1
$\begingroup$

I have been working on deriving the geodesic action via finding the stationary points of the proper-time integral for a massive point particle. Consider a space-time manifold $M$ ($\dim M=4)$ equipped with a metric $g$. The space-time element in coordinates is then $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$ where Einstein convention has been used. Setting $c=1$, the infinitesimal proper time, $d\tau=\sqrt{-ds^2}$, is then $$d\tau=\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$ The proper time elapsed between two points $x(\chi_1)$ and $x_2(\chi_1)$ on a curve parameterized by $x^\rho=x^\rho(\chi)$ can be found by the following integral $$\mathcal{S}=\int_{\chi_1}^{\chi_2}\sqrt{-g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\nu}d\chi$$ where $\dot{x}=\frac{dx}{d\chi}$. We know that a massive particle will take the path that makes this integral stationary to first order, and this path will be a geodesic (this is correct, assuming that the manifold is nice enough, right?).

My question is this: because it is easier to vary just the square of the integrand, is the following correct $$\delta S=\int_{\chi_1}^{\chi_2}\delta\sqrt{-g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\nu}d\chi=\int_{\chi_1}^{\chi_2}\frac{1}{2\sqrt{-g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\nu}}\delta\left( -g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\nu\right) d\chi ~?$$

I have seen this in a document (I don't remember the source) but wasn't sure if this "Chain-rule" for variations/functional derivatices was correct.

I should mention that I am looking for a proof, if possible.

$\endgroup$
  • 1
    $\begingroup$ It is correct if the path is time-like or space-like. $\endgroup$ – Sounak Sinha Jun 28 at 18:35
  • $\begingroup$ Meaning that $ ds^2\neq 0$? Could you offer a proof? $\endgroup$ – Zachary Jun 28 at 18:36
  • $\begingroup$ The explanation is very clear, thank you for that. I don't know if the last equation holds, on the right hand side you are taking the functional derivative of a function, rather than a functional (although you could think about the $\dot x_i$ as functionals of the $x_i$, but I wouldn't know how to calculate the functional derivative in that case). By stationary you mean that it satisfies the Euler-Lagrange equation? $\endgroup$ – S V Jun 28 at 18:40
  • $\begingroup$ Check T. Padmanabhan's book, page 150. Although he has not mentioned it, he has used the chain rule for functional differentiation in equation 4.40. The chain rule works for functional differentiation (check Wikipedia) provided that the quantities we are differentiating are continuous and $C^1$ in the space of functions. $\endgroup$ – Sounak Sinha Jun 28 at 18:41
  • $\begingroup$ Related: physics.stackexchange.com/q/149082/2451 and links therein. $\endgroup$ – Qmechanic Jun 28 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.