I'm trying to solve exercise $19$ in pages $95$ and $96$ in O'Neill's Semi-Riemannian Geometry book. There are four items and I'm having trouble with the last one.

Let $(M,\langle \cdot,\cdot\rangle)$ be a Lorentz manifold and $\alpha$ be a regular curve such that $\alpha''(s) = f(s)\alpha'(s)$.

a) If $\beta = \alpha\circ h$, $\beta$ is pre-geodesic if and only if $h''+ (f\circ h) (h')^2 = 0$.

Easy, just compute $\beta''(s)=( h''(s) + f( h(s)) (h'(s))^2)\alpha'(h(s))$ and use that $\alpha$ is regular.

b) If $\beta$ has unit speed and $\langle \alpha',\alpha'\rangle $ is never zero, then $\beta$ is geodesic.

If $\beta$ has unit speed, then $\langle \beta''(s), \beta'(s)\rangle = 0$ for all $s$, after differentiating once, but this evaluates to $h'(s) ( h''(s) + f( h(s)) (h'(s))^2) \langle \alpha'(s),\alpha'(s)\rangle = 0$.

c) $\langle \alpha',\alpha'\rangle$ is always zero or never zero.

From $\langle \alpha'(s),\alpha'(s)\rangle' = 2f(s)\langle \alpha'(s),\alpha'(s)\rangle$, we have $\langle \alpha'(s),\alpha'(s)\rangle = Ce^{2\int f(s)\,{\rm d}s}$. If that is zero in some point it is always zero by uniqueness of solutions of IVP's, and otherwise it is never zero because the exponential is never zero.

d) If $\langle \alpha',\alpha'\rangle$ is always zero, then $\alpha$ is pre-geodesic.

I'm stuck. The only thing I managed to get is that $\langle \alpha''(s),\alpha''(s)\rangle = 0$. If we could prove that $\alpha''(s)$ is not lightlike, this would imply that $\alpha''(s) = 0 $. Help?

  • $\begingroup$ I may be missing something, but if $0=\langle \alpha^{\prime\prime},\alpha^{\prime\prime}\rangle = f^2 \langle \alpha^{\prime},\alpha^{\prime}\rangle $ and $\langle \alpha^{\prime},\alpha^{\prime}\rangle\neq 0 $ then $f=0$ and so $\alpha^{\prime\prime}= f \alpha^{\prime}=0$. If, on the other hand, $f\neq 0$, then the same equations show $\alpha^{\prime\prime}$ is not lightlike. $\endgroup$ – Thomas Nov 10 '15 at 17:26
  • $\begingroup$ @Thomas, I"m sorry, I copied it wrong, the assumption is that $\langle \alpha',\alpha'\rangle $ is always zero, and I used that to get $\langle \alpha'',\alpha''\rangle = 0 $. $\endgroup$ – Ivo Terek Nov 10 '15 at 18:13

What you have to show in (d) is that if $\langle\alpha',\alpha'\rangle=0$ everywhere, then $\alpha$ is a geodesic up to reparametrization, which is the same as $\alpha''(s)=f(s)\alpha'(s)$ everywhere. This is easy if you have already shown that $\langle\alpha'',\alpha''\rangle=0$ everywhere, for we also have $\langle\alpha',\alpha''\rangle=0$ everywhere. This means that for all $s$ in the domain of $\alpha$ we have that $\alpha''(s)$ is null and orthogonal to $\alpha'(s)$, which is also null. This happens if and only if $\alpha''(s)=f(s)\alpha'(s)$.

  • $\begingroup$ I'm sorry, I don't think I quite get it. Shouldn't I prove that $\alpha'' = 0$? $\endgroup$ – Ivo Terek Nov 10 '15 at 18:56
  • $\begingroup$ No. This would imply that $\alpha$ is a geodesic, which is in general not true. Take for instance a null geodesic $\beta$ and a non-affine reparametrization $g$ (e.g. $g(t)=\tanh t$). Then $\alpha=\beta\circ g$ is a pre-geodesic which is not a geodesic, but still $\langle\alpha',\alpha'\rangle=0$ everywhere by the chain rule. $\endgroup$ – Pedro Lauridsen Ribeiro Nov 10 '15 at 19:29
  • $\begingroup$ This example made sense, thanks. I was thinking that being pre-geodesic + constant speed implied being a geodesic. I don't see how $\alpha''(s) = f(s)\alpha'(s)$ is equivalent as being pre-geodesic, though. And again, it seemed that $\alpha''(s) = f(s)\alpha'(s)$ was the hypothesis right from the start. I am confused. $\endgroup$ – Ivo Terek Nov 10 '15 at 19:38
  • $\begingroup$ The hypothesis $\langle\alpha',\alpha'\rangle=0$ doesn't entail constant speed, as my counterexample above shows. To see the aforementioned equivalence, let $\beta:[a,b]\rightarrow M$ be a geodesic, and $g:[c,d]\rightarrow[a,b]$ a diffeomorphism (i.e. a reparametrization). Then $$(\beta\circ g)''(t)=\frac{g''(t)}{g'(t)}(\beta\circ g)'(t)=(\log g'(t))'(\beta\circ g)'(t)$$ for all $t$ by the chain and Leibniz rules, thus yielding $f(t)=(\log g)'(t)$. Reversing the above formula proves the converse statement. $\endgroup$ – Pedro Lauridsen Ribeiro Nov 10 '15 at 19:48
  • $\begingroup$ In my opinion, item (d) is written somewhat imprecisely. (by the way, it should be $f = (\log g')'$ in my previous comment - unfortunately, I can no longer edit it) $\endgroup$ – Pedro Lauridsen Ribeiro Nov 10 '15 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.