1
$\begingroup$

Einstein's relativity principle states that:

  1. All the laws of physics are the same in every inertial frame of reference

  2. The speed of light is the same in every reference frame.

Why these two statements imply that the space-time interval between two events (defined as $ds^2=c^2dt^2-d\vec{x}^2$ ) must be a relativistic invariant and thus that the Lorentz group is the correct group of transformations that rules the connection between coordinates in different inertial frames?

$\endgroup$
3
  • $\begingroup$ One of the answers to Do we know why there is a speed limit in our universe? should help. $\endgroup$
    – mmesser314
    Jul 11, 2023 at 14:44
  • 1
    $\begingroup$ For a proof that$$ \left[\mathrm ds^2=0=\mathrm ds'^2\right] \Longleftrightarrow \left[\mathrm ds^2=\mathrm ds'^2\right] \tag{a} $$ see Reference : 'Introduction to Special Relativity',by W.Rindler, Edition 1982. Section 6. Derivation of the Lorentz transformation . Equations (6.1) to (6.10). $\endgroup$
    – Frobenius
    Jul 11, 2023 at 20:32
  • 1
    $\begingroup$ This comment proves that Stack Exchange is sometimes more useful than going to university. I bet 99% of people in my department don't even ask if such proof is necessary or not. Thank you $\endgroup$ Jul 11, 2023 at 21:21

1 Answer 1

2
$\begingroup$

I will try to give an answer. First, we use postulate (2). Imagine I am inside a train moving with some velocity and you are watching me from outside. Suppose, I shine a light in a direction perpendicular to the movement of the train. Suppose it takes time $ \Delta t $ to reach to the ceiling of the train. If the distance between the torch and the ceiling is $\Delta x$, the postulate (2) implies

$$ \frac{\Delta x}{\Delta t} = c$$

Now consider the situation in your frame. Now because of the movement of the train, the light doesn't travel purely in the perpendicular direction and is now slanted. Therefore, it travels a distance $\Delta x'> \Delta x$ when it reaches the ceiling. Now for postulate (2) to hold, you need to measure the speed as $c$, i.e

$$ \frac{\Delta x'}{\Delta t'} = c$$

where $\Delta t'$ is the time it took for the light to reach to the top according to you. This immediately implies $\Delta t' \neq \Delta t$. This means, we need to account for difference in time as well as distance when we change frames of reference. Generally (2) implies

$$ c^2dt^2 - d\vec{x}^2 = 0$$

in all frames. Now this alone doesn't show the invariance of the line element $ds^2$ for all space-time events. This is where Postulate (1) comes in. Suppose we have a wave traveling at the speed of light in some frame. It obeys the equation

$$ \left ( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \right)f = 0$$

Now suppose we have another frame. The wave equation should be invariant as the physics shouldn't change by postulate (1). Therefore we must find a transformation that keeps the wave equation intact. This is precisely the Lorentz transformations. Furthermore, it turns out the lorentz transformations leave $dsˆ2$ invariant for $\textbf{all}$ spacetime events.

Hope this helps.

$\endgroup$
2
  • $\begingroup$ Nice try, but you prove that $ds^2$ is zero in all frames of reference, if the initial and end pointr of the space time intervals are considered along the path of a light ray. The problem I have is exactly showing that $ds^2$ between any two events is invariant (even if not zero). $\endgroup$ Jul 11, 2023 at 15:03
  • $\begingroup$ You are right. The proof for all spacetime events is done once you deduce that the proper transformations between frames are lorentz transformations. That is why I considered the wave equation. $\endgroup$
    – emir sezik
    Jul 11, 2023 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.