I would like to recover the (timelike) geodesics equations via the variational principle of the following action:

$$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$

Using an arbitrary auxiliary parameter $\lambda$, then one is able to rewrite the action and obtains: $$ d\tau = \sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ $$ -m \int d\tau = -m \int (d\tau / d\lambda) d\lambda = -m\int\sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$

Now we can vary the paths $x^{\mu}(\tau) \rightarrow x^{\mu}(\tau) + \delta x^{\mu}(\tau)$.

I need the following equation to be true, so i get the geodesics equations, but I don't know how one can say the following: $$ \int d\tau \,\delta g_{\mu\nu}\frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = \int d\tau\, g_{\mu\nu, \rho} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} \delta x^{\rho} $$ I know there is a simpler action, which leads to the same equation of motion, but there one has to do the same manipulation, which I sadly don't understand.

Thanks for helping me!

  • $\begingroup$ The last equation you've written is trivial. $\delta g_{\mu\nu}/\delta x^{\rho} = g_{\mu\nu,\rho} $. $\endgroup$
    – user28355
    Jan 19, 2014 at 15:59
  • $\begingroup$ Related: physics.stackexchange.com/q/67483/2451 $\endgroup$
    – Qmechanic
    Sep 27, 2014 at 14:20

2 Answers 2


Your action is: $$ S[x] = -m\int_{\lambda_0}^{\lambda_1}\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ and you have to impose $\delta S=0$ with the constraints $\delta x(\lambda_0) =\delta x(\lambda_1) =0$, that mean that the considered curves in the domain of $S$ have fixed endpoints.

To compute $\delta S$ you have to replace $x$ for $x+ \epsilon \delta x$ (so $\frac{dx}{d\lambda}$ must be replaced for $\frac{dx}{d\lambda} + \epsilon\frac{d \delta x}{d\lambda}$ ) and finally to compute the derivative respect to $\epsilon$ for $\epsilon=0$.

$$\delta S[x] = \frac{d}{d\epsilon}|_{\epsilon=0} S[x+ \epsilon \delta x]\:.$$

The computation leads to (assuming that $g$ and the curves are $C^1$, these curves defined on the compact $[\lambda_0,\lambda_2]$ one can safely swap the symbol of integral with that of $\epsilon$ derivative, essentially by a known theorem by Lebesgue) $$ \delta S[x] = -\frac{m}{2}\int_{\lambda_0}^{\lambda_1}\frac{- \frac{\partial g_{\alpha \beta}}{\partial x^\delta} \delta x^\delta \tfrac{dx^{\alpha}}{d\lambda}\tfrac{dx^{\beta}}{d\lambda} - 2g_{\alpha\beta} \frac{d \delta x^\alpha}{d\lambda}\frac{d x^\beta}{d\lambda}}{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} d\lambda\:. $$ Notice that $x$ appears in $g_{\mu\nu}=g_{\mu\nu}(x)$, too, and it gives rise to the contribution $\frac{\partial g_{\mu\nu}(x)}{\partial x^\sigma}\delta x^\sigma$ you mentioned in your question.

The denominator in the integral does not vanish as we are varying our curve in the class of timelike curves joining the two fixed endpoints.

Integrating by parts, one gets: $$ \frac{2}{m}\delta S[x] = \int_{\lambda_0}^{\lambda_1}\delta x^\delta\frac{ \frac{\partial g_{\alpha \beta}}{\partial x^\delta} \tfrac{dx^{\alpha}}{d\lambda}\tfrac{dx^{\beta}}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} d\lambda -\int_{\lambda_0}^{\lambda_1} \delta x^\alpha\frac{d}{d \lambda}\frac{2g_{\alpha\beta} \frac{d x^\beta}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} d\lambda + [...]\delta x^\alpha(\lambda_1)-[...]\delta x^\alpha(\lambda_0)\:. $$ The last two terms can be dropped as they vanish by hypothesis. Changing the name of some summed indices we end up with:

$$ \frac{2}{m}\delta S[x] = \int_{\lambda_0}^{\lambda_1}\delta x^\delta\left[\frac{ \frac{\partial g_{\alpha \beta}}{\partial x^\delta} \tfrac{dx^{\alpha}}{d\lambda}\tfrac{dx^{\beta}}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} -\frac{d}{d \lambda}\frac{2g_{\delta\beta} \frac{d x^\beta}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} \right]d\lambda\:. $$ Since the LHS vanishes for every choice of the variation $\delta x^\delta(\lambda)$, we conclude that $\delta S[x]=0$ on a curve $x=x(\lambda)$ is equivalent to the requirement that the said curve verifies: $$\frac{ \frac{\partial g_{\alpha \beta}}{\partial x^\delta} \tfrac{dx^{\alpha}}{d\lambda}\tfrac{dx^{\beta}}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} -\frac{d}{d \lambda}\frac{2g_{\delta\beta} \frac{d x^\beta}{d\lambda} }{\sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}}} =0\:.\quad (1)$$ We can change parameter and use the proper time $d\tau$ so that: $$d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} = d\tau$$ and (1) becomes: $$\frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^\delta} \frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau} -\frac{d}{d \tau}g_{\delta\beta} \frac{d x^\beta}{d\tau} =0\:.\quad (2)\:.$$ Expanding the last derivative changing the name of $\beta$ to $\mu$ in the last term: $$\frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^\delta} \frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau} - \frac{\partial g_{\delta\beta}} {\partial x^\sigma} \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} -g_{\delta\mu} \frac{d^2 x^\mu}{d\tau^2} =0\:.\quad \:.$$ In other words: $$\frac{d^2 x^\mu}{d\tau^2} - g^{\delta\mu} \frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^\delta} \frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau} + g^{\delta\mu} \frac{\partial g_{\delta\beta}} {\partial x^\sigma} \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} =0\:.$$ Renaming some indices: $$\frac{d^2 x^\mu}{d\tau^2} + \frac{1}{2}g^{\mu\delta}\left(2\frac{\partial g_{\delta \beta}}{\partial x^\sigma} - \frac{\partial g_{\sigma\beta}} {\partial x^\delta}\right) \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} =0\:.$$ Eventually, exploiting $g_{\delta \beta}= g_{\beta\delta}$: $$\frac{d^2 x^\mu}{d\tau^2} + \frac{1}{2}g^{\mu\delta}\left(\frac{\partial g_{\delta \beta}}{\partial x^\sigma} + \frac{\partial g_{\beta \delta}}{\partial x^\sigma}- \frac{\partial g_{\sigma\beta}} {\partial x^\delta}\right) \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} =0\:.$$ Now notice that: $$\frac{\partial g_{\delta \beta}}{\partial x^\sigma} \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} = \frac{\partial g_{\delta \sigma}}{\partial x^\beta} \frac{d x^\beta}{d\tau}\frac{d x^\sigma}{d\tau}=\frac{\partial g_{\delta \sigma}}{\partial x^\beta} \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau}$$ so the found identity can be re-written as: $$\frac{d^2 x^\mu}{d\tau^2} + \frac{1}{2}g^{\mu\delta}\left(\frac{\partial g_{\delta \sigma}}{\partial x^\beta} + \frac{\partial g_{\beta \delta}}{\partial x^\sigma}- \frac{\partial g_{\sigma\beta}} {\partial x^\delta}\right) \frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} =0\:.$$ We have found: $$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\sigma_\beta}\frac{d x^\sigma}{d\tau}\frac{d x^\beta}{d\tau} =0\:,$$ as wished.

  • $\begingroup$ Huge thanks for your well explained answer! I totally forgot the dependence $g_{\mu\nu}(x(\lambda))$ and therefore I was stuck. $\endgroup$
    – nerdizzle
    Jan 21, 2014 at 11:29
  • $\begingroup$ Why in the action you used $\frac{dx^{\mu}}{d\lambda}$ what is $\lambda$ here and why didn't we use $\tau$ instead? $\endgroup$ Oct 12, 2015 at 17:00
  • $\begingroup$ Because I cannot fix the range of $\tau$ which depends on the same solution I am looking for! For this reason I use a generic parameter $\lambda$ whose range $[\lambda_0,\lambda_1]$ is defined a priori. $\endgroup$ Oct 12, 2015 at 17:45
  • $\begingroup$ Injecting $d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} = d\tau$ in (1) does not produce (2). A direct replacement produces $$ \frac{\tfrac{ \partial g_{\alpha \beta}}{\partial x^\delta} \tfrac{dx^{\alpha}}{d\lambda}\tfrac{dx^{\beta}}{d\lambda} }{\tfrac{d\tau}{d\lambda}} -\frac{d}{d \lambda} \frac{2g_{\delta\beta} \tfrac{d x^\beta}{d\lambda}}{\tfrac{d\tau}{d\lambda}} =0 $$ Can you produce the missing steps to produce (2) - I am struck otherwise :( ? $\endgroup$
    – Anon21
    Jun 28, 2020 at 16:15
  • 1
    $\begingroup$ Just multipy both sides with $\frac{d\lambda}{d\tau}$ $\endgroup$ Jun 28, 2020 at 16:33

There are two definitions of geodesics here.

  1. locally distance minimizing curves You minimize the action, as you did, \begin{equation} S(\gamma) = \int_a^b \sqrt{ g_{\mu\nu} \dot{x}^{\mu} \dot{x}^{\nu} } dt = \int_a^b L dt \end{equation} The Euler Lagrangian equation associated with this action is \begin{equation} \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}^{\mu}}) - \frac{\partial L}{\partial x^\mu}= 0 \end{equation}

  2. curves on which tangent vector is parallel transported. You minimize the action, \begin{equation} E(\gamma) = \int_a^b \frac{1}{2}g_{\mu\nu} \dot{x}^{\mu} \dot{x}^{\nu} dt = \int_a^b \frac{1}{2}L^2 dt \end{equation} through Euler Lagrangian equation \begin{equation} \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}^{\mu}}) - \frac{\partial L}{\partial x^\mu} = -\frac{1}{L}\frac{\partial L}{\partial \dot{x}^{\mu}} \frac{d}{dt}L \end{equation} and get the solution(after some algebra) \begin{equation} \ddot{x}^{\lambda} + \Gamma^{\lambda}_{\mu\nu} \dot{x}^\mu \dot{x}^\nu = 0 \end{equation} This curve parallel transports the tangent vector.

We notice \begin{eqnarray} \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}^{\mu}}) - \frac{\partial L}{\partial x^\mu} = 0\\ \frac{d}{dt}L = 0 \end{eqnarray} solves Euler-Lagrangian equations in both case. $\frac{d}{dt}L$ just fixes the parameterization. For a Riemannian manifold, the parameter differs the length of the curve by an affine transformation \begin{equation} S(\gamma_{sol}(t)) = \int_a^t L(\gamma_{sol}(\tau) d\tau = (t-a) L(\gamma_{sol}(t=0) ) \end{equation} that's why some textbook directly the length(or proper time) as the parameter in the first place.

The second definition is sometimes called affine geodesics. While in GR, in most cases we are talking about affine geodesics, which is a locally minimizing curve plus an affine parameterization.

  • $\begingroup$ Just a remark: The variational definition (making stationary the distance) can be given for timelike and spacelike geodesics, for lightlike ones it cannot be used, as the functional is singular in that case. In this sense the affine definition is more general even if it constraints the choice of the admitted parametrization class. $\endgroup$ Jan 19, 2014 at 15:39
  • $\begingroup$ Another remark for the OP. Timelike geodesics actually locally maximize the distance, while spacelike ones do not minimize nor maximize (in Lorentzian manifolds, in Riemannian ones they minimize as is well known). $\endgroup$ Jan 19, 2014 at 15:44
  • $\begingroup$ @ValterMoretti, I'm having difficulties to understand what you just said. Could you develop it better? I'd really want to see an answer from you here: physics.stackexchange.com/questions/178695/… $\endgroup$
    – Mr. K
    Apr 27, 2015 at 20:10

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