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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?

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  • $\begingroup$ 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
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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 $\alpha''(s)=f(s)\alpha'(s)$.

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  • $\begingroup$ 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

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