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In general relativity when deriving the geodesic equation $$\ddot{x}^\mu + \Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta = 0\tag{1}$$ from the action $$S = \int d\tau \sqrt{|g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu|}\tag{2}$$ (using (-+++) convention) one assumes that it is possible to choose an affine parametrization $$-g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu = 1.\tag{3}$$
This parametrization implies that the trajectory is timelike. For all other curves (e.g. spacelike curves) an affine parametrization is not possible.

Is it possible to derive that any solution of the geodesic equation (using an arbitrary parametrization) is timelike (probably not because there also exist spacelike geodesics)?

Or am I restricting to timelike curves when deriving the Euler-Lagrange equation out of the action? This is also strange because then I cannot allow for arbitrary $\delta x^\mu$.

Or am I just assuming that the curve is timelike after I derived the geodesic equation, then choosing the affine parametrization and then simplifying the geodesic equation to the form it is usually stated? But then why can I insert this into the action before deriving the Euler-Lagrange equation?

Or is there some other solution?

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  • $\begingroup$ The title question: Why can we choose affine parameterization? seems different from the question in the main body: Is it possible to derive that any solution of the geodesic equation is timelike? $\endgroup$ – Qmechanic Oct 18 '18 at 15:17
  • $\begingroup$ why? do you have a better idea for the title? $\endgroup$ – toaster Oct 18 '18 at 23:17
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As you may know, the geodesic equation, your equation (1), is not obtained as the Euler-Lagrange equations of the curve-length functional (2), but rather as the Euler-Lagrange equations of the energy functional

$$E = \frac12\int d\lambda\, g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu.$$

I'm writing $\lambda$ rather than $\tau$ to avoid the suggestion that this has to be the proper time.

It is not very hard to show that extremals of $E$ are extremals of $L$, but the converse doesn't hold, in fact, length extremizing curves are extrema of $E$ if and only if they are true geodesics, i.e. affinely parameterized.

So, your equation (1) are the Euler-Langrange equations of $E$, whose solutions already are affinely parameterized. Adding (3) to it, the only additional requirement is for the curve to be timelike.

All three classes of geodesics, timelike, spacelike and lightlike, have affine parameterizations. For timelike geodesics proper time can be taken as an affine parameter, for spacelike geodesics proper length can be taken, and for lightlike curves no affine parameter has a special meaning.

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Comments to the post (v2):

  1. Whether the solution(s) to the variational problem is(are) timelike or spacelike is determined by (i) the Dirichlet boundary conditions of the problem and (ii) the causal properties of the Lorentzian manifold $(M,g)$ such that no boundary value problems on $(M,g)$ have both timelike and spacelike solutions. (E.g. we assume that $(M,g)$ does not have closed timelike curves.)

  2. If one assumes affine parametrization (3) off-shell, then the action principle (2) is ill-defined as all virtual paths have the same action value $S=\tau_f-\tau_i$, which is completely fixed by the common boundary conditions. Hence affine parametrization (3) can only be imposed on-shell. (Unless one impose it via an explicit Lagrange multiplier, cf. this related Phys.SE post.)

  3. Solutions to the Euler-Lagrange (EL) equation of the square root action (2) have arbitrary parametrization, while the geodesic equation (1) for affinely parametrized geodesics is the EL equation of the corresponding non-square root action, cf. e.g. this Phys.SE post and user doetoe's answer.

  4. The square root Lagrangian (2) is not differentiable, so it doesn't apply to light-like geodesics. This can however be fixed by introducing an auxiliary einbein field, cf. e.g. this Phys.SE post.

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  • $\begingroup$ 2. I didn't mean that you impose the affine parametrization. I meant that you restrict to timelike trajectories (which is independent of the parametrization). $\endgroup$ – toaster Oct 18 '18 at 13:34
  • $\begingroup$ $ \uparrow$ Ok. $\endgroup$ – Qmechanic Oct 18 '18 at 15:49

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