Definition of entropy and entropy flux of conservation laws: component-wise reasoning Consider the conservation law
$$\DeclareMathOperator{\dvg}{\operatorname{div}}
 \partial_t u(x,t) + \dvg G(u(x,t)) =0, \\
u \in U\subseteq \mathbb R^m, x\in X\subseteq \mathbb R^n, G \subseteq \mathbb M^{n\times m}(\mathbb R).
$$
We usually equip the system with an entropy (which is a scalar) $\eta$ with associate flux $Q \in \mathbb R^n$ related by
$$
DQ_{\alpha} =D\eta \cdot DG_\alpha, \quad\alpha = 1,2,\ldots, n. \label{1}\tag{$\ast$}
$$
Here are my concerns about this definition. In thermodynamics, the entropy flow is simply heat flux divided by temperature. It is NOT defined by the equation stated above via some other quantity $u.$
Therefore, I am puzzled by the following questions:

*

*Is the thermodynamic entropy/entropy flux an entropy in the sense of the definition above? In other words, does the tranditional physical entropy arise from some quantity $u$ via the above equation? If so, what is the physical meaning of $u?$

*Are there other nonequivalent definitions of "mathematical" entropy? After all \eqref{1} is not necessary for $\eta(u)$ to be conserved for smooth solutions. To make sure $\eta(u)$ is conserved for smooth solutions, we only need $\text{div }Q = D\eta \cdot \dvg G(U)$ which is of course a much weaker condition. That leaves the possibility of defining other mathematical objects that looks somewhat like entropy.
 A: Carlo's answer brings partial information. Let me complete it.
First of all, a general gas is not polytropic, thus the pressure is not proportional to $e-\frac12\rho u^2=:\rho\varepsilon$. In other words, there is not such beast as $\gamma$, or it is not constant. Thus the entropy $\eta$ is not given by a log-algebraic formula in terms of $p$ and $\rho$. Instead, it is a function $\eta=-\rho s(\rho,p)$, where $s$ is obtained from the (Gibbs) differential equation
$$\vartheta ds=d\varepsilon+pd\left(\frac1\rho\right)\,.$$
Even the above is not satisfactory, because the temperature $\vartheta$ is an integrating factor, which is far from unique. Actually, any $\bar s=h\circ s$ is another solution of the same equation, with $\bar\vartheta=\frac\vartheta{h'\circ s}$. Thus $\bar\eta=-\rho h\circ s$ is another entropy of the Euler system. This shows that the mathematical notion of entropy is far too wide.
Which one is the correct entropy ? To answer this question, one cannot stay at the level of the Euler equation, but we must take in account a diffusive effect, namely the Fourier law of heat diffusion. This leads us to working with either Euler-Fourier system, or even with Navier-Stokes Fourier. Anyway, the conservation of energy is written as a second-order PDE. The heat diffusion occurs through the right-hand side
$${\rm div}(\kappa\nabla\vartheta).$$
Now, the equations tell us what is the physical temperature, and thus what is the physical entropy. It is a theorem that ${\rm div}(\kappa\nabla\vartheta)$ cannot be recast as ${\rm div}(\bar\kappa\nabla\bar\vartheta)$, unless $h$ is an affine function. Thus every other entropy, admissible from the perspective of Physics, is a linear combination of $\eta$ and $\rho$.
A: Yes, the mathematical entropy is more general than the physical entropy, but in many contexts of physical relevance these turn out to be the same quantities.
In particular, for the one-dimensional Euler equation the conserved quantities $U$ are the mass density $\rho$, momentum density $\rho u$ (with velocity $u$), and the energy density $e$. The corresponding fluxes $G$ are, respectively, $\rho u$, $\rho u^2+p$, and $(e+p)u$, with $p$ the pressure, producing the conservation laws
$$\partial U/\partial t +\partial G/\partial x=0.$$
The entropy $\eta$ with flux $Q=u\eta$ is then given by
$$\eta=\rho\log(p/\rho^\gamma),$$
where the exponent $\gamma$ appears in the equation of state
$$p=(\gamma-1)\big(e-\tfrac{1}{2}\rho u^2\big).$$
That this physical entropy satisfies the condition \eqref{1} in the OP required for a mathematical entropy, is checked for example in these notes (page 12).
