PY1002: Special Relativity

Transcription

1 PY100: Speial Relativity Notes by Chris Blair These notes over the Junior Freshman ourse given by Dr. Barklie in Mihaelmas Term 006. Contents 1 Galilean Transformations 1.1 Referene Frames Galilean Invariane Galilean Transformations Mihelson-Morley Experiment.1 The Ether Brief Outline of The Experiment Lorentz Transformations The Speial Theory Of Relativity Lorentz Transformations Relativisti Effets Length Contration Time Dilation Light Pulse Clok Method Simultaneity Minkowski Diagrams Composition Of Veloities Angle Transformation ( Headlight Effet ) Relativisti Doppler Effet Relativisti Dynamis Relativisti Energy And Momentum Problems Involving Relativisti Dynamis Inelasti ollision Positron-Eletron Collision and Annihilation Collision and Sattering Angle Compton Effet A Appendix: List of Formulae 17 1

2 1 Galilean Transformations 1.1 Referene Frames A key onept in speial relativity is that of a referene frame. A referene frame is basially a set of axes (a o-ordinate system) whih a partiular observer uses to reord events. A referene frame is either inertial (aelerating) or non-inertial (not aelerating). In speial relativity, an event onsists of a partiular set of spae-time o-ordinates of the form (x, y, z, t). 1. Galilean Invariane The priniple of Galilean Invariane states that the basi laws of physis are idential in all inertial referene frames. Essentially, this means that there is no way to tell by experiment if the referene frame you are in is moving or not. 1.3 Galilean Transformations Consider an inertial referene frame S and a seond inertial frame S moving with veloity v away from it along the x-axis. y S S y z x z Suppose the origins of the two frames oinide at time t = t = 0. Then we have the following Galilean transformations for events in the two frames: and the veloity transformations: x = x + vt y = y z = z t = t v x = v x + v v y = v y v x Galilean transformations Galilean veloity transformations v z = v z However these transformations are only orret when the veloities involved are far less than the speed of light,. Mihelson-Morley Experiment.1 The Ether Nineteenth entury physiists believed in the existene of a stationary, all-permeating medium alled the ether whih allowed light to travel through spae.

3 . Brief Outline of The Experiment The movement of the Earth through the ether was believed to ause an ether wind of veloity v. It followed that light travelling in the diretion of the ether wind would have a veloity of v +, and when travelling in the opposite diretion would have a veloity of v. The Mihelson-Morley experiment used a devie alled a Mihelson interferometer to split a beam of light into two beams, that would eah travel an idential distane l but with different veloities with respet to the ether, as a result of the Earth s movement. When reombined, the two split beams would have been travelling for slightly different times, and so would be out of phase with eah other, produing an interferene effet. Rotating the apparatus would produe a seond time differene leading to a different interferene pattern - essentially the idea was to measure the fringe shift aused by this hange. Using the known veloity of the Earth through the ether, this fringe shift ould be alulated theoretially. However, when the experiment was performed the measured fringe shift was far less than the expeted one, and within experimental error orresponded to a omplete absene of a fringe shift. The failure of the Mihelson-Morley experiment implied that the ether did not in fat exist, though physiists at the time were not quik to aept this. It also highlighted the need for a new understanding of spae and time. For a more thorough analysis of this experiment, see just about any physis book that disusses speial relativity. 3 Lorentz Transformations 3.1 The Speial Theory Of Relativity In 1905, Albert Einstein published a paper entitled On The Eletrodynamis of Moving Bodies in whih he developed a theory whereby transformations ould be found that left Maxwell s equations of eletromagnetism invariant when going from a stationary frame to a moving one. This theory, whih has ome to be known as the Speial Theory of Relativity, also aounted for the results of the Mihelson-Morley experiment and did away with the need for the ether. 3

4 The two postulates of Einstein s Speial Theory of Relativity are: 1. The basi laws of physis are idential in all inertial frames.. The speed of light in spae will always be measured to have the same value,, independent of the motion of the light soure. Einstein s paper onerned itself first with the kinemati effets of these two postulates, then with the eletrodynami effets. In what follows, we shall deal almost wholly with the former. 3. Lorentz Transformations The essential transformations that Einstein found, and whih atually had been previously disovered by others, are known as the Lorentz Transformations. Consider two referene frames, as before: z y S S x z Suppose again that the origins oinide at t = t = 0, and that at this time there is a burst of light at the origins. As light travels with speed in both frames, at time t in frame S, the wavefront of the light forms a sphere of radius r = t entred on the origin of S, and in S the wavefront forms a similar sphere of radius r = t. Now, r = x + y + z = t and similarly y v x + y + z = t (3.1) x + y + z = t (3.) We wish to find a transformation between (3.1) and (3.). Starting with the Galilean transformation x = x vt, y = y, z = z, t = t, we find that: x xvt + v t + y + z = t whih does not agree with (3.1). We now try t = t + fx for some onstant f, obtaining: x xvt + v t + y + z = (t + tfx + f x ) and we note that the xvt term on the left anels with the tfx term on the right if f = v. Hene for t = t vx, we get: x xvt + v t + y + z = ( t tvx + v x 4 ) x xvt + v t + y + z = t tvx + v x x (1 v ) + y + z = t (1 v ) 4 x

5 and we an remove the (1 v ) terms by altering our transformations so that: x = ( x vt 1 v t = t vx ( 1 v Letting β = v and γ = 1, we an now write the Lorentz transformations as: (1 β ) 1 And equivalently: x = γ(x vt) ( t = γ t vx ) y = y, z = z x = γ(x + vt ) ) t = γ (t + vx y = y, z = z Note that for v <<, γ 1, and these redue to the Galilean transformations. Also note that γ 1. Lorentz transformations 4 Relativisti Effets 4.1 Length Contration Consider two frames S, S with S moving away from S with veloity v along the x-axis, as before. Let there be a rod of length l 0 in frame S. This rod is at rest with respet to S, but moving relative to S. S S v x 1 x Now, l 0 = x x 1 in S. In S the length l of the rod is the distane between points x 1 and x, reorded at time t 1 = t = t. Using the Lorentz transformations, x 1 = γ(x 1 + vt ) x = γ(x 1 + vt ) x x 1 = γ(x x 1) l 0 = γ l As γ 1, this means we have length ontration when there is relative motion between the length being measured and the observer ompared to when there is no relative motion. ontra- Length tion Lorentz transformations 5

6 4. Time Dilation Consider two frames as before, and an event ouring at x 0 at times t 1, t. The time interval between these two events is known as the proper time interval τ 0 - it is the shortest possible time interval between two events. S (x S 0, t 1 ) (x 0, t ) v x 0 In S using the Lorentz transformations for time, ( t 1 = γ t 1 vx ) 0 ( t = γ t vx ) 0 t t 1 = γ(t t 1 ) τ = γτ 0 where τ is the time interval as measured in S. Thus we have time dilation: time runs slower for moving loks. This has been verified experimentally, by omparing the number of mu-mesons that deay when at rest in the laboratory with the number that deay while moving at lose to the speed of light while travelling through the atmosphere. For the moving mu-mesons, time moves slower meaning that less deay than when they are at rest. (Another experiment involves synhronising two atomi loks and sending one on a plane journey. When the two are ompared afterwords, the one that has been moving is found to have run slower and be behind the other.) Time dilation 4.3 Light Pulse Clok Method Time Dilation We an also derive the expressions for both time dilation and length ontration using the idea of a light pulse lok. Consider a light soure moving with veloity v relative to a frame S along the x-axis. We let S be the frame in whih the light soure is at rest. Now, in S suppose the light soure emits a pulse of light that travels from a point A to a point B and returns to A in a time interval τ 0 = l0 : S B l0 A The situation in S is different: 6

7 S v t 1 l B S B l 0 A A Here the light pulse lok itself travels a distane of vτ where τ is the time interval in S while the light pulse travels from A to B and bak. We an use Pythagoras to show that in S the light pulse travels a total distane of [ l 0 + ( vτ )] 1 τ = [ l 0 + ( vτ )] 1 τ = [ ] 4 l0 + ( v τ 4 ) τ ( 4 v 4 ) = l 0 l 0 τ = ( v = l 0 (1 v τ = γτ 0 Length Contration Consider the same light pulse lok and frames as before, but with the light pulse lok rotated through 90 degrees. In S, we have: S l 0 A B and the time for the light pulse to travel from A to B and bak is again τ 0 = l0. In S we an onsider the journey in the following stages: S A l v 7

8 S v t l A The fixed distane between A and B is l. As the light pulse travels from A to B, the point B moves a distane v t 1 to the right, and as it returns, A moves a distane v t. Hene, we have the folowing two expressions for the distane travelled by the light pulse: l + v t 1 = t 1 and as τ = γτ 0 we have l v t = t τ = t 1 + t = l v + τ = l + v = l l = γ (1 v ) γ l = γ l 0 γl = l 0 l v 4.4 Simultaneity In Speial Relativity, simultaneity is relative. That is, two events that are simultaneous in one frame may not be simultaneous in another. Consider the Lorentz Transformation for time, t = γ(t + vx ) and two events in S, (x 1, t 1) and (x, t ). Then in S, t t 1 = γ(t t 1) + γv (x x 1) So if t = t 1 and x = x 1 then t = t 1 and the events are simultaneous in both frames. If x x 1 then the events are simultaneous in S but not S. 4.5 Minkowski Diagrams Spae-time graphs are a useful aid in speial relativity. In their most basi form, we an represent the one-dimensional motion of a body on a graph of t or t against x. The path of a body is known as its world line. As an example, we an onsider the problem of simultaneity again, and onstrut a spae-time diagram for a frame S in whih the points A, B and C are stationary. The diagram shows a beam of light emitted from the point B at time t = 0 and arriving at A and C at time t 0. S t t 0 A B C x 8

9 If now we let the points A, B and C be moving relative to S we find that the same beam of light does not arrive simultaneously at A and C. t t t 1 x A B C and note that the arrival of the light at A and C is a simultaneous event in some other frame S. The time axis of this frame would be perpendiular to the line joining the points of intersetion of the light beam with A and C. Very often t is hosen instead of t for the time axis. Then both axes express distanes, and then in all frames the world line of a beam of light starting at the origin bisets the t and x axes. (Note that these axes are not always perpendiular to eah other.) Spae-time diagrams are also alled Minkowski diagrams after their inventor, Hermann Minkowski. 4.6 Composition Of Veloities Consider two frames S, S as before, moving relative to eah other with veloity v. Suppose the omponents of the veloity of a body in frame S are (u x, u y, u z ), and those of the body as measured in frame S are (u x, u y, u z). Now, using the Lorentz transformations: and Dividing (4.1) by (4.), we then have: dx = γ(dx + vdt ) dx = γ( dx dt + v)dt = γ(u x + v)dt (4.1) ) dt = γ (dt + vdx dx dt = γ(u x + v)dt ( γ dt + vdx ) u x = u x + v 1 + vu x And starting with dy = dy, we an say that dy = dy dt dt = u ydt, and dividing by 4. as before, we get the transformation: Similarly we have: u y = u y γ 1 + vu x u x = u x v 1 vux (4.) Composition veloities Composition veloities of of 9

10 u y u γ y = 1 vux These are the formulae for the omposition of veloities in speial relativity. Note that it is impossible to measure a veloity to be faster than using these expressions. A body travelling at 0.75 with respet to one frame will have its veloity observed to be 4 5 by an observer in a frame moving towards it at 0.75, rather than the veloity of 1.5 predited by the Galilean veloity transformation. 4.7 Angle Transformation ( Headlight Effet ) Consider a light soure moving with veloity v in frame S. In frame S, whih is moving with veloity v away from S, then light soure is at rest. Consider a photon travelling at angle θ as measured in S. Clearly the x-omponent of the veloity of the photon is u x = os θ. In S, the x-omponent is u x = os θ. From the veloity transformations: u x = u x + v 1 + vu x os θ = os θ + v 1 + v os θ os θ = os θ + v 1 + v os θ os θ = os θ + β 1 + β os θ the angle transformation. A onsequene of this result is the headlight effet: light from a moving soure is observed to not be emitted in all diretions. Rather, it appears to be emitted in a beam (like from a ar headlight). 4.8 Relativisti Doppler Effet Consider a soure of light at rest in a frame S at x = 0. It emits a rest of light at time t = 0 and a seond rest at time t = τ = period. Consider now an observer in a frame S moving with veloity v away from the soure of light along the x-axis, and at x = x 0 at time t = t = 0. We an onstrut a spae-time diagram to illustrate the situation: (x, t ) t (x 1, t 1 ) t = τ transfor- Angle mation t = 0 x 0 x x 1 x 10

11 Now, x 1 = t 1 = x 0 + vt 1 and so and x = (t τ) = x 0 + vt x 0 = t 1 vt 1 = (t τ) vt t 1 ( v) = t ( v) τ t t 1 = τ v x x 1 = v(t t 1 ) x x 1 = vτ v In the frame S, using the Lorentz transformation for time t = γ(t vx ), we have that [ τ = t t 1 = γ t t 1 v ] (x x 1 ) [ τ τ = γ v v τ = γτ v ] vτ v ] [1 v τ = γτ ] [1 1 v v τ = γτ [ 1 β ] = γτ(1 + β) 1 β τ τ(1 + β) τ(1 + β) = = (1 β (1 β (1 + β τ = ( 1 + β τ 1 β the expression for the relativisti Doppler effet hange in period. As T = 1 f expression for the hange in frequeny: f = ( 1 β f 1 + β In the ase that S is moving towards the soure, swith all the signs: τ = f = ( 1 β τ 1 + β ( 1 + β f 1 β we also have the Relativisti Doppler Effet Change in Period Change in Frequeny Relativisti Doppler Effet (Observer Moving Towards) 11

12 5 Relativisti Dynamis 5.1 Relativisti Energy And Momentum The Speial Theory of Relativity neessitated that revisions be made to lassial dynamis. Traditionally, we have Newton s Seond Law, F = dp dt = ma if mass m is onstant. A onsequene of this is that a onstant fore would produe a onstant aeleration, leading to the veloity v of the body inreasing indefinitely to infinite speeds. However, in reality nothing an travel at speeds greater than. This suggests that as the body s speed inreases, its mass does not remain onstant. Indeed, it is found that: m 0 m(v) = ( = m 0 γ 1 v where m 0 = m(v = 0) is the body s rest mass. Now, we an expand γ in a Taylor expansion to get: Relativisti Mass m = m 0 (1 + 1 v + 3 v ) m = m m 0 = 1 v m v m 0... m = 1 m 0v (1 + 3 v 4...) m = 1 m 0v (1 + 3 v 4...) and this is the kineti energy of the body. So: K.E. = m = m m 0 This implies that m = E, leading to: = m 0 (γ 1) E = m where m = m 0 γ. This is the relativisti expression for energy. It introdues a new onept - that of mass-energy equivalene. Mass an be onverted into energy, and vie-versa. The expression for momentum is just: Relativisti Kineti Energy Relativisti Energy Using the above two equations, we note that P = mv = m 0 γv E P = m 0γ 4 m 0γ v = m 0γ 4 (1 v ) 1 But as γ =, we then have the following useful result: (1 v E P = m 0 4 = onstant in all frames Mo- Relativisti mentum An Energy- Momentum Invariant 1

13 For a olletion of partiles, E = the sum of energies, P = vetor sum of momentum, and E P = E 0 where E 0 = energy in a frame where the total momentum is zero. For a photon, E = hf, v = for all observers and the rest mass is zero, so: E p = 0 P = E = hf Momentum of a Photon the expression for the momentum of a photon. Combined with the laws of onservation of energy (more properly mass-energy) and momentum, the above expressions are used to solve problems of relativisti dynamis. 5. Problems Involving Relativisti Dynamis In this setion, we will look at some examples of problems involving relativisti dynamis Inelasti ollision Two idential partiles of rest mass m 0 travelling in opposite diretions, eah with veloity v ollide head on. The partiles stik together after the ollision forming a new partile of rest mass M 0, whih is at rest (thus onserving momentum). m v v m Before: M 0 After: Let m be the relativisti mass of eah partile travelling at veloity v. Then from the onservation of energy we have: m = M 0 m = M 0 Now, m = γm 0 m > m 0 M 0 > m 0. As m = m 0 + K.E., we have that: (m 0 + K.E.) = M 0 M 0 m 0 = K.E. This example illustrates that though energy and momentum are always onserved, rest mass need not be. 5.. Positron-Eletron Collision and Annihilation A positron is a partile with the same rest mass as an eletron but the opposite harge. Here, a positron moving with speed v ollides inelastially with an eletron at rest, and forms a so-alled positronium atom whih reoils freely. The positronium atom subsequently annihilates during the ourse of its motion, forming two γ-ray photons. We wish to work out the speed V of the positronium atom, and the maximum energy a photon so produed may have. We onsider first the ollision of the positron with the eletron. 13

14 m 0 γ 1 v m 0 Before: After: From onservation of momentum: M 0 V m 0 γ 1 v = M 0 γ V (5.1) And from onservation of energy, We then substitute (5.) in (5.1) and obtain: m 0 γ 1 + m 0 = M 0 γ m 0 γ 1 + m 0 = M 0 γ (5.) V = γ 1v γ If we are given the positron s energy/kineti energy rather than its speed, we an work out γ 1 (and hene v) from our expressions for energy. We now onsider the annihilation of the positron into two γ-ray photons. It an be seen (using onservation of momentum and P = E ) that the ase where one of the photons has the maximum possible energy is that in whih the photons move in opposite diretions along the x-axis: We again use onservation of energy: and onservation of momentum: M 0 Before: V After: E E 1 (5.3) M 0 γ = E 1 + E (5.4) M 0 γ V = E 1 E using P = E for the momentum of the photons. We have an expression for M 0 γ in (5.), and so solving (5.4) and (5.5) for E 1 and E is straightforward. A version of this problem has ourred frequently on Junior Freshman examination papers Collision and Sattering Angle In this problem we onsider the ollision between two idential partiles, one of whih is initially at rest with rest energy E 0. The other is moving with total energy E 1 = E 0 + K.E.. After the ollision, the partiles reoil, with θ being the separation angle between them. We onsider this to be the speial ase of a symmetrial ollision and assume both partiles have total energy E after the ollision. (5.5) 14

15 E 1, P 1 E 0 θ E, P E, P If the moving partile was travelling along the x-axis before the ollision, then by the symmetrial nature of the ollision the angle between the veloity of eah partile after the ollision and the x-axis is θ Ḟrom onservation of energy: E 1 + E 0 = E (5.6) and from onservation of momentum: P 1 = P os θ and using E P, Substituting E 1 = E 0 + K.E. into (5.8) we get: P 1 = P 1 P = os θ (5.7) E 1 P 1 = E P = m 0 = E 0 (5.8) ( ( ) E 0 + K.E.) E 0 = K.E. E 0 + K.E. (5.9) And from (5.6) we have E = E1 Dividing (5.9) by (5.10) gives: Using (5.7): + E0 ( P = E 0 + K.E., whih ombines with (5.8) to give: P 1 P ) E 0 = K.E. ( = E 0 + K.E. E 0 + K.E. 4 4 os θ = E 0 + K.E. E 0 + K.E. 4 E 0 + K.E. 4 = 8E 0 + 4K.E. 4E 0 + K.E. ) (5.10) (5.11) os θ = E 0 + K.E. 4E 0 + K.E. Using the trigonometri identity os θ = 1 (1 + os θ) os θ = os θ 1 we get the final result for the sattering angle: os θ = 4E 0 + K.E. 1 4E 0 + K.E. os θ = K.E. 4E 0 + K.E. 15

16 5..4 Compton Effet The Compton effet is the ollision of a photon with a free eletron, suh that the eletron reoils and the photon is sattered with less energy and hene a greater wavelength. E 1, P 1 E, P θ 3 E, P We use the relationship P = E to write the momentum of the photon before and after the ollision as: P 1 = E 1 ˆn 1 and P = E ˆn where ˆn 1 and ˆn are unit vetors along the diretion of travel of the photon. The onservation laws give: E 1 + m 0 = E + E E 1 ˆn 1 = E ˆn + P We solve these for E and P and subsitute into E P = m 0 4, giving: [ (E1 E ) + m 0 ] [E1ˆn 1 E ˆn ] = m 0 4 E1 E 1 E + E + (E 1 E )m 0 + m 0 4 [ E1 E 1 E os θ + E ] = m 0 4 Note the use of the vetor dot produt when squaring the seond braket: ˆn 1 ˆn = os θ. Tidying up the terms, (E 1 E )m 0 E 1 E (1 os θ) = = 1 (1 os θ) E E 1 m 0 having divided aross by E 1 E. We now use the relationships E = hf and = fλ E = h λ Compton effet hange of wavelength: λ λ 1 = to get our final result for the h (1 os θ) (5.1) m 0 16

17 A Appendix: List of Formulae Galilean Transformations x = x + vt v x = v x + v y = y v y = v y z = z v z = v z t = t Lorentz Transformations where or x = γ(x + vt ) ) t = γ (t + vx y = y, z = z 1 γ = ( 1 v 1 γ = (1 β x = γ(x vt) ( t = γ t vx ) β = v Length Contration Time Dilation l 0 = γ l τ = γτ 0 Composition of Veloities u x = u x + v 1 + vu x u x = u x v 1 vux u y = u y γ 1 + vu x u y = u y γ 1 vux Angle Transformation os θ = os θ + β 1 + β os θ 17

18 Relativisti Doppler Effet Observer moving away: τ = Observer moving towards: τ = Mass, Energy, Kineti Energy, Momentum An Energy-Momentum Invariant Momentum of a Photon ( 1 + β 1 β ( 1 β 1 + β m(v) = m 0 γ E = m K.E. = m 0 (γ 1) P = m 0 γv E P = m 0 4 P = E = hf τ, f = τ, f = ( 1 β 1 + β ( 1 + β 1 β f f Compton Effet Change Of Wavelength λ λ 1 = h (1 os θ) m 0 18