The energy momentum tensor for a perfect fluid is defined as:

$$T^{\mu \nu} = (\rho + p)U^\mu U^\nu + p \eta^{\mu \nu}$$

Where $\rho$ is the energy density, $p$ is momentum, $U$ is the four-velocity (tangent vector parameterized by proper time), and $\eta$ is the Minkowski metric.

I've read:

To project $\partial _ \mu T^{\mu \nu}$ along the four-velocity, simply contract it into $U_\nu$. $$U_\nu \partial_\mu T^{\mu \nu}$$

First of all, I believe $\partial_\mu T^{\mu \nu}$ has four components, where each component is the divergence of momentum multiplied by $U^\nu$. (This is the change in momentum flux in the $\nu$ direction?)

So, $U_\nu \partial_\mu T^{\mu \nu}$ is an inner-product between the four-velocity and the momentum-flux, which should result in a single number, correct?

Furthermore, I write a lot of notes in my textbook, and in trying to explain this, I find myself struggling when referring to parts of a tensor. For example, can I say something like "The second index of $T^{\mu \nu}$ represents the directional component of the tensor."

(I am struggling w/ tensors, so any detailed explanation is very much appreciated).


Geometrically, the stress-energy-momentum tensor is the covector-valued 3-form $$ T = T_{\mu\nu} \;\mathrm dx^\mu \otimes \star\mathrm dx^\nu $$ where we have represented the 3-form part via Hodge duality (think normal vector).

This leads to several natural operations, eg integration over 3-surfaces of spacetime (yielding a 4-momentum covector), contraction with a 4-velocity (yielding energy-momentum densities as measured in the corresponding frame), or exterior differentiation (think divergence, which can eg be used to generalize Newton's equations to relativistic continuum mechanics).

What the abstract expression $U_\nu\partial_\mu T^{\mu\nu}$ represents is the combination of the latter two operations, though that's only true under the assumption of a symmetric SEM tensor - by geometric reasoning, we should contract the derivatives with the second index: $$ \mathrm dT = \partial_\lambda T_{\mu\nu}\; dx^\mu \otimes \mathrm dx^\lambda \wedge \star\mathrm dx^\nu = \eta^{\lambda\nu} \partial_\lambda T_{\mu\nu}\; dx^\mu \otimes \mathrm d^4x $$

More pragmatically, the meaning of the individual components of the SEM tensor are

SEM tensor components

When we contract $\partial_\nu$ with the column index $\nu$ of $T^{\mu\nu}$ we get a continuity-equation style expression for each row indexed by $\mu$, ie $$\text{time derivative of density} + \text{divergence of flux}$$

Contraction with $U_\mu$ will then project onto the time-like component (aka energy) as measured by an observer moving with 4-velocity $U^\mu$, ie $$\text{time derivative of observed energy density} + \text{divergence of observed energy flux}$$

Not sure if it will help much, but here's a short explanation how the formalism works for people without a background in differential geometry:

The differentials $\mathrm dx^\mu$ are the basis vectors of momentum space. For example, the momentum of a particle is given by $$ p = p_\mu\mathrm dx^\mu = E\,\mathrm dt + p_i\,\mathrm dx^i $$

In continuum mechanics, we have to operate with spatial densities, so instead of, say, an energy $E$, we need an energy density $\epsilon\,\mathrm dV$.

Naively, the generalization of the momentum 4-vector to continuum mechanics would then be something like $$ \rho_\mu\,\mathrm dx^\mu\otimes\mathrm dV $$ where $\rho_0=\epsilon$ is our energy density and $\rho_i$ are the spatial momentum densities.1

Such an expression should make a relativist's alarm bells ring, because where there's a $$ \mathrm dV = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz $$ we also expect $$ \mathrm dt\wedge\mathrm dx\wedge\mathrm dy \qquad \mathrm dt\wedge\mathrm dy\wedge\mathrm dz \qquad \mathrm dt\wedge\mathrm dz\wedge\mathrm dx $$ to show up.

The correct generalization of the 4-momentum turns out to be the SEM tensor, whose first index labels the directions of momentum space, and its second one the above volume elements.

Finally, for convenience, the Hodge star can be used to represent the volume elements of the 3-dimensional hyperplanes by their normal vectors, eg $$ \star\mathrm dt = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz $$

1 and that's indeed the first column of the SEM tensor

  • $\begingroup$ Thanks for the explanation (the picture and that below it was helpful), but I'm afraid a lot of what you wrote is lost on me. $\endgroup$ – AmagicalFishy May 27 '17 at 14:33
  • $\begingroup$ @AmagicalFishy: I added some parapraphs that might or might not be helpful to you... $\endgroup$ – Christoph May 27 '17 at 15:34

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