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Unformatted text preview: 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 deﬁnite time and place, like a ﬁrecracker going oﬀ. Let us say I assign to it coordinates ( x,t ) and you, moving to the right at velocity u , assign coordinates ( x ,t ). It is assumed that when you, sitting at x = 0 passed me (sitting at x = 0), we set our clocks to zero: t = t = 0. Thus our origins in spacetime coincide. Let us see how the coordinates would have been related in preEinstein days . First, once we synchronize clocks at t = t = 0, they remain synchronized for all future times t = t . This is the notion of absolute time we all believe in our daily life. You will assign to the event the spatial coordinates x = xut. (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 xcoordinate you assign to an event will be less by ut compared to what I assign. You can invert this relation to say x = x + ut. (2) How are these two equations modiﬁed postEinstein? 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 x = xut , you will say that my lengths need to be modiﬁed by a factor γ so that the correct answer is x = γ ( xut ) . (3) x x' S S' ut FIG. 1. The same event (solid dot) is assigned coordinates ( x,t ) by me (S) and ( x ,t ) by you (S’). In the preEinstein days t = t not only initially (when our origins crossed) but always. 2 Likewise when you predict I will say x = x + ut I will say, ”No, your lengths are oﬀ, so the correct result is x = γ ( x + ut ) . ” (4) Note two things. First, I leave open the option that the time elapsed between when we synchronized clocks and when the ﬁrecracker went oﬀ is t for you and t for me, with two times being possibly diﬀerent. 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 ( x + ut ) (5) x = γ ( xut ) . (6) We proceed to nail down γ as follows. The event in question was a ﬁre cracker going oﬀ. Suppose when our origins coincided we sent oﬀ a light pulse that this pulse set oﬀ the ﬁrecracker. Since the light pulse took t seconds to travel x meters according to me and took t seconds to go x 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 x = ct . (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 xx = γ 2 ( xx + xutx utu 2 tt ) , and upon setting x = ct, x = ct we get (8) c 2 tt = γ 2 ( c 2 tt + ucttuct tu 2 tt ) and now upon cancelling tt (9) γ 2 = 1 1u 2 c 2...
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 Fall '08
 RAMAMURTISHANKAR
 Physics, Momentum, Special Relativity, Velocity, Particle, Frame

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