I'm trying to understand how hydrodynamics arise from a precise, mathematical formulation of thermodynamics, learning mostly from Landau's "Hydrodynamics".

So Landau starts from formulating the dynamical equations for the non viscous fluid, 5 equations, because there are 5 variables (pressure, density and a 3-component velocity field). The equations are: the mass continuity equation, the Euler equation (momentum continuity equation) and a statement of the fact, that there is no dissipation of energy, i.e. entropy is constant ($\frac{ds}{dt} = 0$).

Now, when describing a viscous flow we add dissipation through stress $-$ the Euler equation becomes the Navier-Stokes equation. My problem concerns the last equation, which should account for entropy production.

You see Landau doesn't say anything about entropy per se, but just finds an equation for energy transfer analogous to the other continuity equations, and finds a way of describing a lost in kinetic energy of a flow caused by viscosity.

My question is: how to state this equation as a direct generalization of the non viscous case, i.e. in terms of entropy? How to even approach this in terms of thermodynamics? What (thermodynamically speaking) is the process of dissipation? we need to know that, to calculate the $DQ = dE_{kin}$, don't we? Assuming we know the rate of change of the kinetic energy. Trusting Landau on that one, it has the form:

$$ \frac{dE_{kin} }{dt} = -\frac{\eta}{2} \int \left( \frac{\partial v_i}{\partial x_k}+ \frac{\partial v_k}{\partial x_i}\right)^2 dV $$

To get the $\frac{ds}{dt}$, do we just need to divide this by temperature? I'm trying to explain everything here in terms of thermodynamics, because I only know typical, very basic examples of thermodynamics processes and I have a difficulty in interpreting what is truly happening here (in terms of entropy).

  • 2
    $\begingroup$ There is a nice derivation on de Groot's book $\endgroup$
    – Hydro Guy
    Jul 7, 2014 at 3:28
  • $\begingroup$ Viscous motion is a complicated non-equilibrium process. There may not be simple way to apply the concept of thermodynamic entropy with the familiar properties to it. $\endgroup$ Jul 12, 2015 at 3:07
  • $\begingroup$ See Clausius–Duhem inequality, a way of expressing the second law in continuum mechanics (also useful in determining whether the constitutive relation of a material is thermodynamically allowable). $\endgroup$
    – Quillo
    Jul 26 at 15:15

1 Answer 1


Each continuum transport equation is derived from a closed-system counterpart using the Reynolds Transport theorem. In general, the steps to derive an expression for the rate of entropy generation are

  1. Apply the Reynolds Transport Theorem to Conservation of Energy, giving an equation for the rate of change of (internal plus kinetic) energy.
  2. Recognize that $$ \rho\frac{\text{D}}{\text{D}t}\left(\frac{\left| \vec{v} \right|^2}{2} \right) = \vec{v} \cdot \rho\frac{\text{D}}{\text{D}t} \vec{v} $$ and thus an equation for the rate of change of kinetic energy can be derived directly from the momentum equation
  3. Subtract result (2) from result (1) to get an equation for the rate of change of internal energy
  4. Apply the Reynolds Transport Theorem to the Entropy Balance. This gives an equation with both the rate of change of entropy and the rate of generation of entropy.
  5. Apply the Gibbs Relation $$ \text{d} u = T \text{d}s - P \text{d}v $$ to relate the rate of change of entropy to the rate of change of internal energy and specific volume: \begin{align*} \frac{\text{D}}{\text{D}t} u &= T \frac{\text{D}}{\text{D}t}s - P \frac{\text{D}}{\text{D}t}v \\ T \frac{\text{D}}{\text{D}t}s &= \frac{\text{D}}{\text{D}t} u + {P \frac{\text{D}}{\text{D}t}v} \end{align*} Note that I will neglect the rate of change of specific volume because I presume that Landau treats the fluid as incompressible.
  6. Sub result (3) into result (5)
  7. Sub result (6) into result (4)
  8. Break up the total stress tensor $\sigma$ into an isotropic, inward pressure component $-pI$ (where $I$ is the identity matrix) and a viscous component $\tau$, i.e., $\sigma=-pI + \tau$. The pressure component will cancel out, demonstrating that work done by pressure does not generate entropy and is thus reversible (note: this cancellation occurs even if the fluid is compressible - work by thermodynamic pressure is always reversible)
  9. Rearrange for the entropy generation

Assuming that Landau neglects heat transfer, the results of these steps will be

  1. $\rho\frac{\text{D}}{\text{D}t} (u+ke) = \vec{\nabla}\left(\vec{v}\cdot\sigma\right) + \vec{v} \cdot \vec{b}$ where $\vec{b}$ is the body force
  2. $\rho\frac{\text{D}}{\text{D}t} (ke) = \vec{v}\cdot \vec{\nabla}\sigma + \vec{v} \cdot \vec{b}$
  3. $\rho\frac{\text{D}}{\text{D}t} (u) = \sigma:\vec{\nabla}\vec{v}$
  4. $\rho\frac{\text{D}}{\text{D}t} (s) = \rho\dot{s}_\text{gen}$ (other terms would appear here if heat transfer were present)
  5. $\rho\frac{\text{D}}{\text{D}t}s = \frac{1}{T}\left( \rho\frac{\text{D}}{\text{D}t} u \right)$
  6. $\rho\frac{\text{D}}{\text{D}t}s = \frac{1}{T}\sigma:\vec{\nabla}\vec{v}$
  7. $\frac{1}{T}\sigma:\vec{\nabla}\vec{v} = \rho\dot{s}_\text{gen}$
  8. Result is $\frac{1}{T}\tau:\vec{\nabla}\vec{v} = \rho\dot{s}_\text{gen}$ because $I:\vec{\nabla}\vec{v} = 0$
  9. $\rho\dot{s}_\text{gen} = \frac{1}{T}\sigma:\vec{\nabla}\vec{v}$

If viscous stress tensor given by $$ \tau = \frac{\eta}{2}\left( \vec{\nabla}\vec{v} + \left( \vec{\nabla}\vec{v} \right)^\text{T} \right) = \frac{\eta}{2} \left( \frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i} \right) $$ then the important results are:

  1. $\rho\frac{\text{D}}{\text{D}t} (u) = \frac{\eta}{2}\left( \vec{\nabla}\vec{v} + \left( \vec{\nabla}\vec{v} \right)^\text{T} \right)^2 = \frac{\eta}{2} \left( \frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i} \right)^2$
  2. $\rho\dot{s}_\text{gen} = \frac{\eta}{2T}\left( \vec{\nabla}\vec{v} + \left( \vec{\nabla}\vec{v} \right)^\text{T} \right)^2 = \frac{\eta}{2T} \left( \frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i} \right)^2$

In the inviscid case, $\tau = 0$, so these results simplify to

  1. $\rho\frac{\text{D}}{\text{D}t} (u) = 0$
  2. $\rho\dot{s}_\text{gen} = 0$

I notice that the equation that you quote for the rate of change in kinetic energy is closely related to the one that I quote for the rate of change in internal energy. I believe that Landau's expression comes from assuming that the total energy of the system is constant, and thus the change in kinetic energy must equal the negative of the change in internal energy... or something along these lines.

Getting back to your question:

  • The derivation above shows how both the viscous and inviscid results can be derived using the same argument. One can assert that experimental evidence shows that inviscid flow is reversible, and therefore doesn't generate entropy, but the corresponding argument for the viscous case only supports the conclusion that the entropy generation is non-zero (it doesn't generate a specific expression)
  • The expression that you propose for the entropy generation does turn out to take the same form as the one that I derive, but I wouldn't view this as a general result - the entropy generation is not always related to the rate of change in kinetic energy in this way
  • What's happening in terms of entropy is:

    • the pressure component of the stress does work in a reversible fashion, and therefore does not generate entropy
    • the viscous component of the stress (which is present whenever there are velocity gradients in a viscous medium) does work in an irreversible fashion, and therefore generates entropy. We can draw an analogy between heat transfer and viscous dissipation: both generate entropy by causing gradients to flatten out - $T$ gradients for heat transfer, and $\vec{v}$ gradients for viscous dissipation

Note that we can also infer that $\eta \geq 0$, as a negative $\eta$ would give negative entropy generation.


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