# If two points can be joined by a causal curve that is not a null geodesic, can they be joined by a timelike curve?

Suppose $$p$$ and $$q$$ are connected by a causal (i.e. its tangent vectors have non-positive norm) curve $$\gamma$$. If $$\gamma$$ is not a null geodesic, can it be deformed into a smooth, timelike curve still connecting $$p$$ and $$q$$?

This is in Hawking and Ellis The large scale structure of space-time, Proposition 4.5.10. I write the proof below with as many details as I could fill in.

Denote by $$D_t$$ the covariant derivative along $$\gamma$$ induced by the Levi-Civita connection $$\nabla$$. We know $$\gamma$$ is a null geodesic if the acceleration vector $$D_t \gamma '(t) = 0$$ and $$\langle \gamma'(t), \gamma'(t)\rangle = 0$$ everywhere. Thus, if $$\gamma$$ is not a null geodesic, there must exist a point $$t_0$$ where $$D_t\gamma'(t_0) \neq 0$$ or where $$\langle \gamma'(t_0), \gamma'(t_0) \rangle < 0$$. By continuity, if either of these cases hold, then they hold on an open interval $$I = (t_1, t_2)$$. However, if $$\langle \gamma'(t), \gamma'(t) \rangle < 0$$ on $$I$$ then $$\gamma$$ is already timelike, so there is nothing to prove. Consequently, we assume that $$D_t \gamma' \neq 0$$ on $$I$$. Here it is claimed in the book that $$\langle D_t \gamma'(t), \gamma'(t) \rangle = \frac 12 \partial_t \langle \gamma'(t), \gamma'(t) \rangle = 0,$$ but I cannot see why this is so. Can we reparametrize any causal curve so its velocity has constant norm even it at some points the curve becomes null? Taking this for granted, we deduce that $$D_t \gamma'$$ is spacelike and thus $$a(t) ^2 := \langle D_t \gamma'(t), D_t \gamma'(t)\rangle > 0$$ on $$I$$.

We can Fermi transport a vector with positive inner product against $$\gamma'$$ to obtain a vector field $$W$$ along $$\gamma|_I$$ s.t. $$c(t) := -\langle W, \gamma'(t)\rangle > 0$$ on $$I$$ and generate the variation $$\Gamma(s, t) = \exp_{\gamma(t)}(s (x W + y D_t \gamma'(t))), \quad S = \partial_s \Gamma(s, t), \quad T = \partial_t \Gamma(s, t)$$ where $$x$$ and $$y$$ are functions to be chosen momentarily but keeping in mind the requirement that $$x(t_1) = x(t_2) = y(t_1) = y(t_2) = 0$$ to ensure $$\Gamma_s(t_1) = \gamma(t_1)$$ and $$\Gamma_s(t_2) = \gamma(t_2)$$. Then, $$\Gamma_s(t)$$ will be timelike if $$\langle T, T \rangle < 0$$ everywhere, which would hold if the the derivative with respect to $$s$$ at $$s = 0$$ were negative:

\begin{align} - 1 &= \frac{1}{2} \partial_s|_0 \langle T, T \rangle = \langle D_s T, T \rangle \\ &= \langle D_t S, T \rangle & \nabla \text{ is torsion free,} \\ &= \partial_t \langle S, T \rangle - \langle S, D_t T\rangle & \text{metric compatibility,} \\ &= \partial_t (x \langle W, \gamma' \rangle) - x \langle W, D_t \gamma'\rangle - y \langle D_t \gamma', D_t \gamma' \rangle & T|_{s = 0} = \gamma', \\ &= u' + \langle W, D_t \gamma'\rangle c ^{-1} u - y a(t) ^2 & \text{Letting } u = - x c. \end{align} With the integrating factor $$b(t) = -\int_{t_1} ^t \langle W, D_t \gamma'(s) \rangle c(s) ^{-1} \mathrm{d} s$$ we have \begin{align} (u e ^{-b})' e ^{b} = y a(t) ^2 - 1 &\Rightarrow u = e ^b \int_{t_1} ^t e ^{-b(s)} (y(s) a(s) ^2 - 1) \mathrm{d} s \\ &\Rightarrow x = c(t) ^{-1}e ^{b(t)} \int_{t_1} ^t e ^{-b(s)} (1 - y(s) a(s) ^2) \mathrm{d} s. \end{align} To satisfy our earlier constraints, we could let $$y$$ be a parabola with roots ata $$t_1$$ and $$t_2$$. By scaling and applying the intermediate value theorem, we could find an appropriate factor s.t. $$\int_{t_1} ^{t_2} e ^{-b(s)}(1 - y(s) a(s) ^2) \mathrm{d} s = 0$$ because $$a(s)$$ does not vanish. However, in the book, they instead define $$x = c^{-1} e ^b \int_{t_1}^t e^{-b}(1 - a^2 y/2) \text d s$$ and I do not understand where this extra factor of a half is coming from. I think I may be misunderstanding the proof. Could someone point me in the right direction?

• Please give as well the title of the book "The large scale structure of space-time". Commented Sep 24, 2022 at 8:13

The proof is not exactly simple and require a couple of lemmas. I'm not familiar with the proof given in Hawking & Ellis, but you can see Proposition 46 in page 294 of Barrett O'Neill's Semi-Riemannian Geometry with Applications to Relativity. The gist of the argument there is to somehow obtain a vector field $$V$$ along $$\gamma$$ such that $$\langle D_tV,\dot{\gamma}\rangle < 0$$, and then take any variation $$\gamma_s$$ of $$\gamma$$ whose variational vector field is $$V$$ (for $$s$$ non-zero small enough, $$\gamma_s$$ will be timelike). If $$\gamma$$ is timelike somewhere, say at $$\gamma(t_\ast)$$, then let $$V$$ be a suitable function multiple of the (Levi-Civita) parallel transport of $$\dot{\gamma}(t_\ast)$$. When $$\gamma$$ is lightlike, it is also possible to find such $$V$$, but the argument is more elaborate and done in cases.
is false: if the curve is null at some point and you could obtain such a reparametrization, it would be null everywhere. What you can do (and I don't see how that would be useful here) is the following: if $$\gamma$$ is a null curve in a Lorentz manifold and $$D\dot{\gamma}/{\rm d}t$$ is nowhere null, then $$\gamma$$ may be reparametrized so that $$\|D\dot{\gamma}/{\rm d}t\|=1$$, reducing its domain if necessary.
• Thank you! I had looked in Barrett's book but I did not find this lemma. But I think I can complete the one I mentioned by noting that $\langle \gamma', \gamma' \rangle$ vanishing implies it achieves a local maximum an thus has $0$ derivative. Then $D_t \gamma'$ is spacelike at a point and by continuity on a whole interval, which I think suffices to complete the proof. Commented Sep 25, 2022 at 17:36