# Space-time interval Lorentz co/invariant

1. Sep 2, 2013

### fluidistic

1. The problem statement, all variables and given/known data
Hi guys!
I must show by brute force that $\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2$ is invariant under Lorentz transforms.

2. Relevant equations
Lorentz transforms:
$\Delta t' = \gamma \left ( \Delta t -\frac{\vec v \cdot \Delta \vec x}{c^2} \right )$ and $\vec \Delta x'=\gamma (\Delta \vec x - \vec v t)$.
I will use the convention $\Delta \vec x= (\Delta x, \Delta y , \Delta z )$ and $\vec v = (v_x ,v_y, v_z)$

3. The attempt at a solution
Basically I must show that $\Delta s'^2=\Delta s^2$. Where $\Delta s'^2=-c^2\Delta t'^2+\Delta x'^2+\Delta y'^2+ \Delta z'^2$.
I have that $\Delta x'^2+\Delta y' ^2+\Delta z'^2=\gamma ^2[\Delta x^2+\Delta y^2+\Delta z^2 +\Delta t^2 \underbrace{(v_x^2+v_y^2+v_z^2)}_{=\vec v^2}-2\Delta t (\Delta x v_x+\Delta yv_y +\Delta z v_z)]$
While $\Delta t'^2=\gamma ^2 \{ \Delta t^2-\frac{2\Delta t}{c^2}(v_x\Delta x+v_y\Delta y +v_z \Delta z)+\frac{1}{c^4}[v_x^2 \Delta x^2+v_y^2\Delta y^2+v_z^2 \Delta z^2+2(v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z)] \}$.
So then I performed $\Delta s'^2$ and after all the simplifications I could do, I reached that $\Delta s'^2=\gamma ^2 \left [ \Delta t^2 (\vec v^2-c^2) -\frac{\vec v \cdot \vec x}{c^2} +\Delta \vec x +\frac{2}{c^2} (v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z) \right ]$ which does not seem equal to $\Delta s$. I've redone the algebra 4 times (lost hours on this), I reach the same answer every time. So either the expression I found is right but I must simplify it even more and it indeed equals $\Delta s$, or I'm wrong somewhere.
I'd appreciate some help... thank you!

2. Sep 2, 2013

### TSny

For a boost in an arbitrary direction, I think your expression for $\Delta t'$ is ok. But the correct expression for $\vec \Delta x'$ is more complex than what you wrote. See for example

http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_any_direction

Are you sure you can't orient your axes so that the x-axis is parallel to the boost?

3. Sep 2, 2013

### fluidistic

I see, thank you very much.
Due to the complexity of the expression for a boost in any direction, I guess I can orient the axes so that it results to be in the x-direction. I will consider this case only.

4. Sep 2, 2013

### fluidistic

At first glance I get $\Delta s'^2=\Delta t^2 (-c^2\gamma^2+v^2)+\gamma ^2 \left ( \Delta x^2+v^2 \Delta t^2 - \frac{v^2\Delta x^2}{c^2} \right )$.
I guess I'll have to redo the math.
Edit:Yeah I forgot 2 terms. I reach now $\Delta s'^2=\Delta t^2[v^2+\gamma(v^2-c^2)]+\Delta x^2+\Delta y^2+\Delta z^2$. Almost the desired result but apperently I still have some job to do :)

Edit 2:This reduces to $\Delta s'^2=\Delta t^2 (v^2-c^2)+\Delta \vec x$. I'm almost there. But I would reach the result only if v=0... which is not correct.

Last edited: Sep 2, 2013
5. Sep 2, 2013

### vela

Staff Emeritus
Everything multiplying $\Delta t^2$ should be proportional to $\gamma^2$, so your expression in your first edit looks kinda odd.

6. Sep 2, 2013

### TSny

Trace back to see how you got the first term of $v^2$ in the first brackets. I don't think that should be there.

7. Sep 2, 2013

### fluidistic

I've tracked it down (already on my own), at $\Delta x'=\gamma (\Delta x - v \Delta t)$. When I must take the square of this quantity, there's a term $\gamma ^2 v^2\Delta t^2$. I couldn't cancel this term out though I must re-recheck the algebra.

Edit: Nevermind, I made a mistake. I now reach the desired result! Thank you guys.

Last edited: Sep 2, 2013