Proof of Invariance of Spacetime Interval

In summary, the "Proof of Invariance of Spacetime Interval" demonstrates that the spacetime interval between two events remains constant across different inertial reference frames in accordance with Einstein's theory of relativity. This invariance is derived from the Lorentz transformations, which relate the coordinates of events as observed from different frames moving at constant velocities. The proof highlights that despite variations in individual time and distance measurements, the spacetime interval, defined as the square of the time separation minus the square of the spatial separation, retains the same value for all observers, reinforcing the fundamental principles of relativity.
  • #1
PLAGUE
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TL;DR Summary
How to prove it without using flash of light but a particle?
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.
 
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  • #2
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
 
  • #3
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.
 
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  • #4
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?
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
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FAQ: Proof of Invariance of Spacetime Interval

What is the spacetime interval?

The spacetime interval is a measure of the separation between two events in four-dimensional spacetime, combining both spatial distance and time difference. It is given by the formula \( s^2 = -c^2\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \) in special relativity, where \( c \) is the speed of light, and \( \Delta t \), \( \Delta x \), \( \Delta y \), and \( \Delta z \) are the differences in time and spatial coordinates between the events.

Why is the invariance of the spacetime interval important?

The invariance of the spacetime interval is crucial because it ensures that the laws of physics are the same for all observers, regardless of their relative motion. This invariance underpins the principle of relativity, which states that the fundamental laws of physics do not change between different inertial frames of reference.

How is the spacetime interval invariant under Lorentz transformations?

The spacetime interval remains invariant under Lorentz transformations because these transformations mix space and time coordinates in such a way that the interval \( s^2 = -c^2\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \) remains unchanged. Lorentz transformations preserve the form of the interval by properly accounting for the relative motion between observers, ensuring that the interval calculated in one frame of reference is the same as in another.

Can you provide a simple proof of the invariance of the spacetime interval?

A simple proof involves applying the Lorentz transformation equations to the coordinates of two events. For an event with coordinates \((t, x, y, z)\) in one frame and \((t', x', y', z')\) in another frame moving at a constant velocity relative to the first, the Lorentz transformations are:\[ t' = \gamma (t - \frac{vx}{c^2}) \]\[ x' = \gamma (x - vt) \]\[ y' = y \]\[ z' = z \]where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). By substituting these into the spacetime interval formula and simplifying, one can show that the interval \( s^2 \) remains the same in both frames.

What are the physical implications of the invariance of the spacetime interval?

The invariance of the spacetime interval leads to several important physical implications, including time dilation and length contraction. It implies that measurements of time and space are relative to the observer's state of motion

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