How to get inverse Lorentz tranformation from direct Lorentz transformation

In summary: 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...In summary, JesseM said that to invert the Lorentz transformation, you need to solve for x in the direct transformation and then use the inverse transformation.
  • #1
Odyssey
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0
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 (https://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!
 
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  • #2
Odyssey said:
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 (https://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 = [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).
 
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  • #3
JesseM said:
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.
 
  • #4
Thanks guys. It was very clear. Now I get the problem! :)
 
  • #5
meopemuk said:
I think it should be

t = gamma*(t' + vx'/c^2)
Yes, sorry, I mistyped.
 
  • #6
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
Odyssey said:
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 (https://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
 
  • #8
pmb_phy said:
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...!
 
  • #9
Ayame17 said:
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?
 
  • #10
pmb_phy said:
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.
 
  • #11
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...!
 
  • #12
Express v^2/c^2 in terms of gamma.
 
  • #13
Doc Al said:
Express v^2/c^2 in terms of gamma.

It wouldn't have occurred to me to do that, thankyou! :smile:
 
  • #14
Ayame17 said:
It wouldn't have occurred to me to do that, thankyou! :smile:

If you use rapidity, your Euclidean trigonometric intuition would have guided you.
 

1. What is the inverse Lorentz transformation?

The inverse Lorentz transformation is a mathematical operation used in physics to convert coordinates between frames of reference in special relativity. It is the reverse of the direct Lorentz transformation and allows for the transformation of coordinates from a moving frame to a stationary frame.

2. Why is the inverse Lorentz transformation important?

The inverse Lorentz transformation is important because it allows for the accurate description of physical phenomena in different reference frames, particularly in the context of special relativity. It is necessary for understanding concepts such as time dilation and length contraction, which are fundamental to our understanding of the universe.

3. How do you derive the inverse Lorentz transformation?

The inverse Lorentz transformation can be derived from the direct Lorentz transformation using algebraic manipulation and the use of the Lorentz factor, which is a constant value based on the relative velocity between the two frames of reference. It can also be derived geometrically using Minkowski diagrams.

4. Can the inverse Lorentz transformation be applied to any reference frame?

Yes, the inverse Lorentz transformation can be applied to any reference frame, as long as the relative velocity between the two frames is less than the speed of light. It is a fundamental principle of special relativity and is applicable in all inertial reference frames.

5. Are there any practical applications of the inverse Lorentz transformation?

Yes, the inverse Lorentz transformation has many practical applications in fields such as particle physics, astrophysics, and engineering. It is used to calculate the effects of time dilation and length contraction in high-speed travel and to make accurate measurements in experiments involving high-speed particles.

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