0
$\begingroup$

How do you prove $d\tau = dt/\gamma$ is a Lorentz invariant?

$\endgroup$
1
  • 3
    $\begingroup$ How do you prove anything is Lorentz invariant? Apply an arbitrary Lorentz transformation and see if the transformed $d\tau'$ is the same as the original $d\tau$! $\endgroup$ Apr 6, 2022 at 13:36

3 Answers 3

5
$\begingroup$

It is less laborious to prove that $\mathrm{d}s=c\mathrm{d}\tau$ is a Lorentz invariant. We have: \begin{equation} \mathrm{d}s^2 =c^2\mathrm{d}\tau^2 =c^2\mathrm{d}t^2-\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2 =\eta_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu. \end{equation} Taking a Lorentz-Transformation $x'^\kappa=\Lambda_\rho^\kappa x^\rho$ with $\mathrm{d}x'^\kappa=\Lambda_\rho^\kappa\mathrm{d}x^\rho$, we have: \begin{equation} \mathrm{d}s'^2 =\eta_{\kappa\lambda}\mathrm{d}x'^\kappa\mathrm{d}x'^\lambda =\eta_{\kappa\lambda}\Lambda_\rho^\kappa\Lambda_\sigma^\lambda\mathrm{d}x^\rho\mathrm{d}x^\sigma =\eta_{\rho\sigma}\mathrm{d}x^\rho\mathrm{d}x^\sigma =\mathrm{d}s^2. \end{equation}

$\endgroup$
2
$\begingroup$

It might be enlightening to consider the Euclidean-geometry analogue of your question.

How do you prove that $$(\mbox{ hypotenuse} )=\frac{(\mbox{adjacent side})}{\cos\theta},$$ is a Euclidean invariant, where $\theta$ is the angle between the hypotenuse and the adjacent side?

Think of the "adjacent side" as the component of the hypotenuse along an axis of your choice and the angle $\theta$ is the angle from that axis to the hypotenuse. The question is "does the ratio $$\frac{(\mbox{adjacent side})}{\cos\theta}$$ depend on the choice of axis"?

Yes, you may wish to express your answer algebraically, in tensor notation. Maybe you want to use rotation matrices in three dimensions. But the essence of the problem is one from high-school geometry.

$\endgroup$
1
$\begingroup$

$$c^2 \mathrm d\tau^2=c^2\mathrm dt^2-\mathrm d\vec x^2$$ I have used $\vec x$ to represent the vector of x, y and z coordinates.

$$\implies c^2\mathrm d\tau^2=\mathrm dt^2(c^2-(\dfrac{\mathrm d\vec x}{\mathrm dt})^2)$$ $$\implies\mathrm d\tau^2=\mathrm dt^2(1-\dfrac{v^2}{c^2})$$ $$=\dfrac{\mathrm dt^2}{\gamma^2}$$ $$\implies \mathrm d\tau=\dfrac{\mathrm dt}{\gamma}$$

$\endgroup$
1
  • $\begingroup$ Isn't it what you asked for? $\endgroup$ Apr 7, 2022 at 9:44

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