# In general relativity, are light-like curves light-like geodesics?

Just as the title. If a curve is light-like, i.e. a null-curve, is it definitely a null geodesic?

No.

Here is a simple counterexample due to my friend Antonio Russo, a fellow UCLA grad student. Set $c=1$ for convenience, and consider the following curve in $\mathbb R^{2,1}$ (aka three-dimensional flat space with one time dimension $t$ and two space dimensions $x$ and $y$): \begin{align} t(\lambda) = \lambda, \qquad x(\lambda) = R\cos(\alpha\lambda), \qquad y(\lambda) = R\sin(\alpha\lambda). \end{align} This is the trajectory of particle moving along a circle. Notice that \begin{align} \dot x^\mu \dot x_\mu = -1+(R\alpha)^2, \end{align} where the overdot denotes derivative with respect to $\lambda$, and we are using $(-,+,+)$ signature. Therefore this curve is null if we choose $R\alpha = 1$. It is not a geodesic, however, since geodesics in flat space are straight lines, and it is not a straight line.

• I believe it should be $y(\lambda) = R\sin(\alpha\lambda)$.
– Psi
Sep 4, 2013 at 2:10
• Geodesics are a concept from General Relativity and can be curves, not straight lines,no? Sep 4, 2013 at 4:33
• @annav Notice that I wrote "Consider the following curve in $\mathbb R^{2,1}$," which is three-dimensional Minkowski space (flat space). The geodesics in flat space are straight, so this is an example where the set of geodesics in a certain geometry does not coincide with the set of null curves. Of course you can, however, come up with similar examples for non-trivially curved spacetimes as well, but I think that overcomplicates the issue. Sep 4, 2013 at 4:54
• This example is simple enough and illustrates the problem clearly, and I have got the idea. Thanks a lot! Sep 4, 2013 at 7:33
• Dear Anna, geodesics are "curves" because in a general curved spacetime, it isn't even possible to find something that could be called "straight lines". On the other hand, geodesics are closest to this goal because geodesics are, by the definition translated in a way, the "straightest" lines you can find in a general space! So at short distances, when any space becomes nearly flat, geodesics must be straight lines, too. In a curved spacetime, geodesics is whatever you get by connecting these straight infinitesimal pieces i.e. by parallel transport of the forward line segment. Sep 4, 2013 at 7:37

In 3 or more spacetime dimensions the answer is No, as simple counterexamples show already in Minkowski space. However, in 2 spacetime dimensions, it actually holds, cf. the next theorem.

Theorem: In an arbitrary 1+1D Lorentzian manifold $M$, a smooth light-like curve satisfies the geodesic equation.

Proof: Let there be given a parametrized null curve $\gamma: [a,b] \to M$ with zero parameter speed squared

$$\tag{A} g(\dot{\gamma},\dot{\gamma})~=~0,\qquad \dot{\gamma} ~\neq~0.$$

Here dot denotes differentiation wrt. to the parameter $\lambda$. By further differentiation of eq. (A) wrt. $\lambda$, we get that $\nabla_{\dot{\gamma}}\dot{\gamma}$ and $\dot{\gamma}$ are orthogonal/perpendicular

$$\tag{B} g(\nabla_{\dot{\gamma}}\dot{\gamma},\dot{\gamma})~=~0$$

wrt. to the Lorentzian metric $g$. Here $\nabla$ is the Levi-Civita (LC) connection. In eq. (B) we have used the metric property of the LC. connection.

Next recall that in 1+1D, the light-cone

$$\tag{C} \mathbb{R}n_+ \cup \mathbb{R}n_- ~\subset~ T_pM$$

in a point $p\in M$ consists of two null rays, spanned by two non-zero null vectors $n_{\pm}\neq 0$, with the inner products

$$\tag{D} g_p(n_{\pm},n_{\pm})~=~0, \qquad g_p(n_{\pm},n_{\mp})~\neq~0.$$

The pair $n_{\pm}$ also happens to be a basis for the tangent space $T_pM$. In other words, in the point $p\in M$, the vector $\nabla_{\dot{\gamma}}\dot{\gamma}$ must be a linear combination of $n_{\pm}$. Moreover the null vector $\dot{\gamma}$ must be proportional to either $n_+$ or $n_-$. Together with eq. (B), it follows that $\nabla_{\dot{\gamma}}\dot{\gamma}$ and $\dot{\gamma}$ are collinear/parallel/proportional

$$\tag{E} \nabla_{\dot{\gamma}}\dot{\gamma} ~\parallel~ \dot{\gamma}~\neq~0.$$

In other words, there exists a function $f:[a,b]\to \mathbb{R}$ such that

$$\tag{F} \nabla_{\dot{\gamma}}\dot{\gamma}~=~f \dot{\gamma}.$$

Eq. (F) is the eq. for a non-affinely parametrized geodesic. Locally one may reparametrized a non-affinely parametrized geodesic into an affinely parametrized geodesic. An affinely parametrized geodesic $\gamma: [a,b] \to M$ satisfies by definition $$\tag{G} \nabla_{\dot{\gamma}}\dot{\gamma}~=~0.$$ $\Box$

Adding to the other answers, this is just to give an overall explanation.

Case of 1+1 flat spacetime- We can first make the choice of co-ordinates and units such that geodesics are straight lines of slope 1 or -1 on our spacetime diagram. To have a null path, our path must, at all points, have a slope of either 1 or -1. So, there is no way to curve the path smoothly while keeping it a null path.

At best, we can have non-differentiable counter-examples like V-paths or zig-zag paths that always have a slope of either 1 or -1. These are null but aren't geodesics.

Case of n+1 flat spacetime, n>1- In this case, the surface of the light cones has a dimension > 1. So, any non-straight curve on a light cone surface will be null but not a geodesic. One such example is in joshphysics's answer. That curve smoothly spirals over the surface of a light cone.

No. In a Euclidean space, you can have all sort of curves. Not all of them are geodesics. This is also true in Minkowski space.