Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Lagrangian for Euler Equations in general relativity

+ 9 like - 0 dislike
12 views

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)?

This post has been migrated from (A51.SE)
asked Jan 1, 2012 in Theoretical Physics by Ondřej Čertík (55 points) [ no revision ]

1 Answer

+ 9 like - 0 dislike
  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.

This post has been migrated from (A51.SE)
answered Jan 1, 2012 by Luboš Motl (10,178 points) [ no revision ]
Hi Luboš, thanks for the reply. I asked another [question](http://theoreticalphysics.stackexchange.com/questions/771/general-parameters-of-the-stress-energy-tensor-in-local-inertial-frame) about the components of the stress energy tensor. Yes, the geodesic equation follows from the same action if you vary with respect to $x^\mu$, so the action prescribes everything in the case of the dust. You are right. In the case of the perfect fluid, I forgot about the equation of state (which in principle can also follow from some action, but that is model dependent obviously).

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...