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comment Seeking Reference on Transport of Momentum by Diffusion of Mass
I have not looked at the book, but note that in the energy/entropy the leading term is of second order in diffusive fluxes (this follows from the second law of thermodynamics), so corrections beyond Navier-Stokes start at third order.
17h
answered Seeking Reference on Transport of Momentum by Diffusion of Mass
1d
comment Perfect fluid and Cauchy momentum equation
1) The perfect fluid is a systematic approximation (gradient corrections can be made arbitrarily small by considering smooth flows, and can be accounted for order by order in the gradient expansion). 2) $\epsilon=0$ is not a systematic approximation. 3) The only thing I can imagine is that $\epsilon=0$ corresponds to neglecting internal energy compared to rest mass energy density, which is reasonable for a non-relativistic fluid. There is still no godd reason for dropping $\rho\epsilon$, but keeping $P$.
2d
revised Perfect fluid and Cauchy momentum equation
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2d
comment Perfect fluid and Cauchy momentum equation
I think this is not just terminology. $T_{\mu\nu}=(\rho+P)u_\mu u_\nu+Pg_{\mu\nu}$ and $j_\mu=\rho u_\mu$ cannot both be correct. You can consider $\epsilon=0$ (in your notation) but i) there is no such fluid known to man (finite pressure but zero internal energy), ii) the original question is pointless, because we are in the non-relativistic limit from the start.
2d
comment Perfect fluid and Cauchy momentum equation
It has to be energy density, otherwise the formula for the stress tensor is wrong.
May
26
comment Perfect fluid and Cauchy momentum equation
No. If there is a conserved particle number, and if $\rho$ is the density of particles, and if $u_\mu$ is defined to be the velocity of the particles (the so-called Eckart frame), then this equation would be correct. But, as is clear from the expression for $T_{\mu\nu}$, $\rho$ is the energy density.
May
26
revised Perfect fluid and Cauchy momentum equation
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May
26
answered Perfect fluid and Cauchy momentum equation
May
18
comment EFT and Renormalizability
This is just an example of the very poor quality of the physics material on wikipedia. Read ``Is renormalizability necessary'' in Vol I of Weinberg (or any other decent text book on modern QFT or EFT methods).
May
8
comment Energy density and pressure in thermal quantum field theory
The $V(\phi)$ in the operator $T_{ii}$ is the original potential. When we write something like $P=-V_T(\phi_0)$ we define $V_T$ by $\langle T_{ii}\rangle=-V_T(\phi_0)$, which is really an implicit definition because $\phi_0=\langle\phi\rangle$. For an interacting field theory, $Z(T)$ can (usually) not be computed exactly. Various approcimations are discussed in text books on thermal field theory.
May
6
revised Energy density and pressure in thermal quantum field theory
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May
6
answered Energy density and pressure in thermal quantum field theory
Mar
19
comment Nature of particle spectra at ALICE
The difference between bosons (pions etc) and fermions (protons) etc is the standard Bose enhancement vs Pauli blocking (no relation to Fermi theory of beta decay). Also note that there is a trivial phase space effect $d^2p_T=p_Tdp_T$.
Jan
25
answered Are sound waves adiabatic or isothermal?
Oct
12
comment H-theorem and Boltzmann equation applied to Boltzmann distribution
Not sure what you are doing here. As written, $r$ and $s$ must refer to pairs. Then $p_r=f_1f_2$ and $p_s=f_3f_4$, and in equilibrium $f=e^{-\beta\epsilon}$ the condition $p_r=p_s$ is just energy conservation.
Oct
8
comment Hydro regime of strongly coupled field theory, low viscosity
The argument is weak coupling --> large viscosity. Therefore, if low viscosity is observed then the coupling must be large. The statement strong coupling --> low viscosity is not always true.
Jun
11
comment Nonequilibrium thermodynamics in a Boltzmann picture
In simple cases (like phi^4 or the dilute Fermi gas) I can also compare (theoretically) transport coefficients from the BE to calculations based on the Kubo relation and equilibrium Green functions, and the result agrees. Hard to see how that would work if the BE is some kind of phenomenological equation.