Questions regarding projecting the energy-momentum tensor for perfect fluid along the four-velocity

Geometrically, the stress-energy-momentum tensor is the covector-valued 3-form $$ T = T_{munu} ;mathrm dx^mu otimes starmathrm dx^nu $$ 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 $U_nupartial_mu T^{munu}$ 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: $$ mathrm dT = partial_lambda T_{munu}; dx^mu otimes mathrm dx^lambda wedge starmathrm dx^nu = eta^{lambdanu} partial_lambda T_{munu}; dx^mu otimes mathrm d^4x $$

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

When we contract $partial_nu$ with the column index $nu$ of $T^{munu}$ 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 $$text{time derivative of observed energy density} + text{divergence of observed energy flux}$$

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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 $$ p = p_mumathrm dx^mu = E,mathrm dt + p_i,mathrm dx^i $$

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 $$ rho_mu,mathrm dx^muotimesmathrm dV $$ 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 $$ mathrm dV = mathrm dxwedgemathrm dywedgemathrm dz $$ we also expect $$ mathrm dtwedgemathrm dxwedgemathrm dy qquad mathrm dtwedgemathrm dywedgemathrm dz qquad mathrm dtwedgemathrm dzwedgemathrm dx $$ 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 $$ starmathrm dt = mathrm dxwedgemathrm dywedgemathrm dz $$

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