If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic.

Question

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?

Answer 1

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

Answer 2

Problem 19 asks the reader to show that in a semi-Riemannian manifold, a regular curve $\alpha$ with $α''(s)=f(s)α'(s)$ is pregeodesic. (That is, it can be reparametrized to be a geodesic.) This result is needed later in the book in the context of geodesic curves in warped product spaces.

As mentioned in the question, if a reparametrization $h$ satisfies the differential equation in a), it makes $\alpha \circ h$ a geodesic. We can prove the result by solving the differential equation. The complication is that we need $h$ to be defined so that its range is the whole domain of $\alpha$, and it is strictly increasing. (Or strictly decreasing.) Let $(a, b)$ be the domain of $\alpha$. For any point $s$ in $(a, b)$ we can find a solution $h_s$ of the differential equation on an interval including zero with $h_s(0) = s$ and $h_s'(0) = 1$. By shrinking the interval we can have $h_s'(t) > 0$ for all $t$ in the domain of $h_s$. Pick a fixed point $s_0$ in $(a, b)$. Notice that if $h$ is a solution of the differential equation, then any affine reparametrization of $h$ is also a solution. Let $[c,d]$ be any closed interval containing $s_0$ and contained in $(a,b)$. Since $[c,d]$ is compact, the ranges of a finite number of the $h_s$ cover $[c,d]$. Starting with $h_{s_0}$, we can extend this function with affine reparametrizations of $h_s$ from the finite set to get a solution $h_{[c,d]}$ whose range is an open interval containing $[c,d]$, and which is strictly increasing. Since for different $[c,d]$ these solutions agree at $s_0$, any two agree on their common domain. We can then glue them together to get a solution whose range is $(a, b)$. This shows that $\alpha$ is pregeodesic as desired.