Is there a difference, conceptually speaking, between solving the geodesic equations using $\lambda$ as an arbitrary parameter vs substituting a coordinate from the metric in it's place?

For instance, if I have:

$\frac{d^2\theta}{d\lambda^2}=-\Gamma^{\theta}_{t\theta}\frac{dt}{d\lambda}\frac{d\theta}{d\lambda}-\Gamma^{\theta}_{p\theta}\frac{dp}{d\lambda}\frac{d\theta}{d\lambda}$, and I decide $\lambda=\theta$, I get $0=-\Gamma^{\theta}_{t\theta}\frac{dt}{d\theta}-\Gamma^{\theta}_{p\theta}\frac{dp}{d\theta}$, which I can multiply by $d\theta$ and then integrate for $t$ and $p$. But I know the answer must not be true for all geodesics, so I have to ask, what does using $\theta$ as $\lambda$ specify?

By the way, if it would help to explicity see what the metric and Christoffel symbols are, I could include those.

  • $\begingroup$ In general, the geodesic might have multiple places at which $\theta$ has a particular value. You seem to be assuming it’s single-valued and therefore $t$ and $p$ are well-defined functions of $\theta$. $\endgroup$
    – Ghoster
    Commented Nov 30, 2022 at 21:51
  • $\begingroup$ I understood that there are 3 variables $t$, $p$ and $\theta$. In this case the geodesic equation should have much more than 2 terms at the right side. $\endgroup$ Commented Nov 30, 2022 at 22:53
  • $\begingroup$ The metric is $-dt^2+dp^2+(5p^2+4t^2)d\theta^2$, so there are in fact only two Christoffel symbols with an upper index of $\theta$. So I believe the above geodesic equation is correct. There are no killing vectors other than $(5p^2+4t^2)d\theta$, so I don't see a way to solve for geodesics analytically. Do you think that some coordinate transformation could change that? $\endgroup$
    – user345249
    Commented Dec 1, 2022 at 1:07
  • $\begingroup$ By the way, all this comes from the Wikipedia article for Ellis wormholes in the section entitled "Dynamic Ellis Wormhole." $\endgroup$
    – user345249
    Commented Dec 1, 2022 at 1:10

1 Answer 1


When we use the Euler-Lagrange equations to find the stationary action of a path with a given parametrization in a manifold with a metric $g$: $$s_{ab} = \int _a^b \sqrt {\frac{\partial X^i}{\partial \lambda}\frac{\partial X^j}{\partial \lambda}g_{ij}(λ)}dλ$$

The resulting differential equation looks like the geodesic equation in the left side, with the connection defined as the metric one:

$$\frac{\partial u^p}{\partial \lambda} + \frac{1}{2}g^{pk}(\partial_i g_{kj} + \partial_j g_{ki} - \partial_k g_{ij})u^iu^j $$

Where $$u^a = \frac{\partial X^a}{\partial \lambda}$$

But the right side is zero only if $\lambda$ is an affine function of the arc length $s$.

That means: while it is possible to choose a parameter $\lambda = f(s)$, where $f$ is not an affine function, the geodesic in this case is not the shortest length, what (as far as I know) it should be in GR.

There a discussion about the issue on this answer.

  • $\begingroup$ Oh, so $\lambda=as+b$ (where $a$ and $b$ are constants)? $\endgroup$
    – user345249
    Commented Dec 2, 2022 at 4:40
  • $\begingroup$ @user345249 yes. $\endgroup$ Commented Dec 2, 2022 at 18:05

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