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The null condition for a nonzero vector $k$ in a Lorentzian metric is $g(k,k)=0$. Applied to the tangent of a curve it makes the curve a null curve. Nullity alone does not imply the geodesic equation; an additional zero or tangent-parallel covariant acceleration is required.

A null geodesic is both a null curve and a geodesic. In geometric optics, light rays follow null geodesics.

A null pregeodesic has a null tangent and can be parametrized as a null geodesic. In a two-dimensional Lorentzian metric, the acceleration of a regular null curve is orthogonal to its tangent and therefore proportional to it.

Let $U$ be the smooth nonzero tangent along a regular null curve, and $a={\nabla }_{U}U$. Metric compatibility gives $g(U,a)=\frac{1}{2}U[g(U,U)]=0$. In a two-dimensional Lorentzian manifold, the orthogonal complement of a nonzero null vector is its own span; this follows directly in the preceding null basis. Therefore $a=\kappa (t)U$.

Consequently every regular null curve is a geodesic up to reparametrization. To obtain an affine parameter $\lambda$, write $q=dt/d\lambda >0$ and solve $dq/dt+\kappa (t)q=0$. Then ${\nabla }_{qU}(qU)=q^2(\kappa +q^{-1}dq/dt)U=0$. A solution $q=C\exp (-\int \kappa dt)$ exists locally and along any nonsingular parameter interval. The original parameter need not be affine; the literal assertion of zero covariant acceleration for every parameter would be false.

In Minkowski spacetime with metric $-d{t}^2+d{x}^2+d{y}^2+d{z}^2$, takeIts tangent vector is $(1,-\sin t,\cos t,0)$, with metric tensor square $-1+{\sin }^{2}t+{\cos }^{2}t=0$. Its acceleration is $(0,-\cos t,-\sin t,0)$, nonzero and not proportional to the tangent because its time component is zero. Thus it is a regular null curve which is not a geodesic under any reparametrization.