# Showing that null geodesics are incomplete

Given a metric: $$ds^2 = -dt^2 + t dx^2$$

for a manifold $$M = \mathbb{R}^+ \times \mathbb{R}$$. The geodesic equation for null geodesics is $$x = x_0 \pm \log t$$ for some constant $$x_0$$. Now I want to show that these geodesics are not complete, but I'm not exactly sure how to do this. The metric reminds of the Rindler wedge so I assume the manifold is not geodescially complete but I'm not sure how to proceed here.

• Find a null geodesic whose affine parameter is not defined on the whole real axis. – Valter Moretti May 20 at 18:39
• What happens if you switch to double-null coordinates? – Jerry Schirmer May 20 at 21:02
• It seems to me that you do not know the notion of geodesically compleness. What you have to do is easy: compute the null geodesics parametrized with their affine parameters (i.e. solving the geodesical equation written in terms of connection symbols) and prove that the parameter of some of these geodesics is not defined on the complete real axis when the geodesic stays in $M$. – Valter Moretti May 21 at 6:50
• The crucial point here is the correct choice of the parameter. I do not know (without explicit computations) if the parameter $t$ you used is an affine parameter. If it is, then you are done, because $\log t$ is defined for $t>0$ only. – Valter Moretti May 21 at 6:55
• By the way, your solution is wrong! $dx/dt= \pm 1/ \sqrt{t}$ has solutions $x(t) = x_0 \pm 2 \sqrt{t}$... – Valter Moretti May 21 at 11:58

By definition a (metric) geodesic parametrized in terms of an affine parameter is a solution of the equation (in local coordinates) $$\frac{d^2x^a}{d\lambda^2}+ \Gamma^a_{bc}\frac{dx^b}{d\lambda}\frac{dx^c}{d\lambda}=0\:,\tag{0}$$ where the $$\Gamma$$s are the standard Christoffel coefficients of the metric $$g$$.

A (bidifferentiable) change of parameter $$\lambda'=\lambda'(\lambda)$$ preserves the form above of the equation if and only if $$\lambda' = k \lambda + h\tag{1}$$ for some $$k\neq 0$$, $$h \in \mathbb{R}$$.

The affine form of (1) is a justification of the name "affine parameter". It turns out that if the geodesic is not light-like, then the length parameter is an affine parameter in particular.

Geodesic completeness means that all geodesic are defined on the whole real axis when parametrized with their affine parameters $$\lambda$$ (this requirement does not depend on the specific choice of the affine parameter in view of the affine form of (1)).

You found (there is a mistake see may comment) an expression for null geodesics, but you do not know if the coordinate $$t$$ you used as the parameter of the curves belongs to the family of affine parameters.

If it is the case, then you are done since, by definition of $$M$$, $$\lambda =t>0$$. However it could happen that, for instance $$t= e^\lambda$$, where $$\lambda \in (-\infty,\infty)$$. So the issue deserves a better scrutiny.

To solve your problem you have to check if $$t$$ is an affine parameter solving (0), However there is another shorter procedure. It is possible to prove (by direct inspection) that the Hamilton equations of the Hamiltonian $$H := \frac{1}{2}g^{ab}p_ap_b$$ in the cotangent space are identical to (0) when re-written into a second order equations in local coordinates in $$M$$. In particular the time parameter'' of these Hamilton equations is actually an affine parameter $$\lambda$$ of the geodesics.

Since the metric you consider is very simple, and $$H = -\frac{p_t^2}{2} + \frac{p_x^2}{2t}\:,$$ it is convenient to use this approach instead of computing the coefficients $$\Gamma^{a}_{bc}$$.

The Hamilton equations read $$\frac{dt}{d\lambda} = -p_t\:, \quad \frac{dx}{d\lambda} = \frac{p_x}{t}\:, \quad \frac{dp_t}{d\lambda} = -\frac{p_x^2}{2t^2}\:, \quad \frac{dp_x}{d\lambda} = 0\:.$$ The first two equations say that $$p_t,p_x$$ are the covariant components of the tangent vector to the considered geodesic: $$p_a = g_{ab} \frac{dx^b}{d\lambda}$$.

Furthermore, since $$H$$ does not depend explicitly on time, $$H$$ is constant along the solutions of the equations above.

Hence, for a null geodesic, since $$H(\lambda =0)= g^{ab}p_a(0)p_b(0)=0$$, $$-\frac{p_t^2}{2} + \frac{p_x^2}{2t} = 0\:,$$ so that $$p_t = \pm \frac{p_x}{\sqrt{t}}\:.$$ Inserting this identity in the first Hamilton equation, we have $$\frac{dt}{d\lambda} = \mp \frac{p_x}{\sqrt{t}}$$ where $$p_x$$ is constant in view of the last Hamilton equation. We conclude that $$\lambda = C \pm \frac{2}{3p_x}t^{3/2}$$ Notice that $$p_x \neq 0$$ since the geodesic is light-like. This result already permits to conclude.

Since $$t>0$$, the parameter $$\lambda$$ cannot range in the whole line $$(-\infty, +\infty)$$ for every solution of the Hamilton equation with light-like tangent vector, i.e., for every null geodesic.