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-   -   How to get inverse Lorentz tranformation from "direct" Lorentz transformation (http://www.physicsforums.com/showthread.php?t=188338)

 Odyssey Oct1-07 05:40 PM

How to get inverse Lorentz tranformation from "direct" Lorentz transformation

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!

 JesseM Oct1-07 05:58 PM

Quote:
 Quote by Odyssey (Post 1450803) 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...?
It's just algebra, it means solving those two equations (a combined system of equations) for x and t. You have:

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 = $$\frac{1}{(1 - v^2/c^2)^{1/2}}$$ this is the same as:

$$x * (1 - v^2/c^2)^1 = (1 - v^2/c^2)^{1/2} * (x' + vt')$$

So if you divide both sides by (1 - v^2/c^2) you get:

$$x = (1 - v^2/c^2)^{-1/2} * (x' + vt')$$

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).

 meopemuk Oct1-07 06:10 PM

Quote:
 Quote by JesseM (Post 1450828) which should work out to t = gamma*(t' - vx'/c^2).
I think it should be

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.

 Odyssey Oct1-07 06:22 PM

Thanks guys. It was very clear. Now I get the problem! :)

 JesseM Oct1-07 06:45 PM

Quote:
 Quote by meopemuk (Post 1450843) I think it should be t = gamma*(t' + vx'/c^2)
Yes, sorry, I mistyped.

 Ayame17 Oct6-07 07:43 AM

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!

 pmb_phy Oct6-07 08:30 AM

Quote:
 Quote by Odyssey (Post 1450803) 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!
That's a lot of work just to say - Switch the sign on the velocity in the Lorentz transformation and you end up with the inverse Lorentz transformation.

Pete

 Ayame17 Oct6-07 08:44 AM

Quote:
 Quote by pmb_phy (Post 1456496) That's a lot of work just to say - Switch the sign on the velocity in the Lorentz transformation and you end up with the inverse Lorentz transformation. Pete
Yes but then again you can't always do that - I'm working on inverting it mathematically, so we're not allowed to just say that! Unfortunately...!

 pmb_phy Oct6-07 09:52 AM

Quote:
 Quote by Ayame17 (Post 1456501) Yes but then again you can't always do that - I'm working on inverting it mathematically, so we're not allowed to just say that! Unfortunately...!
Why not?

 Ayame17 Oct6-07 10:03 AM

Quote:
 Quote by pmb_phy (Post 1456557) Why not?
Because the question I'm working on says "Mathematically invert equations (1) and (2) [ie, x' and t'] to obtain the inverse transformation"

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.

 Ayame17 Oct6-07 11:19 AM

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...!

 Doc Al Oct6-07 11:25 AM

Express v^2/c^2 in terms of gamma.

 Ayame17 Oct6-07 11:33 AM

Quote:
 Quote by Doc Al (Post 1456633) Express v^2/c^2 in terms of gamma.
It wouldn't have occurred to me to do that, thankyou! :smile:

 robphy Oct6-07 02:18 PM

Quote:
 Quote by Ayame17 (Post 1456638) It wouldn't have occurred to me to do that, thankyou! :smile:
If you use rapidity, your Euclidean trigonometric intuition would have guided you.

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