I was following the book by Landau and Lifshitz, Fluid Mechanics (2nd edition) and got stuck trying to understand one of their arguments. On the chapter about Relativistic Fluid Dynamics it is stated the follwoing thermodynamic relation on Eq. 134.6

$$ d(w/n) = T d(s/n) + (1/n) dP$$

$w = \varepsilon + P$ is the heat function desnity, $s$ is entropy density, $P$ is pressure and $n$ is the particle per volume, and $\varepsilon$ is energy density. What I cannot understand is why there is no chemical potential on this relation.

I tryed to rederive using the Gibbs-Duhem relation. I assume that there is several conserved charges $N_j$ and pick one of them to be my ''particle''. Then, the Gibbs-Duhem divided by the particle per volume

$$ \frac{s}{n_i} dT + \frac{n_j}{n_i} d \mu_i = \frac{1}{n_i} dP $$

Since $w/n_i = (\varepsilon + p)/n_i = (sT + n_j \mu_j)/n_i $, I can take the differential from it and use the above expression to get to a similar expression as the desired one.

$$ d(w/n_i) = T d\left(\frac{s}{n_i}\right) + \frac{s}{n_i} dT + \mu_j d\left(\frac{n_j}{n_i}\right) + \frac{n_j}{n_i} d\mu_j = T d(s/n_i) + \mu_j d(n_j/n_i) + (1/n_i)dP$$

As one can see, the there is an extra term proportional to the chemical potential on my result. So, where is my error? Am I missing some additional assumptions?

Notice that the presence of this term completelly spoils the argument the authors later do that isentropic flow is a sufficient condition for potential flow, which is the result I am interested in understanding.

Edit: After some thought, I found that if $i = j$ then $d(n_{j=i}/n_i) = 0$. Since the book deals with the case of only one conserved charge, then the only term proprtional to the chemical potential is the vanishing one. Although this answers why the term is absent on the book. However, this imply that if there is $N$ conserved charges on the fluid, there will be $N-1$ terms proportional to a chemical potential. In turn this implies that potential flow is impossible if more than one conserved charge is present (unless we explicitly set all chemical potentials to 0). Is this correct?

  • $\begingroup$ I am not an expert in this (hence comment and not answer), but perhaps there is a reason to neglect this term. Do you expect it to be the same or larger order of magnitude than the other terms? $\endgroup$ – Aaron Stevens Aug 8 '18 at 3:16
  • 5
    $\begingroup$ You could include chemical potentials if dealing with multiple components in the fluid but i'm assuming the book only treats a single component fluid hence not considering chemical potentials. $\endgroup$ – nluigi Aug 8 '18 at 8:26
  • $\begingroup$ @nluigi Indeed the chapter is treating of a single componente fluid. Even so, a single fluid could be carrying a number of conserved charges. That is the reason why I included several chemical potentials. Maybe this is the source of my conffusion? $\endgroup$ – WilhelmM Aug 8 '18 at 13:35
  • $\begingroup$ What I don't understand here is the first equation. $w=\epsilon+P$ to me doesn't appear to be homogeneous, if $\epsilon$ is energy per unit mass and $P$ is pressure. Enthalpy is usually written $w=\epsilon+\frac{p}{\rho}$, where $\rho$ is the fluid density. $\endgroup$ – Time4Tea Nov 13 '18 at 16:30
  • $\begingroup$ Since the subject is relativistic fluid dynamics, I think it would not be a good idea to have as one variable energy per unit of mass. We need to consider the possibility of converting part of the mass on energy. Thus, by energy density I mean energy per unit of volume. The masses are computed together with the energy and the densities are not mass densities, but densities of other conserved charges, e.g. eletric charge or baryon number. $\endgroup$ – WilhelmM Nov 20 '18 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.