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Thermodynamic quantities are well defined at equilibrium. a fluid at zeroth order can be approximated by a perfect fluid and conserved currents such as electric current and energy momentum tensor is given by: \begin{equation} \begin{aligned} & j^{\mu}=nu^{\mu}\\ &T^{\mu\nu}=(\epsilon +P)u^{\mu}u^{\nu}+Pg^{\mu\nu} \end{aligned} \end{equation} In the next order in derivative expansion at Landau frame which is define by $ u_{\mu} j^{\mu}=-n $ electric current is as follow: \begin{equation} j^{\mu}=nu^{\mu}-\sigma_Q\mathcal P^{\mu\nu}(\partial_{\nu}\mu-\frac{\mu}{T}\partial_{\nu}T) \end{equation} where the project operator $ \mathcal P^{\mu\nu} $ is defined as follow: \begin{equation} \mathcal P^{\mu\nu}=g^{\mu\nu}+u^{\mu}u^{\nu} \end{equation} According to the definition of the temperature, it is only meaningful at equilibrium, so why do we consider the gradients of temperature and chemical potential?

Arian
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    This question is very confusing. Maybe you can clarify where you use $T$ as temperature or tensor and equally $\mu$ as index chemical potential. Also please note how in physics we distinguish many different equilibria. Someone at the level to use tensor notation should understand that. – AtmosphericPrisonEscape Jan 22 '20 at 22:24
  • Related question (in particular, the background provided by the asker probably answers this question): https://physics.stackexchange.com/questions/330003/assumption-of-local-thermodynamic-equilibrium-in-fluid-dynamics – probably_someone Jan 22 '20 at 22:25
  • So when you have heat conduction along a rod (a non-equilibrium situation), the temperature profile and the rate of heat flow through the rod that you calculate are meaningless? – Chet Miller Jan 22 '20 at 23:58
  • @AtmosphericPrisonEscape. I did not understand what you are mentioned. Everything is clear in my notation. These equations are very well known in the literature. – Arian Jan 23 '20 at 09:07
  • @ChetMiller. no that is not meaningless, but the true notion of temperature is based on equilibrium. of course I should consider the local equilibrium. – Arian Jan 23 '20 at 09:15
  • Yes. When dealing with transport processes, as correctly pointed out by @Thomas, we need to accept the idea that intensive properties such as temperature, specific internal energy, pressure, specific entropy, chemical potential, and velocity can be meaningfully defined locally, as can the fluxes of heat, mass, and momentum. – Chet Miller Jan 23 '20 at 12:14

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Fluid dynamics is an effective theory for the long wavelength, long time behavior of classical or quantum many body systems. In this limit the system reaches approximate local (but not global) equilibrium. This is the case because local equilibration takes place on a microscopic time scale (the collision rate between atoms, or leptons, quarks, etc), whereas global equilibration requires diffusion or propgation of density or pressure disturbances, which takes place on a macroscopic time scale.

In local thermal equilibrium we can use the coarse grained values of the conserved charges $n$, $\epsilon=T_{00}$ and $\pi_i=T_{0i}$ to define local values of the temperature, chemical potential, and fluid velocity. By assumption these are slowly varying, so we can express the fluxes as a gradient expansion in the local thermodynamic variables.

At zeroth order in gradients we obtain ideal (relativistic) fluid dynamics. at first order we encounter three new terms, two that involve derivatives of the fluid velocity, controlled by bulk and shear viscosity, and one that involves gradients of $T,\mu$, controlled by the thermal conductivity.

P.S.: There is an ambiguity (really, "reparametrization invariance" or "frame dependence") in mapping the densities $(n,\epsilon,\pi_i)$ on the thermodynamic variables $(T,\mu,u_\mu)$. This has to do with the definition of the fluid velocity. Roughly, I can use either the energy current (Landau) or the particle currrent (Eckhardt), or anything in between, to define the fluid velocity. This affects $(T,\mu)$ because I use the fluid rest frame to relate the energy and particle density to ($T,\mu)$. It also affects the dissipative terms, because in the Eckhardt frame (for example) the particle current is defined to be $nu_\mu$, so it cannot have any dissipative corrections. The $\nabla T$ term must then appear in the energy current. Of course, physical quantities must be unaffected (to the order in $\nabla$ that we are working) by this choice.

Thomas
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  • Thanks. what I found from your text is that we have not freedom to redefine temperature and chemical potential and everything is obtained from local equilibrium by $ n$, $T_{00} $ and $T_{0i} $, would you please more explain about freedom to redefine temperature and chemical potential. we usually fix a special frame for example Landau frame to fix the first order in derivative expansion. I can not see the role of this freedom in local equilibrium. – Arian Jan 23 '20 at 09:35
  • added a postcript – Thomas Jan 23 '20 at 14:45
  • thanks a lot dear Thomas. I am thankful to you if you also introduce me a reference to understand the subject in detail. – Arian Jan 23 '20 at 15:45
  • This is discussed in standard text books on fluid dynamics (and Landau's book is still among the best and most careful). More sophisticated discsussion of reparametrization invariance can be found in the recent literature, e.g. https://arxiv.org/abs/1511.03646 – Thomas Jan 24 '20 at 17:21
  • thank you so much. – Arian Jan 24 '20 at 19:16