In Landau & Lifshtiz Volume 6 on fluid mechanics we derive the general equation of heat transfer by starting with the expression $$ \partial_t \left( \frac{1}{2} \rho v^2 + \rho \varepsilon \right) = - \vec{\nabla} \cdot \left[ \rho \vec{v} \left( \frac{1}{2} v^2 +w \right) \right] $$ derived from the conservation of energy for an ideal fluid. Here $\rho$ is the density, $v$ the velocity, $\varepsilon$ the internal energy per unit mass and $w$ the enthalpy per unit mass.

We argue that two terms need to be added:

  • $- v_i \sigma_{ij}$ due to flux related to internal friction ($\sigma_{ij}$ is the viscous stress tensor)
  • $q_i = -\kappa \partial_i T$, the heat flux density $q$ with thermal conductivity $\kappa$ and temperature $T$.

This then gives the final equation $$ \partial_t \left( \frac{1}{2} \rho v^2 + \rho \varepsilon \right) = - \vec{\nabla} \cdot \left[ \rho \vec{v} \left( \frac{1}{2} v^2 +w \right) - \vec{v}\cdot \sigma -\kappa \vec{\nabla} T \right] $$

This is fine, however to derive energy flux $\rho \vec{v} \left(\frac{1}{2} v^2 +w \right)$ for the ideal fluid case we assume the general adiabatic equation $$ \partial_t s + v_i \partial_i s = 0 $$ with $s$ denoting the entropy per unit mass. Which requires the absence of heat exchange, i.e. an adiabatic motion of the fluid. Assuming this we are able to cancel terms

  • $+\rho T v_i \partial_i s$ coming from the kinetic energy part after substituting the Euler equation $\rho \partial_tv_i + v_j \partial_j v_i = -\partial _i P$ ($P$ denoting the pressure) and using the thermodynamic relation $\mathrm{d}w=T\mathrm{d}s + 1/\rho \mathrm{d}P$
  • $- \rho T v_i \partial_i s = \rho T \partial_t s$ originating from the internal energy term using $\rho\mathrm{d}\varepsilon=\rho T\mathrm{d}s+P/\rho\mathrm{d}\rho$

If we now add the term for the heat flux density mentioned above, the general adiabatic equation does no longer hold (?!), and thus we cannot cancel the mentioned terms, right? So why are these terms not appearing in the general equation?


1 Answer 1


Alright, turns out I simply missed an important part of the derivative. For completeness and future reference I put the mention what resolved my issue. Apparently the equation I mentioned above is the the total energy flux $$ \partial_t \left( \frac{1}{2} \rho v^2 + \rho \varepsilon \right) - \vec{\nabla} \cdot \left[ \rho \vec{v} \left( \frac{1}{2} v^2+ w - \vec{v} \cdot \sigma - \kappa \vec{\nabla}T \right) \right] $$ which is to be compared with the proper differentiation of the term on the left. This ultimately leaves us with the general equation of heat transfer $$ \rho T \left( \partial_t s + v_i \partial_is \right) = \sigma_{ij} \partial_jv_i + \partial_i \left( \kappa \partial_i T \right) $$


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