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The title basically says it all. Given $\boldsymbol{v}$ and $\boldsymbol{w}$ with contravariant components $v^a$ and $w^a$, I'm asked to show that $v^a w_a$ after being parallel-transported along a curve, and using this result, prove that a timelike (or spacelike or null) geodesic will remain such at all points.

I've tried taking the intrinsic derivative of both the contravariant components of $\boldsymbol{v}$ and the covariant components of $\boldsymbol{w}$ and takind the product between those two, but it turns out really difficult to calculate and I'm not even sure I'm doing it right... Any help will be much appreciated!

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Firstly, a vector $v$ (or any arbitrary tensor) is parallel transported along a curve $X^{\mu}(\lambda)$ if its absolute derivative vanishes along the curve, $$ \tag{1} \frac{D}{D \lambda}v^{\mu} = \nabla_{u} v^{\mu} \equiv u^{\nu} \nabla_{\nu} v^{\mu} = 0 \ , $$ where $u^{\nu} = \frac{d X^{\nu}(\lambda)}{d \lambda}$ is the tangent vector along the curve. Note that the definition of parallel transport makes no reference to geodesics. (A geodesic$^1$ is an example of a curve where the tangent vector is parallel transported with respect to itself.)

Applying this to the norm of two vectors $v$ $w$ we have $$\tag{2} \frac{D}{D \lambda}(v^{\mu}w_{\mu}) = u^{\rho} \nabla_{\rho}(g_{\mu \nu} v^{\mu} w^{\nu}) = g_{\mu \nu} w^{\nu}( u^{\rho}\nabla_{\rho}v^{\mu}) + g_{\mu \nu} v^{\mu} ( u^{\rho}\nabla_{\rho}w^{\nu}) = 0 $$ where we've used $\nabla_{\rho}g_{\mu \nu}=0$ and both remaining terms vanish due to (1).

As previously mentioned, a geodesic is a special case where the vector we're parallel transporting is the tangent vector to the curve itself $u = \frac{d X}{d \lambda}$. Recall the geodesic equation$^{1}$ is just $$ \nabla_{u}u^{\rho}=u^{\sigma}\nabla_{\sigma}u^{\rho}=0 \ , $$ and that we also classify geodesics (and vectors in general) as timelike/null/spacelike by looking at $u^{\nu}u_{\nu}$. In (2) we saw that when parallel transported along a curve, the norm of two vectors is conserved. The expression $u^{\nu}u_{\nu}$ is just a specific example of (2), and hence geodesics that are timelike/null/spacelike must remain so.


$^1$Note that for all the geodesics paths $X^{\mu}(\lambda)$ we've assumed $\lambda$ to be an affine parameter, which can always be found.

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    $\begingroup$ As written, this answer is confusing. The question was about arbitrary parallel transported vectors. You immediately jump to the particular case of the 4-velocity. The link between the two may not be obvious to everyone. $\endgroup$
    – TimRias
    Mar 11, 2021 at 14:35
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    $\begingroup$ @mmeent I rewrote my answer which I hope is now more clear. You're right that my initial one wasn't clear. $\endgroup$
    – Eletie
    Mar 11, 2021 at 15:38
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    $\begingroup$ I think this should be much clearer. $\endgroup$
    – TimRias
    Mar 11, 2021 at 15:55
  • $\begingroup$ Thanks! I think I got it right! This also helped me while following the proof. $\endgroup$ Mar 11, 2021 at 19:01

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