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The derivation of geodesic equation is straight from Padmanabhan's book on General relativity.

Consider the action

$$A = \int d\tau=\int\sqrt{-g_{ab}dx^adx^b}.\tag{4.39}$$

We impose the condition $\delta A=0$.

First we note,

$$\delta (d\tau^2)=\delta(-g_{ik}dx^idx^k)=-dx^idx^k(\partial_l g_{ik})\delta x^l-2g_{ik}dx^id \delta x^k\tag{4.40a}$$

Also,

$$\delta (d\tau^2)= 2d\tau \delta d\tau\tag{4.40b}$$

which implies, $$\delta d \tau=\delta(d\tau^2)/(2d\tau).$$

Substituting $\delta(d\tau^2)$ from above,

$$0=\delta A=-\int\delta d\tau=\int\left[\frac12\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}(\partial_lg_{ik})\delta x^l+g_{ik}\frac{dx^i}{d\tau}\frac{d\delta x^k}{d\tau}\right]d\tau$$
$$=\int\left[ \frac12\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}(\partial_lg_{ik})\delta x^l-\frac{d}{d\tau}\left( g_{ik}\frac{dx^i}{d\tau} \right)\delta x^k \right]d\tau+g_{ik}\left.\frac{dx^i}{d\tau}\delta x^k\right|^{\tau_1}_{\tau_2}.\tag{4.41}$$

(Well, the details are not that important to understand my question.)

Now we discard the boundary term to obtain the geodesic equation, which is the term in the square brackets equated to zero.

Then a few pages further, he goes on to say that this form of the geodesic equation is valid only for affine parametrizations, that is, either $\tau$ is the proper time, or is related to the proper time by a linear transformation.

My question is, where did we assume in the derivation above that the parametrization is affine?

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  • $\begingroup$ The hidden assumption here is that one can use the proper length $\tau$ as parameter. It is an affine parameter for spacelike or timelike curves. For light-like curves that is not possible and the proof is wrong. $\endgroup$ Commented Feb 8, 2022 at 18:34
  • $\begingroup$ At this point, you have not. If you work it out carefully, you'll find that the equation of motion you end up with does not take the standard form of the geodesic equation. The "extra" term that you find can then be set to zero by choosing $\tau$ to be the affine parameter. $\endgroup$
    – Prahar
    Commented Feb 8, 2022 at 18:36
  • $\begingroup$ @Prahar the equation of motion that one ends up with after manipulating the one above does NOT have that 'extra' term you're referring to. $\endgroup$
    – Singh
    Commented Feb 9, 2022 at 3:01
  • $\begingroup$ Ya, you are right. But I now realize that you implicitly assumed that $\tau$ is an affine parameter. $\endgroup$
    – Prahar
    Commented Feb 9, 2022 at 8:53
  • $\begingroup$ Could you spell out explicitly where did I assume it? I am unable to point that out. $\endgroup$
    – Singh
    Commented Feb 9, 2022 at 14:08

2 Answers 2

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I'll just do the Euler-Lagrange equations for the Lagrangian of a free particle here. Consider the Lagrangian $$L=\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}.$$

We find the momentum and force in each direction, $$p_\mu=\frac{\partial L}{\partial\dot x^\mu}=\frac{g_{\mu\alpha}\dot x^\alpha}{L}$$ $$F_\mu=\frac{\partial L}{\partial x^\mu}=\frac{g_{\alpha\beta,\mu}\dot x^\alpha\dot x^\beta}{2L}$$

And write the EL equation $$\dot p_\mu=F_\mu$$ $$\frac{g_{\mu\alpha,\beta}\dot x^\alpha\dot x^\beta+g_{\mu\alpha}\ddot x^\alpha}{L}-\frac{g_{\mu\alpha}\dot x^\alpha}{L}\frac{\dot L}{L}=\frac{g_{\alpha\beta,\mu}\dot x^\alpha\dot x^\beta}{2L}$$

We move terms around, and use some symmetries of $g$ and $\dot x\dot x$ to get $$g_{\mu\alpha}\ddot x^\alpha+\frac{1}{2}(g_{\mu\beta,\alpha}+g_{\alpha\mu,\beta}-g_{\alpha\beta,\mu})=g_{\mu\alpha}\dot x^\alpha\frac{\dot L}{L}$$

Raising the $\mu$ we recognize the connection $\Gamma$ and write $$\ddot x^\mu+\Gamma^\mu_{\alpha\beta}\dot x^\alpha\dot x^\beta=\frac{\dot x^\mu\dot L}{L}$$

The term on the right is what Prahar mentioned in his comment. So if we here assume $L=\mathrm{const},$ we get both the affine parameter condition of $g_{\mu\nu}\dot x^\mu\dot x^\nu=\mathrm{const}$ as well as the standard geodesic equation

$$\ddot x^\mu+\Gamma^\mu_{\alpha\beta}\dot x^\alpha\dot x^\beta=0.$$

It shouldn't be too hard to see where, in the rest of your derivation, this comes about and you do a similar assumption.

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TL;DR: The issue is hidden in eq. (4.40b), which essentially allow Padmanabhan to vary the square of the square-root action (= the non-square-root action), whose EL equations are known to have affine parametrization rather than general parametrization, cf. e.g. my Phys.SE answer here.

In more detail, let us define action $$S[x]~=~\int \! d\lambda \sqrt{L_0},$$ where$^1$ $$L_0(x,\dot{x})~:=~ -g_{ij}(x) \dot{x}^i \dot{x}^j, \qquad \dot{x}^i~:=~\frac{dx^i}{d\lambda},$$ and where $\lambda$ is an arbitrary WL parameter.

If we identify $$d\tau~=~d\lambda \sqrt{L_0}, $$ then the manipulation (4.40b) suggests that $$ \delta(d\tau) ~=~\frac{\delta(d\tau)^2}{2d\tau} ~=~\frac{(d\lambda)^2}{2d\tau}\delta L_0,$$ i.e. Padmanabhan is effectively varying $L_0$ rather than $\sqrt{L_0}$ in eq. (4.41).

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$^1$ Conventions. We use Minkowski signature $(-,+,+,+)$ and work in units where $c=1$.

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