Curve on Lorentz manifold

123 Views Asked by At

If $\alpha : [0,B) \rightarrow M, B \leq \infty$, is an extendible, piecewise smooth (nonspacelike curve) in a Lorentz manifold, then $\alpha$ has a finite length.

Any hints on how to show this? I know what all the words mean, but I have trouble seeing the trees. I tried writing out the arc length integral and playing around with the limits (using extendibillity), but it got me nowhere.

1

There are 1 best solutions below

2
On BEST ANSWER

It is enough to treat the case of finite B, by reparametrizing.

But in this case, extending $\alpha$ past B, you can see that the length of $\alpha$ from $0$ to B is an integral of a continuous function ($\|\alpha'(t)\|$) over the compact set $[0, B]$, which is finite.