Special Cases of Stress-Energy Tensor in GR

[ohwilleke]

TL;DR Summary

Are there are scientific names for the special cases of the stress-energy tensor in GR, for example, where all components of the stress-energy tensor of GR except the mass component, or where all components except the mass and pressure components, are zero?

Background and Motivation

The stress energy tensor of general relativity, as conventionally defined, has sixteen components.

One of those component, conventionally component T₀₀, also called ρ, is mass-energy density, including the E=mc² conversion for electromagnetic fields.

The other components (three each for components in each of the three space dimensions) represent linear momentum, angular momentum, electromagnetic flux, sheer stress, and pressure.

In a static, pressureless case, without electromagnetic fields, however, all of these components except ρ are zero.

There are lots of physical situations where these fifteen components, while not actually zero, are small enough to be ignored in a quite good first order approximation of the gravitational forces that arise, but where it would be desirable to make a relativistic rather than a Newtonian calculation. And, with just one component, you have greatly simplified the computations involved.

Another case which would be considerably simpler and physical relevant for first order approximations in some cases, although not quite as simple, would be the static case without electromagnetic fields, but with both mass-energy density and pressure.

Question

My primary question is whether there is a special technical name or buzzword for this special case that would make it easier to search for and easier to talk about intelligently?

Also, is there a special technical name or buzzword for the slightly less simplified static case without electromagnetic fields, in which all elements not on the diagonal of the stress energy tensor are zero, but ρ and the three pressure components of the stress energy tensor are non-zero?

Finally, are there other special cases of the stress energy tensor similar to these cases that have special technical names?

 

[PeterDonis]

In a static, pressureless case, without electromagnetic fields, however, all of these components except ρ are zero.

This will be true only in coordinates in which the matter in question is at rest. The "buzzword" for this case is "dust". Note that there is actually no requirement that this case be static; for example, FRW spacetime with $p = 0$ is a perfectly good dust solution, but it is not static.

Also, is there a special technical name or buzzword for the slightly less simplified static case without electromagnetic fields, in which all elements not on the diagonal of the stress energy tensor are zero, but ρ and the three pressure components of the stress energy tensor are non-zero?

For the case where all three pressure components are the same, the "buzzword" is "perfect fluid". "Dust" is then a special case of a perfect fluid with zero pressure. (My remarks about static not being required above apply here as well.)

I'm not aware of any special "buzzword" for a diagonal stress-energy tensor where the spatial components are not all the same. I have not seen any real discussion of that case in the textbooks and papers I have read.

 

[ohwilleke]

Thanks! Greatly appreciated. It is hard to pick up things like this with confidence by reading texts and papers alone.

 

[PAllen]

Also, note that this tensor symmetric, so there are at most 10 different components. Further, there are only 6 functional degrees of freedom when diffeomorphisms are taken into account.

 

[pervect]

Generally the buzzwords are coordinate independent. So a dust solution is described in abstract notation as $T = \rho u \otimes u$ where u is a 4-vector, the 4-velocity. $\otimes$ is the tensor product. In abstract index notation this would be $T^{ab} = \rho \, u^a u^b$

If u has the components (1,0,0,0) the stress energy tensor has only one non-zero component, $T^{00}$. Any non-null dust can be described in coordinates where the stress energy tensor has that property, so it would look like this.

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

However, it's still described as a dust even in coordinates that have more non-zero components than the above. For instance, in different coordinates u might be (1, 0.1, 0, 0) rather than (1, 0, 0, 0).

 

[PeterDonis]

Generally the buzzwords are coordinate independent. So a dust solution is described in abstract notation as $T = \rho u \otimes u$ where u is a 4-vector, the 4-velocity. $\otimes$ is the tensor product. In abstract index notation this would be $T^{ab} = \rho \, u^a u^b$

Yes, good point. And a perfect fluid solution, more generally, would be, in abstract index notation, $T^{ab} = \left( \rho + p \right) \, u^a u^b + p \, g^{ab}$

 

[vanhees71]

There are some buzzwords for typical energy-momentum tensors, e.g., "dust"
$$T^{\rho \sigma}=\mu u^{\mu} u^{\nu}.$$
These are just non-interacting particles, described in a continuum mechanical way. $u^{\mu}$ is the four-velocity and $\mu$ the invariant-mass-density (energy density in the local rest frame of the "dust").

For a perfect fluid you have
$$T^{\rho \sigma}=(\epsilon+P) u^{\rho} u^{\sigma}-P g^{\mu \nu},$$
where $\epsilon$ is the internal energy and $P$ the pressure as measured in the local rest frame of the fluid.

The components of the energy-momentum tensor mean energy density ($T^{00}$), energy-current/momentum density ($T^{0j}=T^{j0}$) and the stresses ($T^{jk}$).

 

[PAllen]

Another somewhat less common special case is null dust, a dust of massless particles, which for some purposes only, can be used as a simple model of incoherent light:

https://en.wikipedia.org/wiki/Null_dust_solution

 

[PeterDonis]

For a perfect fluid you have
$$T^{\rho \sigma}=(\epsilon+P) u^{\rho} u^{\sigma}-P g^{\mu \nu},$$
where $\epsilon$ is the internal energy and $P$ the pressure as measured in the local rest frame of the fluid.

I think the sign in front of the second term depends on the metric signature convention; it's a minus sign, as you have here, if you are using $+ - - -$, but it is a plus sign, as I had it in post #6, if you are using $- + + +$.

 

[vanhees71]

Yes, sure. I should have mentioned that I'm in the "west-coast camp".

 

[ohwilleke]

Also, note that this tensor symmetric, so there are at most 10 different components. Further, there are only 6 functional degrees of freedom when diffeomorphisms are taken into account.

The special case is still much simpler to work with mathematically than the general case.