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I am having trouble understanding the "meaning" of the one of the ways to get the geodesic equation in GR. The first way, which seems to me also the most intuitive, is to say that the geodesic is the path with extremise the GR "action" (if I may call it that, it looks like a distance though): $$S = \int(\epsilon g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda})^{1/2}d\lambda$$ Where $\epsilon$ is -1 or 1 for time-like and space-like paths respectively. I am not sure about the case of light-like paths.

I say it is intuitive, because the method closely resemble the way we use to minimize the length of a path on a surface, i.e. we minimize $L = \int ds$.

However, there is also another way to obtain the geodesic equation, and that is to say that the geodesic "path" is the one which parallel transport it's own tangent vector. I am struggling to understand this definition. I know the equations gives the right result, but I'd like to get an intuitive understanding.

Indeed, I think it has something to do with the fact that the parallel transport path is extremal, so that if the tangent vector follows it, it's a geodesic.

I don't know if that's clear, so I'll try to explain it with an analogy. In Euclidean space, the gradient points to the direction of greatest increase of the function. So, given two values of the function, the "fastest" way is to follow the direction of the gradient. Is there a similar interpretation for the definition of the geodesic ?

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    $\begingroup$ For null paths the action is zero, so a null geodesic is defined by the geodesic equation itself. $\endgroup$ – Ryan Unger Feb 23 '17 at 14:05
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In euclidean space, a geodesic is a straight line. If you imagine a straight line, and its tangent vector at a point, then as you move along the straight line, the tangent vector slides about the line in a parallel fashion, right?

There is a subtlety, however, which is relevant to a more "physical" interpretation. A curve is usually defined as a mapping from the real line (or an interval of it) into the manifold: $\gamma:\mathbb{R}\rightarrow M$, so a curve comes naturally with a parametrization.

On the other hand, the image of a curve is not dependent on the parametrization.

Consider the following curves in euclidean 3-space:

$$ \gamma(t)=\mathbf{r}_0+\mathbf{v}t $$ and $$\lambda(t)=\mathbf{r}_0+\mathbf{v}(e^{\alpha t}-1)$$ and we only consider $t\ge0$. Both curves are straight lines, in fact, their image coincides, but if we interpret these curves as a particle's trajectory, and we calculate the acceleration, we get $$ \ddot{\gamma}(t)=\dot{\mathbf{v}}=0 $$ ($\mathbf{v}$ we take as a constant vector), but $$\ddot{\lambda}(t)=(\alpha\mathbf{v}e^{\alpha t})^.=\alpha^2e^{\alpha t}\mathbf{v}. $$ What we see is that for $\gamma$, the acceleration is zero, for $\lambda$, the acceleration is not zero, but is $\alpha\dot{\lambda}(t)$, so the acceleration is parallel (proportional) to the velocity.

Why? The first curve describes a free particle in uniform motion. The second curve describes an accelerating particle that nontheless does one-dimensional motion, eg. its velocity's direction does not change.

Let's move back to general relativity. In special relativity a free particle is such that its 4-acceleration $d^2\gamma^\mu/d\tau^2$ vanishes, right? As usual we follow the principle of minimal substitution as we move from SR to GR, so we take the covariant equivalent of the same statement as the definition of a free particle.

A free particle moves on a path for which $$ \frac{D}{d\tau}\frac{d\gamma^\mu}{d\tau}=\frac{d\gamma^\nu}{d\tau}\nabla_\nu\frac{d\gamma^\mu}{d\tau}=\frac{d^2\gamma^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{d\gamma^\alpha}{d\tau}\frac{d\gamma^\beta}{d\tau}=0. $$ By the usual interpretation of the covariant derivative (a vector is covariantly constant along a curve = the vector is parallel transported along the curve), this means that the curve's tangent vector, $d\gamma^\mu/d\tau$ is parallel transported along the curve (eg. along itself).

Here is a point of note though. Sometimes curves satisfying $$ \frac{D}{d\epsilon}\frac{d\gamma^\mu}{d\epsilon}=\alpha\frac{d\gamma^\mu}{d\epsilon} $$ are also called geodesics. Sometimes some authors call these "pregeodesics".

When we derived the equation of motion for a free particle, we parametrized the curve with proper time. But we have reparametrization freedom, and as I explained with the euclidean example above, if you use a "non-uniform" parametrization, then the curve's tangent vector will not parallel transport along itself, but will change in a way that is proportional to itself.

A geodesic for which $\alpha=0$ is called an "affinely parametrized geodesic', and any "pregeodesic" can be turned into an affinely parametrized geodesic by properly reparametrizing.

Note: Of purely mathematical interest (or in case of extended theories with multiple connections), is the fact that the "curve that parallel transports its tangent vector" definition is often better, because geodesics can be defined for connections that are not associated with metrics too, but then the "path length" interpretation is meaningless.

Of a not-so mathematical interest, for null curves, the path length is zero, but the other definition still gives nontrivial geodesics.

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  • $\begingroup$ Thank you for your clarifications, although I'm not sure this answer my question. How do you justify that the particle follows a curve that satisfy the geodesic equation, without using the action ? I understand that, if you use the free particle action, and find the equation, you get indeed an equation of a curve parallel transporting it's own vector, but I was looking for another way to justify this "fact". Anyway, I'm not sure there is even a "physical" or "intuitive" method of understanding the second definition ! $\endgroup$ – Frotaur Feb 24 '17 at 2:21
  • $\begingroup$ @Frotaur I don't think I even mentioned the action. The usual rule of moving from SR to GR is 1) write SR equation in generally covariant form, 2) assume it is true in GR. This is a consequence of the equivalence principle and the existence of Riemann normal coordinates. (This can fail if second covariant derivatives appear, but they don't in our example.) In our case, in SR we have $d^2x^\mu/d\tau^2=0$, which in generally covariant form is $D/d\tau\ dx^\mu/d\tau=0$. Only one covariant derivative is involved, so this is the GR equation too. $\endgroup$ – Bence Racskó Feb 24 '17 at 8:35
  • $\begingroup$ Ooh ok, that's how you get the equation. Then that kind of answers the question, although I was looking for a answer justifying intuitively as to why is following your tangent vector an extremal path, but I understand it may not be possible to understand it easily since the concepts are 4-dimensional and abstract ! $\endgroup$ – Frotaur Feb 27 '17 at 6:28
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This is a somewhat handwaving answer (you can tell by how long it is): I've held back from giving it in the hope that someone would arrive with a more formal one as I have forgotten a lot of this stuff, but no-one has, so.

Start off in Euclidean space, because it's easy to visualize. Geodesics are just straight lines. Straight lines have two properties:

  1. They are extrema (minima) of distance between any two points on the line: of all the curves connecting $A$ and $B$, the straight line is the shortest.
  2. They parallel-transport their own tangent vectors (and in fact any vector in the tangent space at a point on the curve). What this means is that if you take the tangent vector at any point on the line (in fact: any vector in the tangent space at that point) and drag it along the line so that it remains parallel to itself as it is dragged, then you get another tangent vector (...) at the point where it is dragged to.

(1) is pretty straightforward. (2) is disturbingly vague: I haven't really said what it means to 'parallel transport' a vector or even what a tangent vector / tangent space really is. Well, these can be made precise of course, but I'm going to cop out on that.

An important property of Euclidean space is that these two properties pick out the same curves: if you take, say, an arc of a circle neither the shortest curve between any two points on the arc, and nor does it parallel transport its own tangent vectors -- if you drag a tangent vector half-way-around a circle you'll get something which is $-1$ times the tangent vector at the new point! And in fact straight lines are the only lines which do both things in Euclidean space.

But these two properties are not obviously the same: in particular, if we consider a general manifold there may be no metric: so if we want to define parallel transport we either can't or we can't use the metric to do it.

Well, we can define parallel transport, and we do so using a connection, which is just a mechanism for defining what it means to parallel transport vectors: it defines how you connect the tangent spaces at different points on the manifold. With a connection (or perhaps with a suitably well-behaved one) you can define parallel transport, and geodesics which are now not extrema of length, because there is no length, but are families of curves which parallel transport their own tangent vectors. And you get things like covariant derivatives and so on (in fact I think covariant derivatives and connections are equivalent).

It's important to note that a connection is additional structure on a manifold: it's not something that you can derive from the basic notion of what it means to be a manifold.

So, if we add a metric to the manifold (which is another bit of additional structure) we now have two different notions of what it means to be a geodesic: geodesics can be curves which parallel transport their own tangent vectors, and curves which are extrema of length. These two notions correspond to the two additional bits of structure that we've added, and they do not need to be the same and will not be in general.

But we can choose them so they are. In particular if the connection is such that the inner product of any two vectors that are parallel transported along a curve remains constant, then the connection & the metric are compatible. There are some other definitions which are equivalent to this one: if the covariant derivative (which comes from the connection) of the metric vanishes then this is the same thing, for instance.

In fact there is slightly more that you need to do: a connection and a metric can be compatible but the connection can have torsion. I have never understood torsion very well, but I think it means that, although the inner products of vectors are preserved as you drag them along a curve, they somehow rotate together.

If you insist that the connection is both compatible with the metric and has no torsion then there is (I am almost sure, but possibly with some other sanity conditions added) exactly one such connection, and this is the Levi-Civita connection: it's the unique torsion-free Riemannian connection on a manifold (and Riemannian, to me anyway, means 'metric compatible' although I think there is a little more to it than that).

And for such a connection, then the two notions of geodesic will be the same: geodesics both parallel transport tangent vectors, and are extrema of length.

But none of this answers the question of why we should choose them to be compatible: we don't need to after all. I think that the answers to that are essentially that choosing them to be compatible gets us a spacetime which is 'closer' to Euclidean space, and also for which there are a bunch less parameters which need to be chosen somehow: if we picked them not to be compatible then we'd have to have some set of differential equations which specified how the connection behaved as well as ones which specified how the metric behaved, for instance). And, most compellingly, if we choose them to be compatible then we get a theory of gravitation which works very well.

People have investigated theories of gravity with torsion, for instance.

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  • $\begingroup$ Thank you for the clarification. I must say that I did not consider non torsion-free connections, and/or non-compatible connections. Indeed, looking back to it we see that the geodesic equation involves the Christoffel connection, while the parallel transport equation use the general connexion that has been defined. I must say I'm very intrigued of this correspondence : metric compatible connection $\Leftrightarrow$ geodesic and parallel transport curves coincide. I'll try to find more on the subject. Thanks for the insight. $\endgroup$ – Frotaur Feb 27 '17 at 6:39

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