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Relativity Notes 2006

# Relativity Notes 2006 - 1 Physics 200a Relativity notes...

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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 space-time coincide. Let us see how the coordinates would have been related in pre-Einstein 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 = 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 = x + ut. (2) How are these two equations modiﬁed 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 x = x-ut , you will say that my lengths need to be modiﬁed by a factor γ so that the correct answer is x = γ ( x-ut ) . (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 pre-E-instein 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 be-tween 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 ( x-ut ) . (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 + xut-x ut-u 2 tt ) , and upon setting x = ct, x = ct we get (8) c 2 tt = γ 2 ( c 2 tt + uctt-uct t-u 2 tt ) and now upon cancelling tt (9) γ 2 = 1 1-u 2 c 2...
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Relativity Notes 2006 - 1 Physics 200a Relativity notes...

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