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In Hartle's General Relativity book ("Gravity"), one of the problems (chapter 8 problem 6) is to prove that $g_{\mu\nu}u^\mu u^\nu$ is conserved along geodesics (really not hard to show), where $u^\mu$ is the 4-velocity. My question is: Isn't it true that $g_{\mu\nu}u^\mu u^\nu$ is equal to $-1$ for any timelike curve whether it is a geodesic or not? This follows (I think) from

$$ g_{\mu\nu}u^\mu u^\nu = g_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = \frac{g_{\mu\nu}dx^\mu dx^\nu}{d\tau^2}=\frac{ds^2}{d\tau^2} = \frac{-d\tau^2}{d\tau^2} = -1. $$

Am I wrong about this? Why should we need the geodesic equation to prove this if it's true for any timelike curve?

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Be aware that the normalization depends of the trace of the metric chosen--there are weirdos who use the $\mathop{Tr}(g) = 1$ metric, for instance. –  dmckee Nov 11 '11 at 23:21
    
I'm using the signature (-+++). But my question still applies, changing -1 to +1, in the signature (+---). (Not sure what you mean by $Tr(g)=1$ though, since in either of these two signatures $g^\mu_\mu = \delta^\mu_\mu = 4$) –  Joss L Nov 11 '11 at 23:39
    
@Joss L: dmckee is referring to that some authors use a normalized trace ${\rm Tr}(M)=\frac{1}{n}\sum_{i=1}^n M^i{}_i$. –  Qmechanic Nov 12 '11 at 8:36
    
Oh, okay. Ususlly in general relativity when people say trace I think they mean $M^\mu_{\phantom{\mu}\mu}=g^{\mu\nu}M_{\mu\nu}$. But this is all besides the point. –  Joss L Nov 12 '11 at 17:53

2 Answers 2

Indeed, you are absolutely right, and this is a terrible exercise. The quantity v dot v (using the metric) is the length of the tangent to a curve, and this depends on the parametrization. In an arclength parametrization of timelike curves, as you use to derive the geodesic equation, it is always minus one (in your convention). For null curves, it is zero, for spacelike curves +1. An equivalent statement is that v dot a is zero, the second derivative is relativistically perpendicular to the velocity, and this is just as true in geometry, where in an arc-length parametrization, the second derivative is perpendicular to the first derivative. For null curves, the condition that a is orthogonal to v defines affine parameter, since arclength is zero.

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I am puzzled. Usually Ron's answers contain some strange statements as 'all subsets of the reals are measurable', and they are upvoted. This one is perfectly fine, and is downvoted!!! –  MBN Nov 12 '11 at 15:07
    
It wasn't me that voted it down. It is certainly a better answer than Killercam's. –  Joss L Nov 12 '11 at 17:55
    
@MBN: I figured out why it was voted down--- before I edited it, it said that v dot v is always zero. This is so stupid, I didn't even see the error when I reread it. As for "all subsets of the reals are measurable", this is well known to be true of all subsets which are constructed without the axiom of choice, and it has an intuitive proof which translates to a rigorous forcing construction which is needlessly hidden from non-logicians. –  Ron Maimon Nov 12 '11 at 21:27
    
Well, there is a difference between "all subsets..." and "all subsets such that..." –  MBN Nov 13 '11 at 20:22
    
@MBN: the intuition you have the the notion of "all subsets of R" has an objective meaning which is well defined outside of a given formal mathematical system doesn't withstand scrutiny. The collection of all subsets of R are too big to be objectively defined, they are not listable by a computer program. This means that one must be careful to axiomatize the allowed operations which generate subsets in a model of the real number system. Within ZFC, you are allowed to use choice to select nonmeasurable subsets. In random forcing models, you are allowed to pick numbers at random between 0 and 1. –  Ron Maimon Nov 14 '11 at 3:50
up vote 1 down vote accepted

I emailed my TA and here was his answer, which I think makes sense:

While it is true that a curve which is everywhere timelike can be parametrized so that its tangent vector has unit norm, it is also possible to draw a curve which starts out timelike and then becomes null or spacelike, so its norm won't be the same everywhere. The problem in Hartle simply shows that a geodesic which starts out timelike will always remain timelike (same holds for null or spacelike geodesics).

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A geodesic, being defined as a curve whose tangent vector doesn't change along that curve (parallel transports its own tangent vector), will always return the same norm for that unchanged tangent vector. The same can't be said of other curves, because they don't necessarily have the same tangent vector at different points. –  P O'Conbhui Mar 31 '12 at 0:06

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