other quanttes..3. Momentum Momentum s conserved n relatvstc collsons, provdng we defne t as the product of the apparent mass and the velocty. p mu (7) Notce that when you use momentum conservaton n collsons, you wll have to watch the factors. Snce these are functons of the speed u, they wll change f the speed changes..3. Force The force on a partcle s the tme dervatve of ts momentum. Therefore d d d F p m u u. (8) dt dt dt In the case where the speed s not changng, wll stay constant, and the equaton reduces to the much more straghtforward F=m a. One example s the moton of an electron n a magnetc feld..3.3 Knetc Energy Now that we have an expresson for force, we can ntegrate t wth respect to dstance to obtan the work done n acceleratng a partcle. As shown n secton.., ths wll gve the knetc energy of the partcle. We obtan the result 7 7 If you wsh to derve ths yourself, here are the stages you need. Frstly, dfferentate wth respect to u to convnce yourself that d du 3 u u du c c. 3 u c 3 c d u Usng ths result, the dervaton can be completed (see over the page): Page 9

30 Correctons March 7 c m K. (9) Ths states that the gan n energy of a partcle when accelerated s equal to the gan n mass c. From ths we postulate that any ncrease n energy s accompaned by a change n mass. The argument works backwards too. When statonary, the partcle had mass m. Surely therefore, t had energy m c when at rest. We therefore wrte the total energy of a partcle as. () c m c m K E.3.4 A Relatvstc Toolkt We can derve a very useful relatonshp from (), (7) and the defnton of : 4 4 c m c v c m v c c m c p E. () Ths s useful, snce t relates E and p wthout nvolvng the nasty factor. Another equaton whch has no gammas n t can be derved by dvdng momentum by total energy: c u c m u m E p o, () whch s useful f you know the momentum and total energy, and wsh to know the speed..3.5 Tacklng problems If you have to solve a collson type problem, avod usng speeds at all costs. If you nsst on havng speeds n your equatons, you wll also have gammas, and therefore headaches. So use the momenta and energes of the ndvdual partcles n your equatons nstead. Put more bluntly, you should wrte lots of p s, and E s, but no u s. Use the c m d u c u m d u c u u m d d du u u m dt dt du u m dt dt d u m Fudt Fdx K Page 3

31 Correctons March 7 conservaton laws to help you. In relatvstc work, you can always use the conservaton of E even n non-elastc collsons. The nterestng thng s that n an nelastc collson, you wll fnd the rest masses greater after the collson. To obtan the values you want, you need an equaton whch relates E and p, and ths s provded by (). Notce n partcular that the quantty E p c when appled to a group of partcles has two thngs to commend t. Frstly, t s only a functon of total energy and momentum, and therefore wll reman the same before and after the collson. Secondly, t s a functon of the rest masses (see equaton ) and therefore wll be the same n all reference frames. Fnally, f the queston asks you for the fnal speeds, use () to calculate them from the momenta and energes..4 The Lorentz Transforms The facts outlned above (wthout the dervatons) wll gve you all the nformaton you need to tackle Internatonal Olympad problems. However, you may be nterested to fnd out how the observatons of secton. follow from the general assumptons of secton.. A full justfcaton would requre a whole book on relatvty, however we can gve a bref ntroducton to the method here. We start by statng a general problem. Consder two frames of reference (or co-ordnate systems) Andrew s perspectve (t,x,y,z), and Betty s perspectve (t,x,y,z ). We assume that Betty s shootng past Andrew n the +x drecton at speed v. Suppose an event happens, and Andrew measures ts co-ordnates. How do we work out the co-ordnates Betty wll measure? The relatonshp between the two sets of co-ordnates s called the Lorentz transformaton, and ths can be derved as shown below:.4. Dervaton of the Lorentz Transformaton We begn wth the assumpton that the co-ordnate transforms must be lnear. The reason for ths can be llustrated by consderng length, although a smlar argument works for tme as well. Suppose that Andrew has two measurng stcks joned end to end, one of length L and one of length L. He wants to work out how long Betty reckons they are. Suppose the transformaton functon s T. Therefore Betty measures the frst rod as T(L) and the second as T(L). She therefore wll see that the total length of the rods s T(L) + T(L). Ths must also be equal to T(L+L), snce L+L s the length of the whole rod Page 3

32 Correctons March 7 accordng to Andrew. Snce T(L+L) = T(L) +T(L), the transformaton functon s lnear. We can now get to work. Let us consder Betty s frame of reference to be movng n the +x drecton at speed v, as measured by Andrew. Betty wll therefore see Andrew movng n her x drecton at the same speed. To dstngush Betty s co-ordnates from Andrew s, we gve hers dashes. Gven the lnear nature of the transformaton, we wrte x A t C B x D t where A, B, C and D are functons of the relatve velocty +v (.e. Betty s velocty as measured by Andrew). There must also be an nverse transformaton x D t d C B x A t where d s the determnant of the frst matrx. Now ths second matrx s n tself a transformaton for a relatve velocty v, and therefore should be of a very smlar form to the frst matrx. We fnd that the only way we can ensure that there s symmetry between the two s to make the determnant equal to one (d=). We shall therefore assume ths from here on. Next we consder what happens f x =. In other words we are tracng out Betty s moton as Andrew sees t. Therefore we must have x=vt. Usng the frst matrx, ths tells us that B=-vA. A smlar argument on the second matrx where we must have x =-vt where Betty now watches Andrew s moton [x=], gves Dv = B = -va. Therefore A=D. We now have B and D expressed n terms of A, so the next job s to work out what C s. Ths can be done snce we know that the determnant AD BC =. Therefore we fnd that A C. va Page 3

33 Correctons March 7 Summarzng, our matrx s now expressed totally n terms of the unknown varable A. We may calculate t by rememberng that both Andrew and Betty wll agree on the speed of travel, c, of a ray of lght. Andrew wll express ths as x=ct, Betty would say x =ct, but both must be vald ways of descrbng the moton. Therefore A - D snce A c v A A v c A va c va B Cc B Ac D c Cc D Cc B Ac t ct c t ct D C B A t ct Ths concludes our reasonng, and gves the Lorentz transforms (after a lttle algebra to evaluate C) as: c v c v x t t vt x x c xv t t vt x x. We have not consdered any other dmensons here, however the transformaton here s easy snce Andrew and Betty agree on all lengths n the y and z drectons. In other words y =y, z =z. Ths s a necessary consequence of the prncple of relatvty: the dstance between the ends of a rod held perpendcular to the drecton of moton can be measured smultaneously n all frames of reference. If ths agreed measurement was dfferent to that of an dentcal rod n a dfferent frame, the observers would be able to work out whch of them was movng and whch of them was stll. Page 33

34 Correctons March 7.4. Usng the Lorentz Transforms Havng these transforms at our dsposal, we can now derve the shrnkng rod and slowng clock equatons. Suppose Betty s holdng a stck (of length L) parallel to the x-axs. We want to know how long Andrew thnks t s. To measure t, he wll measure where the ends of the rod are at a partcular moment, and wll then measure the dstance between these ponts. Clearly the two postons need to be measured smultaneously n hs frame of reference, and thus t s the same for both measurements. We know from that Betty thnks t has length L, and therefore x =L. Usng the frst of the Lorentz equatons (the one whch lnks x, x and t), and rememberng that t s the same for both measurements, x x L apparent L. Smlarly we may show how a clock appears to slow down. Betty s carryng the clock, so t s statonary wth respect to her, and x (her measurement of the clock s poston) wll therefore be constant. The tme nterval shown on Betty s clock s t, whle Andrew s own clock wll measure tme t. Here t s the tme Andrew sees elapsng on Betty s clock, and as such s equal to T apparent. Usng the fourth Lorentz equaton (the one wth x, t and t n t), and rememberng that x remans constant, we have t t T apparent T..4.3 Four Vectors The Lorentz transforms show you how to work out the relatonshps between the (t,x,y,z) co-ordnates measured n dfferent frames of reference. We descrbe anythng that transforms n the same way as a four vector, although strctly speakng we use (ct,x,y,z) so that all the components of the vector have the same unts. Three other examples of four vectors are: (c, u x, u y, u z ) s called the four velocty of an object, and s the dervatve of (ct, x,y,z) wth respect to the proper tme. Proper tme s the tme elapsed as measured n the rest frame of the object t=. (mc,p x,p y,p z ) the momentum four vector. Here m s equal to m. Ths must be a four vector snce t s equal to the rest mass Page 34

35 Correctons March 7 multpled by the four velocty (whch we already know to be a four vector). (/c,k x,k y,k z ) the wave four vector, where s the angular frequency of the wave (=f), and k s a vector whch ponts n the drecton the wave s gong, and has magntude /. Ths can be derved from the momentum four vector n the case of a photon, snce the momentum and total energy of a photon are related by E=pc, and the quantum theory states that E=hf=h/ and p=h/=hk/. It also turns out that the dot product of any two four-vectors s framenvarant n other words all observers wll agree on ts value. The dot product of two four-vectors s slghtly dfferent to the conventonal dot product, as shown below: ct, x, y, z ct, x, y, z x y z ct. Notce that we subtract the product of the frst elements. The dot product of the poston four vector wth the wave four vector gves ct x, y, z c, k, k, k k r t,. x y Now ths s the phase of the wave, and snce all observers must agree whether a partcular pont s a peak, a trough or somewhere n between, then the phase must be an nvarant quantty. Accordngly, snce (ct,x,y,z) makes ths nvarant when dotted wth (/c,k x,k y,k z ), t follows that (/c,k x,k y,k z ) must be a four vector too..5 Questons. Work out the relatvstc factor for speeds of %, 5%, 9% and 99% of the speed of lght.. Work out the speeds needed to gve factors of.,.,.,.. 3. A muon travels at 9% of the speed of lght down a ppe n a partcle accelerator at a steady speed. How far would you expect t to travel n s (a) wthout takng relatvty nto account, and (b) takng relatvty nto account? 4. A partcle wth rest mass m and momentum p colldes wth a statonary partcle of mass M. The result s a sngle new partcle of rest mass R. Calculate R n terms of p and M. z Page 35

36 Correctons March 7 5. The prncpal runway at the spaceport on Arcturus-3 has whte squares of sde length m panted on t. A set of lght sensors on the base of a spacecraft can take a pcture of the whole runway at the same tme. What wll the squares look lke n the mage f the spacecraft s passng the runway at a very hgh speed? Each sensor takes a pcture of the runway drectly underneath t, so you do not need to take nto account the dfferent tmes taken by lght to reach the sensors from dfferent parts of the runway. 6. When an electron s accelerated through a voltage V, ts knetc energy s gven by ev where e s the sze of the charge on the electron and s equal to.6 9 C. Takng the mass of the electron to be 9. 3 kg, work out (a) the knetc energy and speed of the electron when V=5kV (b) the knetc energy and speed when V=kV (c) the percentage error n the knetc energy for V=kV when calculated usng the non-relatvstc equaton ½ mu. 7. Prove that the knetc energy of a partcle of rest mass m and speed u s gven by ½ mu f the speed s small enough n comparson to the speed of lght. Work out the speed at whch the non-relatvstc calculaton would be n error by %. 8. Suppose a spacecraft accelerates wth constant acceleraton a (as measured by the spacecraft s onboard accelerometers). At t= t s at rest wth respect to a planet. Work out ts speed relatve to the planet as a functon of tme (a) as measured by clocks on the spacecraft, and (b) as measured by clocks on the planet. Note that the nstantaneous speed of the craft relatve to the planet wll be agreed upon by spacecraft and planet. Page 36

37 Correctons March 7 3 Rotaton Rotatonal moton s all around us [groan] from the acts of subatomc partcles, to the moton of galaxes. Calculatons nvolvng rotatons are no harder than lnear mechancs; however the quanttes we shall be talkng about wll be unfamlar at frst. Havng already studed lnear mechancs, you wll be at a tremendous advantage, snce we shall fnd that each rule n lnear mechancs has ts rotatonal equvalent. 3. Angle In lnear mechancs, the most fundamental measurement s the poston of the partcle. The equvalent base of all rotatonal analyss s angle: the queston How far has the car moved? beng exchanged for How far has the wheel gone round? a queston whch can only be answered by gvng an angle. In mechancs, the radan s used for measurng angles. Whle you may be more famlar wth the degree, the radan has many advantages. We shall start, then by defnng what we mean by a radan. Consder a sector of a crcle, as n the dagram; and let the crcle have a radus r. The length of the arc, that s the curved lne n the sector, s clearly related to the angle. If the angle were made twce as large, the arc length would also double. Can we use arc length to measure the angle? Not as t stands, snce we haven t taken nto account the radus of the crcle. Even for a fxed angle (say 3 ), the arc wll be longer on a larger crcle. We therefore defne Arc length = r f s measured n radans r the angle (n radans) as the arc length dvded by the crcle radus. Alternatvely you mght say that the angle n radans s equal to the length of the arc of a unt crcle (that s a crcle of m radus) that s cut by the angle. Notce one smplfcaton that ths brngs. If a wheel, of radus R, rolls a dstance d along a road, the angle the wheel has turned through s gven by d/r n radans. Were you to calculate the angle n degrees, there would be nasty factors of 8 and n the answer. Before gettng too nvolved wth radans, however, we must work out a converson factor so that angles n degrees can be expressed n radans. To do ths, remember that a full crcle (36 ) has a crcumference or arc length of r. So 36 = rad. Therefore, radan s equvalent to (36/) = (8/). Page 37

38 Correctons March 7 3. Angular Velocty Havng dscussed angle as the rotatonal equvalent of poston, we now turn our attenton to speed. In lnear work, speeds are gven n metres per second the dstance moved n unt tme. For rotaton, we speak of angular velocty, whch tells us how fast somethng s spnnng: how many radans t turns through n one second. The angular velocty can also be thought of as the dervatve of angle wth respect to tme, and as such s sometmes wrtten as, however more commonly the Greek letter s used, and the dot v sn s avoded. To check your understandng of ths, v try and show that rpm (revoluton per mnute) s equvalent to /3 rad/s, whle one cycle per second s equvalent to rad/s. r Now remember the defnton of angle n radans, and that the dstance moved by a pont on the rm of a wheel wll move a dstance s = r when the wheel rotates by an angle. The speed of the pont wll therefore be gven by u = ds/dt = r d/dt = r. For a pont that s not fxed to the wheel, the stuaton s a lttle more complex. Suppose that the pont has a velocty v, whch makes an angle to the radus (as n the fgure above). We then separate v nto two components, one radal (v cos ) and one rotatonal (v sn ). Clearly the latter s the only one that contrbutes to the angular velocty, and therefore n ths more general case, v sn = r. 3.3 Angular acceleraton It should come as no surprse that the angular acceleraton s the tme dervatve of, and represents the change n angular velocty (n rad/s) dvded by the tme taken for the change (n s). It s measured n rad/s, and denoted by or or. For an object fastened to the rm of a wheel, the actual acceleraton round the rm (a) wll be gven by a=du/dt 8 = r d/dt = r, whle for an object not fastened, we have a sn = r. 3.4 Torque Angular Force Before we can start dong mechancs wth angles, we need to consder the rotatonal equvalent of force the amount of twst. Often a twst can be appled to a system by a lnear force, and ths gves us a way n to the analyss. We say that the strength of the twst s called the moment of the force, and s equal to the sze of the force multpled by the dstance from pvot to the pont where the force acts. A complcaton 8 Here we are not ncludng the centrpetal acceleraton whch s drected towards the centre of the rotaton. Page 38

39 Correctons March 7 F arses f the force s not tangental clearly a force actng along the radus of a wheel wll not turn t and so our smple moment equaton needs modfyng. 9 r There are two ways of proceedng, and they yeld the same answer. Suppose the force F makes an angle wth the radus. We can break ths down nto two r sn components one of magntude F cos, whch s radal and does no turnng; and the other, tangental component (whch does contrbute to the turnng) of magntude F sn. The moment or torque only ncludes the relevant component, and so the torque s gven by C = Fr sn. The alternatve way of vewng the stuaton s not to measure the dstance from the centre to the pont at whch the force s appled. Instead, we draw the force as a long lne, and to take the dstance as the perpendcular dstance from force lne to centre. The dagram shows that ths new dstance s gven by r sn, and snce the force here s completely tangental, we may wrte the moment or torque as the product of the full force and ths perpendcular dstance.e. C = F r sn, as before. 3.5 Moment of Inerta Angular Mass Of the three base quanttes of moton, namely dstance, mass and tme, only tme may be used wth mpunty n rotatonal problems. We now have an angular equvalent for dstance (namely angle), so the next task s to determne an angular equvalent for mass. Ths can be done by analogy wth lnear mechancs, where the mass of an object n klograms can be determned by pushng an object, and calculatng the rato of the appled force to the acceleraton t caused: m = F/a. Gven that we now have angular equvalents for force and acceleraton, we can use these to fnd out the angular mass. Thnk about a ball of mass m fxed to the rm of a wheel that s acceleratng wth angular acceleraton. We shall gnore the mass of the wheel tself for now. Now let us push the mass round the wheel wth a F F sn force F. Therefore we calculate the angular mass I by 9 Why force radus? We can use a vrtual work argument (as n secton...4) to help us. Suppose a tangental force F s appled at radus r. When the object moves round by angle, t moves a dstance d = r, and the work done by the force = Fd = Fr = Fr angular force angular dstance. Now snce energy must be the same sort of thng wth rotatonal moton as lnear, the rotatonal equvalent of force must be Fr. Page 39

40 Correctons March 7 I C rf sn a sn r r F a r m where we have used the fact that the mass m wll be the rato of the force F to the lnear acceleraton a, as dctated by Newton s Second Law. Ths formula can also be used for sold objects, however n ths case, the radus r wll be the perpendcular dstance from the mass to the axs. The total angular mass of the object s calculated by addng up the I = m r from each of the ponts t s made from. Usually ths angular mass s called the moment of nerta of the object. Notce that t doesn t just depend on the mass, but also on the dstance from the pont to the centre. Therefore the moment of nerta of an object depends on the axs t s spun round. An object may have a hgh angular nerta, therefore, for two reasons. Ether t s heavy n ts own rght; or for a lghter object, the mass s a long way from the axs. 3.6 Angular Momentum In lnear moton, we make frequent use of the momentum of objects. The momentum s gven by mass velocty, and changes when a force s appled to the object. The force appled, s n fact the tme dervatve of the momentum (provded that the mass doesn t change). Frequent use s made of the fact that total momentum s conserved n collsons, provded that there s no external force actng. It would be useful to fnd a smlar thng for angular moton. The most sensble startng guess s to try angular mass angular velocty. We shall call ths the angular momentum, and gve t the symbol L = I. Let us now nvestgate how the angular momentum changes when a torque s appled. For the moment, assume that I remans constant. dl dt d dt I I d I C dt Thus we see that, lke n lnear moton, the tme dervatve of angular momentum s angular force or torque. Two of the mportant facts that stem from ths statement are:. If there s no torque C, the angular momentum wll not change. Notce that radal forces have C =, and therefore wll not change the angular momentum. Ths result may seem unmportant but thnk of the planets n ther orbts round the Sun. The tremendous force exerted on them by the Sun s gravty s radal, and therefore does not change ther angular momenta even a smdgen. We can therefore calculate the velocty of planets at dfferent parts of ther orbts usng the fact that the angular momentum wll reman the same. Ths prncple also holds when Page 4

41 Correctons March 7 scentsts calculate the path of space probes sent out to nvestgate the Solar System.. The calculaton above assumes that the moment of nerta I of the object remans the same. Ths seems sensble, after all, n a lnear collson, the nstantaneous change of a sngle object s mass would be bzarre, and therefore we don t need to guard aganst the possblty of a change n mass when we wrte F = dp/dt. In the case of angular moton, ths stuaton s dfferent. The moment of nerta can be changed, smply by rearrangng the mass of the object closer to the axs. Clearly there s no external torque n dong ths, so we should expect the angular momentum to stay the same. But f the mass has been moved closer to the axs, I wll have got smaller. Therefore must have got bgger. The object wll now be spnnng faster! Ths s what happens when a spnnng ce dancer brngs n her/hs arms and the correspondng ncrease n revs. per mnute s well known to ce enthusasts and TV vewers alke. To take an example, suppose that all the masses were moved twce as close to the axs. The value of r would halve, so I would be quartered. We should therefore expect to get four tmes larger. Ths s n fact what happens. 3.7 Angular momentum of a sngle mass movng n a straght lne If we wshed to calculate the angular momentum of a planet n ts orbt round the Sun, we need to know how L s related to the lnear speed v. Ths s what we wll now work out. Usng the same deas as n fgure, the velocty v wll have both radal and rotatonal components. The rotatonal component wll be equal to v sn, whle the radal component cannot contrbute to the angular momentum. It s the rotatonal component that corresponds to the speed of a mass fxed to the rm of a wheel, and as such s equal to radus angular velocty. Thus v sn = r. So the angular momentum L I mr v sn mvr sn r Two cautons. Frstly, n a rocket, the mass of the rocket does decrease as the burnt fuel s chucked out the back, however the total mass does not change. Therefore F=dp/dt=ma stll works, we just need to be careful that the force F acts on (and only on) the stuff ncluded n the mass m. A complcaton does arse when objects start travellng at a good fracton of the speed of lght but ths s dealt wth n the secton on Specal Relatvty. Page 4

42 Correctons March 7 s gven by the product of the mass, the radus and the rotatonal (or tangental) component of the velocty. For an object on a straght lne path, ths can also be stated (usng fgure 3) as the mass speed dstance of closest approach to centre. 3.8 Rotatonal Knetc Energy Lastly, we come to the calculaton of the rotatonal knetc energy. We may calculate ths by addng up the lnear knetc energes of the parts of the object as the spn round the axs. Notce that n ths calculaton, as the objects are purely rotatng, we shall assume = /.e. there s no radal moton. K mr I mv m r We see that the knetc energy s gven by half the angular mass angular velocty squared whch s a drect equvalent wth the half mass speed of lnear moton. 3.9 Summary of Quanttes Quantty Symbol Unt Defnton Other equatons Angular velocty rad/s = d/dt r = v sn Angular acceleraton rad/s = d/dt r = a sn Torque C N m C = F r sn Moment of nerta I kg m I = C / I = m r Angular momentum Rot. Knetc Energy L kg m /s L = I L = m v r sn K J K = I / K = ½m (v sn ) Page 4

43 Correctons March 7 3. Rotatonal mechancs wth vectors Ths secton nvolves much more advanced mathematcs, and you wll be able to get by n Olympad problems perfectly well wthout t. However, f you lke vectors and matrces, read on... So far we have just consdered rotatons n one plane that of the paper. In general, of course, rotatons can occur about any axs, and to descrbe ths three dmensonal stuaton, we use vectors. Wth velocty v, momentum p and force F, there s an obvous drecton the drecton of moton, or the drecton of the push. Wth rotaton, the drecton s less clear. Imagne a clock face on ths paper, wth the mnute hand rotatng clockwse. What drecton do we assocate wth ths moton? Up towards o clock because the hand sometmes ponts that way? Towards 3 o clock because the hand sometmes ponts that way? Both are equally rdculous. In fact the only way of choosng a drecton that wll always apply s to assgn the rotaton drecton perpendcular to the clock face the drecton n whch the hands never pont. Ths has not resolved our dffculty completely. Should the arrow pont upwards out of the paper, or down nto t? After thought we realse that one should be used for clockwse and one for ant-clockwse moton, but whch way? There s no way of proceedng based on logc we just have to accept a conventon. The custom s to say that for a clockwse rotaton, the drecton s down away from us, and for antclockwse rotaton, the drecton s up towards us. Varous ades-memore have been presented for ths my favourte s to consder a screw. When turned clockwse t moves away from you: when turned antclockwse t moves towards you. For ths reason the conventon s sometmes called the rght hand screw rule. r come to that later. r sn Wth ths conventon establshed, we can now use vectors for angular velocty, angular momentum L, and torque C. Knetc energy, lke n lnear moton, s a scalar and therefore needs no further attenton. The moment of nerta I s more complex, and we shall Let us consder the angular velocty frst. If we already know and r, what s v, assumng that only rotatonal veloctes are allowed? Rememberng that w must pont along the axs of the rotaton, we may draw the dagram above, whch shows that the radus of the crcle that Page 43

44 Correctons March 7 our partcle actually traces out s r sn where s the angle between r and. Ths factor of sn dd not arse before n ths way, snce our moton was restrcted to the plane whch contaned the centre pont, and thus = / for all our -dmensonal work. Therefore the velocty s equal to w multpled by the radus of the crcle traced out,.e. v = r sn. Ths may be put on a sold mathematcal foundaton usng the vector cross product namely v = r. Ths s our frst vector dentty for rotatonal moton. By a smlar method, we may analyse the acceleraton. We come to the correspondng concluson a = r. Next we tackle torque. Notng our drecton conventon, and our earler equaton C = F r sn, we set C = r F. Smlarly, from L = (mv) r sn = p r sn, we set L = r p. Wth these three vector equatons we may get to work. Frstly, notce: d dt L d dt d dt r p v p r p r mv r ma r F C The tme dervatve of angular momentum s the torque, as before. Notce too that the (v p) term dsappears snce p has the same drecton as v, and the vector cross product of two parallel vectors s zero General Moment of Inerta Our next task s to work out the moment of nerta. Ths can be more complex, snce t s not a vector. Prevously we defned I by the relatonshps C = I, and also used the expresson L = I. Now that C,, L and are vectors, we conclude that I must be a matrx, snce a vector s made when I s multpled by the vectors or. Our am s to fnd the matrx that does the job. For ths, we use our vector equatons v = r and C = r F, we let the components of r be (x,y,z), and we also use the mathematcal result that for any three vectors A, B and C, A B C A C B A B. d dt C We use to represent the angle between r and, to dstngush t from the angle between r and v, whch s of course a rght angle for a strct rotaton. Ths ntentonally does not nclude the centrpetal acceleraton, as before. If you am to calculate ths a from the former equaton v = r, then you get a = dv/dt = d( r)/dt = r + v = r + ( r) = r + (r.) r. The fnal two terms n ths equaton deal wth the centrpetal acceleraton. However n real stuatons, the centrpetal force s usually provded by nternal or reacton forces, so often problems are smplfed by not ncludng t. Page 44

45 Correctons March 7 C r F m m r mr α α m r x m xy xz y z m xy xz r r r a α r r x xy y yz xy z yz r x xz yz z xz yz y α α Ths result looks horrble. However let us smplfy matters by algnng our axes so that the z axs s the axs of the acceleraton. In other words = (,,). We now have xz C m yz x y whch s a lttle better. Notce that t s stll pretty nasty n that the torque requred to cause ths z-rotaton acceleraton s not necessarly n the z- drecton! Another consequence of ths s that the angular momentum L s not necessarly parallel to the angular velocty. However for many objects, we rotate them about an axs of symmetry. In ths case the xz and yz terms become zero when summed for all the masses n the object, and what we are left wth s the mass multpled by the dstance from the axs to the masses (that s x + y ). Alternatvely, for a flat object (called a lamna) whch has no thckness n the z drecton, the xz and yz terms are zero anyway, because z=. At ths pont, you are perfectly justfed n sayng yuk and stckng to twodmensonal problems. However ths result we have just looked at has nterestng consequences. When a 3-d object has lttle symmetry, t can roll around n some very odd ways. Some of the asterods and planetary moons n our Solar System are cases n pont. The moment of nerta can also be obtaned from the rotatonal momentum, however, the form s dentcal to that worked out above from Newton s second law, as shown here. L r p m m r mr ω ω r r mr α v ωr The calculaton then proceeds as before. Page 45

46 Correctons March General Knetc Energy Our fnal detal s knetc energy. Ths can be calculated usng v = r, A B C B C Α. and the vector rule that K mv m v m ω ω ω r r ω L ω r m p v r v v m v ω Iω For the cases where I can be smplfed, ths reduces to the famlar form K = I /. 3. Moton n Polar Co-ordnates When a system s rotatng, t often makes sense to use polar coordnates. In other words, we characterse poston by ts dstance from the centre of rotaton (.e. the radus r) and by the angle t has turned through. Converson between these co-ordnates and our usual Cartesan (x,y) form are gven by smple trgonometry: x r cos y r sn () y θˆ r rˆ x When analysng moton problems, though, there are complcatons f polar co-ordnates are used. These stem from the fact that the ncreasng r and ncreasng drectons themselves depend on the value of, as we shall see. Let us start by defnng the vector r to be the poston of a partcle relatve to some convenent orgn. The length of ths vector r gves the dstance from partcle to orgn. We defne rˆ to be a unt vector parallel to r. Smlarly, we defne the vector θ ˆ to be a unt vector pontng n the drecton the partcle would have to go n order to ncrease whle keepng r constant. Let us now evaluate the tme dervatve of r n other words, let s fnd the velocty of the partcle: Page 46

47 d dt Correctons March 7 d r rˆ dr d rˆ d rˆ r rˆ r r rˆ r, () dt dt dt dt where we have used the dot above a letter to mean tme dervatve of. Now f the partcle does not change ts, then the drecton rˆ wll not change ether, and we have a velocty gven smply by rrˆ. We next consder the case when r doesn t change, and the partcle goes n a crcle around the orgn. In ths case, our formula would say that the d rˆ velocty was r. We know from secton 3. that n ths case, the dt speed s gven by r, that s r, so the velocty wll be r θˆ. In order to make ths agree wth our equaton for dr/dt, we would need to say that d dt rˆ θˆ. (3) Does ths make sense? If you thnk about t for a moment, you should fnd that t does. Look at the dagram below. Here the angle has changed a small amount. The old and new r ˆ vectors are shown, and form two sdes of an sosceles trangle, the angle between them beng. Gven that the sdes r ˆ have length, the length of the thrd sde s gong to be approxmately equal to (wth the approxmaton gettng better the smaller s). Notce also that the thrd sde the vector correspondng to rˆ new rˆ s pontng n the drecton of θ ˆ old. Ths allows us to justfy statement (). r θˆ rˆnew rˆold In a smlar way, we may show that d dt θˆ Rememberng that our velocty s gven by v r r rˆ r θ, ˆ we may calculate the acceleraton as rˆ. (4) Page 47

48 Correctons March 7 rˆ θˆ a v rˆ θˆ θˆ d d r r r r r dt dt rˆ θˆ θˆ θˆ r r r r r rˆ. (5) r r rˆ r r θˆ Now suppose that a force actng on the partcle (wth mass m), had a radal component F r, and a tangental component F. We could then wrte F r m r r F m r r. (6) There are many consequences of these equatons for rotatonal moton. Here are three:. For an object to go round n a crcle (that s r stayng constant, so that r r ), we requre a non-zero radal force F r mr. The mnus sgn ndcates that the force s to be n the opposte drecton to r, n other words pontng towards the centre. Ths, of course, s the centrpetal force needed to keep an object gong around n a crcle at constant speed.. If the force s purely radal (we call ths a central force), lke gravtatonal attracton, then F =. It follows that mr mr mr mrr, (7) d dt mr and accordngly the angular momentum mr mr does not change. Ths ought to be no surprse, snce we found n secton 3.6 that angular momenta are only changed f there s a torque, and a radal force has zero torque. 3. One consequence of the conservaton of angular momentum s the apparently odd behavour of an object comng oblquely towards the centre (that s, t gets closer to the orgn, but s not amed to ht t). Snce r decreases, must ncrease, and ths s what happens n fact the square term causes to quadruple when r halves. We can analyse ths n terms of forces usng (6): F r m r when F =. Snce r s decreasng, whle ncreases, the non-zero value of the mr term gves rse to a non-zero, and hence an acceleraton of rotaton. If you were sttng next to the partcle at the Page 48

49 Correctons March 7 tme, you would wonder what caused t to speed up, and you would thnk that there must have been a force actng upon t. Ths s another example of a fcttous force (see secton..3), and s called the Corols force. It s used, among other thngs, to explan why the ar rushng n to fll a low pressure area of the atmosphere begns to rotate thus settng up a cyclone. Some people have attempted to use the equaton to explan the drecton of rotaton of the whrlpool you get above the plughole n a bath. Put very bluntly the Corols force s the force needed to keep the object gong n a true straght lne. Of course, a statonary observer would see no force after all thngs go n straght lnes when there are no sdeways forces actng on them. The perspectve of a rotatng observer s not as clear and ths Corols force wll be felt to be as real as the centrfugal force dscussed n secton Moton of a rgd body When you are dealng wth a rgd body, thngs are smplfed n that t can only do two thngs move n a lne and rotate. If forces F are appled to postons r on a sold object free to move, ts moton s completely descrbed by a lnear acceleraton gven by mass of the body, and a F M, where M s the total a rotatonal acceleraton gven by α r F I about a pont called the centre of mass, where r s the poston of pont relatve to the centre of mass and I s the moment of nerta of the object about the axs of rotaton. 3 Ths means, among other thngs, that the centre of mass tself moves as f t were a pont partcle of mass M. In turn, f a force s appled to the object at the centre of mass, t wll cause the body to move wth a lnear acceleraton, wthout any rotatonal acceleraton at all. The proof goes as follows. Suppose the object s made up of lots of ponts r (of mass m ) fxed together. It follows that Newton s second law states (as n secton...) 3 Ths assumes that the angular acceleraton s a smple speedng up or slowng down of an exstng rotaton. If and are not parallel, the stuaton s more complex. Page 49

50 Correctons March 7 d d( m u ) d dt dt ( m r ) dt m r F F F Now suppose we defne the poston R such that MR = m r, then t follows that M d R F dt total and the pont R moves as f t were a sngle pont of mass M beng acted on by the total force. Ths poston R s called the centre of mass. Gven that we already know that R does not have any rotatonal moton, ths must be the centre of rotaton, and we can use the equaton from secton 3. to show that the rate of change of angular momentum of the object about ths pont, d(i)/dt, s equal to the total torque (r R) F actng on the body about the pont R. Gven that the masses don t change, we may wrte d dt m r m r m u m a a a F fj j F fj j r F r r F j j f j r f j. The fnal term sums to zero snce f j +f j =, and the nternal forces between two partcles must ether consttute a repulson, an attracton or the two forces must occur at the same place. In any of these cases f j (r rj) =. If we now express the postons r n terms of the centre of mass poston R and a relatve poston r, where r = R + r (so a = A+a ), then Page 5

51 Correctons March 7 m m m m m m m F r a r F r F R a r a R F r F R a r A r a R F r R a A r R snce m r = m (r R) = MR MR =. Now, as shown earler, m m m dt m d a r a r u u u r, and so C F r ω L F r u r I dt d dt d dt d and so the rate of change of angular momentum about the centre of mass s gven by the total moment of the external forces about the centre of mass. 3.3 Questons. A car has wheels wth radus 3cm. The car travels 4km. By what angle have the wheels rotated durng the journey? Make sure that you gve your answer n radans and n degrees.. Why does the gravtatonal attracton to the Sun not change the angular momentum of the Earth? 3. Calculate the speed of a satellte orbtng the Earth at a dstance of 4 km from the Earth s centre. 4. A space agency plans to buld a spacecraft n the form of a cylnder 5m n radus. The cylnder wll be spun so that astronauts nsde can walk on the nsde of the curved surface as f n a gravtatonal feld of 9.8 N/kg. Calculate the angular velocty needed to acheve ths. 5. A televson company wants to put a satellte nto a 4 km radus orbt round the Earth. The satellte s launched nto a crcular low-earth orbt km above the Earth s surface, and a rocket motor then speeds t up. It then coasts untl t s n the 4km orbt wth the correct speed. How fast does t need to be gong n the low-earth orbt n order to coast up to the correct poston and speed? 6. Estmate the gan n angular velocty when an ce-skater draws her hands n towards her body. Page 5

52 Correctons March 7 7. One theory of planet formaton says that the Earth was once a lqud globule whch gradually soldfed, and ts rotaton as a lqud caused t to bulge outwards n the mddle a stuaton whch remans to ths day: the equatoral radus of the Earth s about km larger than the polar radus. If the theory were correct, what would the rotaton rate of the Earth have been just before the crust soldfed? Assume that the lqud globule was suffcently vscous that t was all rotatng at the same angular velocty. Page 5

53 Correctons March 7 4 Vbes, Wggles & Lght 4. Oscllaton Any system n stable equlbrum can be persuaded to oscllate. If t s removed from the equlbrum, there wll be a force (or other nfluence) that attempts to mantan the status quo. The sze of the force wll depend on the amount of the dsturbance. Suppose that the dsturbance s called x. The restorng force can be wrtten 3 F Ax Bx Cx, () where the mnus sgn ndcates that the force acts n the opposte drecton to the dsturbance. If x s small enough, x and x 3 wll be so small that they can be neglected. We then have a restorng force proportonal to the dsplacement x. Just because the system has a force actng to restore the equlbrum, ths does not mean that t wll return to x= mmedately. All systems have some nerta, or reluctance to act quckly. For a lteral force, ths nerta s the mass of the system and we know that the acceleraton caused by a force (F) s gven by F/m, where m s the mass. We can therefore work on equaton () to fnd out more: Ths dfferental equaton has the soluton: d x F m Ax dt. () d x A x dt m x x cos t, (3) A m whch s ndeed an oscllaton. We are usng x to denote the ampltude. Notce that, as we are workng n radans, the cosne functon needs to advance to go through a whole cycle. Therefore we can work out the tme perod (T) and frequency (f): T T. (4) f T Page 53

54 Correctons March 7 Seeng that =f, we notce that s none other than the angular frequency of the oscllaton, as defned n chapter 3. These equatons are perfectly general, and so whenever you come across a system wth a dfferental equaton lke (), you know the system wll oscllate, and furthermore you can calculate the frequency. 4.. Non-lnearty Equaton () has left an unanswered queston. What happens f x s bg enough that x and x 3 can t be neglected? Clearly soluton (3) wll no longer work. In fact the equaton probably won t have a smple soluton, and the system wll start dong some really outrageous thngs. Gven that t has quadratc terms n t, we say t s non-lnear; and a non-lnear equaton wll send most physcs students runnng away, screamng for mercy. Let me gve you an example. There are very nce materals that look harmlessly transparent. However they are desgned so that the nonlnear terms are very mportant when lght passes through them. The result you put red laser lght n, and t comes out blue (at twce the frequency). They are called doublng crystals and are the sort of thng that mght freak out an unsuspectng GCSE examner. Our world would be much less wonderful f t were purely lnear no swrls n smoke, no wave-breakng (and hence surfng), and extremely borng weather not to menton rgd populaton dynamcs. Whle the non-lnear terms add to the spce of lfe, I for one am grateful that many phenomena can be well descrbed usng lnear equatons. Otherwse physcs would be much more frustratng, and brdge desgn would be just as hard as predctng the weather. 4.. Energy Before we move from oscllatons to waves, let us make one further observaton. The energy nvolved n the oscllaton s proportonal to the square of the ampltude. We shall show ths n two ways. Frst: If the dsplacement s gven by equaton (3), we notce that the velocty s gven by u x x sn t. (5) At the moment when the system passes through ts equlbrum (x=) pont, all the energy s n knetc form. Therefore the total energy s x x mu m E K (6) x whch s ndeed proportonal to the ampltude squared. Page 54

55 Correctons March 7 Second: When the dsplacement s at ts maxmum, there s no knetc energy. The energy wll all be n potental form. We can work out the potental energy n the system at dsplacement x, by evaluatng the work done to get t there: E pot Fdx Axdx Ax. (7) Notce that we dd not nclude the mnus sgn on the force. Ths s because when we work out the work done the force nvolved s the force of us pullng the system. Ths s equal and opposte to the restorng force of the system, and as such s postve (drected n the same drecton as x). The total energy s gven by the potental energy at the moment when x has ts maxmum (.e. x=x ). Therefore x x Ax E E pot. (8) Equatons (8) and (6) are n agreement. Ths can be shown by nsertng the value of from equaton (3) nto (6). Whle we have only demonstrated that energy s proportonal to ampltude squared for an oscllaton, t turns out that the same s true for lnked oscllators and hence for waves. The ntensty of a wave (joules of energy transmtted per second) s proportonal to the ampltude squared n exactly the same way. Intensty of a wave s also related to another wave property ts speed. The ntensty s equal to the amount of energy stored on a length u of wave, where u s the speed. Ths s because ths s the energy that wll pass a pont n one second (a length u of wave wll pass n ths tme). 4. Waves & Interference The most wonderful property of waves s that they can nterfere. You can add three and four and get sx, or one, or 4.567, dependng on the phase relatonshp between the two waves. You can vsualze ths usng ether trgonometry or vectors (phasors). However, before we look at nterference n detal, we analyse a general wave. 4.. Wave number Frstly, we defne a useful parameter called the wave number. Ths s usually gven the letter k, and s defned as k, (3) Page 55

56 Correctons March 7 where s the wavelength. If we wrte the shape of a paused wave as y=a cos(), the phase of a wave s gven by kd. (4) We can see that ths makes sense by combnng equatons (3) and (4): D kd. (5) If the dstance D s equal to a whole wavelength, we expect the wave to be dong the same thng as t was at =. And snce cos()=cos(), ths s ndeed the case. A varaton on the theme s possble. You may also see wave vectors k: these have magntude as defned n (3), and pont n the drecton of energy transfer. 4.. Wave equatons We are now n a poston to wrte a general equaton for the moton of a wave wth angular frequency and wave number k: y Acos t kx. We can check that ths s correct, snce f we look at a partcular pont (value of x), and watch as tme passes, we wll pass from one peak to the next when t =f t has got bgger by (.e. t=/f as t should). f we look at a partcular moment n tme (value of t), and look at the poston of adjacent peaks, they should be separated by one wavelength = /k. Now for adjacent peaks, the values of kx wll dffer by accordng to the formula above, and so ths s correct. f we follow a partcular peak on the wave say the one where t kx+f=, we notce that x=(t+f)/k = t/k + constant, and hence the poston moves to ncreasng x at a speed equal to /k, as ndeed t should snce /k = f/(/) = f = v. It follows that a leftwards-travellng wave has a functon whch looks lke y Acos t kx. Page 56

57 Correctons March Standng waves Imagne we have two waves of equal ampltude passng along a strng n the two dfferent drectons. The total effect of both waves s gven by addng them up: y Acos cos t kx Acost kx t coskx At any tme, the peaks and troughs wll only occur at the places where the second cosne s + or, and so the postons of the peaks and troughs do not change. Ths s why ths knd of stuaton s called a standng wave. Whle there s moton, descrbed by the frst cosne, the postons of constructve nterference between the two counterpropagatng waves reman fxed (these are called antnodes), as to the postons of destructve nterference (the nodes). Whle there are many stuatons whch nvolve counter-propagatng waves, ths usually s caused by the reflecton of waves at boundares (lke the ends of a gutar strng). Accordngly, there s nothng keepng the phase constants and the same, and so the standng wave doesn t develop. However f the frequency s just rght, then t works, as ndcated n secton Trgonometrc Interference We are now n a poston to look at the fundamental property of waves namely nterference. Our frst method of analyss uses trgonometry. Suppose two waves arrve at the same pont, and are descrbed by x Acost and x B cost respectvely. To fnd out the resultng sum, we add the two dsturbances together. X x Acost B cos t x Acost B cost cos Bsnt sn A B cos cost Bsn snt X cos cost sn snt X cos t (9) where we defne X cos A B cos B sn A B A B cos X AB cos sn B sn X. () Page 57

58 Correctons March 7 The ampltude of the resultant s gven by X. Notce that f A=B, the expresson smplfes: X A cos A A A cos cos, () 4cos and we obtan the famlar result that f the waves are n phase (=), the ampltude doubles, and f the waves are radans (half a cycle) out of phase, we have complete destructve nterference. Equaton () can be used to provde a more general form of ths statement the mnmum resultant ampltude possble s A-B, whle the maxmum ampltude possble s A+B. Ths statement s remnscent of the trangle nequalty, where the length of one sde of a trangle s lmted by a smlar constrant on the lengths of the other two sdes. Ths brngs us to our second method of workng out nterferences: by a graphcal method Graphc Interference In the graphc method a vector represents each wave. The length of the vector gves the ampltude, and the relatve orentaton of two vectors ndcates ther phase relatonshp. If the phase relatonshp s zero, the two vectors are parallel, and the total length s equal to the sum of the ndvdual lengths. If the two waves are out of phase, the vectors wll be antparallel, and so wll partly (or f A=B, completely) cancel each other out. The dagram below shows the addton of two waves, as n the stuaton above. Notce that snce + =, cos = cos. One applcaton of the cosne rule gves n agreement wth equaton (). X A B AB cos () X B A Page 58

59 Correctons March Summary of Interference Prncples The results of the last secton allow us to determne the ampltude once we know the phase dfference between the two waves. Usually the two waves have come from a common source, but have travelled dfferent dstances to reach the pont. Let us suppose that the dfference n dstances s D ths s sometmes called the path dfference. What wll be? To fnd out, we use the wave number k. The phase dfference s gven by D kd. (4) If the dstance D s equal to a whole wavelength, we expect the two waves to nterfere constructvely, snce peak wll meet peak, and trough wll meet trough. In equaton (5), f D=, then =, and constructve nterference s ndeed obtaned, as can be seen from equaton (). Smlarly, we fnd that f D s equal to /, then =, and equaton () gves destructve nterference Instances of two-wave nterference Young s Two Slt Experment Two cases need to be dealt wth. The frst s known as the two-slt experment, and concerns two sources n phase, whch are a dstance d apart, as shown n the dagram below. The path dfference s gven by D d sn, n the case that d s much smaller than the dstance from sources to observer. Usng the condtons n the last secton, we see that nterference wll be constructve f D d sn n where n s an nteger. To obser vaton pont d D = d sn Thn flms and colours on soap bubbles The second case s known as thn flm nterference, and concerns the stuaton n the dagram below. Here the lght can take one of two routes. Page 59

60 Correctons March 7 B t A C t E The path dfference s calculated: D AC CB AC CE t cos Before we can work out the condtons requred for constructve or destructve nterference, there s an extra cauton to be borne n mnd the reflectons Hard & Soft Reflectons The reflecton of a wave from a surface (or more accurately, the boundary between two materals) can be hard or soft. Hard reflectons occur when, at the boundary, the wave passes nto a sterner materal. At these reflectons, a peak (before the reflecton) becomes a trough (afterwards) and vce-versa. Ths s usually stated as a phase dfference s added to the wave by the cos cos. reflecton. These mean the same thng snce To vsualze ths magne that you are holdng one end of a rope, and a frend sends a wave down the rope towards you. You keep your hand stll. At your hand, the ncomng and outgong waves nterfere, but must sum to zero (after all, your hand s not movng, so nether can the end of the rope). Therefore f the ncomng wave s above the rope, the outgong wave must be below. In ths way, peak becomes trough and vce-versa. Soft reflectons, on the other hand, are where the boundary s from the sterner materal. At these reflectons, a peak remans a peak, and there s no phase dfference to be added. Page 6

61 Correctons March 7 What do we mean by sterner? Techncally, ths s a measurement of the restorng forces n the oscllatons whch lnk to produce the wave the A coeffcents of (). However, the followng table wll help you to get a feel for sternness. Wave From To Reflecton Lght Reflecton off mrror Hard Lght Ar Water / Glass Hard Lght Water / Glass Ar Soft Lght Lower refractve ndex Hgher refractve ndex Hard Sound Sold / lqud Ar Soft Sound Ar Sold / lqud Hard Wave on strng Reflecton off fastened end of strng Hard Wave on strng Reflecton off unsecured end of strng Soft Flm Interference Revsted Gong back to our thn flm nterference: sometmes both reflectons wll be hard; sometmes one wll be hard, and the other soft. The formulae for constructve nterference are: Both reflectons hard, or both soft: D t cos n (6) One reflecton hard, one soft: D t cos n (7) The dfference comes about because of the phase change on reflecton at a hard boundary Standng Waves Equatons (6) and (7) wth = can be used to work out the wavelengths allowed for standng waves. For a standng wave, we must have constructve nterference between a wave and tself (havng bounced once back and forth along the length of the devce). The condtons for constructve nterference n a ppe, or on a strng of length L (round trp total path = L) are Page 6

62 Correctons March 7 L n soft reflectons at both ends L n soft reflecton at one end L n hard reflectons at both ends Two waves, dfferent frequences All the nstances gven so far have nvolved two waves of dentcal frequences (and hence constant phase dfference). What f the frequences are dfferent? Let us suppose that our two waves are descrbed by x Acost and x B cos t, where we shall wrte. When we add them, we get: X x Acost B cost cost Bsnt snt, (8) A B cost cos t Bsnt sn t X x cos t where X cos A B cost B snt A B A B cost X AB cost. (9) We see that the effectve ampltude fluctuates, wth angular frequency. On the other hand, f the two orgnal waves had very dfferent frequences, then ths fluctuaton may be too quck to be pcked up by the detector. In ths case, the resultant ampltude s the root of the sum of the squares of the orgnal ampltudes. Put more brefly f the frequences are very dfferent, the total ntensty s smply gven by the sum of the two consttuent ntenstes. These fluctuatons are known as beats, and the dfference f -f s known as the beat frequency. To gve an llustraton: Whle tunng a voln, f the tunng s slghtly off-key, you wll hear the note pulse: loud-soft-loud-soft and so on. As you get closer to the correct note, the pulsng slows down untl, when the nstrument s n tune, no pulsng s heard at all because f -f = Addng more than two waves Dffracton Gratng The frst case we come to wth more than two waves s the dffracton gratng. Ths s a plate wth many narrow transparent regons. The lght can only get through these regons. If the dstance between adjacent slts s d, we obtan constructve nterference, as n secton 4..4., Page 6

63 Correctons March 7 when d sn n - n other words when the lght from all slts s n phase. The dfference between ths arrangement and the double-slt s that when d sn n we fnd that nterference s more or less destructve. Therefore a gven colour (or wavelength) only gets sent n partcular drectons. We can use the devce for splttng lght nto ts consttuent colours Bragg Reflecton A varaton on the theme of the dffracton gratng allows us to measure the sze of the atom. d The dagram shows a secton of a crystal. Lght (n ths case, X-rays) s bouncng off the layers of atoms. There are certan specal angles for whch all the reflectons are n phase, and nterfere constructvely. Lookng at the small trangle n the dagram, we see that the extra path travelled by the wave bouncng off the second layer of atoms s D d sn. () When D=n, we have constructve nterference, and a strong reflecton. There s one thng that takes great care notce the defnton of q n the dagram. It s not the angle of ncdence, nor s t the angle by whch the ray s deflected t s the angle between surface and ray. Ths s equal to half the angle of deflecton, also equal to /. Usng ths method, the spacng of atomc layers can be calculated and ths s the best measurement we have for the sze of the atom n a crystal Dffracton What happens when we add a lot of waves together? There s one case we need to watch out for when all possble phases are represented wth equal strength. In ths case, for each wave cos t, there wll be an equally strong wave cos t cost, whch wll cancel t out. Page 63

64 Correctons March 7 How does ths happen n practce? Look at the dagram below. Compared wth the wave from the top of the gap, the path dfferences of the waves comng from the other parts of the gap go from zero to Dmax W sn, where W s the wdth of the gap. To obser vaton pont W Dmax = W sn If kd max s a multple of, then we wll have all possble phases represented wth equal strength, and overall destructve nterference wll result. To summarze, destructve nterference s seen for angles, where W sn n. () Make sure you remember that W s the wdth of the gap, and that ths formula s for destructve nterference. Ths formula s only vald (as n the dagram) when the observer s so far away that the two rays drawn are effectvely parallel. Alternatvely the formula works perfectly when t s appled to an optcal system that s focused correctly, for then the mage s at nfnty Resoluton of two objects How far away do you have to get from your best frend before they look lke Cyclops? No offence but how far away do you have to be before you can t tell that they ve got two eyes rather than one? The results of dffracton can help us work ths out. Let s call ths crtcal dstance L. The rays from both eyes come nto your eyeball. Let us suppose that the angle between these rays s, where s small, and that your frend s eyes are a dstance s apart. Therefore tan sn s L. These two rays enter your eye, and spread out (dffract) as a result of passng through the gap called your pupl. They can only just be resolved that s notced as separate when the frst mnmum of one s dffracton pattern lnes up wth the maxmum of the other. Therefore W sn where W s the wdth of your pupl. Page 64

65 Correctons March 7 Pupl of your eye Puttng the two formulae together gves: s sn L W. () sw L For normal lght (average wavelength about 5nm), a 5mm pupl, and a cm dstance between the eyes: you frend looks lke Cyclops f you are more than km away! If you used a telescope nstead, and the telescope had a dameter of cm, then your frend s two eyes can be dstngushed at a dstances up to km The Bandwdth Theorem In the last secton, we asked the queston, What happens when you add lots of waves together? However we cheated n that we only consdered waves of the same frequency. What happens f the waves have dfferent frequences? Suppose that we have a large number of waves, wth frequences evenly spread between f and f+f. The angular frequences wll be spread from to +, where =f as n equaton (4). Furthermore, magne that we set them up so that they all agree n phase at tme t=. They wll never agree agan, because they all have dfferent frequences. The phases of the waves at some later tme t wll range from t to ()t. Intally we have complete constructve nterference. After a short tme t, however, we have destructve nterference. Ths wll happen when (as stated n the last secton) all phases are equally represented when the range of phases s a whole multple of. Ths happens when t =. After ths, the sgnal wll stay small, wth occasonal complete destructve nterference. From ths you can reason (f you re magnatve or trustng) that f you need to gve a tme sgnal, whch has a duraton smaller than t, you must use a collecton of frequences at least =/t. Ths s called the bandwdth theorem. Ths can be stated a lttle dfferently: Page 65

66 Correctons March 7 t f t. (3) f t A smlar relatonshp between wavelength and length can be obtaned, f we allow the wave to have speed c: x f c f x. (4) c x Expressed n terms of the wave number k, ths becomes: k x. (5) k x In other words, f you want a wave to have a pulse of length x at most, you must have a range of k values of at least /x Resoluton of spectra A spectrometer s a devce that measures wavelengths. Equaton (5) can be used to work out the accuracy (or resoluton) of the measurement. If you want a mnmum error k n the wavenumber, you must have a dstance of at least x=/k. But what does ths dstance mean? It transpres that ths s the maxmum path dfference between two rays n gong through the devce and as such s proportonal to the sze of the spectrometer. So, the bgger the spectrometer, the better ts measurements are Doppler Effect Classcal Doppler Effect Suppose a bassoonst s playng a beautful pure note wth frequency f. Now magne that he s practsng whle drvng along a road. A fellow motorst hears the lugubrous sound. What frequency does the lstener hear? Let us suppose that the player s movng at velocty u, and the lstener s movng at velocty v. For smplcty we only consder the problem n one dmenson, however veloctes can stll be postve or negatve. Page 66

67 Correctons March 7 Furthermore, magne that the dstance between player and lstener s L at tme zero, when the frst wave-peak s broadcast from the bassoon. We assume that the waves travel at speed c wth respect to the ground. Ths peak s receved at tme t, where L L vt ct c vt (6) The frst lne s constructed lke ths: The travellers start a dstance L apart, so by the tme the sgnal s receved, the dstance between them s L + vt. Ths dstance s covered by waves of speed c n tme t hence the rght hand sde. The next wave peak wll be broadcast at tme /f one wave cycle later. At ths tme, the dstance between the two muscans wll be L v ut L v u/ f. Ths second peak wll be receved at tme t, where L v u vt f f c u L f ct c vt f. (7) Fnally we can work out the tme nterval elapsed between our lstener hearng the two peaks, and from ths the apparent frequency s easy to determne. t t c v f t t L c v f f c u c u L f c u f c v (8) From ths we see that f v=u, no change s observed. If the two are approachng, the apparent frequency s hgh (blue-shfted). If the two are recedng, the apparent frequency s low (red-shfted) Relatvstc Doppler Effect Please note that f ether u or v are apprecable fractons of the speed of lght, ths formula wll gve errors, and the relatvstc calculaton must be used. For lght only, the relatvstc formula s Page 67

68 Correctons March 7 c u f f, (9) c u where u s the approach velocty as measured by the observer (u s negatve f the source and observer are recedng). The relatvstc form for other waves s more complcated, and wll be left for another day. 4.3 Rays 4.3. Reflecton and Refracton All the dscusson so far has centred on the oscllatory nature of waves. We can predct some of the thngs waves do wthout worryng about the oscllatons lke reflecton and refracton. The dagram below shows both. We refer to a refractve ndex of a materal, whch s defned as Speed of lght n vacuum Refractve Index ( n ). (3) Speed of lght n the materal Ar has a refractve ndex of about.3 4, glass has a refractve ndex of about.5, and water about.3. n n r Frst of all, the angle of reflecton s equal to the angle of ncdence (both were labelled n the dagram). Secondly, the angle of ncdence s related to the angle of refracton r by the formula: sn sn r n n (3) 4 The refractve ndex also gves a measure of pressure, snce n- s proportonal to pressure. Page 68

69 Correctons March 7 Notce that snce the sne of an angle can be no larger than one, f n <n, then refracton becomes mpossble f sn n n. Ths lmtng angle s called the crtcal angle. For greater angles of ncdence, the entre wave s reflected, and ths s called total nternal reflecton. When a wave passes from one materal nto another, the frequency remans the same (subject to the lnearty provsos of secton 4..). Gven that the speed changes, the wavelength wll change too. The wavelength of lght n a partcular materal can be evaluated: c c (3) f nf n where c (and ) represent speed of lght (and wavelength) n vacuum. 4.4 Fermat s Prncple Fermat s prncple gves us a method of workng out the route lght wll take n an optcal system. It states that lght wll take the route that takes the least tme. Gven that the tme taken n a sngle materal s equal to T D nd (33) c c where c s the speed of lght n vacuum; mnmzng the tme s the same as mnmzng the product of dstance and refractve ndex. Ths latter quantty (nd) s called the optcal path. It s possble to prove the laws of reflecton and refracton usng ths prncple Questons. A hole s drlled through the Earth from the U.K. to the centre of the Earth and out of the other sde. All the materal s sucked out of t, and a kg mass s dropped n at the Brtsh end. How much tme passes before t momentarly comes to rest at the Australan end? (NB You may need some hnts from secton..4) +. Repeat q where a straght hole s drlled between any two places on Earth. Assume that the contact of the mass wth the sdes of the hole s frctonless To do ths, magne the plane as a sheet of graph paper, wth the boundary along the x-axs. Suppose that the lght starts at pont (,Y), and needs to get to (X,-Y). Now assume that the lght crosses the x-axs at pont (x,). Work out the total optcal path travelled along the route, and then mnmze t wth respect to x. You should then be able to dentfy sn and sn r n the algebrac soup, and from ths, you should be able to fnsh the proof. Page 69

70 Correctons March 7 3. Show that Fermat s prncple allows you to derve the Law of Reflecton. Assume that you have a mrror along the x-axs. Let lght start at pont (X,Y ) and end at pont (X,Y ). Show that the least-tme reflected route s the one whch bounces off the mrror where angles of ncdence and reflecton wll be equal Show that Fermat s prncple allows you to derve Snell s Law. Assume that you have a materal wth refractve ndex for y> (that s, above the x-axs), and refractve ndex n where y<. Show that the shortest tme route from pont (X,Y ) to (X,Y ), where Y > and Y <, crosses the boundary at the pont where sn / sn r = n You are the navgator for a hkng expedton n rough ground. Your company s very thrsty and tred, and your supples have run out. There s a rver runnng East-West whch s 4km South of your current poston. Your objectve s to reach the base camp (whch s km South and 6km West of your current poston), stoppng off at the rver on the way. What s the quckest route to the camp va the rver? + 6. You are the offcer n charge of a food convoy attemptng to reach a remote vllage n a famne-strcken country. On your map, you see that 5km to your East s a straght border (runnng North-South) between scrub land (over whch you can travel at 5km/hr) and marsh (over whch you can only travel at 5km/hr). The vllage s 4km South-East of your current poston. What s the fastest route to reach the vllage? + 7. In an nterferometer, a beam of coherent monochromatc lght (wth wavelength ) s splt nto two parts. Both parts travel for a dstance L parallel to each other. One travels n vacuum, the other n ar. The beam s then re-combned. If destructve nterference results, what can you say about L, and n ar? 8. Your wnd band s about to play on a pck-up truck gong down a motorway at 3m/s. You want people on the brdges overhead to hear you playng n tune (such that treble A s 44Hz) when you are comng drectly towards them. What frequency should you tune your nstruments to? 9. A polce speed gun uses mcrowaves wth a wavelength of about 3cm. The gun conssts of a transmtter and recever, wth a small mrror whch sends part of the transmsson drectly nto the recever. Here t nterferes wth the man beam whch has reflected off a vehcle. The receved sgnal strength pulsates (or beats). What wll be the frequency of ths pulsaton f the vehcle s travellng towards you at 3mph? +. How far away do you need to hold a ruler from your left eye before you can no longer resolve the mllmetre markngs? Keep your rght eye covered up durng ths experment. Use equaton () to make an Page 7

71 Correctons March 7 estmate for the wavelength of lght based on your measurement. Remember that W s the wdth of your pupl.. A sgnal from a dstant galaxy has one thrd of the frequency you would expect from a statonary galaxy. Calculate the galaxy s recesson velocty usng equaton (9), and comment on your answer. (NB redshfts ths bg are measured wth very dstant astronomcal objects.) Page 7

72 Correctons March 7 5 Hot Physcs Ths secton gves an ntroducton to the areas of physcs known as thermodynamcs and statstcal mechancs. These deal wth the questons What happens when thngs heat up or cool down? and Why? respectvely. We start wth a statement that wll be very famlar but then fnd that t leads us nto new terrtory when explored further. 5. The Conservaton of Energy You wll be used to the dea that energy can nether be used up nor created only transferred from one object to another, perhaps n dfferent forms. For our purposes, ths s stated mathematcally as dq + dw = du, () where dx refers to a small change n X. Put nto words, ths states: Heat enterng object + Work done on the object = the change n ts nternal energy. Internal energy means any form of stored energy n the object. Usually ths wll mean the heat t has, and wll be measured by temperature. However f magnetc or electrc felds are nvolved, U can also refer to electrcal or magnetc potental energy. Gven that the conservaton of energy must be the startng pont for a study of heat, t s called the Frst Law of Thermodynamcs. Equaton () can be appled to any object or substance. The most straghtforward materal to thnk about s a perfect gas, and so we shall start there. It s possble to generalze our observatons to other materals afterwards. Imagne some gas n a cylnder wth a pston of cross-sectonal area A. The gas wll have a volume V, and a pressure p. Let us now do some work on the gas by pushng the pston n by a small dstance dl. The force requred to push the pston F = p A, and so the work done on the gas s dw = F dl = p A dl. Notce that AdL s also the amount by whch the volume of the gas has been decreased. If dv represents the change n volume, dv = -A dl. Therefore dw = -p dv. For a perfect gas n a cylnder (or n fact n any other stuaton), the Frst Law can be wrtten a bt dfferently as: dq = du + p dv () Page 7

73 Correctons March 7 5. The Second Law Whle the Frst Law s useful, there are certan thngs t can never tell us. For example thnk about an ce cube sttng on a dsh n an oven. We know what happens next the ce cube melts as heat flows from oven to ce, warmng t up untl t reaches meltng pont. However the Frst Law doesn t tell us that. As far as t s concerned t s just as possble for heat to flow from the ce to the oven, coolng the ce and heatng the oven. We stumbled across our next law called the second law of thermodynamcs. Ths can be stated n several ways, but we shall start wth ths: Heat wll never flow from a cold object to a hotter object by tself. Ths helps us wth the ce n the oven, but you may be wonderng what the sgnfcance of the by tself s. Actually heat can be transferred from a cold object to a hotter one that s what frdges and ar condtonng unts do. However they can only do t because they are plugged nto the electrcty supply. If you are prepared to do some work then you can get heat out of a cold object and nto a hotter one, but as soon as you turn the power off and leave t to ts own devces, the heat wll start flowng the other way agan. 5.3 Heat Engnes and Frdges Hot reservor - the atmosphere at temperature T(h) dq Frdge Work dw provded by electrc compressor dq Cold reservor - the ce box at temperature T(c) The frdge s shown dagrammatcally above. It s a devce whch uses work dw (usually provded by an electrc compressor) to extract heat dq from the ce-box (coolng t down), and pump t out nto the surroundngs (warmng them up). However, by the conservaton of energy, the amount of energy pumped out dq s bgger than the amount of energy removed from the ce-box. By conventon dq >, and dq <, snce heat flowng n s regarded as postve. The Frst Law therefore states that dq + dq + dw =. Page 73

74 Correctons March 7 The frdge s a devce that uses work to move heat from cold objects to hot. The opposte of a frdge s a heat engne. Ths allows heat to flow ts preferred way namely from hot to cold but arranges t to do some work on the way. Petrol engnes, steam engnes, turbo-generators and jet engnes are all examples of heat engnes. Hot reservor - the fre-box at temperature T(h) dq Engne dq Cold reservor - the atmosphere at temperature T(c) Work dw produced, drves electrcal generator It was Carnot who realsed that the most effcent heat engne of all was a reversble heat engne. In other words one that got the same amount of work out of the heat transfer as would be needed to operate a perfect frdge to undo ts operaton. In order to do ths, t s necessary for all the heat transfers (between one object and another) to take place wth as small a temperature dfference as possble. If ths s not done, heat wll flow from hot objects to cold a process whch could have been used to do work, but wasn t. Therefore not enough work wll be done to enable the frdge to return the heat to the hot object. Carnot therefore proposed that the rato of heat comng n from the hot object to the heat gong out nto the cold object has a maxmum for ths most-effcent engne. Ths s because the dfference between heat n and heat out s the work done, and we want to do as much work as possble. Furthermore, he sad that ths rato must be a functon of the temperatures of the hot and cold objects only. Ths can be stated as dq f ( T, T ) (3) dq where T s the temperature of the hot object, and T s that of the cold object. More lght can be shed on the problem f we stack two heat engnes n seres, wth the second takng the heat dq from the frst (at Page 74

75 Correctons March 7 temperature T ), extractng further work from t before dumpng t as heat (dq 3 ) nto a yet colder object at temperature T 3. The two heat engnes separately and together gve us the equatons: dq dq f ( T, T ) f ( T, T dq dq 3 3 ) f ( T f ( T, T f ( T, T, T 3 ) ) f ( T g( T ) ) g( T ) dq dq 3, T 3 ) f ( T, T 3 ). (4) 5.3. Thermodynamc Temperature However f g(t) s a functon of the temperature alone, we mght as well call g(t) the temperature tself. Ths s the thermodynamc defnton of temperature. To summarze: Thermodynamc temperature (T) s defned so that n a reversble heat engne, the rato of heat extracted from the hot object (Q ) to the heat ejected nto the cold object (Q ): Q T (5) Q T The kelvn temperature scale obtaned usng the gas laws satsfes ths defnton. For ths reason, the kelvn s frequently referred to as the unt of thermodynamc temperature Effcency of a Heat Engne The effcency of a reversble heat engne can then be calculated. We defne the effcency () to be the rato of the work done (the useful output) to Q (the total energy nput). Therefore dw dq. (6) dq dq dq T T Ths, beng the effcency of a reversble engne, s the maxmum effcency that can be acheved. A real engne wll fall short of ths goal. Notce that for a coal-fred power staton, n whch T (the temperature of the boler) s frequently 84K, and T (the temperature of the stream outsde) s 3K; the maxmum possble effcency s 3 64%. 84 Page 75

76 Correctons March 7 In practce the water leaves the turbo generator at 53K, and so the effcency can t go any hgher than 53 37%. 84 The desgn of modern large power statons s such that the actual effcency s remarkably close to ths value. 5.4 Entropy Now we need to take a step backwards before we can go forwards. Look back at the defnton of thermodynamc temperature n equaton (5). It can be rearranged to state dq T dq T dq T dq T REVERSIBLE. (7) Remember that ths s for the deal stuaton of a reversble process as n a perfect frdge or heat engne. Suppose, then, that we start wth some gas at pressure p and volume V. Then we do somethng wth t (squeeze t, heat t, let t expand, or anythng reversble), and fnally do some more thngs to t to brng t back to pressure p and volume V. The lst of processes can be broken up nto tny stages, each of whch saw some heat (dq) enterng or leavng the system, whch was at a partcular temperature T. The only dfference between ths stuaton, and that n (7) s that there were only two stages n the process for the smpler case. The physcs of (7) should stll apply, no matter how many processes are nvolved. Therefore provdng all the actons are reversble we can wrte completecycle dq T dq T REVERSIBLE (8) where the crcle on the ntegral mples that the fnal poston (on a p,v graph) s the same as where the gas started. Now suppose that there are two ponts on the (p,v) graph whch are of nterest to us, and we call them A and B. Let us go from A to B and then back agan (usng a dfferent route), but only usng reversble processes. We call the frst route I, and the second route II. Equaton (8) tells us Page 76

77 Correctons March 7 dq T B A B A B A dq T I dq T dq T I I B A B A A B dq T II dq dq T T II II REVERSIBLE (9) In other words the ntegral of dq/t between the two ponts A and B s the same, no matter whch reversble route s chosen. Ths s a very specal property of a functon we label dq/t as a functon of state, and call t the entropy (S). Ths means that the current entropy of the gas, lke pressure, volume and temperature, s only a functon of the state that the gas s n now and does not depend on the preparaton method. 5.5 Irreversble Processes and the Second Law We must stress that entropy s only gven by dq T when the ntegral s taken along reversble processes n whch there s no wastage of heat. Heat s wasted when t s allowed to flow from a hot object to a cold one wthout dong any work on the journey. Ths would be rreversble, snce you could only get the heat back nto the hot object f you expended more energy on t. Let us make an analogy. Reversble processes are lke a world n whch purchasng prces and sellng prces are the same. If you started wth, and spent t n varous ways, you could sell the goods and end up wth cash at the end. Irreversble processes are lke the real world n that a trader wll want to sell you an apple for more than she bought t for. Otherwse she won t be able to make a proft. If you started wth, and spent t, you would never be able to get the back agan, snce you would lose money n each transacton. You may end up wth worth of goods, but you would have to be satsfed wth a prce lower than f you wanted to sell t all for cash. Let us now return to the physcs, and the gas n the pston. What does rreversblty mean here? We haven t lost any energy the Frst Law has ensured that. But we have lost usefulness. Equaton (8) tells us that f we come back to where we started, and only use reversble processes on the way, the total entropy change wll be zero. There s another way of lookng at ths, from the pont of vew of a heat engne. Page 77

78 Correctons March 7 Let us suppose that the temperature of the boler n a steam engne s T A. In a perfect heat engne, the cylnder wll receve the steam at ths temperature. Suppose Q joules of heat are transferred from boler to cylnder. The boler loses entropy Q/T A, the cylnder gans entropy Q/T A, and the total entropy remans constant. Now let us look at a real engne. The boler must be hotter than the cylnder, or heat would not flow from boler to cylnder! Suppose that the boler s stll at T A, but the cylnder s at T C. We have now let rreversblty loose n the system, snce the heat Q now flows from hot to cooler wthout dong work on the way. What about the entropy? The boler now loses Q/T A to the connectng ppe 6, but the cylnder gans Q/T C from t. Snce T C <T A, the cylnder gans more entropy than the boler lost. Ths s an alternatve defnton of the Second Law. Processes go n the drecton to maxmze the total amount of entropy. 5.6 Re-statement of Frst Law For reversble processes, dw = -p dv, and dq = T ds. Therefore the Frst Law (), can be wrtten as T ds = du + p dv. (9) We fnd that ths equaton s also true for rreversble processes. Ths s because T, S, U, p and V are all functons of state, and therefore f the equaton s true for reversble processes, t s true for all processes. However care must be taken when usng t for rreversble processes, snce TdS s no longer equal to the heat flow, and pdv s no longer equal to the work done. 5.7 The Boltzmann Law The Boltzmann Law s smple to state, but profound n ts mplcatons. Probablty that a partcle has energy E kt e E () 6 What s the ppe got to do wth t? Remember that we sad that change n entropy ds s only gven by dq/t for reversble processes. The passng of Q joules of heat nto the ppe s done reversbly (at temperature T A ), so we can calculate the entropy change. Smlarly the passng of Q joules of heat from ppe to cylnder s done reversbly (at T C ), so the calculaton s smlarly vald at the other end. However somethng s gong on n the ppe whch s not reversble namely Q joules of heat passng from hgher to lower temperature. Therefore we mustn t apply any ds = dq/t arguments nsde the ppe. Page 78

79 Correctons March 7 where k s the Boltzmann constant, and s about.38-3 J/K. We also fnd that the probablty that a system has energy E or greater s also proportonal to e -E/kT (wth a dfferent constant of proportonalty). There s common sense here, because () s sayng that greater energes are less lkely; and also that the hgher the temperature, the more lkely you are to have hgher energes. Let s gve some examples: 5.7. Atmospherc Pressure The pressure n the atmosphere at heght h s proportonal to the probablty that a molecule wll be at that heght, and s therefore proportonal to e -mgh/kt. Here, the energy E, s of course the gravtatonal potental energy of the molecule whch has mass m. The proof of ths statement s n several parts. Frstly we assume that all the ar s at the same temperature. Ths s a dodgy assumpton, but we shall make do wth t. Next we dvde the atmosphere nto slabs (each of heght dh and unt area), stacked one on top of the other. Each slab has to support all the ones above t. From the Gas Law (pv = NkT where N s the number of molecules under consderaton), and the defnton of densty (Nm=V) we can show that =pm/kt. Furthermore, f you go up by a small heght dh, the pressure wll reduce by the weght of one slab namely gdh. Therefore dp dh p pmg g kt mgh p exp kt () where p s the pressure at ground level. We see that the Boltzmann Law s obeyed Velocty dstrbuton of molecules n a gas The probablty that a molecule n the ar wll have x-component of ts velocty equal to u x s proportonal to exp(-mu x /kt). Here the energy E s the knetc energy assocated wth the x-component of moton. From ths statement, you can work out the mean value of u x, and fnd t to be: u x x exp mx kt dx exp mx kt dx kt kt m m 3 / / kt m () mu x kt Page 79

80 Correctons March 7 The mean knetc energy s gven by K 3 u u kt mu m u (3) x y z so the nternal energy of a mole of gas (due to lnear moton) s U 3 3 N K N kt RT (4) A A where R N Ak s the gas constant. From ths t follows that the molar heat capacty of a perfect gas 7 3, C V R Vapour Pressure The probablty that a water molecule n a mug of tea has enough (or more than enough) energy to leave the lqud s proportonal to exp(- E L /kt) where E L s the energy requred to escape the attractve pull of the other molecules (latent heat of vaporzaton per molecule) Justfcaton of Boltzmann Law In ths secton, we ntroduce some statstcal mechancs to gve a taste of where the Boltzmann law comes from. Suppose that you have N atoms, and P packets of energy to dstrbute between them. How wll they be shared? In statstcal mechancs we assume that the energy wll be shared n the most lkely way. In a smple example, we could try sharng 4 unts of energy (P=4) among 7 atoms (N=7). Because the ndvdual energy unts are ndstngushable (as are the 7 atoms), the possble arrangements are: atom wth 4 energy unts, 6 atoms wth none atom wth 3 energy unts, wth, 5 wth none atoms wth energy unts, 5 wth none atom wth energy unts, wth, 4 wth none 4 atoms wth energy unt, 3 wth none. These are sad to be the fve macrostates of the system. We can work out how lkely each one s to occur f the energy s dstrbuted randomly by countng the ways n whch each macrostate could have happened. 7 Ths s the heat capacty due to lnear moton. For a monatomc gas (lke helum), ths s the whole story. For other gases, the molecules can rotate or vbrate about ther bonds as well, and therefore the heat capacty wll be hgher. Page 8

81 Correctons March 7 For example, n the frst case (all of the energy s gven to one of the atoms), there are seven ways of settng t up because there are seven atoms to choose from. In the second case, we have to choose one atom to take 3 unts (7 to choose from), and then choose one from the remanng sx to take the remanng unt. Therefore there are 76 = 4 ways of settng t up. Smlarly we can count the ways of rearrangng for the other macrostates 8 : atom wth 4 energy unts, 6 atoms wth none atom wth 3 energy unts, wth, 5 wth none atoms wth energy unts, 5 wth none atom wth energy unts, wth, 4 wth none 4 atoms wth energy unt, 3 wth none. 7 ways 4 ways ways 5 ways 35 ways We can see that there s one macrostate clearly n the lead where 4 atoms have no energy, atoms have one energy unt, and atom has two energy unts. Ths macrostate s nterestng because there s a geometrc progresson n the number of atoms (4,,) havng each amount of energy. It can be shown that the most lkely macrostate wll always be the one wth (or closest to) a geometrc progresson of populatons. 9 In other 8 The calculaton s made more straghtforward usng the formula W N n! n!!! n where W s the number of ways of settng up the macrostate, n s the number of atoms wth no energy, n s the number of atoms wth one unt of energy, and so on. Here s one justfcaton for ths formula: N! gves the total number of ways of choosng the atoms n order. The dvson by n! prevents us overcountng when we choose the same atoms to have zero energy n a dfferent order. A smlar reason holds for the other terms on the denomnator. Alternatvely, W = the number of ways of choosng n atoms from N the number of ways of choosng the n from the remanng (Nn ) the number of ways of choosng the n from the remanng (Nn n ) and so on. Thus W C N n n! C N n n N! N n C! n! N nn n N n! N n n! n! N n n! N! N n n n! n! n! n! 9 To show ths, start wth a geometrc progresson (n other words, say n Af where f s some number), and wrte out the formula for W. Now suppose that one of the atoms wth unts of energy gves a unt to one of the atoms wth j unts. Ths means that n and n j have gone down by, whle n and n j+ have each gone up by one. By comparng the old and Page 8

82 Correctons March 7 words, assumng that ths macrostate s the true one (whch s the best bet), then the fracton of atoms wth n energy unts s proportonal to some number to the power n. The actual fracton s gven by the n formula P N N P. Assumng that the mean number of energy unts per atom s large (so that we approxmate the contnuous range of values that the physcal energy can take), ths means that N P N P e, and so the probablty that an atom wll have n unts of energy s proportonal to Nn P e. Suppose that each packet contans joules of energy. Then the energy of one atom (wth n packets) s E = n, and the probablty that our atom NE P wll have energy E as e. The mean energy per atom s P. Now we have seen that the average energy of an atom n a system s about kt, where T s the temperature n kelvns, and so t should not seem odd E kt that the Boltzmann probablty s e where we replace one expresson for the mean energy per atom P, wth another kt. 5.8 Perfect Gases All substances have an equaton of state. Ths tells you the relatonshp between volume, pressure and temperature for the substance. Most equatons of state are nasty, however the one for an deal, or perfect, gas s straghtforward to use. It s called the Gas Law. Ths states that p V = n R T (8) p V = N k T (9) where p s the pressure of the gas, V ts volume, and T ts absolute (or thermodynamc) temperature. Ths temperature s measured n kelvns always. There are two ways of statng the equaton: as n (8), where n represents the number of moles of gas; or as n (9), where N represents the number of molecules of gas. Clearly N=N A n where N A s the Avogadro number, and therefore R=N A k. You can adjust the equaton to gve you a value for the number densty of molecules. Ths means the number of molecules per cubc metre, and s gven by N/V = p/kt. The volume of one mole of molecules can also be worked out by settng n= n (8): new values of W, you can show that the new W s smaller, and that therefore the old arrangement was the one wth the bggest W. The bet gets better as the number of atoms ncreases. The combnaton (4,,) was the most popular n our example of N=7, P=4, however f you repeated the exercse wth N=7 and P=4, you would fnd a result near (4,,) almost a certanty. In physcs we deal wth huge numbers of atoms n matter, so the gamblng pays off. Page 8

83 Correctons March 7 RT V m. () p You can adjust ths equaton to gve you an expresson for the densty. If the mass of one molecule s m, and the mass of a mole of molecules (the R.M.M.) s M, we have Mass Volume N N A A mp kt M RT mp kt p Mp RT. () Please note that ths s the deal gas law. Real gases wll not always follow t. Ths s especally true at hgh pressures and low temperatures where the molecules themselves take up a good fracton of the space. However at room temperature and atmospherc pressure, the Gas Law s a very good model Heat Capacty of a Perfect Gas We have already shown (n secton 5.7.) that for a perfect gas, the nternal energy due to lnear moton s 3 RT per mole. If ths were the only consderaton, then the molar heat capacty would be there are two complcatons 3 R. However The condtons of heatng In thermodynamcs, you wll see molar heat capactes wrtten wth subscrpts C P and C V. They both refer to the energy requred to heat a mole of the substance (M klograms) by one kelvn. However the energy needed s dfferent dependng on whether the volume or the pressure s kept constant as the heatng progresses. When you heat a gas at constant volume, all the heat gong n goes nto the nternal energy of the gas (dq V = du). When you heat a gas at constant pressure, two thngs happen. The temperature goes up, but t also expands. In expandng, t does work on ts surroundngs. Therefore the heat put n s ncreasng both the nternal energy and s also dong work (dq P = du + p dv). Gven that we know the equaton of state for the gas (8), we can work out the relatonshp between the constant-pressure and constant-volume heat capactes. In these equatons we shall be consderng one mole of gas. Page 83

84 Correctons March 7 C C V P dq dt dq dt C V V P du dt du dt p d dt dv p dt RT p C V R () The type of molecule Gas molecules come n many shapes and szes. Some only have one atom (lke helum and argon), and these are called monatomc gases. Some gases are datomc (lke hydrogen, ntrogen, oxygen, and chlorne), and some have more than two atoms per molecule (lke methane). The monatomc molecule only has one use for energy gong places fast. Therefore ts nternal energy s gven smply by 3 kt, and so the molar nternal energy s can show that C V 3 3 U RT. Therefore, usng equaton (), we 5 R and C C R R. P V A datomc molecule has other optons open to t. The atoms can rotate about the molecular centre (and have a choce of two axes of rotaton). They can also wggle back and forth stretchng the molecular bond lke a rubber band. At room temperature we fnd that the vbraton does not have enough energy to kck n, so only the rotaton and translaton (the lnear moton) affect the nternal energy. Each possble axs of rotaton adds kt to the molecular energy, and so 5 7 we fnd that for most datomc molecules, C V R and C P R Thermodynamc Gamma It turns out that the rato of C P CV crops up frequently n equatons, and s gven the letter. Ths s not to be confused wth the n relatvty, whch s completely dfferent. Usng the results of our last secton, we see that =5/3 for a monatomc gas, and =7/5 for one that s datomc Pumpng Heat If a healthy examner expects you to know about deal gases and thermodynamcs you can bet that he or she wll want you to be able to do thermodynamcs wth an deal gas. In ths secton we show you how to turn a perfect gas (n a cylnder) nto a reversble heat engne, and n dong so we wll ntroduce the technques you need to know. Page 84

85 Correctons March Isothermal Gas Processes As an ntroducton, we need to know how to perform two processes. Frstly we need to be able to get heat energy nto or out of a gas wthout changng ts temperature. Remember that we want a reversble heat engne, and therefore the gas must be at the same temperature as the hot object when the heat s passng nto t. Any process, lke ths, whch takes place at a constant temperature s sad to be sothermal. The Gas Law tells us (8) that pv=nrt, and hence that pv s a functon of temperature alone (for a fxed amount of gas). Hence n an sothermal process pv const. ISOTHERMAL (3) Usng ths equaton, we can work out how much we need to compress the gas to remove a certan quantty of heat from t. Alternatvely, we can work out how much we need to let the gas expand n order for t to absorb a certan quantty of heat. These processes are known as sothermal compresson, and sothermal expanson, respectvely. Suppose that the volume s changed from V to V, the temperature remanng T. Let us work out the amount of heat absorbed by the gas. Frst of all, remember that as the temperature s constant, the nternal energy wll be constant, and therefore the Frst Law may be stated dq=pdv. In other words, the total heat enterng the gas may be calculated by ntegratng pdv from V to V : Q nrt V pdv dv nrt lnv nrt ln. (4) V V V Ths equaton descrbes an sothermal (constant temperature) process only. In order to keep the temperature constant, we mantan a good thermal contact between the cylnder of gas and the hot object (the boler wall, for example) whle the expanson s gong on Adabatc Gas Processes The other type of process you need to know about s the adabatc process. These are processes n whch there s no heat flow (dq=), and they are used n our heat engne to change the temperature of the gas n between ts contact wth the hot object and the cold object. Sometmes ths s referred to as an sentropc process, snce f dq= for a reversble process, TdS=, and so ds= and the entropy remans unchanged. V Whle the terms sentropc and adabatc are synonymous for a perfect gas, care must be taken when dealng wth rreversble processes n more advanced systems. In ths context dq s not equal to TdS. If dq=, the process s sad to be adabatc: f ds=, the process s Page 85

86 Correctons March 7 Before we can work out how much expanson causes a certan temperature change, we need to fnd a formula whch descrbes how pressure and volume are related n an adabatc process. Frstly, the Frst Law tells us that f dq=, then = du + p dv. We can therefore reason lke ths for n moles of gas: du pdv nc dt pdv V Now for a perfect gas, nrt=pv, therefore may contnue the dervaton thus: nrdt pdv Vdp. So we nc V CV dt pdv Vdp R CV pdv Vdp pdv R C ( pdv Vdp) RpdV C V p pdv C pdv Vdp dv V dp p V Vdp. (5) Integratng ths dfferental equaton gves lnv ln p C pv pv e C const. ADIABATIC (6) Equaton (6) s our most mportant equaton for adabatc gas processes, n that t tells us how pressure and volume wll be related durng a change. We now come back to our orgnal queston: what volume change s needed to obtan a certan temperature change? Let us suppose we have a fxed amount of gas (n moles), whose volume changes from V to V. At the same tme, the temperature changes from T to T. We may combne equaton (6) wth the Gas Law to obtan: sentropc. Clearly for a complex system, the two condtons wll be dfferent. Ths arses because n these systems, the nternal energy s not just a functon of temperature, but also of volume or pressure. Page 86

87 Correctons March 7 pv T T pv V nrtv V V const const const. (7) A Gas Heat Engne We may now put our sothermal and adabatc processes together to make a heat engne. The engne operates on a cycle:. The cylnder s attached to the hot object (temperature T hot ), and sothermal expanson s allowed (from V to V ) so that heat Q s absorbed nto the gas.. The cylnder s detached from the hot object, and an adabatc expanson (from V to V 3 ) s allowed to lower the temperature to that of the cold object (T cold ). 3. The cylnder s then attached to the cold object. Heat Q s then expelled from the cylnder by an sothermal compresson from V 3 to V Fnally, the cylnder s detached from the cold object. An adabatc compresson brngs the volume back to V, and the temperature back to T hot. Applyng equaton (4) to the sothermal processes gves us Q Q nrt nrt hot cold V ln V V ln V 4 3. (8) Smlarly, applyng equaton (7) to the adabatc processes gves us V V 3 T T V 4 Thot V T cold V3 V4 V V V V hot cold 3 4 V V. (9) Combnng equatons (8) and (9), gves us Page 87

88 Correctons March 7 Q Q Q Q T T T T hot cold hot cold, (3) where the mnus sgn remnds us that Q <, snce ths heat was leavng the gas. To summarze ths process, we have used a perfect gas to move heat from a hot object to a colder one. In dong ths, we notce less heat was deposted n the cold object than absorbed from the hot one. Where has t gone? It materalzed as useful work when the cylnder was allowed to expand. Had the pston been connected to a flywheel and generator, we would have seen ths n a more concrete way. We also notce that we have proved that the kelvn scale of temperature, as defned by the Gas Law, s a true thermodynamc temperature snce equaton (3) s dentcal to (5). 5.9 Radaton of Heat And fnally... there s an extra formula that you wll need to be aware of. The amount of heat radated from an object s gven by: P AT 4 (3) P s the power radated (n watts), A s the surface area of the object (n m ), and T s the thermodynamc temperature (n K). The constant s called the Stefan-Boltzmann constant, and takes the value of W/(m K 4 ). The amount radated wll also depend on the type of surface. For a perfect matt black (best absorber and radator), the object would be called a black body, and the emssvty would take the value. For a perfect reflector, there s no absorpton, and no radaton ether, so =. 5. Questons. Calculate the maxmum effcency possble n a coal fred power staton, f the steam s heated to 7 C and the rver outsde s at 7 C.. Mechancal engneers have been keen to buld jet engnes whch run at hgher temperatures. Ths makes t very dffcult and expensve to make the parts, gven that the materals must be strong, even when they are almost at ther meltng pont. Why are they makng lfe hard for themselves? Page 88

89 Correctons March 7 3. Two nsulated blocks of steel are dentcal except that one s at C, whle the other s at C. They are brought nto thermal contact. A long tme later, they are both at the same temperature. Calculate the fnal temperature; the energy change and entropy change of each block f (a) heat flows by conducton from one block to the other, and f (b) heat flows from one to the other va a reversble heat engne There s a rule of thumb n chemstry that when you rase the temperature by C, the rate of reacton roughly doubles. Use Boltzmann s Law to show that ths means the actvaton energy of chemcal processes must be of order 9 J The amount of energy taken to turn kg of lqud water at C nto kg of steam at the same temperature s.6 MJ. Ths s called the latent heat of vaporzaton of water. How much energy does each molecule need to free tself from the lqud? 6. By defnton, the bolng pont of a lqud s the temperature at whch the saturated vapour pressure s equal to atmospherc pressure (about kpa). Up a mountan, you fnd that you can t make good tea, because the water s bolng at 85 C. What s the pressure? You wll need your answer to q Estmate the alttude of the mountaneer n q5. Assume that all of the ar n the atmosphere s at C Use the Gas Law to work out the volume of one mole of gas at room temperature and pressure (5 C, kpa). 9. What fracton of the volume of the ar n a room s taken up wth the molecules themselves? Make an estmate, assumng that the molecules are about m n radus.. Estmate a typcal speed for a ntrogen molecule n ntrogen at room temperature and pressure. On average, how far do you expect t to travel before t hts another molecule? Agan, assume that the radus of the molecule s about m. ++. The fracton of molecules (mass m each) n a gas at temperature T mu / kt whch have a partcular velocty (of speed u) s proportonal to e, as predcted by the Boltzmann law. However the fracton of molecules mu / kt whch have speed u s proportonal to u e. Where does the u come from? ++. One ltre of gas s suddenly squeezed to one hundredth of ts volume. Assumng that the squeezng was done adabatcally, calculate the work done on the gas, and the temperature rse of the gas. Why s the adabatc assumpton a good one for rapd processes such as ths? Page 89

90 Correctons March 7 3. A water rocket s made usng a ltre plastc drnks bottle. An amount of water s put nto the bottle, and the stopper s put on. Ar s pumped nto the bottle through a hole n the stopper. When the pressure gets to a certan level, the stopper blows out, and the pressure of the ar n the bottle expels the water. If the bottle was standng stopper-end downwards, t fles up nto the ar. If you neglect the mass of the bottle tself, what s the optmum amount of water to put n the bottle f you want your rocket to (a) delver the maxmum mpulse, or (b) rse to the greatest heght when fred vertcally. ++ Page 9

91 Correctons March 7 6 Sparks & Generaton 6. Electrostatcs when thngs are stll The fundamental fact of electrostatcs wll be famlar to you opposte charges attract: lke charges repel. As a physcst t s not enough to know ths, we also want to know how bg the force s. It turns out that the equatons descrbng the force, energy, potental and so on are very smlar mathematcally to the equatons descrbng gravtatonal attracton. Usng the symbols F for force, U for potental energy, V for potental and E for feld, we have the equatons: F E r r 4 Qq R R Qq U 4 R Q V 4 R 4 Q () Notce that the symbols are slghtly dfferent for gravty feld s now gven E rather than g, and n consequence we have to use another letter for energy hence our choce of U. Here we have put a charge Q at the orgn, and we measure the quanttes assocated wth a small test charge q at dstance R. Notce that we do not have a mnus sgn n front ths allows lke charges to repel rather than attract (whereas n gravty, postve mass attracts postve mass [and we have never found any lumps of negatve mass f we dd ths would upset a lot of our thnkng]). Also, n place of the G of gravty, we have the constant 4 whch s about tmes bgger. No wonder the theoretcans talk of gravty as a weaker force! Now, you may wonder, why the factor of 4? Ths comes about, because the equatons above are not the ncest way of descrbng electrostatc forces. They are based on the Coulomb Force Law, whch s the frst equaton n () however there s another, equvalent, way of descrbng the same physcs, and t s called Gauss Law of Electrostatcs. Ths Gauss Law s on the Olympad syllabus, and you wll fnd t useful because t wll smplfy your electrcal calculatons a lot. Page 9

92 Correctons March Gauss Law of Electrostatcs Frstly, let us say what the law s. Then we wll descrbe t n words, and then prove that t s equvalent to the Coulomb Law. Fnally, we wll show ts usefulness n other calculatons. S Q E ds () What does ths mean? Frstly let us look at the ndvdual symbols. S s a closed surface (that s what the crcle on the ntegral sgn means) lke the outer surface of an apple, a table, or a doughnut but not the outer surface of a bowl (whch ends at a rm). ds s a small part of the surface, wth area ds, and s a vector pontng outward, perpendcular to the surface at that pont. Q s the total electrc charge contaned nsde the surface S. Fnally, the vector E s the electrc feld (n volts per metre) where the vector ponts n the drecton a postve charge would be pulled. The odd lookng ntegral tells us to ntegrate the dot product of E wth ds (a normal vector to the surface) around the complete surface. Ths may sound very foregn, strange, and dffcult, but let us gve some examples. Frst of all, suppose S s a sphercal surface of radus R, wth one charge (+Q) at the centre of the surface. Assumng there are no other charges nearby, the feld lnes wll be straght, and wll stream out radally from the centre. Therefore E wll be parallel to ds, and the dot product wll smply be E ds the product of the magntudes. Now the sze of E must be the same all round the surface, by symmetry. Therefore E ds EdS E ds E 4R (3) S S S snce the surface area of a sphere s gven by 4R. Now by Gauss Law, ths must equal Q. Puttng the two equatons together gves us 4R Q Q E E (4) 4 R n agreement wth Coulomb s Law. Smlarly we may fnd the feld at a dstance R from a wre that carres a charge per metre spread evenly along the wre somethng whch Coulomb s Law could do, but would need a horrendous ntegral to do t. Ths tme, our surface s a cylnder, one metre long, wth the wre runnng down ts axs. The cylnder has radus R. Frst, notce that the feld lnes wll run radally out from the wre. Ths has the consequence that the two Page 9

93 Correctons March 7 flat ends of the cylnder do not count n the ntegral. Thnk about ths for a moment, because ths s mportant. The vector ds for these ends wll pont normal to the surface that means parallel to the wre. The vector E, on the other hand ponts outward. Therefore E s perpendcular to ds, and the dot product s zero. Only the curved surface counts. Agan, E wll have the same magntude at all ponts on t because of symmetry, and E wll be parallel to ds on ths surface. Once agan, we have S E ds EdS E ds E RL (5) curved S curved S where L s the length of the cylnder, and hence RL s ts curved surface area. By Gauss Law, ths must be equal to Q L, and so E. (6) R We notce that for a lne charge, the nverse square of the Coulomb Law has become an nverse, not squared law. Before movng on, let us make two more ponts. Frstly, we have not proved the equvalence of Gauss Law and Coulomb s Law we have only shown that they agree n the case of calculatng the feld around a fxed, pont charge. However the two can be proved equvalent but the proof s a bt nvolved, and s best left to frst (or second) year unversty courses. Secondly, let us thnk about a surface S whch s entrely nsde the same pece of metal. E wll be zero wthn a metal, because any non-zero E (.e. voltage dfference) would cause a current to flow untl the E were zero. Therefore a surface entrely wthn metal can contan no total charge! Impossble, I hear you cry! Let us take a hollow metal sphere, wth a charge +Q at the centre of the cavty. How can the enclosed charge be zero surely t s +Q! Oh, no t sn t. Actually the total enclosed charge s zero, and ths enclosed charge s made up of +Q at the mddle of the cavty, and Q nduced on the nsde wall of the hollow sphere! If the sphere s electrcally solated, and began lfe uncharged, there must be a +Q charge somewhere on the metal, and t sts on the outer surface of the sphere. 6.. Capactors A capactor s a devce that stores charge. To be more precse, a capactor conssts of two conductng plates, wth nsulatng space Page 93

94 Correctons March 7 between them. When the postve plate carres charge +Q, an equal amount of negatve charge s stored on the other. If certan nsulatng materals are used to separate the plates (nstead of ar or vacuum), the amount of charge that can be stored ncreases consderably. The charge stored s proportonal to the potental dfference across t, and we call the constant of proportonalty the capactance. Gauss Law gves us a wonderful way to calculate the capactance of smple capactors, and we wll look at the calculaton for a parallel plate capactor Parallel Plate Capactor At ts smplest, a capactor s shown n the fgure below. The two plates are square, and parallel. Each has area A and the dstance between them s denoted L. Let s work out the capactance. To do ths, we frst suppose that there were a charge Q stored. In other words, there s a charge Q on the top plate, and +Q on the bottom plate. We can work out the electrc feld n the gap usng Gauss Law. We draw a rectangular box-shape surface, wth one of ts faces parallel to, but bured n the bottom plate, and the opposte face n the mddle of the gap. Area A Separaton L Voltage dfference V When workng out the surface ntegral S E ds, only ths face n the mddle of the gap counts. The face bured n the metal of the plate has E=, whle the other four surfaces normals are perpendcular to the feld. Therefore Q S E ds AE. (7) We next work out the voltage. Ths s not hard, as by analogy from gravtatonal work (chapter, equaton 6) Page 94

95 Correctons March 7 Feld = - d(potental)/d(dstance) E = -dv/dx (8) Here, E s constant and unform throughout the nter-plate gap, and so V=Ex+c where c s a constant of ntegraton. Thus the voltage dfference between the plates can be calculated; and from ths the capactance can be worked out. QL Q A V EL C (9) A V L We ought to gve a word of cauton at ths pont. In a real parallel plate capactor, the feld near the edge of the plates wll not pont drectly from one plate to the other, but wll bow out a bt. Therefore the equaton gven above s only true when these edge effects are gnored. It turns out that the equaton s pretty good provdng that L s much smaller than both of the lnear dmensons of the plates. The equaton also allows you to see the effects of wrng capactors n seres or parallel. When two dentcal capactors, each of capactance C, are connected together n parallel, the overall area A s doubled, so the capactance of the whole arrangement s C. On the other hand f the capactors are connected n seres, the result s one capactor wth twce the gap thckness L. Therefore the overall capactance s C/ Decay of Charge on Capactor We next come to the case where a capactor s charged to voltage V (that s, a voltage V across the plates), and then connected n a smple crcut wth a resstor R. How long wll t take to dscharge? To work ths out, we need to use our characterstc equatons for capactor and resstor. For the capactor V=Q/C, for the resstor V=IR. To solve the crcut we need to clarfy the relatonshp between Q and I. I +Q -Q Here we need to take care. Dependng on how the crcut has been drawn, ether I=dQ/dt or I=-dQ/dt. That s why t s essental that you Page 95

96 Correctons March 7 nclude n your crcut dagrams arrows to show the drecton of current flow for I>, and whch plate of the capactor s the +ve one. Here a postve I wll dscharge the capactor, so I=dQ/dt. The voltage across the capactor s the same as that across the resstor, so V dq Q IR R dt C. () dq Q dt RC Ths dfferental equaton can be solved wth an exponental soluton: Q t t RC ( ) Q e () where Q was the ntal charge on the capactor after t was charged up. Gven that the voltage across the capactor V(t)=Q(t)/C, the tme dependence of the voltage obeys a smlar equaton. The constant RC s known as the tme constant, and t s the tme taken for the voltage (or charge) to fall by a factor e (approx.7) Energy consderatons Next, we need to know how much energy has been stored n a capactor, f ts voltage s V and ts capactance C. The energy s actually stored n the electrc feld between the plates but more of ths later. To work the energy out, we charge a capactor up from scratch (ntal charge = ), and contnually measure the current flowng, and the voltage across t. The energy stored must be gven by U VC dv dt P dt dt VI dt V CVdV C dq dt dt VdV C V dt dt V CV d( CV ) () Gven that the energy stored must be zero when V=, and there s no electrc feld n the gap, ths fxes the constant of ntegraton as zero, and we obtan the fact that energy stored = half the capactance the square of the voltage across the gap. You could equally well say that the energy s gven by half the charge multpled by the voltage. Before we leave ths formula, let us do some conjurng trcks wth ths energy, assumng that the capactor s a smple parallel plate devce: U CV AV L A EL L E AL (3) Page 96

97 Correctons March 7 thus the energy stored per unt volume of gap s E. Although we have only shown ths to be true for a perfect parallel plate capactor, t s possble to make any electrc feld look lke rows and columns of parallel plate capactors arranged lke a mosac, and from ths the proof can be generalzed to all electrc felds Polarzaton When we ntroduced capactors, we mentoned that the capactance can be rased by nsertng nsulatng stuff nto the gap. How does ths work? Look at the dagram below. The stuff n the mddle contans atoms, whch have postve and negatve charges wthn them When the plates are charged, as shown, ths pulls the nucle of the stuff to the rght, and the electrons of the atoms to the left. The left plate now has a blanket of negatve charge, and the rght plate has a blanket of postve charge. Ths reduces the overall total charge on the plates, and therefore reduces the voltage across the capactor. Of course the crcut can t remove the polarzaton charges n the blankets as that would requre the chemcal breakdown of the substance. So we have stored the same crcut charge on the capactor for less voltage, and so the capactance has gone up. The rato by whch the capactance ncreases s called the relatve permttvty of the substance (t used to be called the delectrc constant), and s gven the symbol r. In the presence of such a materal, the of all the equatons derved so far n ths chapter needs to be multpled by r. 6. Magnetsm when thngs move So far, we have just consdered electrc charges at rest. Our next job s surely to look at electrc charges that have gone roamng, and then to study magnetsm two thngs stll to do? No. Actually we only have one job, because magnetsm s all about movng charges. We ought to gve one warnng, though. Just because magnetsm s about movng charges, we can t derve ts formulae smply from Coulomb s Law and Classcal mechancs. We need Relatvstc mechancs! That s actually one route nto relatvty t s the knd of mechancs needed f electrcty and magnetsm are to be descrbed together. Put another way, your nearest pece of evdence for specal relatvty s not n a partcle accelerator or arborne atomc clock, but n your wrst-watch (f t has hands), credt card, vacuum cleaner, frdge, CD player, prnter, hard dsk drve, or wherever your nearest magnet s. Page 97

98 Correctons March 7 We shall demonstrate ths at the end of the chapter. However, for the moment, let us stck to what we need for the Olympad, and let us carry on callng t magnetsm as opposed to relatvstc electrcty. 6.. Magnetc Flux Densty If there s a magnetc feld, there must be a measurement of the feld strength, and we call ths the flux densty, and gve t the symbol B. The feld has a drecton (from North to South), and s therefore a vector. The fundamental fact of magnetsm can be stated n two ways:. If a wre of length L s carryng current I, and the wre s n a magnetc feld B, t wll experence a force F, where F = L I B. Wrtten wthout the vector cross product, ths s F = B I L sn, where s the angle between the drecton of the current, and the drecton of the magnetc feld.. If a charged partcle, of charge Q s movng n a magnetc feld B, and t has velocty u, t wll experence a force F, where F = Q u B. Wrtten wthout the vector cross product, ths s F = Q u B sn where s the angle between the drecton of moton, and the drecton of the magnetc feld. You can show that these descrptons are equvalent, by magnng the wre contanng N charges (each Q coulombs) per metre. If the wre has length L, the total charge s Q TOT =NQL, and ths moves when the current s flowng. If the current s I, ths means that the charge passng a pont n one second s I, and hence I=NQu, where u s the speed. Therefore F = L I B = L (NQu) B = (NQL) u B = Q TOT u B. (4) 6.. Dong the Corkscrew Now that we have an expresson for the force on a charge movng n a magnetc feld, we can work out the moton f a charge s thrown nto the vcnty of a magnet. The most mportant fact s that the force s always at rght angles to the velocty. Therefore t never does any work at all, and t never changes the knetc energy (hence speed) of the object. The next mportant fact s that f the velocty s parallel to the magnetc feld, there s no force and the partcle wll just carry on gong: as f the magnet weren t there at all. Page 98

99 Correctons March 7 Ths second fact s useful, because any velocty can be broken down (or resolved) nto two components one parallel to the magnetc feld (u cos), and one perpendcular to t (u sn). The component parallel to the feld wll be unchanged by the moton t wll stay the same, just as Newton s Frst Law requres. We next need to calculate what the effect of the other component wll be. Ths wll cause a force perpendcular to both the velocty and magnetc feld, and we already know from classcal mechancs that when a force consstently remans at rght angles to the moton, a crcular path s obtaned. We can calculate the radus from the equatons of crcular moton: Magnetc Force = Mass Centrpetal Acceleraton m u sn mu sn BQu sn R (5) R BQ Another useful measurement s the tme taken for the partcle to go round the loop once. Snce ts rotatonal speed s u sn, the tme taken to go round a R crcumference s T R u sn. We can also work out the angular velocty: u sn R BQ m (6) The overall moton s therefore a helx, or corkscrew shape, wth the axs of the corkscrew parallel to the magnetc feld. The radus of the helx s gven by R n equaton (5), and the ptch (that s, the dstance between successve revolutons), s equal to D = T u cos. Helx, or corkscrew moton of an electron n a magnetc feld. Please notce that all these formulae reman vald when the partcle starts gong very quckly. The only correcton that specal relatvty requres s that we use the enhanced mass m m. No further rest Page 99

100 Correctons March 7 correcton s needed, because the speed remans constant, and therefore does not change Calculaton of magnetc feld strengths So far, we have just thought about the effects of a magnetc feld. However before you can study what an electron wll do n such a feld, you have to make the feld! How do you do that? At ts smplest, magnetc felds are caused by electrc currents. The bgger the current: the bgger the feld. These may be real currents of electrons n wres, or they can be the effectve currents of electrons orbtng ther nucle n atoms. The latter s responsble for the permanent magnetc property of ron (and some other metals) however the process s qute nvolved and needs no further consderaton for the Olympad. We do need to worry about the magnetsm caused by wres, however, although only a bref descrpton s necessary. The Olympad syllabus precludes questons that nvolve large amounts of calculus (hence ntegraton), and most feld calculatons requre an ntegral. Therefore f you need to know how bg a magnetc feld s, you wll probably be gven the equaton you need. Nevertheless t wll help to see how the calculatons are done. There s a method akn to Coulomb s Law, and an alternatve called Ampere s Law. We shall ntroduce both Magnetc Coulomb The Bot Savart Law To use ths method, the wre (carryng current I) s broken down nto small lengths (dl), lnked head-to-tal. To work out the sze of the magnetc feld at pont r, we sum the effects of all the current elements. Let us take one current element at pont s. Its contrbuton to the B feld at r s: B r I r s dl 4 r s d 3 (7) where (r-s) s the vector that ponts from the bt of wre to the pont at whch we are calculatng B. To work out the total feld B(r), we ntegrate the expresson along the wre. In most cases, ths can be put more smply as I sn db dl (8) 4 d Page

101 Correctons March 7 where d s the dstance from wre element to pont of measurement, and s the angle between the current flow and the vector from wre to measurement pont. It s possble to use ths expresson to calculate the magnetc feld B on the axs of a col of wre wth N turns, radus R, carryng current I, f we measure the feld at a dstance D from the central pont: INR B D R 3 (9) The Ampere Law The alternatve method of calculatng felds s called the Ampere Law. You remember that Gauss electrostatc law nvolved ntegratng a dot product over a surface? Well, the Ampere Law nvolves ntegratng a dot product along a lne: L B dl I () In other words, f we choose a loop path, and ntegrate the magnetc flux densty around t, we wll fnd out the current enclosed by that loop. Let us gve an example. Ths formula s very useful for calculatng the magnetc feld n the vcnty of a long straght wre carryng current I. Suppose we take path L to be a crcle of radus R, wth the wre at ts centre, and wth the wre perpendcular to the plane of the crcle, as shown n the dagram. Current I Integraton path round the wre, keepng constant dstance from t. R We know that B ponts round the wre, and therefore that B s parallel wth dl (dl beng the vector length of a small part of the path). Page

102 Correctons March 7 Furthermore, B= B must be the same all round the path by symmetry. Therefore: L B dl L B dl B L I B R dl RB I () Another splendd use of the Ampere Law s n calculatng the magnetc feld wthn the mddle of a long solenod wth n turns per metre. Ths tme we use a rectangular path, as shown n the dagram. Integraton path B-feld drecton Only one of the sdes of the rectangle counts n the ntegraton the one completely nsde. Of the other three, one s so far away from the col that there s no magnetc feld, and the other two are perpendcular to the B-feld lnes, so the dot product s zero. The rectangle encloses nl turns, and hence a current of nli. Therefore, Ampere s Law tells us: BL nli B ni () Before leavng Ampere s Law, we ought to gve a word of cauton. The Ampere Law s actually a smplfed form of an equaton called the Ampere-Maxwell Law ; and the smplfcaton s only vald f there are no For the curous, the full Ampere-Maxwell Law states B dl I L S E ds where the surface S s any surface that has as ts edge the loop L. Ths reduces to equaton () when E s constant. rel Page

103 Correctons March 7 changng electrc felds n the vcnty. Therefore you would be on dodgy ground usng Ampere s Law near a capactor that s chargng up! 6..4 Flux, Inductance & Inductors To sum up the last secton: f there s a current, there wll be a magnetc feld. Furthermore, the strength of the magnetc feld s proportonal to the sze of the current. It turns out, however, to be more useful to speak of the magnetc flux. Ths s the product of the feld strength B and the cross sectonal area of the regon enclosed by the magnetc feld lnes. We vsualze ths as the total number of feld lnes made by the magnet. We then wrte LI. (3) The total amount of magnetc feld (the flux) s proportonal to the current, and we call the constant of proportonalty L the self-nductance (or nductance for short). Any col (or wre for that matter) wll have an nductance, and ths gves you an dea of how much magnetc feld t wll make when a current passes. You mght thnk of an analogy wth capactors the capactance gves a measure of how much electrc feld a certan charge wll cause (snce C - = V/Q and V s proportonal to E). Now, ths magnetc feld s mportant, because a changng current wll cause a changng magnetc feld, and ths wll generate (or nduce) a voltage, and therefore upset the crcut t s n. Ths s somethng we need to understand better but before we do so, let us remnd ourselves of the laws of electromagnetc nducton: 6..5 Generators & Inducton If a wre thnks t s movng magnetcally, a voltage s nduced n t. It doesn t matter whether the wre s stll, and the magnetc feld s movng or changng; or whether the wre s movng and the magnetc feld s stll. However for the two stuatons, dfferent equatons are used. The two equatons are equvalent however t s easer to remember them both than to prove the equvalence Statonary feld: Movng wre moton of wre electrons n wre pushed to rght Magnetc feld B down nto paper. Page 3

104 Correctons March 7 Here the voltage s easy to calculate. The electrons n the wre must move wth the wre, and f the wre s n a magnetc feld, they wll experence a force pushng them along the wre. Ths causes them to bunch up at one end of the wre whch n turn sets up an electrc feld whch dscourages further electrons to jon the party. Assumng that the wre has length L, and s beng moved at speed u through a unform magnetc feld B (perpendcular to u and L f B s not so nclned, take only the component of B whch s), once equlbrum s establshed Electrc force balances magnetc force qe qub V ub L V LuB (4) Statonary wre: Changng feld Here the voltage nduced across the ends of a crcular loop of wre (complete crcut, apart from the small gap) s gven by V d d B ds dt (5) dt S Equvalence? The two expressons are very smlar, as we shall see when we look at the frst from a dfferent perspectve. Look at the dagram we have completed a loop by usng a very long wre. Regon of magnetc feld B, pontng down nto paper L End wre moved n drecton of whte arrow. Dstance moved equals ut, where u s speed. In tme t, the area s reduced by Lut, so the magnetc flux enclosed by the wre s reduc ed by BLut. The rate of change of flux = nduced voltage = BLu. Page 4

105 Correctons March 7 The total flux enclosed n the loop s equal to B Area. In one second, the Area changes (decreases) by Lu, and hence the flux changes by BLu, and so we see that the frst equaton s a specal case of the second. The second s better as t also allows us to calculate what wll happen f the wre s statonary Drecton of Induced Voltage When equaton (5) s wrtten down, t s customary to put a mnus sgn n front of the dervatve. The sgnfcance of ths negaton was Lenz s dscovery. When the voltage s nduced n a complete crcut, t wll try to (and succeed n) drvng a current. Ths current wll produce a magnetc feld. Lenz postulated that ths produced magnetc feld always opposed the change beng made. Let us have an example. Imagne a large col of wre (say, n a motor), wth a decent szed current flowng n t. Now let us try and lower the current by reducng the voltage of the supply. Ths causes a reducton n the magnetc feld, whch n turn nduces a voltage n the wre, whch pushes a current n a desperate attempt to keep the orgnal current gong. On the other hand, f I were to try an ncrease the current n the motor (by ncreasng the supply voltage), the opposte would happen: the greater current causes the magnetc feld to grow, whch nduces a voltage, whch pushes a current to oppose ths ncrease. Ths s the orgn of the phrase back emf to refer to the voltage nduced across an nductor. Now for a word about the mnus sgn. Yes, the voltage does go n opposton to the change n current, so I suppose one ought to wrte equaton (5) wth a mnus sgn. However f you do, please also wrte Ohm s law as V = IR, snce the voltage opposes the current n a resstor. I would prefer t, however, f you used common sense n applyng your notaton and were not stuck n ruts of always or never usng the mnus sgn. We all remember whch way V and I go n a resstor wthout beng nagged about conventons, so I hope that there s no need for me to nag you when nductors come on the scene Inductors n crcuts Just as a capactor requres an energy flow to change the voltage across t, an nductor requres an energy flow to change the current through t. It doesn t gve n wthout a fght. Let me llustrate ths wth a demonstraton or at least the story of one. A nasty physcs teacher (yes, they do exst...) asked a pupl to come and hold one wre n hs left hand, and one n hs rght completng the crcut wth hs body. He grasped the frst wre, and then the second steelng hmself for the shock whch never came. The teacher then stalled hm wth questons, and kept hm there, whle he surrepttously, slowly ncreased the current n the crcut, whch also ncluded an nductor. Page 5

106 Correctons March 7 Fnally, the teacher sad, OK, now you can go back to your seat. The unsuspectng pupl let go of the wres suddenly, and was very surprsed when the nductor ndgnant to have ts current shut off so quckly made ts dspleasure known wth an arc from the wre to the pupl. It s foolsh n the extreme to starve an nductor of ts current. Its revenge wll be short, but not sweet. To see ths, let us combne equatons (3) and (5) V d d di LI L (6) dt dt dt Equaton (6) s the defntve equaton for nductors, just lke Q=CV was for capactors, and V=IR s for a resstor. Agan, we need to bear n mnd the comments above, that the voltage s n the drecton needed to oppose the change n current. You wll often see (6) wth a mnus sgn n t for that reason. Let us now calculate how much energy s stored n the devce (actually n ts magnetc feld) U di Pdt VI dt L I dt L I di LI (7) dt Snce t seems sensble for the devce to hold no energy when there s no current and no feld, we take the constant of ntegraton to be zero Relatvty and Magnetsm At the begnnng of the chapter we stated boldly that magnetsm could be derved entrely from electrostatcs and specal relatvty. Now s the tme to justfy ths. We shall do so by dervng the same result two ways once usng magnetsm, and once usng relatvty. The phenomenon we choose s the mutual attracton of two parallel wres carryng equal current n the same drecton, and we shall calculate the attractve force per metre of wre. Page 6

107 Correctons March 7 Both wres carry current I Force on rght hand wre Drecton of B- feld due to left hand wre R A Classcal Magnetc calculaton Equaton () tells us that the magnetc feld at a dstance R from a straght nfnte wre s I B R Therefore, the attractve force experenced by one metre of parallel conductor also carryng current I s F L I IB (8) R A Relatvstc calculaton Each wre contans postve on cores (say Cu + for a copper wre), and free electrons. Let us magne the stuaton n the dagram below, wth conventonal current flowng downwards n both wres. The on cores are statonary, whle the electrons move upwards. If the free (electron) charge per metre of cable s called, then the current s related to the electron speed u by the equaton I= u. Page 7

108 Correctons March 7 Dstance between on cores Dstance between electrons appears contracted From perspectve of ons n second wre, frst wre appears negat vely charged. Wres attract. Let s magne the stuaton as perceved by an on core n the second (rght hand) wre. It sees the on cores n the other wre statonary, wth charge densty and fnds them repulsve. However t also sees the electrons on the other wre, and s attracted by them. The electrons are travellng, and therefore our observant on core sees the length between adjacent electrons contracted. Therefore as far as t s concerned, the electron charge densty s hgher than the on core charge densty by factor u c. Therefore ts overall mpresson s attracton wth a total effectve electrc feld (as derved n equaton 6) Total feld = Feld due to on cores + Feld due to loose electrons E R R R u c R (9) where the fnal stage has made use of a Bnomal expanson of (-) to frst order n u/c. Now the total charge of on cores experencng ths feld per metre s of course. Therefore the total attracton of the on cores n the second wre to the frst wre (electrons & on cores) s F ons u QE c I 4c R R (3) Page 8

109 Correctons March 7 Ths s the force experenced by the on cores n the rght hand wre. By an exactly equvalent argument, the electrons n the rght hand wre see ther counterparts n the left wre as statonary, and see the left hand on cores bunched up, and therefore more attractve. Therefore the electrons n the rght hand wre experence an equal attracton, and the total attractve force between the wres s twce the fgure n equaton (3). Fnally, f you get a book of physcal constants and a calculator, you wll dscover that c. Therefore the total force agrees exactly wth our magnetc calculaton n equaton (8). 6.3 Crcuts puttng t together In ths secton, we look at combnng resstors, capactors and nductors n electrcal crcuts. There are two reasons for dong ths. Frstly, once you have left school, you wll be faced wth complcated electronc networks, and you need to be able to analyse these just as well as the smple seres and parallel arrangements you dealt wth n the classroom. Secondly, engneers frequently use electrc crcuts as models or analoges for other systems (say, an oscllatng brdge or the control of the nervous system over the muscles n a leg) the better you understand electrc crcuts, the better you wll understand any lnked system Crcut Analyss Our am here s to be able to solve a crcut lke the one below. The crcles represent constant-voltage sources (a bt lke cells or batteres) and the lnked crcles represent constant-current sources. Our am s to fnd voltage dfference across each component, and also to work out the current n each resstor. A B R C D R R 3 V R 4 I V E In order to solve the crcut, we use two rules the Krchoff Laws. Krchoff s st says that the total current gong nto a juncton s equal to the total current leavng t. Therefore, at B n the crcut below, we would say that I BE = I AB + I CB, where I BE means the current flowng from B to E (through R 4 ). Krchoff s nd Law s that voltages always add up correctly. In other words, no matter whch route we took from E to B, say, we would agree on the voltage dfference between E and B. In symbols, f V BE means Page 9

110 Correctons March 7 the dfference n potental (as measured by a voltmeter) between B and E, then we have V BE = V AE + V BA. Ths s bascally the same thng as the law of conservaton of energy. The voltage (or, more strctly, the potental) at B, V B, s the energy content of one coulomb of charge at B. In travellng to E, t wll lose V B -V E joules, rrespectve of the route taken. 3 In fact, we assume the truth of Krchoff s nd Law whenever we say, let s call the voltage at A V A, for we are assumng that the voltage of A does not depend on the route used to measure t. Usng these two rules, and the equaton for the current through a resstor (for example, V BA = I BA R ), we may wrte down a set of equatons for the crcut. Notce that because currents are sad to go from + to, ths means that f V AB (the voltage of A, measured relatve to B) s postve, then V A s bgger than V B, and hence I AB wll be postve too. To make the notaton easer we wll take the potental at E to be zero. In symbolc form, ths means that we shall call V BE (that s, V B V E ) V B for short. Krchoff s Frst Law: I EA = I AB ; I BE = I AB +I CB ; I = I CB +I CD ; I CD = I DE Krchoff s Second Law: V B = I BE R 4 = V A +V BA = V I AB R = V D + V CD + V BC = V + I CD R 3 I CB R After elmnaton, the equatons reduce to two: V I AB R = V + I R 3 (R 3 + R ) I CB = (I AB + I CB ) R 4, and from these the currents I AB and I CB can be found (after a bt of messy algebra). After ths, the remanng currents and voltages are straghtforward to determne. These prncples can be used to solve any crcut. However, as networks get bgger, t s useful to fnd more prescrptve methods of soluton, whch could be used by a computer. We shall cover two methods here for certan problems, they may be more effcent than the drect applcaton of Krchoff s Laws. 3 To see why the Law of Conservaton of Energy s nvolved, let us suppose that our coulomb of charge would lose 5J gong from B to E va A, whereas t would lose 3J n gong drectly. All t would have to do s go drect from B to E, then back to B va A and t would be back where t started, havng ganed J of energy! Ths s not allowed. Page

111 Correctons March Method of Superposton The method of superposton reles on the fact that for a smple resstor, the current s proportonal to the voltage. It follows that f current I causes a voltage dfference of 3V, and current I causes a voltage dfference of 5V, then current I +I wll cause an 8V p.d. across the component. Here s the procedure: Choose one of the supply components. Remove the other supply components from the crcut. Replace voltage sources wth drect connectons (short crcuts), and leave breaks n the crcut where the current sources were (open crcuts). Calculate the current n each wre, and the voltage across each component. Repeat the procedure for each supply component n turn. The current n each wre for the orgnal (whole) crcut s equal to the sum of the currents n that wre due to each supply unt. The voltage across each component n the orgnal (whole) crcut s equal to the sum of the voltages across that component due to each supply unt. Let s use ths method to analyse the crcut above. We start by consderng only source V. Removng the other supply components gves us a crcut lke ths. A R B R C R 3 D V R 4 E Ths crcut s easer to analyse as t only has one supply. Supply V feeds a crcut wth resstance R + {R 4 // (R +R 3 )} R4 R R 4 R R3 R R 3 where // means n parallel wth. Accordngly, the current suppled by V (and the current through R whch s n seres wth t) s equal to V dvded by ths resstance. The voltages of ponts B, C and D can be calculated, as can the current n each wre. We make a note of the values, and add to them the results of analyses of crcuts only contanng I and only contanng V. Page

112 Correctons March 7 You may fnd ths method good n the sense that you only have to deal wth one supply component at a tme and therefore all you need to know s how to combne resstors (somethng you ve done before). Havng sad that, we end up analysng three crcuts rather than one, so t s more tme consumng. Before leavng the method, you may be curous why voltage sources were replaced wth short crcuts, and current sources wth open crcuts. Here s the reason. A voltage source does not change the voltage across ts termnals, no matter what the current s (d Voltage / d Current = R equvalent = ). The only type of resstor whch behaves lkewse s a perfect conductor (). Smlarly, a current source does not change ts current, no matter what the voltage (d Current / d Voltage = / R equvalent = ). The equvalent resstor n ths case s a perfect nsulator ( ) whch lets no current through ever Method of Loop Currents Here we break the crcut down nto the smallest loops t contans. Here there are three loops: A B R C D R R 3 V I L R 4 I L I I L3 V E E to A to B and back to E (loop ), E to B to C to E (loop ), and E to C to D to E (loop 3). We call the current n loop loop current number one (I L ), wth I L and I L3 representng the currents n the other two loops. We then express all other currents n terms of the loop currents. Clearly, I AB = I L, snce R s n the frst loop alone. Smlarly, I BC = I L, and I CD = I L3. The current through R 4 s more complex, snce ths resstor s part of two of the loops. We wrte I BE = I L I L. Here I L s postve, snce I BE s n the same drecton as I L, whereas I L (whch goes from E to B then on to C) s n the opposte drecton. These desgnatons automatcally take care of Krchoff s Frst Law. Notce that by ths method, I = I L3 I L. Each loop now contrbutes one equaton Krchoff s nd law around that loop. Clearly, f you go all the way round the loop, you must return to the voltage you started wth. Takng the frst loop as an example, we have: = V AE + V BA + V EB Page

113 Correctons March 7 = V I AB R I BE R 4 = V I L R + (I L I L ) R 4. In a smlar way, we wrte equatons for each of the other two loops 4. We then have three equatons n three unknowns (the three loop currents), whch can be solved. The end result s the same as for a drect sledgehammer approach wth Krchoff s Laws but the method s more organzed Alternatng Current Havng looked at crcuts wth resstors n them, we next turn our attenton to crcuts wth nductors and capactors as well. For a drect current, the stuaton s easy. After a bref perod of settlng down, there s no voltage drop across an nductor (because the current sn t changng), and a capactor doesn t conduct at all. For alternatng currents the stuaton s more complcated. Let us suppose that the supply voltage s gven by V=V cos t. It turns out that the crcut wll settle down to a steady behavour (called the steady state). Once ths has happened, the voltage across each component (and the current through each component) wll also be a cosne wave wth frequency, however t may not be n phase wth the orgnal V Resstor, capactor and nductor We start wth the three smplest crcuts the lone resstor, the lone capactor and the lone nductor, each suppled wth a voltage V=V cos t. For the resstor, I=V/R, so the result s straghtforward. For the capactor, Q=VC, and f we take I as postve n the drecton whch charges the capactor, then dq d I CV cost dt dt CV snt CV cos t CV cos t (3) For the nductor, V L di dt, so 4 It may help when wrtng the equatons to notce the pattern: voltage sources count postvely f you go through them from to +, but negatvely f you go from + to. The voltages across resstors (e.g. V BA = I R ) count negatvely f you go through them n the same drecton as the current, and postvely f you go through them the opposte way to the current. Page 3

114 Correctons March 7 di V cost dt L V I snt L V cost L (3) where we have taken the constant of ntegraton to be zero. Falure to do so would lead to a non-zero mean current, whch s clearly mpossble as the mean supply voltage s zero Reactance and Impedance For resstors, the current and voltage are proportonal, and consequently are n phase one peaks at the same tme as the other. For the other two components, ths s not the case. The voltage s / radans (or 9 ) out of phase wth respect to the current. Inductor currents peak 9 later than the voltage (the current lags the voltage), whereas capactor currents peak 9 before the voltage (the current leads the voltage). Nevertheless, the ampltude of the voltage s stll proportonal to the ampltude of the current, and we call the rato of the ampltudes the reactance (X). X X L C L C (33) By conventon, we take reactance to be postve f the current lags the voltage by 9, and negatve f t leads by 9. For capactors and nductors n seres, the total reactance s equal to the sum of the ndvdual components reactances just as resstances add n seres. Smlarly, the formula for combnng reactances n parallel s the same for that used for the resstance of resstors wred n parallel. When a crcut s constructed wth resstors, capactors and nductors, then we need a way of analysng a crcut wth both resstances and reactances. We vsualse the stuaton usng a D (phasor) dagram. For any component or crcut, both voltage and current are represented by vectors. The length of the lnes gves the ampltude, and the angle between the vectors gves the phase dfference. By conventon, we magne the vectors to rotate about the orgn n an antclockwse drecton (once per tme perod of the alternatng current). The vectors for a resstor, capactor and nductor are shown below. Page 4

115 Correctons March 7 V=I X L Orgn I Orgn I V=IR Orgn I V=I X C Resstor Inductor Capactor As the arrows rotate antclockwse, for the nductor, V comes before I. Wth the capactor, I comes before V. Ths accurately represents the phase relatonshps between voltage and current for these components. For a set of components n seres, the current I wll be the same for all of them. We usually draw the current pontng to the rght. Voltages across nductors wll then pont up, those across resstors pont rght, and those across capactors pont down. By addng these voltages vectorally, we arrve at the voltage across the set of components and can calculate ts ampltude and phase relatonshp wth respect to the current. Smlarly, for components n parallel, the voltage wll be the same for each. We thus put voltage pontng to the rght. Currents n capactors now pont up, currents n resstors pont rght, and currents n nductors pont down. The total current s gven by the vector sum of the ndvdual currents. In all cases, we call the rato of the voltage ampltude to the current ampltude the mpedance (Z) rrespectve of the phase dfference between the current and voltage. 5 In general the mpedance of a component s related to resstance and reactance by Z = R + X Complex Numbers and Impedance If you are famlar wth complex numbers, there s an easer way of descrbng all of ths, usng the Argand dagram n place of -dmensonal vectors. The mpedance Z s a now a complex number Z = R + X, wth R as ts real part and X as ts magnary part. The complex mpedances of a resstor, capactor and nductor are accordngly wrtten as R, /C and L respectvely. The mpedance of a set of components n seres s gven by the sum of the ndvdual mpedances. For a parallel network, Z of the network s gven by the 5 In other words, a resstance s a specal knd of mpedance wth zero phase dfference, and a reactance s a specal knd of mpedance when the phase dfference s 9. Page 5

116 Correctons March 7 sum of Z for each component, where nverse (or recprocal ) s calculated n the usual way for complex numbers Root Mean Square values You wll also need to remember the defnton of RMS voltage and current n an a.c. crcut. For a resstor, remember that the RMS supply voltage s the d.c. voltage whch would supply the same mean power to the devce. Vrms P R V V rms cos t V R R V (34) Resonance One further crcut needs a menton, and that s the smple crcut of an nductor and a capactor connected together, as shown n the dagram below. Both the voltage and current for the two components must be the same, and so wth the sgn conventons chosen n the dagram: Drecton of postve current I C +Q -Q L Q di d Q L L C dt dt d Q Q dt LC (35) Ths s an equaton of smple harmonc moton wth angular frequency, where LC. Ths crcut can therefore oscllate at ths frequency, and ths makes t useful n rado recevers for selectng the frequency (and hence rado staton) whch the lstener wants to detect. 6.4 Questons. Calculate the sze of the repulson force between two electrons.nm apart. Page 6

117 Correctons March 7. In ths queston, you wll make an estmate for the sze of a hydrogen atom. Suppose an electron moves n a crcular path around the proton, wth radus r. Calculate, n terms of r, the potental energy of the atom (t wll be negatve, of course), the speed of the electron, and ts knetc energy. Now wrte down an expresson for the total energy of the electron. Fnd the value of r whch mnmzes ths total energy, and compare t to the measured radus of a hydrogen atom, whch s about 5 m. 3. What fracton of the electrons n the solar system would have to be removed n order for the gravtatonal attractons to be completely cancelled out by the electrostatc repulson? 4. A cloud of electrons s accelerated through a kv potental dfference (so that ther knetc energy of each coulomb of electrons s kj). Calculate ther speed. 5. A beam of kv electrons s travellng horzontally. An expermenter wshes to bend ther path to make them travel vertcally (at the same speed) usng a regon wth a unform electrc feld. Ths regon s square wth sde length 5cm. Calculate the sze and drecton of electrc feld needed to do ths. What would happen to a beam of kv electrons passng ths regon? 6. A dfferent expermenter wshes to bend the beam of kv electrons usng a magnetc feld. She chooses to bend the beam round a crcular path of radus 3cm. What magntude and drecton of magnetc feld s needed? What would happen to a beam of kv electrons passng ths regon? 7. I wsh to make a T magnetc flux densty nsde a long col (or solenod) wth radus 5mm. I use wre whch can carry a current of 4A. How many turns per metre of col are needed? 8. Clamp ammeters used by electrcans can measure the current n a wre wthout needng to break the wre. A metal loop encloses the wre, and the magnetc feld around the wre s measured. If the loop s crcular, wth radus 3cm, and s centred on the wre, calculate the magnetc flux densty measured when the current n the wre s A. 9. Calculate the mpedance of a resstor wred n seres wth a 3mH nductor when fed wth alternatng current of 5Hz. A capactor wred n parallel wth ths combnaton causes the overall reactance of the crcut to become zero at 5Hz (n other words, the voltage s n phase wth the total current). Calculate the capactance of the capactor. Page 7

118 Correctons March 7 Fnally, to convert from klograms to joules, multply by c (the speed of lght squared). A partcular case of a nuclear reacton s the annhlaton reacton, n whch a partcle and ts antpartcle (say an electron and a postron) react together. The matter vanshes, and the energy appears n the form of two gamma rays Questons. Calculate the wavelength and frequency of the quantum assocated wth a 6g ball travellng at 4m/s. Why don t we observe nterference effects wth balls such as ths?. Blue lght has a wavelength of approxmately 4nm, whle red lght has a wavelength of approxmately 65nm. Calculate the energes of photons of blue and red lght (a) n joules (b) n electron-volts (ev). One electron-volt s equal to.6 9 J. 3. Work out the wavelengths of lght emtted when electrons from the n=5, 4, and 3 levels descend to the n= level. Why do you thnk that these transtons were more mportant n the hstorcal development of atomc theory than the more fundamental transtons gong down to the n= level? 4. Calculate the energes of the n=,, 3, 4 and levels for an onzed helum atom (a helum nucleus wth a sngle electron). 5. Calculate the sze of a muonc hydrogen atom n comparson wth a normal hydrogen atom. A muonc hydrogen atom has a muon rather than an electron movng near a proton. The muon has a charge equal to that of an electron, but ts mass s 7 tmes greater. 6. In ths queston, you wll make an estmate for the sze of a hydrogen atom. Suppose that the atom s radus s r. Then the uncertanty n the electron s poston s r. Use the uncertanty prncple to work out the uncertanty n ts momentum, and from ths work out ts typcal knetc energy, n terms of r. The electron s typcal electrostatc energy s e /4 r. Fnd the value of r whch mnmzes the total energy of the electron Calculate the energy lberated n the fuson reacton H H He n. The masses of the partcles are gven n the table n unfed mass unts (u). u = kg. 8 When analysng the collson, you fnd that you can not satsfy momentum and energy conservaton at the same tme f only one photon s produced. Page 4

119 Correctons March 7 7 Small Physcs The rules, or laws, of classcal mechancs break down typcally n three cases. We have seen that when thngs start gong quckly, we need to take specal relatvty nto account. Another form of relatvty the general theory s needed when thngs get very heavy, and the gravtatonal felds are strong. The thrd excepton s very mysterous and occurs often when we deal wth very small objects lke atoms and electrons. Ths s the realm of quantum physcs, and many of ts dscoverers expressed horror or puzzlement at ts conclusons and phlosophy. Havng sad that, there s no need to be frghtened. Whle there s much we do not understand, a set of prncples have been set up whch allow us to perform accurate calculatons. Furthermore, those calculatons agree wth experment to a hgh degree of accuracy. The development of the transstor, hosptal scanner, and many other useful devces testfy to ths. The stuaton s analogous to a lon-tamer who can get the lon to jump through a hoop, though she doesn t know what s gong on nsde the lon. 7. Waves and Partcles Quantum objects, lke electrons and photons (packets of lght) are dffcult to descrbe. As physcsts, we have two models, or descrptons, whch we are comfortable usng the wave and the partcle. Waves can nterfere, they have a wavelength, frequency and ntensty, and they carry energy by means of fluctuatons n a medum. The ntensty s contnuous t can take any value. Partcles on the other hand, are lumps. They possess ndvdual masses, energes and momenta. They most certanly do not nterfere f you add apple to apple, you always get apples. Fnally they only come n nteger numbers. You can have one, or two, or ; but you can t have half. The electron fts nether descrpton. Lght fts nether descrpton. The descrptons are too smplstc. However there are nstances when the partcle descrpton fts well but t doesn t always ft. There are also nstances when the wave descrpton fts well but t doesn t always ft. Gven that a partcular electron beam may behave lke partcles one mnute, and waves the next, we need some knd of phrase book to convert equvalent measurements from one descrpton to the other. Quantum theory mantans that such a phrase book exsts. Page 8

120 Correctons March 7 The total number of partcles (n the partcle pcture) s related to the ntensty (n the wave pcture). The exact converson rate can be determned usng the prncple of conservaton of energy. The energy per partcle (n the partcle pcture) s related to the frequency (n the wave pcture) by the relatonshp Energy of one partcle (n J) = h Frequency of wave (Hz), () where h s the Planck constant, and has a value of Js. You may also come across the constant h-bar h, whch can be used n place of h f you wsh to express your frequency as an angular frequency n radans per second. The momentum per partcle (n the partcle pcture) s related to the wavelength of the wave (n the wave pcture) by the relatonshp h Momentum of one partcle (kg m/s). () Wavelength (m) 7. Uncertanty The brdge between wave and partcle causes nterestng conclusons. We have seen n the chapter on Waves that a wave can have a welldefned frequency or duraton (n tme), but not both. Ths was expressed n the bandwdth theorem: f t. When combned wth our wave-partcle translaton, we obtan a relatonshp between energy and tme: E t h. (3) In other words, only somethng that lasts a long tme can have a very well known energy. Let us have an example. Suppose a nucleus s unstable (radoactve), wth a half-lfe T. Seeng as the emsson of the radaton s a process that typcally takes a tme T, the energy of the alpha partcle (or whatever) has an nherent uncertanty of E h T. If we were watchng a spectrometer, montorng the radaton emtted, we would expect to see a spread of energes showng ths level of uncertanty. The bandwdth theorem also has somethng to say about wavelength: x. Page 9

121 Correctons March 7 Ths has the quantum consequence: p x h. (4) Ths s frequently stated as, You can t know both the momentum and the poston of a partcle accurately. It mght be better stated as, Snce t s a bt lke a wave, t can not have both a well defned poston and momentum. We can use ths to make an estmate for the speed of an electron n an atom. Atoms have a sze of about - m. Therefore, for an electron n an atom, x m. So, usng equaton (4), p 3 kg m/s. Gven the electron mass of about -3 kg, ths gves us a speed of about 7 m/s about a tenth the speed of lght! Cauton: Please note that we haven t defned precsely what we mean by uncertanty (). That s why we have only been able to work wth approxmate quanttes. In more advanced work, the defnton can be tghtened up (to mean, say, standard devaton). However t s better for us to leave thngs as they are. In any case, t s never wse to state uncertantes to more than one sgnfcant fgure! 7.3 Atoms Puttng thngs classcally for a moment, the electron orbts the nucleus. Whle a quantum mechanc thnks ths descrpton very crude, we shall use t as a startng pont. Now, let s magne the electron as a wave. For the sake of vsualzaton, thnk of t as a transverse wave on a strng that goes round the nucleus, at a dstance R from t. If the electron wave s to make sense, the strng must jon up to form a complete crcle. Therefore the crcumference must contan a whole number of wavelengths. R n R nh p nh pr n L n (5) The concluson of ths argument s that the angular momentum of the electron, as t goes round the nucleus, must be n the ħ-tmes table. The argument s smplstc, n that the quantum pcture does not nvolve a lteral orbt. However, the quantum theory agrees wth the reasonng above n ts predcton of the angular momentum. Page

122 Correctons March 7 Gven that the angular momentum can only take certan values (we say that t s quantzed t comes n lumps), we conclude that the electron can only take certan energes. These are called the energy levels, and we can work out the energes as follows: 6 L n Knetc Energy I mr Ze Potental Energy 4 R, (6) where Z s the number of protons n the nucleus. We are, of course, gnorng the other electrons n the atom hence ths model s only drectly applcable to hydrogen. 7 Next, we use the relatonshps derved n secton.., where we showed that for a Coulomb attracton, Potental energy = - Knetc Energy Potental energy = E (7) Knetc energy = -E where E s the total energy of the electron. We may use ths nformaton to elmnate the radus n equatons (6), obtanng: 4 Z e m E n 4. (8) When dealng wth atoms, the S.I. unts can be frustratng. A more convenent unt for atomc energes s the electron-volt. Ths s the energy requred to move an electron through a potental dfference of one volt, and as such t s equal to about.6-9 J. In these unts, equaton (8) can be re-wrtten: Z E 3.6 ev. (9) n Ths form should be remembered. It wll help you to gan a feel for the energes an electron can have n an atom, and as a result, t wll help you spot errors more quckly. 6 The knetc energy s calculated usng the relatonshps derved n chapter 3. If you do not wsh to go n there, a smpler dervaton can be employed. L=mvR, where v s the speed. Therefore the knetc energy mv / = L /(mr ). 7 Hydrogen, that s, and hydrogen-lke ons: whch are atoms that have had all the electrons removed apart from one. Page

123 Correctons March 7 When an electron moves from one level (n value) to another, energy s ether requred or gven out. Ths s usually n the form of a photon of lght that s absorbed or emtted. The energy of the photon s, as usual, gven by the Planck constant, multpled by the frequency (n Hz) of the lght. If an electron moves from orbt n to n, where n <n, the frequency of photon emtted s therefore gven by: 5 f Z 3.9 Hz. () n n Smlarly, the formula gves the frequency of photon requred to promote an electron from n to n. The frequency of photon requred to remove the electron completely from the atom (f t starts n level n ) s also gven by equaton (), f n s taken as nfnte. 7.4 Lttle Nuts As far as Romans were concerned, the stones n the mddle of olves were lttle nuts or nucle. We shall thus turn our attenton to nutty physcs. The nuclear topcs requred for the Internatonal Olympad are common to the A-level course. In ths book we shall merely state what knowledge s needed. You wll be able to fnd out more from your school textbook Types of radaton Alpha decay: n whch a helum nucleus (two protons and two neutrons) s ejected from the unstable nucleus. Beta decay: n whch some werd nucleonc processes go on. In all beta decays, the total number of nucleons (sometmes called the mass number) remans constant. In - decay (the most common), a neutron turns nto a proton and an electron. The electron s ejected at speed from the nucleus. There are two other forms of beta radaton. In + decay, a proton turns nto a neutron and an ant-electron (or postron). The postron fles out of the nucleus, and annhlates the nearest electron t sees. The annhlaton process produces two gamma rays. The other permutaton s electron capture () n whch an electron s captured from an nner (low n) orbt, and reacts wth a proton to make a neutron. Ths phenomenon s detected when another electron descends to fll the gap left by the captve and gves out an X-ray photon as t does so. Page

124 Correctons March 7 Gamma decay: n whch the nucleus re-organzes tself more effcently, leadng to a drop n ts nternal potental energy. Ths energy s released as a burst of electromagnetc radaton a gamma ray photon. By conventon hgh energy photons are called X-rays f they come from the electrons n an atom, and gamma rays f they come from a nucleus Radoactve decay It s beyond the wt of a scentst to predct when a partcular nucleus wll decay. However we have so many radonucldes n a sample that the average behavour can be modelled well. The rate of decay (number of decays per second) s proportonal to the number of nucle remanng undecayed. Ths rate of decay s called the actvty, and s measured n Becquerels (Bq). We defne a parameter to be the constant of proportonalty: I I N N N dn dt e t N t e () I e t where N s the ntal number of radonucldes, and I s the ntal actvty. The half-lfe (T) s the tme taken for the actvty (or the number of undecayed nucle) to halve. Ths s nversely proportonal to, as can be seen: exp T ln T ln T ln T. () If a half-lfe s too long to measure drectly, the value of can be determned f I and N are known separately. I would be measured smply by countng the decays n one year (say), whle N would be measured by puttng a fracton of the sample through a mass spectrometer Nuclear Reactons Now for the fnal technque: You wll need to be able to calculate the energy released n a nuclear reacton. For ths, add up the mass you started wth, and add up the mass at the end. Some mass wll have gone mssng. Rememberng that mass and energy are bascally the same thng the lost mass s the energy released from the nucle. Page 3

125 Correctons March 7 H.4 3 H He 4.64 n.8665 Page 5

126 Correctons March 7 8 Practcal Physcs 8. Errors, and how to make them 9 Every dog has ts day, every slver lnng has ts cloud, and every measurement has ts error. If you doubt ths, take (sorry borrow wth permsson) a school metre stck, and try and measure the length of a corrdor n your school. Try and measure t to the nearest centmetre. Then measure t agan. Unless you cheated by choosng a short corrdor, you should fnd that the measurements are dfferent. What s gone wrong? Nothng has gone wrong. No measurement s exact, and f you take a seres of readngs, you wll fnd that they cluster around the true value. Ths spread of readngs s called random error and wll be determned by the nstrument you use and the observaton technque. To be more precse and polte, ths knd of error s usually called uncertanty, as ths word doesn t mply any mstake or ncompetence on the part of the scentst. So, whenever you wrte down a measurement, you should also wrte down ts uncertanty. Ths can be expressed n two ways absolute and relatve. The absolute uncertanty gves the sze of the spread of readngs. You mght conclude that your corrdor was (.3±.)m long. In other words, your measurements are usually wthn cm of.3m. In ths case the absolute uncertanty s cm. The absolute uncertanty only gves part of the story. A cm error n the length of a curtan track mples sloppy work. A cm error n the total length of the M motorway s an mpressve measurement. To make ths clearer, we often state errors (or uncertantes) n percentage form and ths s called relatve uncertanty. The relatve uncertanty n the length of the corrdor s 9 A mathematcan would probably be appalled at some of the statements I make. The study of errors and uncertantes s embedded n statstcs, whch s a well-establshed dscplne. There are many refnements to the results I quote whch are needed to satsfy the rgour of a professonal statstcan. However, the thng about uncertantes n measurements s that quotng them to more than one sgnfcant fgure s mssng the pont, and therefore our methods only need to be accurate to ths degree. If you are dong statstcs and you want to take thngs more serously, then you wll understand () from the addton of varances; and you wll realse that n 8.. we really ought to be addng varances not errors. You wll also apprecate that () ought to have an (n-) n the denomnator to take nto account the dfference between populaton and sample statstcs, and that our secton 8.. s a form of the Bnomal theorem to frst order. Page 6

127 Correctons March 7 Abs. Uncertanty. m Relatve Error.6% %. () Measurement.3 m Notce the roundng off at the end. It s usually pontless to gve uncertantes to more than one sgnfcant fgure. Every measurement has ts uncertanty, and the only way of determnng ths s to take more than one measurement, and work out the standard devaton to measure the spread. In practce the spread can be eyeballed rather than calculated. If the measurements were 54.5cm, 54.7cm and 54.3cm, then there s no need to use a calculator and the techncal defnton of devaton. The observaton that the spread s about ±.cm s perfectly good enough. Notce that the more readngs you take, the better dea you get of the spread of the measurements and hence the better estmate you can make for the mddle, whch s ndcatve of true value. Therefore we fnd, from statstcs, that f you take n measurements, and the absolute uncertanty s x, then the uncertanty of the mean of those measurements s approxmately: 3 x Uncertant y of mean. () n Therefore, the more measurements you take, the more accurate the work. Notce that f you wsh to halve the uncertanty, you need to take four tmes as many readngs. Ths s subject to one provso: Measurements also have a resoluton. Ths s the smallest dstngushable dfference that the measurng devce (ncludng the technque) can detect. For a smple length measurement wth a metre ruler, the resoluton s probably mm. However f, by years of practce wth a magnfyng lens, you could dvde mllmetres nto tenths by eye, you would have a resoluton of.mm usng the same metre stck. That s why we say that the resoluton depends on the technque as well as on the apparatus. The uncertanty of a measurement can never be less than the resoluton. Ths s the provso we mentoned below equaton (). Why should ths be the case? Let us have a parable. Many years ago, the great naton of Chna had an emperor. The masses of the populaton were not permtted to see hm. One day, a ctzen had 3 Ths result wll be proved n any statstcs textbook. To gve a bref justfcaton the more readngs you take, the more lkely you are to have some hgh readngs cancellng out some low readngs when you take the average. Page 7

128 Correctons March 7 the sudden desre to know the length of the emperor s nose. He could not do ths drectly, snce he was not permtted to vst the emperor. So, usng the apparatus of the mperal admnstraton, he asked all the regonal mandarns to ask the entre populaton to make a guess. Each person would make some guess at the mperal nasal length and the error of each guess would probably be no more than ±cm snce nose lengths tend not to vary by more than about 4cm. However, the mean would be a dfferent matter. Averaged over the mllon measurements, the error n the mean would be.7m. So the emperor s nose had been measured ncredbly accurately wthout a sngle observaton havng been made! The moral of the story: uncertantes are reduced by repeated measurement, but the error can never be reduced below the resoluton of the technque here cm snce gnorance can not be crcumvented by poolng t wth more gnorance. 8. Errors, and how to make them worse Errors are one thng. The trouble s that usually we want to put our measurements nto a formula to calculate somethng else. For example, we mght want to measure the strength of a magnetc feld by measurng the force on a current-carryng wre B F IL. If there s a 7% uncertanty n the current, % n the force and % n the length what s the uncertanty n the magnetc feld? There are two rules you need: 8.. Rule Addng or subtractng measurements If two measurements are added or subtracted, the absolute uncertanty n the result equals the sum (never the dfference) of the absolute uncertantes of the ndvdual measurements. Therefore f a car s (3.±.)m long, and a caravan s (5.±.)m long, the total length s (8.4±.3)m long. Smlarly f the heght of a two-storey house s (8.3±.)m and the heght of the ground floor s (3.±.)m, the heght of the upper floor s (5.±.3)m. Even n the second case, we do not subtract the uncertantes, snce there s nothng stoppng one measurements beng hgh, whle the other s low. 3 3 Of course, there s a good chance that the errors wll partly cancel out, and so our method of estmatng the overall error s pessmstc. Nevertheless, ths knd of error analyss s good enough for most experments after all t s better to overestmate your errors. If you want to do more careful analyss, then you work on the prncple that f the absolute uncertantes n a Page 8

129 Correctons March Rule Multplyng or dvdng measurements If two measurements are multpled or dvded, the relatve uncertanty n the result equals the sum (never the dfference) of the relatve uncertantes of the ndvdual measurements. Therefore f the speed of a car s 3mph ± %, and the tme for a journey s 6 hours ± %, the uncertanty n the dstance travelled s %. Notce that one consequence of ths s that f a measurement, wth relatve uncertanty p% s squared (multpled by tself), the relatve error n the square s p% -.e. doubled. Smlarly f the error n measurement L s p%, the error n L n s p n%. Notce that whle a square root wll halve the relatve error, an nverse square (n= ) doubles t. All the mnus sgn does s to turn overestmates nto underestmates. It does not reduce the magntude of the relatve error. 3 Now we can answer our queston about the magnetc feld measurement at the begnnng of the secton. All three relatve errors (n length, force and current) must be added to gve the relatve error n the magnetc feld, whch s therefore %. 8.3 Systematc Errors All the errors mentoned so far are called random, snce we assume that the measurements wll be clustered around the true value. However often an oversght n our technque wll cause a measurement to be overestmated more often than underestmated or vce-versa. Ths knd of error s called systematc error, and can t be reduced by averagng readngs. The only way of spottng ths knd of error (whch s a true error n that there s somethng wrong wth the measurement) s to repeat the measurement usng a completely dfferent technque, and compare the results. Just thnkng hard about the method can help you spot some set of measurements are A, B, C, then the absolute uncertanty n the sum (or n any of the dfferences) s gven by A B C. Ths result comes from statstcs, where we fnd that the varance (the square of the standard devaton) of a sum s equal to the sum of the varances of the two measurements. 3 The conclusons of ths paragraph can be justfed usng calculus. If measurement x has n absolute uncertanty x, and y (a functon of x) s gven by y Ax, then we fnd that the relatve error n y s gven by: y dy x n x y x x An x n n, y dx y y x y x that s n multpled by the relatve error n x. Page 9

130 Correctons March 7 systematc errors, but t s stll a good dea to perform the experment a dfferent way f tme allows. 8.4 Whch Graph? You wll often have to use graphs to check the functonal form of relatonshps. You may also have to make measurements usng the graph. In order to do ether of these, you usually need to manpulate the data untl you can plot a straght lne. A straght lne s conclusve proof that you have got the form of the formula rght! The gradent and y-ntercept can then be read, and these enable other measurements to be made. For example, your am may be to measure the acceleraton due to gravty. You may plot velocty of fallng aganst tme, n whch case you wll need to fnd the gradent of the lne. At ts most general, you wll have a suspected functonal form y=f(x), and you wll need to work out what s gong on n the functon f. Notce that our experment wll gve us pars of (x,y) values what s not known are the parameters n the functon f. We fnd them by manpulatng the equaton: y f ( x). g( x, y) A h( x, y) B We can then plot g(x,y) aganst h(x,y), and obtan the parameters A and B from the gradent and ntercept of the lne. Furthermore, the presence of the straght lne on the graph assures us that our functon f was a good guess. We shall now look at the most common examples Exponental growth or decay Here we have the functonal form y Ae determned. We manpulate the equaton: y Ae Bx ln y ln A. Bx Bx, where A and B need to be So we plot (ln y) on the vertcal axs, and (x) on the horzontal. The y- ntercept gves ln A, and the gradent gves B Logarthmc growth or decay Here we have the functonal form y A B ln x, and agan, we need to work out the values of A and B. Ths equaton s already n lnear form we plot y on the vertcal, and (ln x) on the horzontal. The y-ntercept gves A, and the gradent gves B. Page 3

131 Correctons March Power laws Ths covers all equatons wth unknown powers: B y Ax. The manpulaton nvolves logarthms: y Ax ln y ln A ln x ln y ln A B ln x B Here we plot (ln y) aganst (ln x), and fnd the power (B) as the gradent of the lne. The A value can be nferred from the y-ntercept, whch s equal to ln A Other forms Even hdeous lookng equatons can be reduced to straght lnes f you 3 crack the whp hard enough. How about y A x Bx? Is t tasty enough for your breakfast? Actually t s fne f dgested slowly: B. y A x Bx y A Bx x 3 5. Ths looks even worse, doesn t t? But remember that t s x and y that are known. If we plot ( y x ) on the vertcal, and (x 5/ ) on the horzontal, a straght lne appears, and we can read A and B from the y-ntercept and gradent respectvely. 8.5 Questons. Work out the relatve uncertanty when a 5V battery s measured to the nearest.v.. If I don t want to have to correct my watch more than once a week, and I never want my watch to be more than s from the correct tme, calculate the necessary maxmum relatve uncertanty of the electronc oscllator whch I can tolerate. 3. My two-storey house s 7.5±.m tall. The ground floor s 3.±.m tall. How tall s the frst floor? 4. I want to measure the resstance of a resstor. My voltmeter can read up to 5V, wth an absolute uncertanty of.v. My ammeter can read up to A wth an absolute uncertanty of.a. Assumng that my resstor s approxmately Ω, calculate the absolute uncertanty of the resstance I measure usng the formula R=V/I. Assume that I choose the current to make the relatve uncertanty as small as possble. Page 3

132 Correctons March 7 9 Appendx 9. Multplyng Vectors Physcs s rddled wth quanttes whch have both magntude and drecton velocty, acceleraton, dsplacement, force, momentum, angular velocty, torque, and electrc feld to name but eght. When descrbng these, t s very useful to use vector notaton. At best ths saves us wrtng out separate equatons for each of the components. Whle the addton and subtracton of vectors s reasonably straghtforward (you add, or subtract, the components to get the components of the result), multplcaton s more trcky. You can thnk of a vector as a lttle arrow. You can add them by stackng them nose-to-tal, or subtract them by stackng them nose-to-nose. But how do you go about multplyng them? It s not obvous! To cut a long story short, you can t do t unambguously. However there are two vector operators whch nvolve multplcaton and are useful n physcs. Ordnary multplcaton s commonly wrtten wth ether the cross ( ) or dot ( ), so when t comes to vectors we call our two dfferent multplcaton processes the dot product and cross product to dstngush them. These are the closest we get to performng multplcaton wth vectors. 9.. The Dot product (or scalar product) A ton of brcks s lfted a metre, then moved horzontally by m. How much work s done? Work s gven by the product of force and dstance, however only the vertcal lftng (not the horzontal shufflng) nvolves work. In ths case the work s equal to the weght (about 9.8kN) multpled by the vertcal dstance (m). Ths gves us one useful way of multplyng vectors namely to multply the magntude of the frst, by the component of the second whch s parallel to the frst. If the two vectors are A and B, wth magntudes A and B, and wth an angle between them, then the component of B parallel to A s Bcos. Therefore the dot product s gven by ABcos. A B AB cos () Notce that the dot product of two vectors s tself a scalar. Note that when we talk about the square of a vector, we mean ts scalar product wth tself. Snce n ths case, =, ths s the same as the square of the vector s magntude. Page 3

133 Correctons March 7 The dot product s also commutatve, n other words, the order of the two vectors A and B does not matter, snce A B=B A. The dot product of two vectors wrtten usng Cartesan co-ordnates s partcularly easy to calculate. If we use, j, and k to represent the unt vectors pontng along the +x, +y and +z axes, then a bj ck u vj wk au av j aw k buj bvj j bwj k cuk cvk j cwk k au bv cw () 9.. The Cross product (or vector product) If the dot product produced a scalar, what are we to do f a vector s needed as the result of our multplcaton? Answer: a cross product. Our frst dlemma s to choose the drecton of the result. Gven that the vectors wll, n general, not be parallel or antparallel, we can t choose the drecton of one of them that would not be far! The two vectors wll usually defne a plane, so perhaps we could use a vector n ths plane as the result? No, that wouldn t do ether there s stll an nfnte number of drectons to choose from! A soluton s presented f we choose the vector perpendcular to ths plane. Ths narrows the choce down to two drectons and we use a conventon to choose whch. Notce that the result of the cross product must be zero f the two vectors are parallel, snce n ths case we can t defne a plane usng the vectors. It follows that the cross product of a vector wth tself s zero. Ths means that we aren t gong to be nterested n the component of the second vector whch s parallel to the frst when calculatng the product. On the contrary, t s the perpendcular component whch matters. The cross product of two vectors s defned as the magntude of the frst, multpled by the component of the second whch s perpendcular to the frst. The product s drected perpendcular to both vectors. To be more precse, magne a screw attached to the frst vector. The cross product goes n the drecton the screw advances when the frst vector s twsted to lne up wth the second. The cross product of a vector lyng along the +x axs wth one lyng along the +y axs s one lyng along the +z axs. The cross product of up wth forwards s left. A B AB sn (3) Wth a defnton as obtuse as ths, you could be forgven for wonderng whether t had any practcal use at all! However they turn out to be very useful n physcs especally when dealng wth magnetc felds and rotatonal moton. Page 33

134 Correctons March 7 Notce that, where the dot product was commutatve, the cross product s antcommutatve. In other words, A B= B A, so make sure you don t swap the vectors over nadvertently. The vector product of vectors wrtten n Cartesan form can also be calculated: a bj ck u vj wk au av j aw k buj bvj j bwj k cuk cvk j cwk k avk awj buk bw cuj cv bw cv cu awj av bu a u where the most convenent way of rememberng the result s as the determnant of the 3 3 matrx shown. j b v k c w k (4) Page 34

135 Correctons March 7 9. Dmensonal Analyss If you look back at the flow equaton (4 n secton.3.3), you wll see somethng nterestng about the unts. Current (A) = Charge densty (C/m 3 ) Area (m ) Speed (m/s) If we do algebra wth the unts on the rght hand sde, we get C m C m A, 3 m s s and ths agrees wth the unts of the left hand sde. Now ths may all seem pretty obvous, but t gves us a useful procedure for checkng whether our workng s along the rght lnes. If, durng your calculatons, you fnd yourself addng a charge of 3C to a dstance of 6m to get a result of 9N; or you multply a speed of 3m/s by a tme of 4s and get a current of 5A; then n ether case you must have made a mstake! We can also use the prncple that unts must balance to guess the form of an equaton we do not know how to derve. For example, you may guess that the tme perod of a smple pendulum mght depend on the length of the pendulum L, the strength of the local gravty g and the mass of the pendulum bob m. Now L s measured n m g s measured n N/kg or m/s m s measured n kg, and we want a tme perod, whch wll be measured n s. The only way these measurements can be combned to make somethng n seconds s to take L, dvde t by g (ths gves somethng n s ) and then take the square root. Therefore, wthout knowng any physcs of the smple harmonc oscllator, we have shown that the tme perod of a pendulum s related to L g and wll be ndependent of the mass m. Smlarly, notce what happens f you multply ohms by farads: V C C F s. A V A Yes, you get seconds. Therefore, t should come as lttle surprse to you that f you double the resstance of a capactor-resstor network, t wll take twce as long to charge or dscharge. Furthermore, you have worked ths out wthout recourse to calculus or the tedous electrcal detals of secton 6... Page 35

136 Correctons March 7 Of course, one drawback of the method presented here s that some quanttes have two dfferent unts. For example, gravtatonal feld strength could be N/kg or m/s. Electrc feld strength could be N/C or V/m. How do you know whch to choose? The answer s that f you restrct yourself to usng the mnmum number of unts n your workng, and express all others n terms of them, you wll not have any dffcultes. Usually people choose m, s, kg and A but any other combnaton of ndependent unts 33 wll do equally well. 34 In books you may see folk use L, T, M, I and to represent the dmensons of length, tme, mass, electrc current and temperature. Ths s just a more formal way of dong what we have done here usng the S.I. unts. In these books, the dmenson of speed would be wrtten as [speed] = L T, and the dmensons of force would be wrtten [force] = M L T, where the square brackets mean dmensons of. Techncally, ths s more correct than usng the S.I. unts, because some quanttes are dmensonally the same, but have very dfferent meanngs (and hence unts). For example, torque and energy have the same dmensons, but you wouldn t want to rsk confusng them by usng the same unt for both. Smlarly an angle n radans has no dmensons at all (beng a an arc length n metres dvded by a radus n metres), but we wouldn t want to confuse t wth an ordnary number lke By ndependent we mean that no one unt can be derved entrely from a combnaton of the others. For example, m, s, kg and J would be no good as a set of four snce J can already be expressed n terms of the others J = kg (m/s), and hence we have ambguty arsng as to how we express quanttes. 34 OK, f you want to do work where there are electrc currents and temperatures as well as mechancal quanttes, you mght need to go up to fve (an extra one for temperature). 35 Indeed, quanttes wth no unts are usually sad to have the dmensons of the number one. Thus [angle] =. It follows that angular velocty has dmensons of [angle][tme] = T = T. Ths comes from the property has n beng the unty operator for multplcaton. Page 36