Chestermiller said:
The fluid mechanics textbooks we engineers us have no such defonition.
Interesting. Maybe that's for books on hydrodynamics of liquids (water), where you can assume incompressibility, but I'm sure that you need an equation of state and the fact that ideal fluid-flow is adiabatic to close the equations.
In relativistic fluid dynamics the usual derivation is via the energy-momentum-stress tensor for ideal fluid dynamics and energy-momentum conservation. If I think about it, that's an approach which is even easier to understand than the usual way used in non-relativistic fluid-dynamics textbooks, how and where the enthalpy and internal energy enter the equations. The ideal-fluid ems tensor is easily derived. Consider a fluid cell. In the local rest frame of this cell you have
$$\bar{T}^{\mu \nu}=\mathrm{diag}(\epsilon,P,P,P),$$
where ##\epsilon## is the internal-energy density and ##P## the pressure of the gas. For an arbitrary frame you get the tensor components by introducing the fluid-four-velocity vector (I use units with ##c=1## to save some work), which in the local rest frame is
$$\bar{u}^{\mu}=(1,0,0,0).$$
In this way we get
$$T^{\mu \nu} = (\epsilon+P) u^{\mu} u^{\nu}-P \eta^{\mu \nu}$$
with the Minkowski pseduo-metric tensor ##\eta^{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)##. Note that ##\epsilon## and ##P## are still defined as their values in the local rest frame. This has the advantage that they are all Lorentz scalar fields.
If you now apply energy-momentum conservation (not considering external forces) you get the energy-balance and the relativistic version of the ideal-fluid Euler equation
$$\partial_{\mu} T^{\mu \nu}=0.$$
Now we have to introduce some conserved particle-number like quantity. In relativistic plasma physics one usually uses the electric charge in relativistic heavy-ion collisions the net-baryon number (i.e., the quantity (number of baryons) minus (number of antibaryons), which is conserved under the strong interaction). Then we have the continuity equation
$$\partial_{\mu} (n u^{\mathrm{\mu}})=0,$$
where ##n## is the electric-charge or net-baryon-number density in the local restframe of the fluid cell. Note that we here tacitly assume that the net-charge flow and the energy-momentum flow are both determined by the same four-velocity ##u^{\mu}##. That's also only true for ideal fluids. For non-ideal (viscous) fluids you have to distinguish these four-velocities, and it's a matter of the physical situation which you define as the fluid four-velocity. If you use the energy-momentum flow velocity it's called the Landau framework if you choose a conserved-charge flow it's the Eckart framework. But here let's stick to the ideal fluid, where all the flow velocities are the same.
Then you can write the thermodynamic identities which are for a fixed quantity of substance for the enthalpy (because that's the quantity ##h=\epsilon+p## which occurs in the ideal-fluid ems-tensor above)
$$\mathrm{d}(h/n)=T \mathrm{d}(s/n) + \frac{1}{n} \mathrm{d} P.$$
Here ##s## is the entropy density (as measured in the local restframe of the fluid, i.e., a Lorentz scalar field again) and ##T## the temperature (also measured in the local fluid restframe and thus a scalar).
Writing out the equation for energy-momentum conservation,
$$\partial_{\mu} T^{\mu \nu}= u_{\nu} \partial_{\mu}(h u^{\mu}) + h u^{\mu} \partial_{\mu} u^{\nu} + \partial^{\nu} P=0.$$
Multiplying with ##u^{\mu}## (with the Einstein summation convention effective) we find
$$\partial_{\mu}(h u^{\mu})-u^{\mu} \partial_{\mu} P=0,$$
where we have used ##u_{\nu} u^{\nu}=1##, i.e., ##u_{\nu} \partial_{\mu} u^{\nu}=0##.
Using in the above equation ##h=n (h/n)## and also the continuity equation ##\partial_{\mu} (n u^{\mu})=0## we get
$$u^{\mu} \left (\partial_{\mu}(h/n)-\frac{1}{n} \partial_{\mu}P \right)=0.$$
Using the above thermodynamic relation, we can write this equation as
$$u^{\mu} \partial_{\mu} (s/n)=0,$$
i.e., indeed the entropy per charge is conserved along the stream-worldlines, and this means the fluid flow is adiabatic.
