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Jun
23
comment Why is the chiral symmetry only $SU(3) \times SU(3)$ and not $SU(6)$?
Then you would still have to write something like $\sigma_2u_R^\dagger$ instead of $u_L$ in your 6-component vector. In any case, you violate either $SU(3)_c$ (and baryon number $U(1)_V$) or Lorentz invariance.
Jun
23
comment How do Cooper pairs form?
Yes, of course. I meant ingredients beyond BCS, that are required to get condensation in the presence of the repulsive Coulomb force.
Jun
23
comment Why is the chiral symmetry only $SU(3) \times SU(3)$ and not $SU(6)$?
$L$ and $R$ are in the same representation of $SU(3)_c$, but in different representations of the Lorentz group (there is something wrong with your Dirac operator, by the way, now that you wrote in 2-component notation), so your proposed flavor symmetry mixes different representations of $SO(3,1)$.
Jun
22
comment Why is the chiral symmetry only $SU(3) \times SU(3)$ and not $SU(6)$?
$\bar{q}_Li\gamma^\mu D_\mu q_R=0$. Your symmetry does not act on the fields.
Jun
16
comment Superfluid Fountain
At any temperature below the lambda point (2.2K) liquid helium is a mixture of a superfluid and a normal fluid. The superfluid fraction goes to zero as $T\to T_c$. If you set up a temperature gradient, then the normal fluid flows to low T, and the superfluid flows to high T. This is why the superfluid flows into the heated pipe. The seal ensures that the normal fluid cannot get out, and you get the fountain effect.
Jun
7
comment Incompressible Navier-Stokes pressure solve in simulations
Your second equation is right (I expanded my answer).
Jun
6
comment Incompressible Navier-Stokes pressure solve in simulations
Not sure what you mean by "what to do". Obviously, hundreds of text books have been written about analytical solutions to these equations, and about methods for solving them numerically. Many codes can be downloaded, optimized for all sorts of conditions.
Jun
6
comment Incompressible Navier-Stokes pressure solve in simulations
Sorry, I fixed the index structure. This is just the divergence of the NS equation, $\partial_t u_i+u_j\nabla_ju_i=-1/\rho\nabla_iP +\nu\nabla^2u_i$, using the fact that $\nabla_i u_i=0$.
May
31
comment In a fluid, why are the shear stresses $\tau_{xy}$ and $\tau_{yx}$ equal?
I tried to be a little more explicit.
May
30
comment Gauge potential for locomotion at low Reynolds number
The paper by Shapere and Wilczek has a number of worked examples. Do those not answer your question?
May
28
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.
May
27
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 $P$.
May
27
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.
May
26
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
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.
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 $d^2p_T=p_Tdp_T$.
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.