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I am following an introductory course on General Relativity based on the work of Sean Carroll in: Spacetime and Geometry.
After a lot of trouble we get to the following differential equation: $$\frac{d ^2 x^\mu}{d \lambda ^2}+\Gamma ^{\mu}_{\rho \sigma}\frac{dx^\rho}{d\lambda}\frac{dx^\sigma}{d\lambda}=0 \ \ \ \ \ \ \ \ (1)$$ this of course is the geodesic equation; my understanding is that any curve $x^\mu (\lambda)$ such as to be a solution of $(1)$ is a geodesic. This statement and its proof (based on the definition of a geodesic as a curve along which the tangent vector is parallel-transported) are completely fine for me.
However: at page 109 Carroll states that:

What was hidden in our derivation of (1) was that the demand that the tangent vector be parallel-transported actually constraints the parameterization of the curve, specifically to one that is the proper time or an affine parameter.

I cannot see why this statement has to be true. I get that in GR the proper time is an extremely good parameter for a curve, and that indeed using the proper time as a parameter we get the four-velocity as the tangent vector, which is undoubtedly nice; but still I cannot see why the derivation of the geodesic equation force us to choose a specific parameter.

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The condition is hidden in the statement that a curve that parallel transports its own tangent vector must satisfy

$$\frac{D}{d\lambda} \frac{dx^\mu}{d\lambda} = 0. \tag{1}$$

See, the issue is that this equation (which is just the usual geodesic equation) is not reparametrization invariant. If you use a new parameter $\mu = g(\lambda)$, you will in general find that

$$\frac{D}{d\mu} \frac{dx^\mu}{d\mu} \neq 0$$

unless $g$ is an affine function (hence the name affine parameter).

There are two possible paths we can take here, and both are used in the literature. One is to declare that $(1)$ is the only geodesic equation; that is, a curve $x^\mu(\lambda)$ is a geodesic if and only if it satisfies $(1)$. This means that your definition of geodesic depends on the parameter: the same geometric curve (as in, 1-dimensional subspace) could simultaneously be a geodesic and fail to be one, depending on how you choose to parametrize it. In many situations, it's useful to make sure your geodesics are affinely parametrized.

The other option is to require the condition of geodesic to be reparametrization invariant. We do that by recognizing that in the definition of parallel transport we don't strictly need the acceleration (the left hand side of $(1)$) to be zero; we just need it to point in the same direction as the velocity. This leads us to a "generalized" geodesic equation

$$\frac{D}{d\lambda} \frac{dx^\mu}{d\lambda} = f(\lambda) \frac{dx^\mu}{d\lambda}, \tag{2}$$

where $f$ is an arbitrary function. This equation is reparametrization invariant; changing the parameter just changes $f$. If you try to solve it for a given metric you will get an undetermined function in the solution; this function represents your freedom to choose the parameter however you want. It can be shown that the parameters that make $f=0$ are affine rescalings of proper time (or arclength in general); but note that proper time has no meaning for null curves, so having $f=0$ (that is, satisfying equation $(1)$) is the only definition of affine parameter for them.

This second equation is what you get if you work with the definition that a geodesic is an extremum of length and allow any parameter; as such, it's a bit more intuitive geometrically, if you will, but actually working with it can get a bit gnarly, which is why we almost always use affine parameters and equation $(1)$ as our geodesic equation. This is also why Carroll explicitly chooses proper time as a parameter when deriving the "shortest length" geodesic equation, which also explains why $f$ is nowhere to be found in his book.

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  • $\begingroup$ This is a really useful answer, thanks a lot! But when the proper time comes into play? I mean: why do we specifically choose the proper time as the parameter? Carroll states that we are forced to choose it as the parameter, and seems to me that you left this bit out of your explanation. $\endgroup$
    – Noumeno
    Jul 2 '20 at 20:01
  • $\begingroup$ And another thing: a geodesic should be also the shortest path between two points, this property should be parameterization independent. How can it be when we say that indeed the geodesic is parameterization dependent? $\endgroup$
    – Noumeno
    Jul 2 '20 at 20:24
  • $\begingroup$ @Noumeno I didn't show it, but proper time (or any affine rescaling of it), also known as arclength for non-null curves, is the parameter that makes $f=0$ in $(2)$. That's what sets it apart. $\endgroup$
    – Javier
    Jul 2 '20 at 22:23
  • $\begingroup$ And the "shortest length curve" approach leads to equation $(2)$, unless you do what Carroll does, which is to explicitly use proper time as a parameter. $\endgroup$
    – Javier
    Jul 2 '20 at 22:25
  • $\begingroup$ Wonderful. Thanks! Can I suggest you to edit your answer? It's such a beautiful answer and I feel that leaving this two important details in the comments kinda ruins it a bit. If it is too much work I fully understand. Thanks again! $\endgroup$
    – Noumeno
    Jul 2 '20 at 23:24

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