Classical Mechanics (CM):

Transcription

1 Classical Mechanics (CM): We ought to have some backgound to aeciate that QM eally does just use CM and makes one slight modification that then changes the natue of the oblem we need to solve but much of the hysics emains the same. Outline of This ae:. Defining a Hamiltonian in CM (which then QM uses as the basis of all mechanics). The consevation of enegy, and how that leads to Newton's Laws of motion. 3. Showing alications of diffeent tyes of otentials in CM and solving the motion. 4. Relating the momentum to the velocity. 5. Deiving the Classical Viial Theoem which elates the Kinetic and Potential Enegies. 6. Discussing the mechanics of a chaged aticle going aound a nucleus. Pat : The fist oint is that we need in both QM and CM a statement about the enegy. This statement is used in both foms of mechanics with a Hamiltonian. The Hamiltonian is the total enegy (the sum of the kinetic and otential enegies) witten in tems of the osition (q o x) and momentum () of the aticle. Hamilton fomulated his ideas fo classical mechanics. They ae the bases fo quantum mechanics as well. Basically the classical Hamiltonian is just the sum of the kinetic and otential enegies. Geneally the kinetic enegy (T) deends only on the momentum of the aticle, and the otential enegy (V) deends only on the osition of the aticle. The Hamiltonian is H = T + V. Fo most situations: T = mv = mv v =, and V is just the otential enegy as a m esult of the envionment and theefoe deends only on the osition of the aticle V = V( q), and V V(, q). The dot" notation is quite convenient and is just dq shothand fo taking the time deivative. The velocity (v) is: v vq q. The otential enegies we will conside will be: V = 0 fee sace V = mgx gavity on the suface of the eath V = kx Hooks Law (sings) Hamonic Motion Ze V = Coulomb's Law mm V = gavity in geneal Pat : Thee ae a coule of imotant ideas that hel us undestand the oint of Hamilton s exession. The fist is: The enegy is conseved. Which means it does not change in

2 time. Classically if an object moves into a egion of lowe otential it seeds u so that the total enegy of the aticle is constant. Theefoe: dh = 0 This is vey imotant. The consequence is that we can deive Newton s laws of motion fom this simle equiement: H = T + V T = V = V ( ) m dh dt dv 0 = = + Using these definitions of T and V and the chain ule we have: dt d dv d dv ( ) = = m d d Then combining with the definition of momentum: = yields: m dt dv 0 = + d d dv ( ) 0 = + m d d dv ( ) 0 = + m d The at in the bace is set to zeo: This is Newton s Law of motion. Each tem is elated to the foce on the aticle. The foce comes fom the otential (secifically, the dv ( ) gadient of the otential) F = and the foce causes the aticle momentum to d d change in time F =. A system is said to be "conseved" (enegetically) when thee ae no extenal foces acting on it. Then only the intenal foces ae used to change aticle momentum: d dv ( ) = d d dv ( ) F = = d This exession, combined with the definition of the momentum give us two equations and two unknowns (x,) that can be detemined: Newton's Equations d = m d dv ( ) = d

3 Pat 3: Alications of Newton's Laws Let s see how these (Newton's) equations ae used to solve fo the osition, and momentum (x,) of a aticle as a function of time. Case : Fee sace: V = 0 Because thee is no otential thee is no foce on a aticle, so the momentum is d conseved. 0 =, which imlies ( t) = ( 0) o, a constant. We can now solve fo the osition given a constant momentum fom the definition of the momentum: d m = = o As is a constant this equation is solved by integating (both sides of the equation) fom time 0 to t: t t d m = o = ot t = 0 t = 0 m () t m ( 0) = ot () t = o + m ot So the aticle moves at a constant velocity and the osition changes linealy with time deending only on whee it stated and how much momentum it had initially (in the thee diections). Case : Gavity on the suface of the eath: V = mgx This is a simlified gavity otential whee x is the distance off the suface of the eath: x = o, and the otential enegy is that elative to the eath's suface. The two fundamental equations of motion ae: dx d dv ( x) = m and = dx Fo this alication the secific equations of motion ae that the momentum is subjected to a constant foce: dx d = m and = mg The equations can be integate u, to obtain the momentum (o velocity), using the second equation, and then integated to get the osition, using the fist equation. d Integate = mg () t ( 0) = mgt dx Integate m = = o mgt mx() t mxo = o t mgt x() t = xo + vot gt Notice the momentum become moe negative as time inceases, the object is falling. The osition has a quadatic deendence on time; the aticle is aoaching eath faste and faste. x() t gets smalle in time. Thee is a catch to this that is not shown diectly, and that is that x cannot be negative. Once the aticle hits the gound the equations sto alying, unless the aticle is going down a mine shaft o a fault in the eath, then things 3

4 change slightly. But assuming the otential only alies when the aticle is in the ai, eveything is OK. Thee ae two abitay constants: whee the aticle stats and how fast it is going at the stat ( x, = mv. These ae bounday conditions. Each fist ode diffeential o o o) equation is solved in tun hee but one could wite a second ode D.E. diectly fo x (but not fo ): d d x mg = = m d x = g Notice that the single equation can be solved on its own, but equies two bounday conditions. Also notice the change in osition of the aticle is the same egadless of mass. Often this is the method of solution shown in intoductoy hysics; I think it is moe confusing that the method used above, but it is oula as it gets id of an intemediate vaiable (the momentum). I disagee with this aoach and uge you to focus on how the oblem was solved above. Afte all we ae inteested in the momentum as much as the osition. The thid case: V = kx The Hamonic Oscillato Fo this examle the two fundamental equations become: dx m = d dv ( x) = = kx dx These equations ae quite symmetic in x and. We can combine the two to get a second ode D.E. in tems of eithe x o. d x m d = dx m = d d dx = kx = k d x k d k = x = m m Notice this shows you can think about the equations of motion in eithe x o, indeendently. So eithe way woks fine. The two equations (in the boxes hee) ae identical in fom, just x and ae intechangeable symbols. This means that the same solution alies equally well to eithe quantity. This might seem contadictoy to the idea that the momentum is ootional to the time change of x, but we will see that that is not a oblem In Q.M. one says that eithe a "momentum sace" o "osition sace" 4

5 eesentation of the oblem is equally valid. Unlike the evious two oblems this cannot be integated to obtain eithe o x as a function of time. One must solve eithe the two couled fist ode D.E.s o eithe of the second ode D.E. (in the boxes). The solution is well known, because the second deivative of a cosine o sine function gives that function back (times a constant). Theefoe these ae the two indeendent solutions. = t = Asin ωt+ φ d ( ) ( ) = ω Theefoe the fequency, ω, of oscillation is tied to the sing foce constant and the k mass: ω =. [FN: This is an eigenvalue oblem whee is the eigenfunction.] A, φ o m ae constants one can set: Fom the aoiate exession the osition can be detemined: d ω x= = Acos( ωt+ φo ) k k A x= cos( ωt+ φo ) km The Hamiltonian o enegy is athe inteesting. We said at the beginning that the equations of motion should show that the Hamiltonian is indeendent of time. So let's substitute ou solutions to (x,) and evaluate the Hamiltonian as a function of time: H = + x k m A km H = sin ( ωt+ φo) + cos ( ωt+ φo) m ( km ) A H = m As one knew at the beginning the Hamiltonian must be a constant of the motion, so it is indeendent of time. The enegy (which is the Hamiltonian afte it has been evaluated as a function of time) is ootional to the squae of the amlitude of the motion. Let's conside the hysical inteetation of these equations. Set φ o = 0 o. In this case the bounday conditions (o the values of x and at time zeo) tell us that the sing has been ulled to its maximum extension to the left, negative x. The extension ( x o ) is: A xo =. As the aticle is eleased (at time zeo) it oceeds to the ight (inceasing x, km i.e. a smalle negative numbe), and the momentum inceases. When the aticle cosses π the y axis at x = 0, the time is ω t =, o 90degees, it has gone /4 of its cycle. At this oint the momentum is maximal (and contains all the enegy), and = max = A, and we max find that the maximum in x is: xmax = km 5

6 Pat 4: The momentum and the velocity. We ae all etty well conditioned to accet that the momentum is just the mass times the velocity. This is not always the case. The definition of momentum needs to go beyond what we aleady know. The eason the momentum is associated with a coodinate is that it is deived fom the enegetics of the system in tems of that coodinate's velocity. Recall that momentum is actually call the "conjugate momentum", meaning a elation to a coodinate velocity. So hee is how the momentum is officially defined. [FN: This is intended to exlain why the usual definition of momentum and why we need to do moe.] The full definition of momentum: The kinetic and otential enegies ae witten in tems dq of coodinates, q, and velocities: q. Both T and V can be functions of both{ qq, }. The official definition of momentum conjugate to q is: d( T V) = q = dq This may seem like it came out of thin ai, but we eally do not want to develo the equation that leads to this. Even though this is the eally comlete definition, fo ou uoses, the otential is not a function of the velocity (which only haens when magnetic fields ae esent) and so the moe tyical definition is: dt q = dq The deivative, w..t. the velocity, by the way, means to take the deivative but conside the coodinate (q) to not be a function of vq q. The kinetic enegy, T, is always the same, egadless of oblem but sometimes it is witten in diffeent coodinate systems, and so it can look a little diffeent. Let's stat with Catesian coodinates: T = m{ vx + vy + vz} vq = q o goued as a vecto v = The deivative of T w..t. any one of the thee coodinate-velocities then gives the momentum conjugate to that coodinate. The esult is the usual esult: dt d{ v } x + vy + vz x = = m = mv x dvx dvx The same holds fo y o z. This is the definition of the momentum and its elation to the velocity. Now T and V can be ewitten in tems of the momentum and osition athe than velocities and ositions. T = m Then the Hamiltonian can be fomed: H=T+V, whee T and V ae witten in tems of momentum and osition. That seems staightfowad, and gives the momentum we all know. So why do we need to go though all of this? Well, if we have a oblem of a stationay obit (the electon aound the nucleus o the eath aound the sun) the otential is a cental foce otential and witten in tems of the distance between the aticles, so sheical coodinates ae moe aoiate because then V deends only on one of the thee coodinates (the ). 6

7 So we need to know what the kinetic enegy looks like when we wite it in tems of sheical coodinates. Fom the definition of the Catesian coodinates in tems of sheical coodinates we can wite the velocity and then the kinetic enegy. We use the above definition to find the momentum conjugate to the adius and the angles. The elation between Catesian and ola coodinates: x sinθ cosφ = y = sinθ sinφ z cosθ The velocity is x sinθ cosφ cosθ cosφ sinφ v = = y = sinθ sinφ + θ cosθsinφ + sinθφ cosφ z cosθ sinθ 0 The kinetic enegy is found now fom the inne oduct of the velocity vectos: v v = sinθ cosφ sinθ cosφ cosθ cosφ cosθ cosφ sinφ sinφ sinθ sinφ sinθsinφ + ( θ) cosθsinφ cosθ sinφ + ( sinθφ ) cosφ cosφ cosθ cosθ sinθ sinθ CossTems T = ( ) ( ) mv v = m + θ + sinθφ ( ) The coss tems all vanish and the inne oducts of the vectos with angle ats all give. So the answe is not too bad given the messy intemediate matix multilication. Fom hee now we can obtain the momentum conjugate to{, θ, φ}. This is whee the moe geneal definition of momentum comes in handy. T = m + θ + sinθφ ( ( ) ( ) ) dt dt dt θ ( sinθ) φ d d θ d φ = = m θ = = m φ = = m ( θθ φφ) T = + + Notice that we have now defined the momenta conjugate to each of the velocities and the kinetic enegy can be witten in a fom like that fom Catesian coodinates: T = v. We can finish u by witing T fully in tems of the momenta, using the conventional definition of the moment of inetia, I: I = m T = m + I θ + sin φ θ 7

8 The momenta ae not nealy as tidy as they wee in Catesian coodinates. Also, the kinetic enegy now deends on the coodinates not just the velocities. In aticula θ = I θ, which looks like the angula analogue of linea momentum, but I contains the coodinate (not ). So this makes T moe comlicated and causes us to ethink how we handle the mechanics of motion going aound a cental foce. Pat 5: The viial theoem The oblem we would like to do is the gavitational attaction o chage attaction (Coulomb s law) of one aticle obiting aound anothe. These two oblems ae identical as the otential is ootional to in both cases. The otential is also negative in both cases which lead to bound obits. Howeve, is eally too tedious and time consuming fo us. We can do the obit oblem a diffeent way. And that is to fist ove the vial theoem fo classical mechanics. This is not too had to do fo the cases we ae consideing: The viial (G(t)) is x time (o time = ). We follow this oduct in time: ( ) () G t = dg t d d = + These two tems can each be evaluated (using Newton's Laws, above) in tems of T and V: d m d = = = T m m d dv = d n Fo cases whee the otential enegy is ootional to to some owe V = a we have: n d dv d = = a = nv d d Combining all of the ats, we can elate the viial to T and V: dg ( t) = T nv Now aveage ove time (fom 0 to time T). The angula baces,, mean the same as aveaging. These baces will be used a lot in this couse. t d dg ( t ) G ( t) G ( 0 G ) = T n V t = = t t= 0 If we aveage ove a time such that the system etuns to whee it was at time zeo: Then G t G 0 = 0, and we have the elation between the kinetic and the otential. () ( ) So fo any system that is eiodic, ove a eiod of the motion we have the elation: T = n V 8

9 Fo the Hamonic oscillato n= and we have that the mean kinetic and mean otential enegies ae the same, and fo the bound electon, whee n=-, the elation is: T = V. Pat 6: Motion in a cental foce Fom the viial theoem we can obtain some insight into motion of a aticle in an obit of constant adius. Above we wote the kinetic enegy in tems of ola coodinates. We can simlify the kinetic enegy exession by saying the aticle is confined to a secific adius (so = 0 ), and is otating in the x-y lane (so θ = 0 and sinθ = ), in a stationay obit. The kinetic enegy exession simlifies in this case to: T = m φ = I φ ( ) And the angula velocity, ω = φ, is constant (time indeendent). Then (both T and V ae always constant): T = V Ze Iω = 3 Ze ω = m The oblem is solved by equating all the constants of the motion. This ties togethe the adius to the angula velocity: as the adius of a stationay obit inceases the angula velocity must slow down. All quantities ae constants of motion but the constants ae not indeendent of each othe. This is tue fo lanets as well. The futhe out lanets must obit the sun at slowe ates, and they exactly follow this law. Howeve, showing that one is allowed to say that the angula velocity is a constant of the motion takes a bit of wok. This eally is a good lace to sto. 9

10 Beyond hee lie Dagons: Pat 7 The mechanics of motion in a cental foce Fo the moe geneal case we will need to use Hamilton s equations of motion (which ae a vaiant on Lagange s equations), which educe to Newton s laws fo the simle cases we have develoed. We need, fo late, the idea of the angula momentum and how that elates to the conjugate mometa we defined in at 4. So the definition of angula momentum, witten as a vecto, initially in Catesian coodinates is: L = m v = This is vey useful fo a aticle taveling in a cental foce, because: L = m v+ = mv v V = 0 The angula momentum is a constant of the motion because the two tems vanish. The fist tem vanishes identically because the coss oduct of a vecto with itself must be zeo. The second tem vanishes because the otential only deends on. So the gadient of the otential is in the same diection as (oints outwad, adially only) and so that coss oduct must vanish, as well. To show this exlicitly: = x + y + z ( ) / d 3/ 3/ ( ) ( ) d d x = = = x= 3 dx dx dx = 3 This makes the angula momentum a constant of the motion. Now we need to wite L out in Catesian and ola coodinates and show that: φ L L = θ + sin θ Fom this it follows that the Kinetic Enegy is: T = + L Finding L in ola coodinates: m I 0

11 L = m v sinθ cosφ sinθ cosφ cosθ cosφ sinφ = m sinθsinφ sinθsinφ + θ cosθsinφ + sinθφ cosφ cosθ cosθ sinθ 0 m sinθ cosφ cosθ cosφ sinφ sinθsinφ θ cosθsinφ + sinθφ cosφ cosθ sinθ 0 = m sinφ cosθ cosφ θ cosφ sinθφ cosθ sinφ 0 sinθ = And we can summaize this as: L = Iω, which gives the full definition of the angula velocity in tems of the ola angles. The z comonent of L is, igoously: Lz = Isin θ φ =. φ Now get the squae of the angula momentum: sinφ sinφ cosθ cosφ cosθ cosφ L L I θ cosφ cosφ ( sinθφ) = + cosθ sinφ cosθ sinφ + CossTems 0 0 sinθ sinθ The coss tems cancel and the diect vecto oducts ae : L L = I θ + sinθφ { ( ) } I sin θφ φ L L = ( I θ ) + = θ + sinθ sin θ Conside the aticle to be otating in the x-y lane (sinθ = ). The Kinetic enegy is: T = + φ m I The angula momentum is the momentum conjugate to the angula vaiable, and the angula velocity is ω = φ. Hamilton s equations of motion ae: H H qk = and k = k qk The fist equation is the definition of the conjugate vaiable and the second equation is Newton s law, fo a foce. Now let's aly these equations to see how to solve fo the motion of a aticle constained to be in the x-y lane unde the influence of a cental foce. The fist equation is that fo the hi-otation: φ = 0 o φ = Iω is a constant of the motion. Notice the angula velocity may change in time, we don't know what that does

12 yet, as the distance changes. Now the adial momentum equation of motion is moe comlex and moe inteesting (I suose): The fist equation is just the elation between the adial velocity and the momentum. The second one gives a descition of the foces involved: H H = and = T V T φ ( ) m I m + = = = + = T + V φ Ze = = I If the adial distance does not change we have: φ Ze 0 = = I Ze mω = At this oint we have a fist ode diffeential equation fo the change in the adial comonent and eveything else is a constant: φ Ze m = I 3 φ m = Ze m