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How to get inverse Lorentz tranformation from direct Lorentz transformation 
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#1
Oct107, 05:40 PM

P: 88

Hello, I am having trouble on deriving the inverse Lorentz transformation from the direct Lorentz transformation. I looked at some threads here and I found in here (http://www.physicsforums.com/showthread.php?t=183057) that all I need to do is to "combine" the equation for x and t and I will get the inverse equation...but I don't really know what does it mean to "combine" the equation...? I also found in textbooks that to get the inverse transformation I just need to solve for x in the direct transformation. However, when I do it...it doesn't give the inverse transform equation. Can anybody give me some help here? I'll greatly appreciate it.
Thanks! 


#2
Oct107, 05:58 PM

Sci Advisor
P: 8,470

x'=gamma*(x  vt) and t'=gamma*(t  vx/c^2) So, with the first one you can do: x' = gamma*x  gamma*vt x' + gamma*vt = gamma*x x'/gamma + vt = x And with the second one: t' = gamma*t  gamma*vx/c^2 t' + gamma*vx/c^2 = gamma*t t'/gamma + vx/c^2 = t Then substitute this expression for t into the earlier equation x = x'/gamma + vt, which gives you: x = x'/gamma + v(t'/gamma + vx/c^2) = x'/gamma + vt'/gamma + xv^2/c^2 and if you subtract xv^2/c^2 from both sides, you get: x(1  v^2/c^2) = x'/gamma + vt'/gamma Now since gamma = [tex]\frac{1}{(1  v^2/c^2)^{1/2}}[/tex] this is the same as: [tex]x * (1  v^2/c^2)^1 = (1  v^2/c^2)^{1/2} * (x' + vt')[/tex] So if you divide both sides by (1  v^2/c^2) you get: [tex]x = (1  v^2/c^2)^{1/2} * (x' + vt')[/tex] which is just x = gamma*(x' + vt'), the reverse transformation for x in terms of x' and t'. Then you can plug this into t = t'/gamma + vx/c^2 and get the reverse transformation for t in terms of x' and t', which should work out to t = gamma*(t' + vx'/c^2). 


#3
Oct107, 06:10 PM

P: 1,746

t = gamma*(t' + vx'/c^2) The direct and inverse transformations should differ only by the sign of velocity v. Direct: x'=gamma*(x  vt) t'=gamma*(t  vx/c^2) Inverse: x=gamma*(x' + vt') t=gamma*(t' + vx'/c^2) Eugene. 


#4
Oct107, 06:22 PM

P: 88

How to get inverse Lorentz tranformation from direct Lorentz transformation
Thanks guys. It was very clear. Now I get the problem! :)



#5
Oct107, 06:45 PM

Sci Advisor
P: 8,470




#6
Oct607, 07:43 AM

P: 44

I've been following this, and I can see how to get to x, but am having trouble with t...I've got up to t = t'/gamma + (gamma*x'v)/c^2 + (gamma*t'v^2)/c^2
I just can't see where to go from there! 


#7
Oct607, 08:30 AM

P: 2,954

Pete 


#8
Oct607, 08:44 AM

P: 44




#9
Oct607, 09:52 AM

P: 2,954




#10
Oct607, 10:03 AM

P: 44

Which means you can't just look at it from a physics point of view, you have to show it through the method that JesseM said above. 


#11
Oct607, 11:19 AM

P: 44

Like I said above (sorry for reposting,, feared that it got lost in the much quoting above):
I've been following this method, and I can see how to get to x, but am having trouble with t...I've got up to t = t'/gamma + (gamma*x'v)/c^2 + (gamma*t'v^2)/c^2 I can't see how to make it into the inverse Lorentz from there! Have tried rearranging but just can't make it look right...! 


#13
Oct607, 11:33 AM

P: 44




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