In the following, as usual, a spacetime is a pseudo-Riemannian manifold $M$ with metric $g$ compatible with the Levi-Civita connection $\nabla$. If an affinely-parametrized curve is a geodesic, then the squared length of its tangent vector is constant: $$\nabla_{V} g(V,V) = (\nabla g)(V,V) + g(\nabla_V V, V) + g(V,\nabla_V V) = 0 \implies g(V,V) = c\in\mathbb{R}.$$ We have used the auto-parallel transport condition $\nabla_V V = 0$, only true for affinely-parametrized geodesics.

It is, however, not true that any curve $\gamma \subset M$ with a tangent vector of constant squared length is a geodesic. So how come often in the physics literature, one sees geodesics, say for example the timelike ones of the Schwarzschild spacetime, found by a claim like the following: $$\forall \lambda \in I \subset \mathbb{R} : -1 = g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}.$$

Using Killing vector fields to find conserved quantities along geodesics, the above condition can be reduced to an ODE for one of the coordinates, traditionally $r$.

But I am still not confident that the curves so found are actually geodesics: 3 of the degrees of freedom are those of a geodesic, but why should this guarantee we have actually found a geodesic? Does this reduction to one unknown guarantee this? More generally, is the existence and uniqueness theorem of geodesics at play here: providing but a single tangent vector to $M$ will, at least locally, give a geodesic.


2 Answers 2


For lightlike geodesics it is true in $2D$: those curves can be re-parametrized to geodesics in standard form. For geodesics of other types it is false in general: think of a circle im Minkowski spacetime in the rest space of a Minkowski reference frame.

Regarding lightlike geodesics, if $g(\dot{\gamma}(s),\dot{\gamma}(s))$ is constant, we have that $g(\nabla_{\dot{\gamma}}\dot{\gamma}(s),\dot{\gamma}(s))=0$. Since the tangent vector is lightlike, the only possibility is that $$\nabla_{\dot{\gamma}}\dot{\gamma} = \lambda(s)\dot{\gamma}.$$ At this point, it is possible to reparametrize the curve in order to obtain $$\nabla_{\dot{\gamma}(u)}\dot{\gamma}(u)= 0.$$

  • $\begingroup$ I don't think the primary claim here is true: a null curve need not be reparameterized geodesic, as demonstrated in the counterexample in the linked question physics.stackexchange.com/a/76173/153305. The argument given here fails in that while the orthogonal space to a null vector includes the span of the null vector, it is not only the span of the null vector-- it is a codimension one lightlike subspace of the tangent space--, so $\nabla_{\dot \gamma} \dot \gamma$ is not forced to be parallel to $\dot \gamma$. $\endgroup$
    – jawheele
    Feb 19, 2023 at 16:34
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    $\begingroup$ You are right, I was referring to the 2D case. I corrected my answer. $\endgroup$ Feb 19, 2023 at 17:54

I have a $90\%$ complete answer for you, modulo one technical detail (maybe I’m overlooking something trivial to get around it). First, let us note the following lemma:


Let $(M,g)$ be a pseudo-Riemannian manifold, and $\xi$ a Killing vector field. Then, for any smooth curve $\gamma:I\subset\Bbb{R}\to M$, we have $g(\dot{\gamma},\nabla_{\dot{\gamma}}\xi)=0$ along $\gamma$. As a consequence (by differentiating) we conclude that $g(\dot{\gamma},\xi)$ is constant along $\gamma$ if and only if $g(\nabla_{\dot{\gamma}}\dot{\gamma},\xi)=0$ along $\gamma$.

I’ll leave it to you to prove this fact; it’s a simple application of the product rule and the Killing condition on $\xi$. Next, we come to the following theorem:


Let $(M,g)$ be an $n$-dimensional pseudo-Riemannian manifold, and suppose we have a set of $n-1$ smooth, pointwise linearly independent Killing vector fields $\{\xi_1,\dots,\xi_{n-1}\}$. Let $\gamma:I\to M$ be a smooth curve. Then,

  • If $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ along $\gamma$, then the quantities $g(\dot{\gamma},\xi_i)$ for $i\in\{1,\dots, n-1\}$ and $g(\dot{\gamma},\dot{\gamma})$ are constant along $\gamma$.
  • As a partial converse, if the tangent vector to $\gamma$ is never in the span of these Killing fields and $g(\dot{\gamma},\xi_i)=c_i$ and $g(\dot{\gamma},\dot{\gamma})=\kappa$ are constants, then we have $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ along $\gamma$.

The first bullet point I think you already know the proof for. For the second bullet point, note that by working in a neighborhood of each point, we may without loss of generality assume there is a smooth vector field $R$ (not necessarily Killing) defined on $M$ such that $\{\xi_1,\dots,\xi_{n-1},R\}$ forms a basis for each tangent space. Since we have a basis, let us write the velocity to the curve as \begin{align} \dot{\gamma}&=a^i\xi_i+bR, \end{align} for some smooth functions $a^i,b:I\to\Bbb{R}$; and in fact out assumption of not lying in the span of the Killing fields means that the function $b$ is nowhere-vanishing. Now, differentiating the equation $g(\dot{\gamma},\xi_i)=c_i$, and using the lemma, we find that $g(\nabla_{\dot{\gamma}}\dot{\gamma},\xi_i)=0$ along $\gamma$. Finally, differentiating the equation $g(\dot{\gamma},\dot{\gamma})=\kappa$ (and dividing by the factor of 2) gives \begin{align} 0&=g(\nabla_{\dot{\gamma}}\dot{\gamma},\dot{\gamma})=a^i g(\nabla_{\dot{\gamma}}\dot{\gamma},\xi_i)+bg(\nabla_{\dot{\gamma}}\dot{\gamma},R)=0+bg(\nabla_{\dot{\gamma}}\dot{\gamma},R). \end{align} Since $b$ is nowhere-vanishing, it follows $g(\nabla_{\dot{\gamma}}\dot{\gamma},R)=0$. Therefore, by non-degeneracy of the metric, and the fact that $\{\xi_1,\dots, \xi_{n-1},R\}$ forms a basis, we find that $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ along $\gamma$, as desired.

So, this theorem is telling us that if we find sufficiently many Killing vector fields, then the information in the geodesic equation is equivalent to the conservation laws and the parametrization constraint on the curve. I’m not sure if one can weaken the assumption in the converse too much. We can definitely weaken the assumption slightly, such as “the tangent vector to $\gamma$ is not in the span of the Killing fields for a dense subset of parameter values in $I$”. Because then, we can repeat the argument above to conclude that for these particular values of $\lambda$, we have $\nabla_{\dot{\gamma}(\lambda)}\dot{\gamma}=0$, and hence, by continuity of $\nabla_{\dot{\gamma}}\dot{\gamma}$ on $I$, it vanishes everywhere. So, this is saying that if occasionally the curve $\gamma$ lies in the span of the Killing fields, then that’s not a big deal, and the result is still a geodesic.

In Schwarzschild specifically, we have a whole bunch of Killing fields, $\{T,\Omega_1,\Omega_2,\Omega_3\}$, where the $\Omega_i$ are the Killing fields for the spherical symmetry.

The usual way we do the computations is that we use spherical symmetry so we use that initially to consider only curves in the $\theta=\pi/2$ ‘plane’ (we can justify this initially at $\lambda=0$, and then for all $\lambda$ by existence and uniqueness, and spherical symmetry, as explained in textbooks). So, we have cut down one dimension; we’ll now use the above theorem with $n=3=2+1$. We have the timelike Killing field $T$ (in Schwarzschild coordinates $(t,r,\theta,\phi)$, it is $\frac{\partial}{\partial t}$… as you can immediately recognize since the metric components do not depend on $t$ in this coordinate system), and the angular Killing field $\Phi$ (in Schwarzschild coordinates, it is $\frac{\partial}{\partial\phi}$). These give us the two conserved quantities of ‘energy’ and ‘angular momentum’, and we take $R$ to be any vector field transversal to these guys, for instance the radial vector field of Schwarzschild coordinates, $\frac{\partial}{\partial r}$. In fact, in Schwarzschild, these vector fields are orthogonal, so the algebra is simplified slightly.

  • 2
    $\begingroup$ Also, in Schwarzschild, say, we often look for radial geodesics, so the $R$-component being non-vanishing isn’t that big of a deal. If it does vanish, then I think in this special case, one can verify the last remaining component of the geodesic equation directly (I remember thinking about this once, but I lost my calculations lol, and don’t feel like repeating them… but it shouldn’t be too hard). $\endgroup$
    – peek-a-boo
    Feb 19, 2023 at 20:29

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