# Show that a geodesic remains of the same type at all points by first proving that $v^a w_a$ remains constant when parallel-transported

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!

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.

• 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. Commented Mar 11, 2021 at 14:35
• @mmeent I rewrote my answer which I hope is now more clear. You're right that my initial one wasn't clear. Commented Mar 11, 2021 at 15:38
• I think this should be much clearer. Commented Mar 11, 2021 at 15:55
• Thanks! I think I got it right! This also helped me while following the proof. Commented Mar 11, 2021 at 19:01