- #1

fluidistic

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## Homework Statement

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.

## Homework 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)##

## 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!