# Spacetime Inverval Invariance using Lorentz Transformations

1. Aug 27, 2011

### maccyjj

1. The problem statement, all variables and given/known data
Prove that the spacetime interval
-(ct)$^{2}$ + x$^{2}$ + y$^{2}$ + z$^{2}$
is invariant.

[/itex]$2. Relevant equations Lorentz transformations [itex]\Delta$$x' = \gamma(\Delta$$x-u\Delta$$t)$
$\Delta$$y' = \Delta$$y$
$\Delta$$z' = \Delta$$z$
$\Delta$$t' = \gamma(\Delta$$t-u\Delta$$x/c^{2})$

3. The attempt at a solution
I have tried to prove that $\Delta S = \Delta S'$
So first I said that $\Delta S' = - \Delta (ct')^{2} + \Delta (x')^{2} + \Delta (y')^{2} + \Delta (z')^{2}$

And inserted all the Lorentz Transformations above into the above formula.

I end up simplyfying it to get

$\gamma^{2} (x^{2} + u^{2}t^{2} - c^{2}t^{2} - \frac{u^{2}x^{2}}{c^{2}}) + y^{2} + z^{2}$

How does this equal $S = - \Delta (ct)^{2} \Delta (x)^{2} + \Delta (y)^{2} + \Delta (z)^{2}$ ? I can't see a way to get rid of the extra terms to get this simple function.

Any help would be really really great!

2. Aug 27, 2011

### vela

Staff Emeritus
If you collect the terms, you'll see that the coefficient of x2 is $\gamma^2(1-u^2/c^2)$. Use the definition of $\gamma$ to simplify that.

3. Aug 27, 2011

### maccyjj

Oh of course! How did I miss that?

Thank you so much I got it out now!