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The stress energy tensor for relativistic dust $$ T_{\mu\nu} = \rho v_\mu v_\nu $$ follows from the action $$ S_M = -\int \rho c \sqrt{v_\mu v^\mu} \sqrt{ -g } d^4 x = -\int c \sqrt{p_\mu p^\mu} d^4 x $$ where $p^\mu=\rho v^\mu \sqrt{ -g }$ is the 4-momentum density. One uses the formula: $$ T_{\mu\nu} = - {2\over\sqrt{ -g }}{\delta S_M\over\delta g^{\mu\nu}}\quad\quad\quad\quad\quad\quad\quad(1) $$ And the derivation is given for example in this question that I asked before (or in the Dirac book cited therein): http://physics.stackexchange.com/questions/17604/lagrangian-for-relativistic-dust-derivation-questions

Question 1: is there any Lagrangian (or action) that would give the following stress energy tensor for a perfect fluid: $$ T_{\mu\nu} = \left(\rho+{p\over c^2}\right) v_\mu v_\nu + p g_{\mu\nu} $$ ?

Question 2: What about Navier Stokes equations?

Question 3: If the answer is no to any of the questions above, can the equation (1) still be used as the definition of the stress energy tensor? Or should rather one use a definition that the stress energy tensor is whatever appears on the right hand side of the Einstein's equations (even if it can't be derived from an action)?

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  1. Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor depends on 6 parameters, 4 those in $v_\mu$ and $p$ and $\rho$, which is most of the 10 general parameters in a generic stress-energy tensor. So what you wrote is just a slightly special, but not too special, subset of stress-energy tensors. However, when you write an action for some variables, the least-action principle always immediately tells you how all the degrees of freedom evolve, and you haven't specified the equations, so you can't write an action. For the dust, this problem doesn't really arise because the dust moves along geodesics, which is what probably follows from $\delta S=0$, too.

  2. There's no Lagrangian for the Navier-Stokes equations because they include viscosity, i.e. energy dissipation, and for such irreversible systems with friction-like terms, one can't write down a fundamental description based on the action. However, one may find a generalized description of this sort, "stochastic least-action description", which has some extra integration over random variables, see e.g. http://arxiv.org/abs/0810.0817

  3. Your two "definitions" of the stress-energy tensor are completely equivalent. If you derive Einstein's equations from the principle of least action, and the dependence on the metric tensor's derivatives is just a multiple of $R$, then the right hand side of the Einstein's equations exactly contains the variation of $S$ with respect to the metric tensor. You can't say that one of them is better than the other: they're the same thing whenever an action exists. When an action doesn't exist, well, you may still define the stress-energy tensor as the right hand side of Einstein's equations but the absence of the "variation formula" for the tensor is mostly just due to ignorance because there exists an underlying Lagrangian for any interesting theory (matter coupled to gravity) in $d=4$. Let me also mention that there is a different definition of a stress-energy tensor, one obtained as the (covariantized) Noether current from spacetime translational symmetries in the limit of vanishing gravity i.e. a definition related to the conservation law and symmetries. They're typically the same thing as the variation you wrote whenever the variation is well-defined.

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Note: the Noether canonical stress energy tensor for electromagnetic field is not symmetric and one needs to add a total derivative term $\partial_\alpha K^{\alpha\mu\nu}$ with $K^{\alpha\mu\nu} = F^{\mu\alpha}A^\nu$ (so the new tensor is still conserved), then one obtains the exact same tensor as from the equation (1). But this trick seems quite arbitrary to me, so I personally like the formula (1) the most, which is the right hand side of the Einstein's equation (assuming the Hilbert action, as you wrote), as long as we know the action for the matter. –  Ondřej Čertík Jan 1 '12 at 23:17
    
Question: if we add the ideal gas equation of state, does there exist an action? I can only formulate the nonrelativistic equations: $p = \rho R T$, where $e= T c_v$, where $E=\rho e + {1\over2}\rho v^2$. In here $e$ is the internal energy, $E$ is the total energy (not counting rest mass energy), $T$ is temperature, $R$ the gas constant and $c_v$ is the specific heat capacity at constant volume. –  Ondřej Čertík Jan 1 '12 at 23:38
    
Dear Ondřej, with the ideal gas equation, you added one new variable and one new condition, so you didn't change anything. You may view it as a definition of $T$ and it doesn't change that there are undetermined things. If you made $T$ increase with local friction etc., then you face the same irreversibility problems as those with viscosity, and there won't be any "ordinary" action. Your preference for the variation-defined tensor is legitimate although it is a subjective choice. The Noether currents may still be the primary things in various contexts and the symmetry of the tensor secondary. –  Luboš Motl Jan 2 '12 at 8:57
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