# Spacetimes, metrics and symmetries in the theory of relativity?

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Suekdccia
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Spacetimes, metrics and symmetries in the theory of relativity?
I was discussing this paper with a couple of physicists colleagues of mine (https://arxiv.org/abs/2011.12970)

In the paper, the authors describe "spacetimes without symmetries". When I mentioned that, one of my friends said that no spacetime predicted or included in the theory of relativity could break the Lorentz and diffeomorphism symmetries...

However, the other one didn't seem to agree, as he asked what he meant exactly by Lorentz and diffeomorphism symmetry:

"Are you talking about a symmetries of the theory? The theory of general relativity is diffeomorphism (and so Lorentz) invariant.
Are you talking about a symmetry of a spacetime, solution of the equations of motion of general relativity (the Einstein equations)? Defined how? Is it a change of coordinates and the corresponding change of metric components? In this case, a spacetime is trivially invariant (a change of coordinates cannot have any physical effect).
Or is it an active diffeomorphism in which one keeps using the same coordinate system but moves points to new points and then compares the spacetime metric on the old set of points with the spacetime metric on the new set of points? This is the kind of symmetry the authors mention in their paper. When the authors say "spacetime without symmetries", they mean without this latter kind of symmetries.
For example, Minkowski (flat) spacetime has Poincaré symmetries of this kind (e.g. it is translation invariant, but also rotation and boost invariant). The curvature at any point of the spacetime is the same.
Now, imagine bending Minkowski spacetime in an arbitrary way. The result, in general, will not have translation, rotation and boost symmetry, and the curvature at any 2 arbitrary points will be, in general, different.
"

The thing is that I'm having trouble trying to understand how is the Poincaré/Lorentz and diffeomorphism symmetries crucial for General Relativity while there can be solutions to it corresponding to spacetimes which violate them at the same time...

Homework Helper
Gold Member
what geometrical objects are being subject to various symmetries or not?

It seems to me that your issue isn't really a relativity issue.

Are you referring to points in the manifold?
Or directions in the tangent space at a point?
Or something else?

Consider two [gaussian] surfaces: one a sphere and one some irregular surface.
A rotation that maps a sphere to itself won't generally map the irregular surface to itself.
However, that rotation won't change the distance between any two marked points on each surface.

Mentor
no spacetime predicted or included in the theory of relativity could break the Lorentz and diffeomorphism symmetries...
The first statement is correct, but only when properly qualified: any spacetime that is considered in GR must have local Lorentz symmetry (note the "local"--strictly speaking this is a symmetry of the tangent space at each event). This is built into the theoretical framework of GR.

The second statement makes no sense, since diffeomorphism "symmetry" is a relationship between different mathematical descriptions of the same physical spacetime, not a condition on physical spacetimes.

there can be solutions to it corresponding to spacetimes which violate them at the same time...
There aren't. See above.

atyy, vanhees71 and Omega0
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2022 Award
That's a very important point! It took Einstein at least 3 years from his correct idea that his relativistic theory of gravitation should be formulated in general covariant form via some detour, because at the time the notion of "gauge invariance" was not as advanced as today, since we are now used to the finding that on a fundamental level the fundamental interactions are all described by gauge theories. Indeed, the "general covariance" is a gauge theory, and it's a gauge theory in the modern sense, because you can get to general relativity by the idea to make Poincare symmetry (i.e., the global spacetime symmetry of special relativity ##\text{ISO}(1,3)^{\uparrow}##) local. As long as you avoid fields spin 1/2 you get to GR. This restriction is at the moment not very severe, since in the (mostly astrophysical) applications of GR only scalar fields (to describe matter in terms of continuum mechanics) and the electromagnetic field plays a significant role, and this "matter content" leads to a torsion-free pseudo-Riemannian spacetime manifold and thus GR.

dextercioby
Mentor
This restriction is at the moment not very severe, since in the (mostly astrophysical) applications of GR only scalar fields (to describe matter in terms of continuum mechanics) and the electromagnetic field plays a significant role
I'm not sure this is entirely correct as regards matter; matter in many applications of GR is described by a perfect fluid stress-energy tensor, which is not the same as a scalar field, but also does not require dealing with any fermionic degrees of freedom; the details of the internal constitution of the matter are simply not included in the model.

vanhees71
Gold Member
2022 Award
Exactly this was what I meant with my argument: A (perfect) fluid can be described by three scalar field quantities, and scalar fields##^*## (as well as the electromagneticc field) don't lead to torsion, i.e., the corresponding spacetime is described by standard GR with the spacetime manifold being a pseudo-Riemannian space.

Usual matter, of course is composed of fermions, but in the description of macroscopic matter effectively this plays no role, because it's usually in a spin-saturated state, and you describe it anyway by (ideal) hydro, with no need to do "spin hydro".

##^*## See the description for the SR formulation in Sec. 3.5 of

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

For a more conventional approach and to the extension to GR, see

L. Rezzolla and O. Zanotti, Relativistic hydrodynamics,
Oxford University Press, Oxford (2013),
https://dx.doi.org/10.1093/acprof:oso/9780198528906.001.0001

Mentor
A (perfect) fluid can be described by three scalar field quantities, and scalar fields (as well as the electromagneticc field) don't lead to torsion
Ah, I see. The term "scalar field" is somewhat ambiguous. If by "scalar field" you just mean the energy density and pressure in the fluid's rest frame (not sure what the third scalar would be), these are just scalar functions on spacetime that define the components of the stress-energy tensor in the fluid's rest frame (with its components in any other frame being obtained by the usual process of applying coordinate transformations). The form of the stress-energy tensor is fixed by the statement that it is a perfect fluid. Further, if you know the equation of state of the fluid, that tells you the pressure in terms of the density, so there is really only one independent scalar in the stress-energy tensor.

In field theory terms, however, a "scalar field" is a field on spacetime whose stress-energy tensor is obtained from its field Lagrangian in the usual manner. This stress-energy tensor could, I suppose, be described as a perfect fluid, but it is one with the very unusual equation of state ##p = - \rho##. Further, the form of the stress-energy tensor is not defined to be that of a perfect fluid in this case; it just happens to work out that way when you derive it from the field Lagrangian.

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What I mean is the treatment of relativistic hydro as shown in my manuscript for SR

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I got this description from the great book by

D. E. Soper, Classical field theory, Dover Publications,
Minneola, New York (2008).

It's a not so well-known approach, which I find however very illuminating, because it treats the fluid analogous to the mechanics of point particles. Of course at the end it's equivalent with the standard approach, leading directly to the equations of motion by using the (local) conservation of energy-momentum and particle number (or conserved charges).

The idea is to start with a "Lagrangian description", i.e., you define the fluid as the motion of "fluid elements". In a "standard configuration". For a fluid it's most simple to take the initial positions ##(\xi^A)##, with ##A \in \{1,2,3 \}## of the fluid elements. You can take the ##\xi^A## to denote the Cartesian coordinates for the position with respect to an arbitrary (inertial) reference frame. Then the fluid in the initial state is described by a particle-number density ##\tilde{n}(\vec{\xi})##.

The motion of the fluid elements is then simply described by ##x^{\mu}(s,\vec{\xi})##, where ##s## is an arbitrary world-line parameter, which can be conveniently chosen as the proper time of the fluid element, i.e.,
$$\partial_s x^{\mu} \partial_s x_{\nu}=u^{\mu} u_{\mu}=1.$$
Of course also the coordinate time ##t## can be used instead of ##s##, and thus one can define the fluid motion as well by the Eulerian description, i.e., you define the ##\xi^A## as functions of ##x=(x^{\mu})##: ##\xi^{A}(x)## then are the ##\xi^{A}## of the fluid element in the standard configuration, which at time ##t## is at position ##\vec{x}##. Now the ##\xi^{A}(x)## are three scalar fields.

For the further treatment of fluid dynamics, using this "kinematics" in the Lagrangian action principle, see the above quoted manuscript (Sect. 3.5).