# Eliminating x in Lorentz Transformation

1. Oct 2, 2011

### ZedCar

1. The problem statement, all variables and given/known data

http://www.cv.nrao.edu/course/astr534/LorentzTransform.html

It states;
Lorentz factor:
* y =gamma

How did they eliminate x?

3. The attempt at a solution

I'm taking the x value which is y(x'+vt') and substituting this into the x value in the x' equation.

So I'm getting;

x' = y^2(x'+vt')-yvt

Then I've separated the equation so that t is on LHS;

t = [y^2(x' + vt')] / yv

If anyone knows how this is done, I'd be very interested in learning! Thank you.

2. Oct 2, 2011

### cepheid

Staff Emeritus
Your last expression is wrong, it should be:

$$t = \frac{(\gamma^2 - 1)x^\prime + \gamma^2 vt^\prime }{\gamma v}$$

Now:

$$\gamma^2 = \frac{1}{1-v^2/c^2}$$

$$= \frac{c^2}{c^2 - v^2}$$

so that:

$$\gamma^2 - 1 = \frac{c^2}{c^2 - v^2} - \frac{c^2 - v^2}{c^2 - v^2}$$

$$= \frac{v^2}{c^2 - v^2} = v^2 \frac{\gamma^2}{c^2}$$

Therefore, the above expression becomes:

$$t = \frac{v^2\gamma^2 x^\prime}{\gamma vc^2} + \frac{\gamma^2 vt^\prime}{\gamma v}$$

$$= \frac{\gamma v x^\prime}{c^2} + \gamma t^\prime$$

$$= \gamma(t^\prime + vx^\prime / c^2)$$

3. Oct 4, 2011

### ZedCar

Cheers very much for that.

I understand it now!

Thank you