Proof of Invariance of Spacetime Interval

I was going through Spacetime Physics by Taylor and Wheeler and came to a point where they showed a proof of Invariance of Spacetime Interval. You can find the proof Here and Here is the second part of that proof.

They used an apparatus that flies straight "up" 3 meters to a mirror. There it reflects straight back down to the photodetector. This happens in the rocket frame.

In the laboratory frame, the path of the flash appears as that of a tent. So, light travels a greater distance in this frame. But since the speed of light is constant, the time of the travel is also greater.

Then they use simple geometry and Invariance of Spacetime Interval is proved.

But what if we used a particle instead of a flash of light? Say the particle would keep bouncing between two points? The speed of the particle is definitely not a constant. Wouldn't it take the same time to cross the tent like structure and the straight path in two of the respective frames? In that case, how can one prove Invariance of Spacetime Interval?

Here is the full book.

PLAGUE said:

But what if we used a particle instead of a flash of light? Say the particle would keep bouncing between two points? The speed of the particle is definitely not a constant. Wouldn't it take the same time to cross the tent like structure and the straight path in two of the respective frames? In that case, how can one prove Invariance of Spacetime Interval?

You would need to calculate how the velocity of the particle transforms between reference frames. This is something called relativistic velocity transformation (or addition).

https://phys.libretexts.org/Bookshe...ty/5.07:_Relativistic_Velocity_Transformation

PLAGUE said:

TL;DR Summary: How to prove it without using flash of light but a particle?

In that case, how can one prove Invariance of Spacetime Interval?

As @PeroK said, you can prove it in that case by using the relativistic velocity addition formula. However, once you step away from light it makes more sense to prove it in general by using the Lorentz transform. Once you have a general proof then you do not have to worry about proving it for each case.

The general proof is super straightforward. Simply start with the formula for the spacetime interval $$ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$$ Then substitute the inverse Lorentz transform formulas $$t=\gamma \left( t'+\frac{vx'}{c^2}\right)$$$$x=\gamma\left(x'+vt'\right)$$$$y=y'$$$$z=z'$$Then simplify to get $$ds^2=-c^2 dt'^2+dx'^2+dy'^2+dz'^2=ds'^2$$

This shows that the spacetime interval formula is the same in all reference frames. That is what is meant by it being invariant.

PeroK said:

You would need to calculate how the velocity of the particle transforms between reference frames. This is something called relativistic velocity transformation (or addition).

https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/05:__Relativity/5.07:_Relativistic_Velocity_Transformation

Isn't relativistic velocity is for, in such cases, horizontal motion? As I understand, our particle have no horizontal motion, but only vertical motion. Or am I getting it wrong?

PLAGUE said:

Isn't relativistic velocity is for, in such cases, horizontal motion?

I believe that the special theory of relativity permits motion in all three spatial dimensions.

PLAGUE said:

Isn't relativistic velocity is for, in such cases, horizontal motion?

The simple formula is, yes. You'll need to find a general formula. Dale's approach is easier.

The above libretexts link gives the transformation of all three velocity components for an arbitrary boost in the x-direction. That covers most cases you are likely to need.