Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 36D (General Relativity)

At $p$, choose local normal coordinates with $g_{\alpha \beta }(p)=\mathrm{diag}(-1,1)$. Let $e_0,e_1$ be their coordinate basis. The vectors $V=e_0+e_1$ and $W=e_0-e_1$ have zero metric tensor square, and their component matrix has determinant $-2\ne 0$. Thus two independent null vectors form a basis of the tangent space. Normal coordinates are only needed at the chosen point, not on a whole neighbourhood with flat metric.

Write the nonzero tangent vector as $U=aV+bW$. Since $g(V,V)=g(W,W)=0$ and $g(V,W)=-2$, nullness gives $g(U,U)=-4ab=0$. Thus exactly one coefficient can be nonzero, proving $\boxed{U\parallel V\text{ or }U\parallel W}$. This pointwise argument does not require the curve to be a geodesic.

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.