I'm working on a derivation of the geodesic equation from the action functional. In special relativity, specifically for flat spacetime, we assume that the metric tensor is constant (not necessarily that it's not a function of the coordinates, but that it traces out constant distances between any two points in space since the metric is what keeps track of distances in geometry; maybe saying "it's constant" is wrong in this case).

Anyhow, starting from the spacetime interval

$$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$

with signature $(+,-,-,-)$, the lorentz invariant action in flat spacetime is given by

$$S[x^{\mu}] = \int ds = \int |{\frac{ds}{d{\lambda}}| d{\lambda}}=\int d\lambda \sqrt{g_{\mu\nu} \dot{dx^{\mu}} \dot{dx^{\nu}}}.$$


$$S[x^{\mu}] = \int d\lambda L(x^{\mu}, \dot{x^{\mu}}),$$

the stationary action principle yields Euler-Lagrange equations nearly identical to those from classical, non-relativistic mechanics

$${\frac{\partial L}{\partial x^{\mu}} - {\frac{d}{ds}} {\frac {\partial L}{\partial \dot{x^{\mu}}} } = 0}.$$

The action is essentially the arclength of the worldline.

Now, every derivation I've seen online to derive the geodesic equation begins with assuming that the metric is describing non constant distances and is a function of the coordinates $x^{\mu}({\lambda})$, for scalar parameter ${\lambda}$. Then, most sources take the flat spacetime action as given above and take it's variation. The geodesic equation falls out upon taking the derivatives in the Euler-Lagrange equation properly (and this is commonly how the Christoffel symbol can be introduced).

My problem is, doesn't the term $ds/d{\lambda}$ get defined differently? Why is the derivative with respect to the parameter lambda the same if the metric is non-constant as a function of the coordinates (which are a function of the parameter). I think I'm missing something about the definition of arclength.


2 Answers 2


The derivative $\frac{ds}{d\lambda}$ gives a vector field which is tangent to the world line at every point of the world line. This tangent vector does not depend on how space behaves away from the world line, and therefore, $\frac{ds}{d\lambda}$ has the same meaning no matter how the manifold in which we embed our world line curves. The curvature of space only enters the picture in the form of the metric when it comes to evaluating the length of the tangent vector at a specific point, and subsequently, the arc length along the world line. Arc length on a manifold is defined exactly the way you wrote it: as the integral over the square root of the inner product of two tangent vectors to the curve.

Also, the curve parameter does not have an intrinsic physical meaning. One can (possibly after an appropriate reparametrization) take it as the proper time measured by an observer going along the world line, but mathematically, it does not matter, as the tangent always points in the same direction.


The action between 2 values $a$ and $b$ of a parameter $\lambda$ is: $$s_{ab} = \int_a^b \sqrt{\frac{\partial X^{\mu}}{\partial \lambda}\frac{\partial X^{\nu}}{\partial \lambda}g_{\mu\nu}} d\lambda$$ where in general not only the derivatives but also the $g$'s are function of $\lambda$. Note that it is a sum of integrals, and in the particular case when the metric is constant, the $g$'s can be placed out of the integrals.

The final result of the Euler-Lagrange equations presents in the left side the familiar equation of the geodesic, where the term multiplying the product of the first derivatives is the metric connection or symbol of Christoffel: $$\frac{d^2X^{\sigma}}{d{\lambda^2}} + \frac{1}{2}g^{\sigma k}\left(\frac{\partial g_{k\nu}}{\partial X^{\mu}} + \frac{\partial g_{k\mu}}{\partial X^{\nu}} - \frac{\partial g_{\mu\nu}}{\partial X^k}\right)\frac{\partial X^{\mu}}{\partial \lambda}\frac{\partial X^{\nu}}{\partial \lambda}$$

But the right side is not zero, so it is not the geodesic equation yet, except if the parameter $\lambda$ is chosen to be the arc length $s$! Only then the right side is zero.


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