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The geodesic equation in general relativity is famously invariant under affine reparametrization, i.e., under the reparametrization $\tau \to a\tau + b$ where $\tau $ is the proper time. This can be read off directly from the geodesic equation \begin{align*} \frac{d^2 x^\lambda}{d\tau^2}+\Gamma^{\lambda}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}&=0\text{ (where }d\tau^2\equiv g_{\mu\nu}dx^\mu dx^\nu\text{)}\\ \end{align*} One can reinterpret the affine reparametrization as a transformation $x^\lambda\to ax^\lambda$ where $a$ is a constant. One also observes that the geodesic equation is invariant under the scale transformation $x^\lambda \to a x^\lambda$. Let's just consider the transformation $x^\lambda \to a x^\lambda$. Now, under this transformation, $d\tau \to a d\tau$. The first term in the geodesic equation picks up a factor of $1/a$ ($a$ from the numerator and $a^2$ from the denominator). The second term in the geodesic equation also picks up a factor of $1/a$ due to the one partial derivative in each of the terms in the Christoffel symbols. So, the overall factor of $1/a$ cancels out. Thus, the geodesic equation is invariant under this scale transformation. This is a somewhat weird invariance in my estimation for the following reasons:

  • It is not generically an isometry, because the metric is not necessarily invariant under this transformation.
  • Consequently, the action $S=-m\int d\tau$ is not generically invariant under this transformation. So, we have an equation of motion invariant under a transformation without the action being invariant under the same transformation.

So, I am not sure how to think about this invariance. Is it an artifact but not a real symmetry (like the diffeomorphism invariance of GR which is not generically an isometry)? But if so, how exactly? Because unlike in the case of diffeomorphisms, we are not actually transforming the metric (or anything else) precisely in such a way that it cancels out the factors arising from the transformation of coordinates.

It is also somewhat curious that the geodesic equation is scale-invariant even in the presence of matter when the Einstein-Hilbert action is not scale-invariant. Of course, that is not a contradiction but still, it seems somewhat interesting and I would be interested in finding out if it associates to some nice physical implication.


I should emphasize the point that I made in the comments:

The geodesic equation is invariant under π‘₯β†’π‘Žπ‘₯ not only when I consider π‘₯β†’π‘Žπ‘₯ as a diffeomorphism (which would be unsurprising) but also when I consider it as a "real" transformation, i.e., I don't covariantly change the metric accordingly. The Christoffel symbols do need to change here, but not as they would change under a diffeomorphism. They only pick up an overall factor of 1/π‘Ž from the partial derivatives picking up a factor of 1/π‘Ž as a direct consequence of π‘₯β†’π‘Žπ‘₯.

And yet, this is not an isometry as I pointed out earlier.


To put it very simply, I don't know what to do with an invariance that is coming neither from an isometry nor from a diffeomorphism.

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  • $\begingroup$ "One can reinterpret the affine reparametrization as a transformation π‘₯πœ†β†’π‘Žπ‘₯πœ† where π‘Ž is a constant." -- Do you mind providing more explanation of this step? This is not obvious to me. $\endgroup$
    – Andrew
    Dec 5 '20 at 1:16
  • $\begingroup$ @Andrew I should have made it more explicit because perhaps it is not really a reinterpretation of just the affine reparametrization. But the geodesic equation is indeed invariant under the scale transformation $x\to ax$. [...] $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:22
  • $\begingroup$ [...] So, let's just consider the transformation $x\to ax$. Now, under this transformation, $d\tau \to ad\tau $. The first term in the geodesic equation picks up a factor of $1/a$ ($a$ from the numerator and $a^2$ from the denominator). The second term in the geodesic equation also picks up a factor of $1/a$ due to the one partial derivative in each of the terms in the Christoffel symbols. So, the overall factor of $1/a$ cancels out. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:22
  • $\begingroup$ @Andrew Regarding your deleted post to which I shouldn't be addressing as you deleted it, but still, any scale invariance is essentially a change of units, right? But not all theories are invariant under such a change. For example, EH action in the presence of matter, or Maxwell's theory in the presence of charges/currents, etc. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:24
  • $\begingroup$ Scale invariance usually means scaling the coordinates, not the parameter. Your question is still valid, but it's a different kind of scale invariance than the one we usually talk about. $\endgroup$
    – Javier
    Dec 5 '20 at 1:53
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I think this symmetry is a bit of a red herring. All that is really being expressed is that you can measure proper time along the world line in different units -- seconds, hours, months, etc, are all related by a rescaling of the time parameter.

Let's write the action as

\begin{equation} S = m \int {\rm d} \tau \sqrt{- g_{\mu\nu}(x) \frac{{\rm d}x^\mu}{d\tau}\frac{{\rm d} x^\nu}{{\rm d}\tau}} \end{equation} Let's consider this action as a theory living in 1 dimension, parameterized by $\tau$, with 4 dynamical degrees of freedom given by $x^\mu(\tau)$. In this form, it is perhaps clear that the metric $g_{\mu\nu}(x)$ is playing the role of a fixed set of functions of the dynamical variables $x^\mu$. These function can contain dimensionful parameters reflecting the fact that the geometry may have some scale, for instance if one considers the metric for Schwarzschild which has a scale associated with the mass $M$ \begin{equation} ds^2 = -\left(1-\frac{2GM}{r}\right) {\rm d}t^2 + \frac{{\rm d}r^2}{\left(1-\frac{2GM}{r}\right)} + r^2 {\rm d}\Omega_s^2 \end{equation} You can see that the action won't be invariant under rescaling $x^\mu$, because of the presence of $M$ in the metric.

However, a spacetime like Minkowski will not contain any scales, and the action will be scale invariant in the non-trivial sense that the dynamics do not depend on any scales.

One point you raise is:

Consequently, the action 𝑆=βˆ’π‘šβˆ«π‘‘πœ is not generically invariant under this transformation. So, we have an equation of motion invariant under a transformation without the action being invariant under the same transformation.

In fact (at least classically), if the action changes by an overall constant under a given transformation, $S\rightarrow \lambda S$ for some $\lambda$, then the equations of motion will be invariant under the transformation.

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    $\begingroup$ Thanks, this is helpful! So, you are saying that the geodesic equation is not generically invariant under $x \to ax$, right? This makes sense to me because I think my reasoning in concluding that it is invariant implicitly assumed that the metric is invariant $g_{\mu\nu}(x/a)=g_{\mu\nu}(x)$ which is not true unless it is an isometry. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 2:40
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    $\begingroup$ Yes, that's exactly right. $\endgroup$
    – Andrew
    Dec 5 '20 at 2:42
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This is not a symmetry in the sense that physicists normally think of symmetries, where an observable is invariant under a transformation, and this invariance usually implies a conservation law. It also has nothing to do with the metric or with coordinates.

Let's take the case of a timelike geodesic. Then an affine parameter is the reading on a clock that free-falls along this geodesic. Rescaling the affine parameter simply means that you're changing the units of the clock.

It's true but not obvious that you can define affine parameters without having to have any metric geometry at all. All you need is a knowledge of what curves are geodesics.

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  • $\begingroup$ Thanks for your answer. I realize that the affine reparametrization simply refers to a change of units. However, here, we are really talking about the scale transformation $x\to ax$ and not just the change of the affine reparametrization. See my edited version of the question in case you wrote the answer before that. Also, physicists think of symmetries as invariance of equations of motion (or of action) and not as invariance of quantities. One gets conserved quantities under invariance, not necessarily invariant quantities. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 2:07
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Let me say a couple things about the situation here because there are two types of invariances and a bit of a detail that often gets overlooked.

The Lagrangian for the geodesic equation is invariant under two different transformations, with a caveat. The first, which has no caveat, is general reparameterization invariance. Essentially, the Lagrangian $d\tau$ is defined independent of the parametrization scheme. The form of the geodesic equation may not be preserved for general reparameterizations, but the same set of solutions will always be determined. So it's as good as being invariant in many respects.

The second invariance is general coordinate transformations, known as diffeomorphisms. The caveat is that the metric must change accordingly. That means your Christoffel symbols must change with the change in coordinates as well.

In fact, the geodesic equation is not invariant under the coordinate transformation ($x^\mu\rightarrow ax^\mu$) unless you transform the Christoffel symbols appropriately. With this in mind, yes, the geodesic equation is indeed invariant under this coordinate rescaling. But only because coordinate rescalings are a special case of a general coordinate transformation.

If I can take a guess at what you might find more satisfying as an answer: rescalings of the coordinates are defined often called dialtions and have a close relationship to the conformal group, which plays a large role in certain areas of modern high energy theory. The conformal group is defined to be the subgroup of diffeomorphisms which leave the components of the metric invariant up to an overall scale factor.

There are many sources on the conformal group. For example the book "Conformal Field Theory" by Di Francesco and others, and the paper (a small book, really, but it's on the arXiv) on conformal field theory by Ginsparg.

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  • $\begingroup$ Thank you for your answer, I have the following questions: 1. I am not sure how I see that the geodesic equation is invariant under any reparametrization. It is only invariant under affine reparameterizations as far as I can see. For example, if I take $\lambda = \tau^2$, I will not get $\frac{d^2x}{d\lambda^2}+\Gamma\frac{dx}{d\lambda}\frac{dx}{d\lambda}=0$. [...] $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:38
  • $\begingroup$ [...] 2. The geodesic equation is invariant under $x\to ax$ not only when I consider $x\to ax$ as a diffeomorphism (which would be unsurprising) but also when I consider it as a "real" transformation, i.e., I don't covariantly change the metric accordingly. The Christoffel symbols do need to change as you point out, but not as they would change under a diffeomorphism. They only pick up an overall factor of $1/a$ from the partial derivatives picking up a factor of $1/a$ as a direct consequence of $x\to ax$. [...] $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:39
  • $\begingroup$ [...] 3. This is not a dilation in the usual sense where we would have a Killing equation for the generators of the dilation, because the metric is not generically invariant here. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 1:42
  • $\begingroup$ @DvijD.C. I will note that you will not have a Killing equation for general dilations as the whole point is that the metric is not invariant. You will instead have a so-called conformal Killing equation. For reparametrization, look at the Lagrangian: $d\tau$ is, by definition, independent of the parametrization of the path. This is the simplest way to see it. The form of the geodesic equation may change I suppose, but it is the same equation in the sense that it determines the same set of solutions. $\endgroup$ Dec 5 '20 at 2:12
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    $\begingroup$ And sure, you can parameterize a curve however you want but the geodesic equation will not be "preserved" under a reparametrization that is not affine. $\endgroup$
    – Dvij D.C.
    Dec 5 '20 at 2:19

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