# Total derivative of the Lagrangian with respect to the action?

One usual trick for deriving the geodesic equation is that one can parametrize the Lagrangian such that $\frac{dL}{ds}=0$.

First I tought that the parameter $s$ was the action... but now I am confused, if the parameter is arbitrary, then why would that equality in general be true?

PS: I already saw other related posts, but none of them really explains this.

• Why did you think $s$ was the action, when action itself is the time integral of Lagrangian? According to my understanding, $s$ is the proper time (for massive particles). – Sucheta Aug 27 '16 at 20:03
• The type of parameter you should learn about is called an "affine" parameter. en.wikipedia.org/wiki/Geodesic#Affine_and_projective_geodesics You may also find this answer useful: physics.stackexchange.com/a/149087/19976 – joshphysics Aug 27 '16 at 20:09
• The action is $S$, not $s$; as others have said, $s$ is something else. – J.G. Aug 27 '16 at 22:20
• @Sucheta I don't want to confuse things, but I have often seen $ds$ as the arc length, distance along a curve, rather than time. – user108787 Aug 27 '16 at 23:37
• In this context, $s$ is an affine parameter as @joshphysics said, which could be the proper time $\tau$ (when the path is parametrized as $x(\tau)$, say). – Sucheta Aug 29 '16 at 5:45

I came to a conclusion here by applying the Euler-Lagrange operator to the given Lagrangian:

$L^{2}=g_{ij}(x)\dot{x}^{i}\dot{x}^{j}$

Doing the algebra I find that:

$\ddot{x}^{i}+\Gamma^{i}_{jk}\dot{x}^{j}\dot{x}^{k}=\frac{dL}{ds}\dot{x}^{i} \qquad(1)$

Now, according to this webpage the condition for a vector to be transported along a curve without changing its direction is:

$\nabla_{x}V^{i}=\lambda(\tau)V^{i} \qquad (2)$

where $\nabla_x$ is the covariant derivative. Moreover, if the transported vector keeps the same magnitude, then the condition holds for $\lambda(\tau)=0$, which is only true if one chooses $\tau$ to be an affine parameter.

If I take the covariant derivative of $\dot{x}^{i}$, then:

$\nabla_{x}\dot{x}^{i}=\dot{x}^{k}\bigg(\frac{\partial \dot{x}^{i}}{\partial x^{k}}+\Gamma^{i}_{jk}\dot{x}^{j}\bigg)=\ddot{x}^{i}+\Gamma^{i}_{jk}\dot{x}^{j}\dot{x}^{k}$

which is the same as the left-hand side of the equation $(1)$. According to the parallel transport condition $(2)$, if I am choosing an affine parameter $s$, it must be true that:

$\ddot{x}^{i}+\Gamma^{i}_{jk}\dot{x}^{j}\dot{x}^{k}=0$

This is the same as equation (1) only if $\boxed{\frac{dL}{ds}=0}$.

This means that such condition for the Lagrangian must be true only if one chooses $s$ to be an affine parameter.