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Question

What does it mean for the metric to be scale invariant in curved spacetime (in the sense when I say a property is scale invariant in thermodynamics)? I'm confused as to how to define this. It seems to be either by means of a Weyl scaling or conformal transformation where the scaling factor is a constant? I suspect the correct way would be via means of coordinate transformations? Is there some nice mathematical condition such a metric would satisfy?

Motivation

Consider the stress energy tensor for a perfect fluid:

$$T^{\mu \nu} = \left(\rho + \frac{p}{c^2} \right) U^{\mu} U^\nu + p g^{\mu \nu}, $$

Now keeping our notation ambiguous:

$$g^{\mu \nu} \to \lambda^2 g^{\mu \nu}$$

But $$g^{\mu \nu} g_{\mu \nu} = 4$$

Thus

$$ g_{\mu \nu} \to \frac{1}{\lambda^2}g_{\mu \nu} $$

We also know:

$$ g_{\mu \nu} U^{\mu} U^\nu = c^2 $$

Thus,

$$ U^{\mu} U^\nu \to \lambda^2 U^{\mu} U^\nu $$

Thus we have effectively done the following:

$$ T^{\mu \nu} \to \lambda^2 T^{\mu \nu} $$

More Anonymous
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  • You want your metric to be scale invariant...but which quantity do you want to rescale in the first place so that your metric remains scale invariant? Can you elaborate on that? – KP99 Jan 31 '22 at 15:52
  • I 'd like the stress energy tensor to be scale invariant. If that is scale invariant I suppose all my properties in thermodynamics become extensive? – More Anonymous Jan 31 '22 at 16:05
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    I am not familiar with relativistic thermodynamics, but I guess that the thermodynamics of the matter part will become extensive, but there could be a change in entropy due to the coupled gravitational field. Conformal rescaling of metric will rescale the Weyl curvature, which encodes the gravitational entropy – KP99 Jan 31 '22 at 18:21
  • "If that is scale invariant I suppose all my properties in thermodynamics become extensive?" I don't understand where this is coming from. $\rho$ is the density of an extensive quantity no matter what the metric is. Given an equation of state you can extract densities of other extensive quantities like entropy and conserved charges from the energy momentum tensor too. None of this has to do with scale invariance (do you want to talk about a conformal fluid?) – octonion Feb 02 '22 at 12:10
  • @octonion This is what I had in mind: If I (say) half my volume then even all other intensive properties will half while my extensive properties will remain the same. Note in the above the volume element goes to $V \to \frac{V}{\lambda^2}$. But if take scale the volume of integration ($\int_{x_0}^{x_1} dx \to \int_{kx_0}^{kx_1} dx$) as well then the intensive properties are indeed halved. I might be mixing this with thermodynamics where we can define the extensive property as the ratio of intensive properties :/ – More Anonymous Feb 02 '22 at 12:26
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    @MoreAnonymous, The usual way to treat thermodynamics in relativistic fluids is you effectively trade in all your extensive quantities for densities of extensive quantities (like $\rho$). A good review is Andersson, Comer (https://arxiv.org/abs/2008.12069) – octonion Feb 02 '22 at 12:43
  • Also I have to say I think that "gravitational entropy" comment sounds good but it's a complete red herring – octonion Feb 02 '22 at 12:45
  • I see ... Feel free to ignore it if you answer the question in that case. I don't think I can edit the bounty comment – More Anonymous Feb 02 '22 at 12:46

1 Answers1

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You can't say that your metric is scale "invariant". What you perhaps meant is to say that the theory is scale invariant and the metric is scale covariant.

Assuming you mean the latter, scale invariance of a theory simply implies that if I scale the metric, velocities, stress tensor, etc. as you have shown, then the physics of the system remains unchanged.

A Weyl transformation is $$ g_{ab}(x) \to \Omega^2(x) g_{ab}(x) $$ Here, $\Omega(x)$ is an arbitrary positive function. When $\Omega(x)$ is a constant, we call this a scale transformation.

A conformal transformation is a diffeomorphism $x^a \to x'^a(x)$ such that the metric transforms as $$ g_{ab}(x) \to g'_{ab}(x) = \Omega(x)^2 g_{ab}(x) $$ Here, the Weyl factor $\Omega(x)$ is NOT an arbitrary function. Rather it is related to the diffeomorphism $x'^a(x)$ [which is also not arbitrary].

Prahar
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  • Yes but I kept my notation abstractand I'm uncertain if it represents Weyl or a conformal transformation? – More Anonymous Feb 05 '22 at 10:56
  • @MoreAnonymous I edited the answer for you. To be honest though, I am not entirely certain what you are looking for. – Prahar Feb 06 '22 at 14:19