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In Hobson's General Relativity for physics, a geodesic is described as a curve satisfying $\frac{d\boldsymbol{t}}{du}=f(u)t$, where $u$ is a general parameter, $f(u)$ is some function of $u$ and $\boldsymbol{t}$ is the tangent vector to the geodesic curve $x^a(u)$. Now, the curve it's said to be affinely parametrised if $f(u)=0$, and so $\frac{d\boldsymbol{t}}{du}=0$.

At this point, it's assumed in the book that $\frac{d\boldsymbol{t}}{du}=0$ implies $\frac{D{t^a}}{Du}=0$, where $D$ stands for the intrinsic (or absolute) derivative, and I guess this is happening because in a general coordinate system, the proper derivative to take is not the usual one, but the intrinsic one. Anyway, the problem comes a couple sections later, when an alternative form of the geodesic equation is derived: $$\dot{t}^a=\frac{1}{2}(\partial_ag_{cd})t^ct^d$$

The procedure is clearly detailed, so I have no problem following it, but at the begining of it, it's stated:

For a geodesic, we have $\frac{d\boldsymbol{t}}{du}=0$. This implies that, in some coordinate system, we may write this equation in terms of the intrinsic derivative of the covariant components of the tangent vector as $\frac{Dt_a}{Du}=0$.

From here, the equation follows easily, but I don't understand how this last statement holds, and that actually made me question why did $\frac{D{t^a}}{Du}=0$ follow from $\frac{d\boldsymbol{t}}{du}=0$ instead of $\frac{Dt_a}{Du}=0$, or is it that we can relate both the intrinisic derivatives of the covariant and contravariant components of the tangent vector in such a way that $\frac{Dt^a}{Du}=0$ implies $\frac{Dt_a}{Du}=0$?

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I ended up finding out myself. After doing some research, I got to check that the covariant derivative of the metric tensor is null, so the operation of raising and lowering indexes with the metric tensor can be exchanged with a covariant derivative. This is: $$\nabla_cv^a=\nabla_c(g^{ab}v_b)=(\nabla_cg^{ab})v_b+g^{ab}\nabla_c(v_b)=g^{ab}\nabla_c(v_b)$$

From this reasoning, and writing the intrinsic derivative as $\frac{Dv^a}{Du}=(\nabla_cv^a)\frac{dx^c}{du}^*$, it's clear that $\frac{Dv^a}{Du}=0$ necessarily implies $\frac{Dv_a}{Du}=0$, since you can perform an index rising operation on it.

$^*$Note that, if $\boldsymbol{v}$ is only defined along the curve, it's covariant derivative wouldn't be formally defined and this wouldn't be 'true', though it holds for the sake of our argument.

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In Euclidean space, in an inertial frame, objects without a resultant force acting on them are either at rest or move in straight lines. This is Newtons first law.

Since geodesics are the straight lines in non-Euclidean geometry then a similar statement should also be available for them. It turns out that this is the case:

We can always derive a path on a manifold and this gives us the velocity field which is a vector field along the path. Now, to derive the velocity field to give us the acceleration field we can't simply derive like we did the path, this requires a connection. For any pseudo-Riemannian manifold there is a canonical connection called the Levi-Civita connection. When we derive the velocity field with this connection we get the acceleration field.

Thus, geodesics, that is straight lines, on pseudo-Riemannian manifolds are paths which particles follow when they are unaccelerated. In particular, this includes Lorentzian manifolds and hence the manifolds on which GR is predicated upon.

(Notice that for general manifolds there are many different accelerations, one for every connection. It's only in the pseudo-Riemannian case, and hence the Lorentzian case, we get to pick a canonical one).

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    $\begingroup$ Small comment, it's not really correct to call geodesics 'straight lines in non-Euclidean geometry': straight lines in flat space are geodesics, but geodesics generally aren't straight lines. But you can say they're analogous to, or a generalisation of, straight lines for non-Euclidean spaces. $\endgroup$
    – Eletie
    Mar 14, 2021 at 12:43
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    $\begingroup$ @Eletie: That's being a little pedantic to my mind ... the whole answer is drawing out an analogy between Newtonian Mechanics and Einsteinian mechanics. I think anyone who understands what I've written will understand what I mean; after all, you understood it, even though you're against interpreting it in this fashion. $\endgroup$ Mar 14, 2021 at 12:47
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    $\begingroup$ I don't see the need to write things that are incorrect if they're not even conceptually simpler. The OP is dealing with GR and nowhere implies they need simplifications about geodesics. A key feature of curved space is that there are no straight lines. $\endgroup$
    – Eletie
    Mar 14, 2021 at 13:01
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    $\begingroup$ @Eletie: He's asking for "An alternative derivation of the geodesic equations". I suggested a mechanical analogy. Geodesics are conceptually the straight lines in Non-Euclidean geometry. For example, in the synthetic approach to non-Euclidean geonetry, they say for spherical geometry, all lines intersect - that is all great circles. Great circles are curves, yet they call them lines ... $\endgroup$ Mar 14, 2021 at 13:07
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    $\begingroup$ Your point about drawing a straight line vs a squiggly line is irrelevant, if somebody asks what curve locally minimises the distance between two points on a sphere and you say a straight line that's incorrect. Also the OP was asking about a specific set of equations, perhaps you misread the title (they weren't asking for an alternative derivation). The answer given doesn't seem to actually address the question. $\endgroup$
    – Eletie
    Mar 14, 2021 at 13:23

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