Invariants of the stress energy tensor

[Dale]

Does anyone know of a set of invariants for the stress energy tensor? In particular, I would like to know if there is a small set of linearly independent invariants, each of which (or at least some of which) have a clear physical meaning.

 

[vanhees71]

It depends on which situation you look at.

Take an ideal fluid. It's characterized by an equation of state and a fluid four-velocity field $u^{\mu}$. The energy-momentum tensor is defined through 2 invariants (or rather scalar fields): the internal-energy density and pressure in the local rest frames of the fluid cells, $\epsilon$ and $P$. At one space-time point in the restframe the components read
$$T^{* \mu \nu}=\mathrm{diag}(\epsilon,P,P,P).$$
Since $u^{* \mu}=(1,0,0,0)$ and $\eta^{*\mu \nu}=(1,-1,-1,-1)$ you can write this in manifestly covariant form as
$$T^{* \mu \nu}=(\epsilon+P) u^{*\mu} u^{* \nu}-P \eta^{* \mu \nu}.$$
Since $u^{\mu}$ are four-vector components, and $\eta^{* \mu \nu}=\eta^{\mu \nu}$ are invariant tensor components (under Lorentz boosts), the equation holds in any frame,
$$T^{\mu \nu} =(\epsilon+P) u^{\mu} u^{\nu} - P \eta^{\mu \nu}.$$
The invariants (scalar fields) in this case are
$$u_{\mu} u_{\nu} T^{\mu \nu}=\epsilon$$
and
$$\eta_{\mu \nu} T^{\mu \nu}=\epsilon-3P.$$
Of course the latter scalar ("the trace") is one you can define for any energy-momentum tensor. For a free electromagnetic field or a fluid of massless particles it vanishes (in the classical-field theory approximation) because of the scale invariance of free electromagnetic fields and massless particles making up an ideal fluid.

 

[Dale]

Do you know if there are other invariants for a more general stress energy tensor or only for a perfect fluid?

 

[Orodruin]

Like any (1,1) tensor, the invariants should just be the eigenvalues or combinations thereof? This is reflected in #2 where the assumption of an ideal fluid makes three of the eigenvalues equal so that you only have two independent ones.

 

[Ibix]

Does $\nabla_\mu T^{\mu\nu}=0$ count? It's the local conservation of energy, and is four invariants constructed from the stress-energy tensor.

 

[Dale]

Like any (1,1) tensor, the invariants should just be the eigenvalues or combinations thereof?

Is that true in all coordinate systems or just in locally inertial coordinates?

 

[pervect]

I rather thought the invariants would be $T^u{}_u$ and perhaps $*T^u{}_u$, where $*T^{cd} = \epsilon^{abcd}T_{ab}$. But that was mainly from a discussion of an anti-symmetric tensor, I'm not sure what difference symmetry might make.

 

[Orodruin]

I rather thought the invariants would be $T^u{}_u$ and perhaps $*T^u{}_u$, where $*T^{cd} = \epsilon^{abcd}T_{ab}$. But that was mainly from a discussion of an anti-symmetric tensor, I'm not sure what difference symmetry might make.

Well, to start $T$ is not a 2-form. Since it is symmetric, $*T$ would be zero trivially.

The trace of $T$ is indeed an invariant as it is the sum of eigenvalues.

Is that true in all coordinate systems or just in locally inertial coordinates?

Eigenvalues of a (1,1) tensor do not depend on the coordinates. The coordinate independent eigenvector equation is $T(V)=\lambda V$.

 

[Dale]

The coordinate independent eigenvector equation is $T(V)=\lambda V$.

So in coordinate notation that is $g_{\nu\xi}T^{\mu\nu}V^{\xi}=\lambda V^{\mu}$

 

[Orodruin]

So in coordinate notation that is $g_{\nu\xi}T^{\mu\nu}V^{\xi}=\lambda V^{\mu}$

Yes.