We are all familiar with the kinetic energy dissipation and how it is converted into heat which can either be radiated away or go into the internal energy of the system. In the enstrophy transport equation:

\begin{align} \frac{\partial\Omega^2}{\partial t} + u_j \frac{\partial\Omega^2}{\partial x_j} & = \omega_i S_{ij} \omega_j + \nu \frac{\partial^2\Omega^2}{\partial x_j\partial x_j} - \Phi_0 \\ \Omega^2 & = \frac{1}{2} \omega_i \omega_i \\ \Phi_0 & = \nu \frac{\partial\omega_i}{\partial x_j} \frac{\partial\omega_i}{\partial x_j} \\ S_{ij} & = \frac{1}{2} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) \end{align}

there is a dissipation term, $\Phi_0$, very similar to that in the kinetic energy equation. Is there some mechanism or "place" where the dissipated enstrophy goes similar to the KE? Does enstrophy have to be conserved in the same sense that the total energy of a system (KE + PE + IE, etc.) has to be conserved?

Some people have explained it to me that since vorticity is a mathematical construct, then there is no "place" that the dissipated energy has to go. But you can describe velocity in the same sense as it is a construct that we created to represent particle motion in space.

Since the vorticity field is directly related to the velocity field (via the curl operator), then does that mean that the dissipated enstrophy is directly related to the dissipated kinetic energy? I'm currently attempting to reform and rewrite the enstrophy equation in terms of KE ($1/2 \times U_i U_i$) and see if there is any direct relation.


It is possible to rewrite both dissipation terms in terms of the strain rate and rotation rate tensor. This gives a slightly better picture of what's going on though it still doesn't answer my question.

\begin{align} \omega_i = -\epsilon_{ijk} R_{jk} \\ \frac{\Phi_0}{\nu} = \epsilon_{ijk} \epsilon_{inp} \frac{\partial R_{jk}}{\partial x_l} \frac{\partial R_{np}}{\partial x_l} = (\delta_{jn} \delta_{kp} - \delta_{jp} \delta_{kn})\frac{\partial R_{jk}}{\partial x_l}\frac{\partial R_{np}}{\partial x_l} = 2\frac{\partial R_{jk}}{\partial x_l}\frac{\partial R_{jk}}{\partial x_l} \\ \frac{\Phi_{KE}}{\nu} = \frac{\partial u_i}{\partial x_j}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}) = (S_{ij} + R_{ij})(2S_{ij}) = 2S_{ij}S_{ij} \\ \end{align}

  • $\begingroup$ Just as with momentum, this is only one of a set of coupled equations. You are missing the total energy equation expressed in terms of enstrophy. In there, you should find a production term that matches your enstrophy dissipation term, indicating that just like the dissipation of momentum turns into internal energy, the dissipation of enstrophy turns into internal energy. $\endgroup$
    – tpg2114
    Commented Feb 5, 2015 at 23:20
  • $\begingroup$ What is the total energy equation in terms of enstrophy? Is there even such a conservation law? The equation that I posted does have a production term, namely the first term on the RHS, but does this exactly balance the energy that's lost? $\endgroup$
    – Kimusubi
    Commented Feb 5, 2015 at 23:25
  • $\begingroup$ See this paper for example. I don't work with the enstrophy formulation so I can't provide a full answer. But generally speaking, enstrophy is directly related to the amount of kinetic energy in the flow. So it is natural that as kinetic energy is dissipated, it has to change into internal energy. There must be a conservation expression that contains internal energy and enstrophy in some form. $\endgroup$
    – tpg2114
    Commented Feb 5, 2015 at 23:30
  • 1
    $\begingroup$ Hi @Kimusubi I wrote your equations using the MathJax formatting we encourage here. You should check to make sure they're still right (I couldn't tell if that was a nu or a v, and I'm not sure on the subscript on $\Phi$). If you want a more thorough guide to latex-style MathJax, see here. $\endgroup$
    – user10851
    Commented Feb 5, 2015 at 23:39
  • $\begingroup$ More on enstrophy: physics.stackexchange.com/search?q=is%3Aq+enstrophy $\endgroup$
    – Qmechanic
    Commented Feb 5, 2015 at 23:42

2 Answers 2


I'd say part of the answer must be that whatever dynamic variable you use, like Enstrophy, Vorticity, their potential analogues, etc. those are always 'filtered' fields.

Filtered in the sense, that you start with the velocity field $\vec v = \sum u_i \vec e_i$ that has full information over the dynamics and then apply some operators (integration and differentiation mostly) on top of that to generate your dynamic variable of interest.

Usually information is lost through that process. Sometimes, you can reconstruct $\vec v$ from the vorticity $\vec \omega$ in the incompressible fluid-case, as an example.

However my point here is, that the dissipation of those constructed variables, is always in the end the expression of dissipation of linear momentum, and therefore generation of heat, just filtered through the construction operator.

  • $\begingroup$ A very interesting viewpoint. Would it be possible to elaborate a bit more on how the filter works? $\endgroup$
    – Amey Joshi
    Commented Feb 6, 2015 at 4:11
  • $\begingroup$ I think the simplest argument for that is that, if you take a function, put a derivative on it and want to get back over the integral, then you get an integration constant. So in a sense you could say you have lost information where the zero-level was after the derivative. Thats why there is not necessarily a 1:1 correspondence for $\vec v$-field to vorticity $\vec \omega$, or Enstrophy $\Omega$. You loose ofc even more information with $\Omega$, as it's a scalar. $\endgroup$ Commented Feb 6, 2015 at 5:37
  • $\begingroup$ @AtmosphericPrisonEscape - I really like your filter explanation. That's pretty much what's happening, but I was hoping for a more rigorous explanation. I'm currently trying to recast the energy equations in different forms in the hopes of explaining the equivalence between the KE dissipation and enstrophy dissipation. I made a small edit to my original post that shows the dissipation terms in different forms that helps clarify the mechanisms behind them. $\endgroup$
    – Kimusubi
    Commented Feb 6, 2015 at 5:55
  • $\begingroup$ @Kimusubi: But then from here on it's easy isn't it? There are popular forms of the energy equation containing the momentum flux tensor explicitly, w/o newtonian modelling. Should be something like $\partial_t v_i^2 /2 = v_i \partial_j \Pi_{ij}$ $\endgroup$ Commented Feb 6, 2015 at 6:16

Wondering if the above contributions are making things rather too complicated.

  1. The notion of filtering is relevant to any computation or measurement, but not to the basic equations (unless I mistake your meaning, in which case, please do explain)
  2. One basic definition of vorticity states it is a measure of local solid-body motion of the fluid. Thus, destruction of enstrophy should relate to a cessation of the relative motion associated with local solid-body rotation. Though he does not mention enstrophy by name, BR Morton ("The generation and decay of vorticity" Geophys. Asotrphys. Fluid Dynamics, 1984, vol. 28, 277-308) states plainly that "the only means of decay or loss of vorticity is by cross-diffusion and annihilation of vorticity of opposite signs." Since enstrophy is a measure of the intensity of that local rotation rate, we might say that enstrophy destruction arises from this mechanism.

  3. So, where does the "destroyed" enstrophy go (or better(?), transform into)? The question presumes enstrophy to be a conserved quantity (like energy or mass - but NOT momentum). The enstrophy equation itself belies this idea: if enstrophy were conserved, we could simply write d(enstrophy)/dt = 0.

Perhaps I'm oversimplifying. But a return to basic definitions is a good place to start. Would be grateful for feedback on this!


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