Before going any further, I should emphasize that I know we cannot use the action principle for locally dissipative systems or even Noether's theorem for that matter. There are plenty of stackexchange articles discussing the subject, but I precisely want to understand where it is that the theory breaks down in the mathematics and what happens to the Noether' current when it does not satisfy the continuity equation.

Let $(\mathcal{M},\eta)$ be a 4d Minkowski spacetime and with a boundary $\partial\mathcal{M}$. Consider a Lagrangian density of the usual form $\mathcal{L}(\phi_i,\partial_a\phi_i)$ in a flat background. The action of the Lagrangian on the spacetime manifold is then given by $$\mathcal{S}[\phi] = \int_{\mathcal{M}}\mathcal{L}(\phi_i,\partial_a\phi_i)\text{d}^4x.$$ The variation $\delta \mathcal{S}$ gives rise to the following integral $$\delta\mathcal{S}[\phi]=\delta\int_{\mathcal{M}}\mathcal{L}\text{d}^4x=\int_{\mathcal{M}}\left\{\left[\frac{\partial\mathcal{L}}{\partial\phi_i}-\partial_a\left(\frac{\partial\mathcal{L}}{\partial(\partial_a\phi_i)}\right)\right]\delta\phi_i + \partial_a\left[\frac{\partial\mathcal{L}}{\partial(\partial_a\phi_i)}\delta\phi_i\right]\right\}\text{d}^4x,$$ here we have defined the deformations field $\delta\phi_i$ according to joshphysics' excellent reply to these (1) and (2) stackexchange posts. With regards to Qmechanic's notation in the same articles, we are considering both a vertical and horizontal deformations for complete generality.

For conservative systems, Noether's theorem states that every differentiable symmetry which leaves the action invariant gives rise to a conserved quantity. In such systems, one often considers the Lagrangian changing by a total derivative, therefore we may write $$\delta\int_{\mathcal{M}}\mathcal{L}\text{d}^4x = \int_{\mathcal{M}}\partial_a f^a\text{d}^4x,$$

and by a simple rearrangement, we would find that

\begin{align} \int_{\mathcal{M}}{\partial_a \left(f^a - \frac{\partial\mathcal{L}}{\partial(\partial_a\phi_i)}\delta\phi_i\right)}\text{d}^4x &= \int_{\mathcal{M}}\left\{\left[\frac{\partial\mathcal{L}}{\partial\phi_i}-\partial_a\left(\frac{\partial\mathcal{L}}{\partial(\partial_a\phi_i)}\right)\right]\delta\phi_i \right\}\text{d}^4x.\tag{1} \end{align}

The term on the left hand side is usually called the Noether current, $j^a$, which satisfies the continuity equation $$\partial_aj^a = 0$$ when the Euler-Lagrange equations are satisfied.

Now that the framework is mostly set up, I have a few questions. Suppose the system was dissipative locally but not globally. By this I mean the following. We define an open thermodynamic system as an open 4d region $\Omega\subset\mathcal{M}$ with boundary $\partial\Omega$. Note, $\Omega$ is not compact. Consider placing a patch of heat within the interior of $\Omega$. Though the flux of heat through $\partial\Omega$ is non-zero, it is zero on $\partial\mathcal{M}$.

Question 1: What is the connection between the boundary and the bulk with regards to Noether's theorem? Does Noether's theorem hold with respect to the boundary on $\mathcal{M}$ even if it does not hold locally in the bulk of $\mathcal{M}$? (My thinking here is that the energy is always fixed globally even if it changes locally).

Question 2: Since the system is dissipative with respect to $\Omega$, is it possible to prove that the Lagrangian does not change by a total derivative?

$$\delta\int_{\mathcal{\Omega}}\mathcal{L}\text{d}^4x \ne \int_{\mathcal{\Omega}}\partial_a f^a\text{d}^4x,$$

Question 3: In the case that it does not change by a total derivative, what is the physical interpretation of the divergence of the vector field given by

\begin{align} {X^a = \left(\frac{\partial\mathcal{L}}{\partial(\partial_a\phi_i)}\delta\phi_i\right)}? \end{align}

Question 4: Lastly, does $\partial_a X^a \ne 0$ everywhere in $\mathcal{M}$ (not on the boundary $\partial\mathcal{M}$) or just across $\Omega$.

  • $\begingroup$ Comment to the post (v3): OP already knows that (i) Noether's theorem assumes a stationary action principle, and (ii) dissipative systems don't have a stationary action principle. Nevetheless, in the paragraph just before the four questions, OP seems to inconsistently assume the existence of a dissipative action formulation in the $\Omega$ region. $\endgroup$
    – Qmechanic
    Sep 26, 2018 at 21:52
  • $\begingroup$ @Qmechanic, I dispute your (ii) as the paper I cited is a self-consistent (and in agreement with the canonical examples of dissipative systems) framework of an action principle for dissipative systems. Thus, my original assumption is consistent and it is possible to answer my original questions in the $\Omega$ region. $\endgroup$
    – Dhuality
    Dec 9, 2018 at 22:51

1 Answer 1


After a few months of searching and thinking I believe I have an answer. I recently came across a paper by Galley, Tsang, and Stein who are tackling the issue of formulating an action principle for nonconservative systems. They start with a variational calculus formulation which is consistent with the initial value formalism and thereby allowing for dissipative effects. Using their results, it is possible to answer my original questions.

(1) With regards to Noether's theorem, when dissipative effects are taken into account the usual notion of conserved quantities break down. Therefore, Noether's theorem does not hold on the boundary or locally. Instead one finds an equation of motion for the Noether current $X^a$ such that $$\text{div}\cdot X = \left[\partial_a\left(\frac{\partial{\mathcal{L}}}{\partial{(\partial_a\phi_i)}}\right)-\frac{\partial{\mathcal{L}}}{{\partial\phi_i}}\right]\delta\phi_i.$$ The degrees of freedom, which escape from the bulk to the boundary, affect the motion by removing the energy necessary for motion. Thus Noether's theorem will not hold either in the interior or the boundary.

(2) If the system is dissipative , in $\Omega$, it is possible to prove that the Lagrangian does not change by a total derivative. This follows from the fact that any continuous symmetry of the action of a non-conservative system cannot be written as a total derivative. If it could be written as a total derivative, then one could in principle integrate out the total derivative and, with appropriate boundary conditions, eliminate the total derivative thus leaving the equations of motion unchanged. However, for dissipative systems, the equations of motion do not follow from the ordinary variational principle. These two facts lead to a contradiction. If a system is dissipative, then the Lagrangian does not change by a total derivative.

(3) This is a more difficult question to straightforwardly answer. If the divergence of a vector field is given, then given the appropriate volume form, one has $$(\text{div}\cdot X)\ \mu = (\mathcal{L}_X \mu)$$ where $\mathcal{L}_X$ denotes the Lie derivative of the volume form $\mu$ along the Noether current. In dynamical systems, one imposes $\mathcal{L}_X \mu =0$. This is known as a measure preserving dynamical system and is associated with conservative systems. If the measure is not preserved, this is an indication of a dissipative system. The answer to my question is that the non-zero divergence of the Noether current is directly responsible for the transfer of the degrees of freedom from the bulk to the boundary.

(4) The degrees of freedom travel across the boundary of $\Omega$ and flow towards the boundary. Therefore $\partial_a X^a \ne 0$ across $\Omega$ and decays at the spacetime infinities, i.e., on the boundary of $\mathcal{M}$.


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