Inferring coordinate change from the form of the metric

[Kostik]

TL;DR

Dirac infers the form of a change in coordinates from the fact that $g_{\mu\nu}$ remains a function of $l_\sigma x^\sigma$ only ... how?

In Dirac's discussion of gravitational waves ("GTR", Chap. 33), he is working in the case where $g_{\mu\nu}$ are plane waves: waves moving in one direction only. In this case, $g_{\mu\nu}$ is a function of the single variable $l_\sigma x^\sigma$.

Here $l_\sigma$ is the wave vector, and one can show that $l_\sigma x^\sigma$ is a scalar.

Dirac claims that if, under a change of coordinates $x^\mu \rightarrow x^{\mu'}$, the transformed metric tensor $g_{\mu'\nu'}$ remains a function of the single variable $l_\sigma x^\sigma$, then the coordinate transformation must be of a certain form.

To be clear, the metric tensor transforms $$g_{\mu'\nu'} = {x^\rho}_{,\mu'}{x^\sigma}_{,\nu'} g_{\rho\sigma} \, .$$ Dirac states that if the metric tensor is a function of $l_\sigma x^\sigma$ only: $g_{\mu\nu} = g_{\mu\nu}(l_\sigma x^\sigma)$, and if likewise $$g_{ {\mu'}{\nu'} } = g_{ {\mu'}{\nu'} }(l_{\sigma'} x^{\sigma'}) = g_{ {\mu'}{\nu'} }(l_{\sigma} x^{\sigma})$$ then the coordinate transformation must be of the form $$x^{\mu'} = x^\mu + b^\mu$$ where $b^\mu$ is a function of $l_{\sigma} x^{\sigma}$ only. How does he know that?

 

[Ibix]

Isn't that just the only thing it can be? In general I suppose he could multiply by some function of $l_\sigma x^\sigma$, but he can use his freedom to rescale coordinates to undo that.

 

[Kostik]

Any coordinate transformation that satisfies ${x^\rho}_{,\mu'}={x^\rho}_{,\mu'}(l_\sigma x^\sigma)$ would work ... surely that's a much wider class of transformations than the one specified?

For example, $x^{\mu'} = 2x^\mu$? This does not match the type of transformation specified by Dirac.

 

[Ibix]

Yes to your second paragraph, but there I think Dirac is simply exercising his freedom to rescale coordinates and ignore that change.

For more general transforms, isn't this weak field? I might be missing something, but don't we Taylor expand the transform function and ignore anything that's stronger than linear in the coordinate derivatives anyway?

 

[Kostik]

You are correct that the setting here is a weak field where the $g_{\mu\nu}$ are approximately constant. My question is: how does the form of the coordinate transformation follow from the fact that $g_{\mu'\nu'}$ is a function of the single variable $l_\sigma x^\sigma$?

 

[Ibix]

Let me go and re-read that section.

 

[Kostik]

Let me go and re-read that section.

Thank you. Please also note Dirac's comment: "With the restriction that we have waves moving only in one direction, gravitational energy can be localized." This is difficult to understand. It is well-understood that the equivalence principle prohibits the idea of local energy-momentum of a gravitational field in a tensorial (coordinate-free) way. What does he mean by "gravitational energy can be localized"?

 

[ergospherical]

Thank you. Please also note Dirac's comment: "With the restriction that we have waves moving only in one direction, gravitational energy can be localized." This is difficult to understand. It is well-understood that the equivalence principle prohibits the idea of local energy-momentum of a gravitational field in a tensorial (coordinate-free) way. What does he mean by "gravitational energy can be localized"?

Not for general gauge transformations but only for the family of gauge transformations along the propagation direction (in which case the pseudo-tensor transforms like a tensor, c.f. 33.9).

 

[Kostik]

Well, if it can be localized within some restricted family of coordinate transformations, but not all (obviously including transforming to a locally inertial system of coordinates), then I'd say it cannot be localized, period. I think Dirac expressed himself poorly here.

 

[ergospherical]

I'll let him know.

(P.S., localization of energy in GR is still an open problem. It's not worth to get too bogged-down in word-play: the game is to look for something that resembles a stress energy tensor with a local conservation law. It's true that there's no such gauge invariant object, but if you drop that requirement then there are many possible ideas and interpretations out there).

 

[PeterDonis]

if you drop that requirement

The problem is that dropping that requirement means attributing physical meaning to things that are frame dependent, which goes against the principle of relativity.