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Relativity Notes: Shankar, 2006 [PDF] 1 Physics 200a Relativity notes Shankar (2006) Let us go over how the Lorentz transformation was derived and what it represents. An event is something that happens at a de¯nite time and place, like a ¯recracker going o®. Let us say I assign to it coordinates (x; t) and you, moving to the right at velocity u, assign coordinates (x0; t0). It is assumed that when you, sitting at x0 = 0 passed me (sitting at x = 0), we set our clocks to zero: t = t0 = 0. Thus our origins in space-time coincide. Let us see how the coordinates would have been related in pre-Einstein days. First, once we synchronize clocks at t = t0 = 0, they remain synchronized for all future times t = t0. This is the notion of absolute time we all believe in our daily life. You will assign to the event the spatial coordinates x0 = x ¡ ut: (1) This relationship is easy to understand: your origin is ut meters to the right of mine (Figure (2)) because you have been moving to the right at velocity u for time t so the x-coordinate you assign to an event will be less by ut compared to what I assign. You can invert this relation to say x = x0 + ut: (2) How are these two equations modi¯ed post-Einstein? If the velocity of light is to be same for both you and me, it is clear we do not agree on lengths or times or both. Thus if I predict you will say the event is at x0 = x ¡ ut, you will say that my lengths need to be modi¯ed by a factor γ so that the correct answer is x0 = γ(x ¡ ut): (3) S S' ut x x' FIG. 1. The same event (solid dot) is assigned coordinates (x; t) by me (S) and (x0; t0) by you (S'). In the pre-E- instein days t = t0 not only initially (when our origins crossed) but always. 2 Likewise when you predict I will say x = x0 + ut0 I will say, "No, your lengths are o®, so the correct result is x = γ(x0 + ut0):" (4) Note two things. First, I leave open the option that the time elapsed be- tween when we synchronized clocks and when the ¯recracker went o® is t0 for you and t for me, with two times being possibly di®erent. Next, the "fudge factor" for converting your lengths to mine and mine to yours are the same γ. This comes from the postulate that both observers are equivalent. So let us look at the equations we have: x = γ(x0 + ut0) (5) x0 = γ(x ¡ ut): (6) We proceed to nail down γ as follows. The event in question was a ¯re cracker going o®. Suppose when our origins coincided we sent o® a light pulse that this pulse set o® the ¯recracker. Since the light pulse took t seconds to travel x meters according to me and took t0 seconds to go x0 meters according to you and we both agree on the value of c, it must be true for this particular event that x = ct and x0 = ct0: (7) Let us multiply the LHS of Eqn 5 by the LHS of 6 and equate the result to the product of the RHS's to get xx0 = γ2(xx0 + xut0 ¡ x0ut ¡ u2tt0); and upon setting x = ct; x0 = ct0 we get (8) c2tt0 = γ2(c2tt0 + uctt0 ¡ uct0t ¡ u2tt0) and now upon cancelling tt0 (9) 2 1 γ = u2 (10) 1 ¡ c2 1 γ = r : (11) u2 1 ¡ c2 Note that once we have found γ it does not matter that we got it form this speci¯c event involving a light pulse. It can applied to Eqns. (5,6) valid for a generic event. Putting γ back into Eqn. (6) we obtain x ¡ ut x0 = r : (12) u2 1 ¡ c2 3 If we now go to Eqn 5 and isolate t0 we ¯nd ux t ¡ 2 t0 = r c (13) u2 1 ¡ c2 upon remembering that 1 u2 1 ¡ = : (14) γ2 c2 To summarize, if I am frame S and you are S', and you are moving to the right (increasing x direction) at speed u and my coordinates for an event are (x; t) and yours are (x0; t0) the Lorentz transformation tells us that x ¡ ut x0 = q (15) 1 ¡ u2=c2 u t ¡ 2 x t0 = q c (16) 1 ¡ u2=c2 If we consider two events labelled 1 and 2, then the coordinate di®erences obey ¢x ¡ u¢t ¢x0 = q (17) 1 ¡ u2=c2 u ¢t ¡ 2 ¢x ¢t0 = q c (18) 1 ¡ u2=c2 where ¢x = x2 ¡ x1 etc., and di®erences are not necessarily small. If you want to get my coordinates in terms of yours, you need to invert Eqns. (15) and 16. The answer is we get the same equations but with u !¡u. The same goes for the di®erences. As a result the di®erences will be related as follows; ¢x0 + u¢t0 ¢x = q (19) 1 ¡ u2=c2 0 u 0 ¢t + 2 ¢x ¢t = q c (20) 1 ¡ u2=c2 Now for the velocity transformation law . Let us follow a particle as it moves by an amount ¢x in time ¢t according to me and ¢x0 in time ¢t0 according to you. Let us agree that velocities are de¯ned as follows once and for all: 4 ¢x v = according to me (21) ¢t ¢x0 w = according to you (22) ¢t0 u = your velocity relative to me (23) In the above all ¢0s better be in¯nitesimals going to zero, as we are talking about instantaneous velocities and the derivative needs to be taken. Suppose I see a particle with velocity v. What is the velocity w according to you? To get this we take the ratio of the equations (37 -38) that give the primed coordinate di®erences in terms of unprimed ones: ¢x0 ¢x ¡ u¢t v ¡ u w = 0 = u = uv (24) ¢t ¢t ¡ c2 ¢x 1 ¡ c2 where in the last step we have divided top an bottom by ¢t. It is good to check that for small velocities (dropping the terms that go as 1=c2) we get results agreeing with common sense. Let us get used to going from your description to mine. Suppose you think a particle has velocity w. What will I think its velocity is? Now we use Eqns.(19-20). Taking the ratios as before and recalling the de¯nition of w we get w + u v = vw ; (25) 1 + c2 which just has the sign of u reversed in Eqn. (24) as expected and w $ v. Suppose you see an object moving at w = 3c=4 and you yourself are moving relative to me at u = 3c=4. In the old days I would expect that object to be moving at 1:5c. However the correct answer is 3c 3c 4 + 4 24 v = 9 = c: (26) 1 + 16 25 It is interesting to see that if you choose to apply this to a light pulse seen by you (w = c) the speed I will ¯nd (for any relative velocity u between us) c + u v = = c: (27) 1 + u=c 5 Time dilatation. Let me carry a clock with ticks ¢t = ¿0 seconds apart. If the two events are two successive ticks, the spatial separation is ¢x = 0 since the clock is at rest for me. You think the time di®erence between ticks is u2 ¿0 ¡ 0 2 ¿0 ¢t0 = q c = q (28) 1 ¡ u2=c2 1 ¡ u2=c2 Thus you say my clock is slow. This is what I will hear from anyone moving relative to me. Let us rederive this using Eqns (19-20). Thus set 0 u 0 ¢t + c2 ¢x ¢t = ¿0 = q (29) 1 ¡ u2=c2 and note that the other equation says ¢x0 + ut0 ¢x = 0 = q (30) 1 ¡ u2=c2 which means ¢x0 = ¡u¢t0. (This just means you think I and my clock are moving to the left at speed u) Feeding this into Eqn.29 to get ¢t0(1 ¡ u2=c2) ¿0 = q (31) 1 ¡ u2=c2 which agrees with Eqn. 28. Length contraction Suppose you are carrying a rod of length L0. This is its rest length. Suppose you are moving relative to me at velocity u. How long will I say it is? To ¯nd its length I have two of my co-moving assistants measure the location of its front and back ends at the same time. These two measurements, which can be seen as two events, have ¢x = L, the length according to me, and ¢t = 0 since the measurements of a moving rod's ends have to be simultaneous. (What happens if I measure one end, take a break and them measure the other?) The separation in space between the two events is the rest length L0 of the rod according to you. Thus q L ¡ 0 2 2 L0 = q or L = L0 1 ¡ u =c : (32) 1 ¡ u2=c2 q that is I will say your meter stick is actually 1 ¡ u2=c2 meters long, i.e., shortened. 6 Likewise youq will see my meter sticks are contracted. Suppose the contrac- tion factor 1 ¡ u2=c2 = :5. I say your meter sticks are :5m long. You say yours are ¯ne and mine are :5m long. How do you understand a guy with a shortened meter stick accusing you of having a shortened meter stick? Should I not say your are 2m long? The answer is this : You will say that I measured the leading end of your meter stick ¯rst and then after a delay , during which time the stick slid by, I measured the other end.
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