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In classical General Relativity (meaning not modified) one can think of geodesics in two ways.

  1. One way is to say that a geodesic is the curve which is the straightest (in analogy with the flat case) among all curves. More or less the story goes like this (correct me if I'm wrong): in the flat case geodesics are of the form $x^\mu(s)=st^\mu +b^\mu$ where $t$ and $b$ are constant vectors and $s$ is the curve parameter. The curve tangent vector is $\frac{dx^\mu}{ds}=t=const$ so $$\frac{dt^\mu}{ds}=t^\nu\partial_\nu t^\mu=0.$$ This is a tensor equation and by general covariance it holds true in a general spacetime, mutatis mutandis: $$t^\nu\nabla_\nu t^\mu=0.$$

  2. Another way is to obtain the equation by finding the minimum of the length functional $$L=\int \sqrt{g(t,t)} \,\, d\tau,$$ we get $$\frac{dt^\mu}{d\tau}+\Gamma^\mu_{\alpha\beta}t^\alpha t^\beta=0.$$ The above equation turns out to be just $t^\nu\nabla_\nu t^\mu=0$, with a certain parametrization choice.

The two formulations are thus equivalent. My question is when and why is it so? Is there a deep reason?

Looking up Wald's book I found the following, unclear to me, argument:

"On a manifold with a Riemannian metric, one can always find curves of arbitrarily long length connecting any two points. However, the length will be bounded from below, and the curve of shortest length connecting two points (assuming the lower bound in length is attained) is necessarily an extremum of length and thus a geodesic. Thus, the shortest path between two points is always a straightest possible path."

Later in the book he also says something about conjugate points.

I always thought it had to do with torsion: If we relax the condition $\Gamma^\mu_{\alpha\beta}-\Gamma^\mu_{\beta\alpha}=0$ then the connection is not Levi-Civita, hence $\delta_g L=0$ remains unchanged but $t^\nu\nabla_\nu t^\mu=0$ does and gives a different geodesic equation.. so Straightest $\neq$ Shortest anymore.

Can anyone clarify things to me?

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    $\begingroup$ You're asking for a "deep reason" why in flat space the shortest distance between two points is a straight line. I don't think there's a "deeper" reason for this than for the fact that $2+2=4$. $\endgroup$ – tparker Jul 4 '17 at 0:40
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    $\begingroup$ The two definitions are rarely equivalent globally, the shortest path is always a geodesic, but a geodesic may not be the shortest path. The "deep reasons" are that the equation derived by varying the length functional is the geodesic equation, and a theorem that solutions to it are minimizing length unless they pass through conjugate or cut points, see cut locus. But if the connection is not Levi-Civita then it is not compatible with a metric, so it is unclear how you are defining the length functional. $\endgroup$ – Conifold Jul 4 '17 at 3:30
  • $\begingroup$ @tparker two words: surface geometry. $\endgroup$ – The Great Duck Jul 4 '17 at 6:11
  • $\begingroup$ I think it is actually a geometric axiom that geodesics be the straightest and shortest path. In essence, the definition of straightness changes. After all, straight on a sphere is far curvier than on a plane. $\endgroup$ – The Great Duck Jul 4 '17 at 6:12
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    $\begingroup$ @ChrisBecke one geodesic would be the shortest path, but the other geodesic would not be the longest path. On a sphere, if you fix 2 endpoints, you can produce a smooth curve joining them of arbitrarily large length. For example, you can always spiral around as many times as you like, to get longer and longer paths. $\endgroup$ – Malkoun Jul 4 '17 at 13:27
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The two formulations are thus equivalent.

This is false.

In the case of spacelike geodesics, your definition of $L$ gives an imaginary number. The complex numbers are not an ordered field, so there is no such thing as "shortest."

There is a similar problem for null geodesics. A null geodesic has $L=0$, and perturbations of a null geodesic may make $L$ either real or imaginary.

Even in the timelike case, it can happen that a geodesic is not a maximal-time curve. There is a discussion of this in Misner, Thorne, and Wheeler, p. 318.

The only general definition of a geodesic that works is that it parallel-transports its own tangent vector, i.e., it's the straightest path.

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    $\begingroup$ Can't you get around this complex-number issue by requiring that the spacetime interval $\int d\tau\, ds^2$ be extremized instead of the proper length $\int d\tau\, \sqrt{ds^2}$? $\endgroup$ – tparker Jul 4 '17 at 14:51
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Though I cannot explain it better than Penrose in The Road to Reality, as Ron Gordon wrote, which is a very nice book with very nice figures, I thought I'd provide some remarks here.

Regarding your paragraph which starts "I always thought it had to do with torsion", I have a few (hopefully helpful) comments. First any affine connection on a manifold gives a notion of parallel transport (and I note that Ron Gordon referred the OP to the parallel transport page on wikipedia already), and so you can define geodesics as smooth curves for which the tangent vector is parallel. Note that I did not mention any metric, and indeed, one can talk about geodesics while just having a connection.

But what if the manifold has a Riemannian metric $g$? The Levi-Civita connection is a torsion-free $g$-compatible connection. I like to think of it as being in some sense canonically associated to $g$ (since given a metric, it exists and it is unique). In this case, you can also define geodesics as being smooth curves which are critical points of the length functional, with their endpoints held fixed. One can also define them as critical points of an Energy functional, rather than the length functional, thus eliminating the square root involved in the length. This is similar to considering the Polyakov action rather than the Nambu-Goto one in String theory.

My final remark is this. While a geodesic on a Riemannian manifold always locally minimizes the length, it may not always globally minimize the length, with its 2 endpoints held fixed. For example, consider a sphere, and say you start from the north pole N, and travel along a great circle until you reach the South pole S, and then continue a little more on the same great circle, a little past S. This is a geodesic, but it is not the shortest path between the 2 endpoints. Indeed, one can start and N, and go in the opposite direction as the initial path, and reach the endpoint in a shorter geodesic path.

Edit: my remarks assumed a Riemannian metric, meaning Euclidean signature, rather than Lorentzian. There are other issues arising in Lorentzian signature as some users (in particular Ben Crowell) have correctly pointed out.

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References: Penrose pg 294, Ch 14 of The Road to Reality provides a very approachable explanation of parallel transport using geodesics to find "short" paths and describes how to solve the problem of path dependence. He avoids rigorous development to focus on the rationale.

https://en.wikipedia.org/wiki/Parallel_transport The mathematical development for parallel transport in Reimmann Geometry as well as other treatments to resolve the parallel path dependence problem.

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    $\begingroup$ $\uparrow$ Reimmann? $\endgroup$ – Qmechanic Jul 4 '17 at 3:26
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Let there be given a pseudo-Riemannian manifold $(M,g)$ with a connection $\nabla$ that is compatible with the metric $g$ but not necessarily torsion-free. Let $\nabla^{LC}$ denote the Levi-Civita connection for $g$. Then the geodesic equation $\nabla^{LC}_{\dot{\gamma}}\dot{\gamma}=0$ and the auto-parallel equation $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ are not necessarily the same. They are the same iff the torsion tensor is totally antisymmetric. See e.g. this Phys.SE post for details.

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I don't really understand your question, but I think this is what you're asking:

"Geodesics can defined in two different ways: (a) trajectories that parallel-transport their tangent vector, or (b) trajectories that maximize the proper time along all possible paths between two points. [Note than the former definition is local and the second is global. As Ben Crowell points out, the indefinite signature of the metric leads to subtleties with the second definition, so for simplicity it's easiest to only consider timelike trajectories, which of course means that this definition only makes sense if the endpoints are causally connected. The cases of spacelike or null paths end up being very similar.] The first definition only references the connection (not the metric) and the second definition only references the metric (not the connection). For what relationships between the metric and the connection are these definitions equivalent?"

Carroll discusses both definitions on pgs. 106-108 of this GR texbook: "these two concepts coincide if and only if the connection is the Christoffel connection ... On a manifold with a metric, extremals of the length functional are curves that parallel transport their tangent vectors with respect to the Christoffel connection associated with that metric [emphasis added]. It doesn't matter if any other connection is defined on the same manifold."

As you say, if the connection has torsion (or is not metric-compatible), then the two concepts are no longer equivalent. In alternatives to GR that consider connections with torsion, free-particle paths obey the first equation rather than the second, so the first definition is more fundamental in the context of gravitational physics.

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  • $\begingroup$ A bit unrelated maybe: Do the connection "choices" that differ from the Christoffel connection fit well with the Equivalence principle? I tend to think there is no choice in the connection if one has to follow the Equivalence principle. This viewpoint, in my opinion, is somewhat supported by Weinberg's treatment. He doesn't assume anything about the torsion apriori and (using the Equivalence principle) proves that the connection turns out to be the Christoffel connection. $\endgroup$ – Dvij Mankad Oct 11 '17 at 15:26

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