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I have a question about an exercise in Misner, Thorne, and Wheeler's Gravitation. On page 321, exercise 13.5 says

Show that a geodesic of spacetime which is timelike at one event is everywhere timelike. Similarly, show that a geodesic initially spacelike is everywhere spacelike, and a geodesic initially null is everywhere null. [Hint: This is the easiest exercise in the book!]

Ok so the idea is that timelike means that the timelike geodesic have a negative-length 4-velocity, a spacelike would have a positive-length 4-velocity and a null would have a 0-length 4-velocity. So my idea was to use the definition of the geodesic to show that the sign of the length of the 4-velocity doesn't change. The definition of the geodesic translates to a curve that parallel transport conserves the tangent vector to the curve.

My idea was to say that the tangent vector of a spacelike is always in the space parts of spacetime. But it doesn't seem true at all. Finally the Hint doesn't help me much.

I assume there are two kinds of solutions for this problem: one purely abstract with words and one with a more mathematical approach of the concept of geodesics and tangent vectors.

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  • $\begingroup$ Just calculate $D(u_\mu u^\mu ) /ds$. $\endgroup$ – A.V.S. Jul 2 at 16:58
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    $\begingroup$ What does “a negative 4-velocity” mean? Vectors aren’t positive or negative. Various components can be positive or negative, and the “length” can be positive or negative. $\endgroup$ – G. Smith Jul 2 at 17:01
  • $\begingroup$ I really don't understand the last sentence. A purely abstract answer with words is less mathematical? $\endgroup$ – Fabio Di Nocera Jul 3 at 15:30
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Let $u^\mu = \frac{d x^\mu}{d\tau}$ be the 4-velocity of a geodesic where $\tau$ is the affine parameter along a geodesic. It satisfies the geodesic equation $$ \frac{du^\lambda}{d\tau} + \Gamma^\lambda_{\mu\nu} u^\mu u^\nu = 0 . $$ Then, \begin{align} \frac{ d }{d\tau} ( g_{\mu\nu} u^\mu u^\nu ) &= u^\mu \left( u^\nu \frac{ d }{d\tau} g_{\mu\nu} + 2 g_{\mu\nu} \frac{ d }{d\tau} u^\nu \right)= u^\mu \left( u^\nu u^\rho \partial_\rho g_{\mu\nu} + 2 g_{\mu\nu} \frac{ d }{d\tau} u^\nu \right) \end{align} Now, recall the definition $$ \Gamma^\lambda_{\mu\nu} = \frac{1}{2} g^{\lambda \rho} \left( \partial_\mu g_{\nu\rho} + \partial_\nu g_{\mu\rho} - \partial_\rho g_{\mu\nu} \right) $$ Then, $$ 2 g_{\lambda \rho} \Gamma^\lambda_{\mu\nu} = \partial_\mu g_{\nu\rho} + \partial_\nu g_{\mu\rho} - \partial_\rho g_{\mu\nu} $$ Interchanging $\mu$ and $\rho$, we find $$ 2 g_{\lambda \mu} \Gamma^\lambda_{\rho\nu} = \partial_\rho g_{\mu\nu} + \partial_\nu g_{\mu\rho} - \partial_\mu g_{\nu\rho} $$ Adding the two equations above and then rearranging the indices, we find $$ \partial_\rho g_{\mu\nu} = 2 g_{\lambda(\mu}\Gamma^\lambda_{\nu)\rho} $$ Substituting this into the equation, we find \begin{align} \frac{ d }{d\tau} ( g_{\mu\nu} u^\mu u^\nu ) &= 2 g_{\mu\lambda} u^\mu \left( u^\nu u^\rho \Gamma^\lambda_{\nu\rho} + \frac{ d u^\lambda }{d\tau} \right) = 0. \end{align} Thus, $u^2$ is a constant along the geodesic. An immediately corollary of this result is the statement in the question.

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  • $\begingroup$ Trust me, this is the easiest calculation you can hope for in GR. $\endgroup$ – Prahar Jul 2 at 17:06
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Just use the geodesic equation to evaluate the derivative along the curve of the "length" of the tangent to the curve. You'll find that it is zero, so the "length" does not change along the curve. In particular, the sign of the "length" does not change.

The geodesic equation is \begin{align} T^a \nabla_a T^b = 0, \end{align} where $T^a$ is the tangent to the curve, in whatever affine parameterization you want. The norm of the tangent (which determines whether it's timelike, spacelike, or null) is given by $T^b T^c g_{bc}$. The quantity $T^a \nabla_a$ is just the derivative along the curve, so the derivative along the curve of the norm of the tangent is \begin{align} T^a \nabla_a \left( T^b T^c g_{bc} \right) = 0. \end{align} Since the covariant derivative of the metric is zero, we can pull it out of the derivative, and expand this as \begin{align} T^a \nabla_a \left( T^b T^c g_{bc} \right) &= g_{bc} T^a \nabla_a \left( T^b T^c \right) \\ &= g_{bc} \left[ \left( T^a \nabla_a T^b \right) T^c + T^b \left( T^a \nabla_a T^c \right) \right] \\ &= 0 \end{align} To get to zero, we just apply the geodesic equation twice in that second-to-last line. Again, since this derivative along the curve is zero, the norm does not change — so the sign of the norm does not change.

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A geodesic is a curve of auto-parallel transport, then the tangent vectors are transported parallely.

In GR one usually uses Levi-Civita connection, so the scalar product of transported vectors is conserved.

So the scalar product of the tangent vector with himself (its norm) is conserved, so if it is less than zero in one point it is less than zero everywhere on the geodesic.

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  • $\begingroup$ A beautiful and elegant answer! $\endgroup$ – G. Smith Jul 2 at 21:46

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