
#1
Mar1210, 11:41 PM

P: 6

It seems that the common approach to obtain the equations for the Lorentz transformations is to guess at its form and then, by considering four seperate situations, determining the values for the constants. From these equations, things like time dilation and length contraction can be worked out. Now, my goal was to go the other way around: starting from time dilation and length contraction, arrive at the Lorentz equations.
Suppose that in the reference frame O, the reference frame O' is moving at a speed v in the xdirection, with their origins coinciding at t = 0. An event E occurs at (x, y, z, t) in the O frame. It's straightforward to show that y' = y and z' = z. Next I considered x'. In the O frame, the distance between O' and E is x  vt. Because the ruler O' uses is shortened by a factor of γ, she will then measure the distance x  vt as being greater and O measures it, by a factor of γ. Thus, x' = γ(x  vt) (if this approach is incorrect, let me know!). However, I'm having trouble with t'. I know that t' = γ(t  vx/c2). I assume that the vx/c2 term comes from that, because O' believes that she's at rest, when the light emitted from the event reaches her, she doesn't treat herself as moving into the light, and thus there's a discrepancy as to how long before the light reaches O' did the event actually occur. Unfortunately, I can't arrive algebraically at this term. Finally, the gamma factor. I assume this comes from time dilation, but the wouldn't O' 's clock be running slower? So wouldn't the term have to be 1/γ (because less time transpires on her clock)? Can someone please tell me how to get the final Lorentz term this way? I know there are probably other, easier routes, but for personal reasons I would like to know how to do it this way. Thanks a lot! 



#2
Mar1310, 12:01 AM

Sci Advisor
P: 8,470

Remember that you have to include the possibility that clocks which are synchronized in one frame will be outofsync in another (relativity of simultaneity). In post #14 of this thread I derived the Lorentz transformation using an approach like the one you're suggesting, by putting in variables for length contraction, time dilation and the amount by which two clocks would be outofsync, and then solving for these variables:
http://www.physicsforums.com/showthread.php?t=180578 



#3
Mar1310, 03:26 PM

P: 6

Hey thanks a lot for the reply. I read through it, and I seem to understand it (later tonight I'll go through it more thoroughly). Some questions: first, if instead of considering there to be outofsync clocks in the O' frame, would taking into account the fact that, in the O frame, light hitting O' will take less time than O' thinks it should (because she's moving into the beam of light in the O' frame) be the same thing (as that, I believe, is the cause of loss of simultaneity)? Because if that's true, then (for personal tastes; I believe this to be more pleasing a route) how would you be able to account for this difference? It should turn out to be vx/c^2, but I can't seem to get the algebra to work. I know that the many clocks method is similar, but I would like to be able to do it this way too. Thanks!




#4
Mar1310, 07:05 PM

P: 231

Derivation of Lorentz transformations
JT7, Look at Einsteins original derivation in his 1905 paper 'On the electrodynamics of moving bodies' ( in the Dover book The Principle of Relativity). There is no guesswork and the use of the Postulate of Constant Light Speed is shown.
JM 



#5
Mar1310, 07:37 PM

Sci Advisor
P: 8,470

By the way, I realized in retrospect that what I proved there was not quite the same as what you were askingI showed that starting from the two basic postulates of SR, one could derive the familiar formulas for length contraction and time dilation. But you're asking how, given length contraction and time dilation, we can the derive the Lorentz transformation equations. OK, suppose an event E happens at (x_{1},t_{1}) in the O frame. Since the ruler used to mark position in the O' frame is moving at speed v, any point on the O' ruler will move a distance of vt_{1} between t=0 and t=t_{1}. So, the marking M on the O' ruler that's at position x=x_{1} at time t=t_{1} in the O frame must have been at position x_{1}  vt_{1} at time t=0 in the O frame. And we know that at time t=0 the marking x'=0 on the O' ruler coincided with position x=0 in the O frame, so in the O frame the distance between M and the x'=0 mark on the O' ruler must be x_{1}  vt_{1}. But since we know that in the O frame the O' ruler is shrunk by a factor of sqrt(1  v^{2}/c^{2}), that means that in the O' frame the distance between M and x'=0 is larger than this by a factor of 1/sqrt(1  v^{2}/c^{2}) = gamma, so the mark M must have a reading of gamma*(x_{1}  vt_{1}) on the O' ruler. Now let's say the observer O' sits at position x'=0 on the O' ruler, and at time t=0 in the O frame this was at x=0, after which she moved in the positive x direction with speed v. Suppose that when the event E happens at x=x_{1} and t=t_{1} in the O frame, it sends a beam of light towards O'. In the O frame, O' will be at position x=vt_{1} at time t_{1}, so the initial distance between O' and the event E is x_{1}  vt_{1} if E happened further in the +x direction than O' was at that point, or vt_{1}  x_{1} if O' was further in the +x direction at that time. If E happened further in the +x direction, then O' is moving at v in the +x direction while the light is moving at c in the x direction, so the distance between them is shrinking at a speed of (c + v), meaning the light will take a time of (x_{1}  vt_{1})/(c + v) to reach O' after being emitted at E. And since it was emitted at t_{1}, the time t that it reaches O' as seen in the O frame will be: [t_{1}] + [(x_{1}  vt_{1})/(c + v)] = [(ct_{1} + vt_{1})/(c + v)] + [(x_{1}  vt_{1})/(c + v)] = (x_{1} + ct_{1})/(c + v) = (c  v)*(x_{1} + ct_{1})/(c^{2}  v^{2}) = (c  v)*(x_{1} + ct_{1})/((c^{2})*(1  v^{2}/c^{2})) On the other hand, if O' was further in the +x direction, then O' is moving at v in the +x direction while the light is moving at c in the +x direction too trying to catch up with O' from behind, so in this case the distance between them is only shrinking at a speed of (c  v), so the light will take a time of (vt_{1}  x_{1})/(c  v) to reach O' after being emitted at E. And again, it was emitted at t_{1}, so the time t it reaches O' as seen in the O frame will be: [t_{1}] + [(vt_{1}  x_{1})/(c  v)] = [(ct_{1}  vt_{1})/(c  v)] + [(vt_{1}  x_{1})/(c  v)] = (ct_{1}  x_{1})/(c  v) = (c + v)*(ct_{1}  x_{1})/(c^{2}  v^{2}) = (c + v)*(ct_{1}  x_{1})/((c^{2})*(1  v^{2}/c^{2})) Now the clock of O' is slowed down by a factor of sqrt(1  v^{2}/c^{2}) in the O frame, and her clock read t'=0 at time t=0 in the O frame, so at any later time t in the O frame her clock reads t*sqrt(c^{2}  v^{2})/c. So in the first scenario where E was further in the +x direction, the light reached O' when her own clock read: sqrt(1  v^{2}/c^{2})*(c  v)*(x_{1} + ct_{1})/((c^{2})*(1  v^{2}/c^{2})) = (c  v)*(x_{1} + ct_{1})/((c^{2})*sqrt(1  v^{2}/c^{2})) In the second scenario where O' was further in the +x direction, the light reached O' when her own clock read: sqrt(1  v^{2}/c^{2})*(c + v)*(ct_{1}  x_{1})/((c^{2})*(1  v^{2}/c^{2})) = (c + v)*(ct_{1}  x_{1})/((c^{2})*sqrt(1  v^{2}/c^{2})) In both cases, if the event E happened at the x' = gamma*(x_{1}  vt_{1}) mark on her ruler, she must subtract a time of gamma*(x_{1}  vt_{1})/c from the time she observed light from E to get the actual time of E in her frame (that's if the mark is at a positive value of her x' ruler, if it's at a negative value then she would subtract a time of gamma*(x_{1}  vt_{1})/c. It will be at a positive value in the first scenario where the event E happened further in the +x direction, and at a negative value in the second scenario where O' was further in the +x direction when E occurred). And gamma*(x_{1}  vt_{1})/c = c*(x_{1}  vt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})). So if we subtract this from the observed time in the first scenario to get the actual time of E in the O' frame, we get: [(c  v)*(x_{1} + ct_{1})/((c^{2})*sqrt(1  v^{2}/c^{2}))]  [c*(x_{1}  vt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2}))] = (cx_{1} + c^{2}*t_{1}  vx_{1}  cvt_{1}  cx_{1} + cvt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})) = (c^{2}*t_{1}  vx_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})) = (t_{1}  vx_{1}/c^{2})/sqrt(1  v^{2}/c^{2}) And if we subtract c*(x_{1}  vt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})) from the observed time in the second scenario to get the actual time of E in the O' frame, we get: [(c + v)*(ct_{1}  x_{1})/((c^{2})*sqrt(1  v^{2}/c^{2}))]  [c*(x_{1}  vt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2}))] = (c^{2}*t_{1}  cx_{1} + cvt_{1}  vx_{1} + cx_{1}  cvt_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})) = (c^{2}*t_{1}  vx_{1})/(c^{2}*sqrt(1  v^{2}/c^{2})) = (t_{1}  vx_{1}/c^{2})/sqrt(1  v^{2}/c^{2}) So, in both scenarios we find that the time of E in the O' frame is gamma*(t_{1}  vx_{1}/c^{2}). 



#6
Mar1410, 01:02 AM

Sci Advisor
P: 8,470

Incidentally, there was a lot of algebra in the above derivation, but it does become easier if you think in terms of synchronized clocks rather than in terms of a light signal sent to O'. I showed in the derivation on the other thread that:
So suppose we have some event E that occurs at x_{1} and t_{1} in the frame of O, and that frame O has a ruler of length x_{1} with one end at x=0 and the other at x=x_{1}, with two synchronized clocks at either end. So, the end of this ruler at x=x_{1} is at the position of E when it happens, and the clock there reads t=t_{1}. Now consider how things must look in the the frame of O', who sees this ruler moving at speed v in the x' direction. If x_{1} is a positive number, then x_{1} is the length of the ruler in its own frame, and the end of the ruler at x=x_{1} must be the "back end" as seen in O', so if the clock at this end shows a time of t_{1} the clock the the "front end", x=0, must be behind by vx_{1}/c^2 (using the equation at the end of the previous paragraph), showing a time of t_{1}  vx_{1}/c^2. On the other hand, if x_{1} is a negative number then the length of the ruler in its own frame is x_{1}, and the end of the ruler at x=x_{1} must be the "front end", so if the clock at this end shows a time of t_{1} the clock at the "back end", x=0, must be ahead by v*(x_{1})/c^2, showing a time of t_{1} + v*(x_{1})/c^2 = t_{1}  vx_{1}/c^2. So either way, in the O' frame at the moment the event E is happening the clock at the spatial origin of O reads a time of t_{1}  vx_{1}/c^2. We also know that the clocks at the origins of each frame both read 0 at time t=0 and t'=0 in each frame, and that in frame O' the clock at the origin of O is running slow by a factor of 1/gamma, so the at the moment the clock at the origin of O reads t=t_{1}  vx_{1}/c^2, the time in O' must be t'=gamma*(t_{1}  vx_{1}/c^2). And we know that the event of the clock at the origin of O shows this reading at the same time as event E in the O' frame! So, the event E must also have a time coordinate of t'=gamma*(t_{1}  vx_{1}/c^2) in the O' frame. 



#7
Mar1410, 12:04 PM

P: 6

Thanks for the detailed response, I understand now. Cheers!



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