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

Question

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

Answer 1

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.

Answer 2

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 reparametrize a non-affinely parametrized geodesic into an affinely parametrized geodesic, cf. e.g. my Phys.SE answer here. An affinely parametrized geodesic $\gamma: [a,b] \to M$ satisfies by definition $$\tag{G} \nabla_{\dot{\gamma}}\dot{\gamma}~=~0.$$ $\Box$

Answer 3

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.