According to Noether's theorem, every continuous symmetry (of the action) yields a conservation law.

In fluid, there is a local particle number conservation law, which is $$\partial{\rho}/\partial{t}+\nabla \cdot \vec{j} ~=~0,$$ where $\rho$ and $\vec{j}$ is the density and current respectively. I just wonder is there any symmetry associated with this conservation law?

  • $\begingroup$ I think your conservation law is backwards, it should be $\partial_t\rho+\nabla\cdot\vec{j}=0$. $\endgroup$
    – Kyle Kanos
    Dec 13, 2013 at 17:03
  • $\begingroup$ Thank you. You mean the particle number conservation is the symmetry? $\endgroup$
    – Blue
    Dec 13, 2013 at 17:06
  • $\begingroup$ @KyleKanos If so, is it possible to write it down mathematically in the language of field theory? $\endgroup$
    – Blue
    Dec 13, 2013 at 17:09
  • $\begingroup$ No, gauge transformations are the symmetry: $\mathbf{A}'=\mathbf{A}+\nabla\lambda$ and $\varphi'=\varphi+\partial_t\lambda$ $\endgroup$
    – Kyle Kanos
    Dec 13, 2013 at 17:21
  • $\begingroup$ @KyleKanos Suppose there is no electromagnetic field, the field is not coupled to $\vec{A}$. In this situation, gauge transformation does not make sense, I think. $\endgroup$
    – Blue
    Dec 13, 2013 at 18:14

2 Answers 2


Noether's theorem in its usual form assumes that the system (in this case a fluid) is governed by an action principle. We assume for simplicity that the fluid consists of just one type of fluid particles.

I) In the Lagrangian fluid picture, the (local) conservation of fluid particles is manifest from the onset, since the dynamical variables are the labels ${\bf a}$ of the fluid particles.

We will assume that the labels are chosen such that the mass density in label ${\bf a}$-space (as opposed to position ${\bf r}$-space) is a constant. Then particle conservation is the same as mass conservation

$$\tag{1} \frac{D\rho }{Dt} +\rho {\bf \nabla} \cdot {\bf u} ~\equiv~ \frac{\partial \rho }{\partial t} + {\bf \nabla} \cdot (\rho {\bf u})~=~ 0.$$

II) In the Eulerian fluid picture, the mass density $\rho$ is a dynamical field. The mass conservation (1) is imposed by the Euler-Lagrange equation for the unpaired variable $\phi$ in the Clebsch velocity potential

$$\tag{2} {\bf u}~=~{\bf \nabla}\phi +\ldots. $$

The corresponding global symmetry is $\phi \to \phi+ \text{const}.$


  1. R. Salmon, Hamiltonian Fluid Mechanics, Ann. Rev Fluid. Mech. (1988) 225. The pdf file can be downloaded from the author's webpage.

I don't think that you should think of particle conservation as a conservation law in the context of classical physics. As Qmechanic says in a classical system, the particle number is the number of degrees of freedom and is fixed as a definition of the problem. Once we have decided how many particles will be there we can write a Lagrangian, investigate its symmetries and find conserved charges.

In the case of fluid dynamics there is an infinite number of particles. The density which takes the place of the number of particles is a thermodynamic quantity. It's dynamics (and associated conservation laws) is then fixed by the choice of the equation of state. We could imagine a fluid composed of particles that disintegrate when they come in close contact. Then raising the density would cause more disintegration which would in turn lower the density again. The (equilibrium) equation of state would forbid high densities and the continuity equation would have an additional loss term.

Hydrodynamics is basically a list of conservation laws. It however does not know where these come from. In order to write hydrodynamic flow equations we assume that there is an underlying thermodynamic system of particles and that it is described by some Hamiltonian. Every fluid element is however in thermal equilibrium such that there are infinitely many degrees of freedom that are not described by the hydrodynamic equations and can absorb (or emmit) energy, momentum and (if you cook up an appropriate example) particles. That is the origin of viscosity and the reason why we need an additional equation of state to close the hydrodynamic equations. Then we know that the energy and momentum are conserved (because there is an underlying Hamiltonian and we control the forces applied to the system externally) so that we can simply write the conservation laws in terms of the thermodynamics quantities: density, local average velocity, etc. If these quantities are not conserved it is necessary to take this into account by hand by adding proper sources and sinks such as external forces, a rising local entropy density, particle disintegration rates, etc.

The situation is however different in quantum field theory. There particle may be converted to energy and back and we consider systems where the particle number is not fixed. In this case there is a symmetry that generates particle conservation: overall phase rotations. In quantum physics the degrees of freedom are complex numbers. Their overall phase however is not physically observable and does not affect the dynamics. If all the objects of the theory are multiplied by the same complex number of modulus equal to one, the Lagrangian is unaffected. We have a symmetry and the corresponding conservation law is particle conservation (or charge conservation if the particles are charged).

Note that I have not read 'R. Salmon, Hamiltonian Fluid Mechanics, Ann. Rev Fluid. Mech. (1988) 225'. (See the answer of Qmechanic.) I don't know if (and how) it is possible to write a Hamiltonian for fluid dynamics nor if it yields conservation laws. However, even if this is possible, I prefer to think of hydrodynamics as a phenomenological theory which is "guessed" by writing down conservation laws.


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