# Conservation laws and continuity equations

I'm a bit messed up with conservation laws and continuity equations.

This is my understanding:

• A conservation law describes that a physical quantity $G$ is conserved with time. It does not prevent "quantity teleportation", as long as at a given time, the created quantity and the disappeared quantity are equal. In Wikipedia's wordings: "For example, it is true that "the total energy in the universe is conserved". But this statement does not immediately rule out the possibility that energy could disappear from Earth while simultaneously appearing in another galaxy."

• A continuity equation is stronger: It implies that there is no "quantity teleportation".

• A global conservation law describes that globally, a physical quantity $G$ is conserved with time: $$\dfrac{\mathrm{d}G}{\mathrm{d}t}=0$$ Using mathematics, this can be written as the sum of an integral over a volume and an integral over a surface, or as a single integral over a volume using Stokes theorem (and introducing a divergence)

• A local conservation law is the result of writing that the integrand of the global law are equal.

Questions:

1. Is "quantity transportation" possible in a local conservation law? If not is there any difference between the two?
2. In the equation (i.e. mathematically), where do you see the differences between continuity equations and conservation laws?
• 1. Do you know about flux? And how it comes into these kind of equations? 2. I would say, focus on the physical meaning of the terms in the equations, rather than the equation itself. Commented Sep 5, 2014 at 18:50
• @Bernhard: 1. Yes. I thought, but that might be where I'm missing something about continuity. 2. When I do so, I don't see how a local conservation law could include "quantity teleportation". Commented Sep 5, 2014 at 18:54
• Is it correct that writing the local conservation law requires continuity? Continuity would make Stokes theorem applicable, and transform the flux density with a divergence. Commented Sep 5, 2014 at 18:57
• I don't really understand what you mean with "quantity teleportation". Also, do you have a specific field in mind? Commented Sep 5, 2014 at 20:05
• @Bernhard By "quantity teleportation" I mean the following: Consider the case when the quantity (momentum, thermal energy, mass, etc.) we're looking at $G$ in a given domain at $t$ and at $t+\mathrm{d}t$ is the same: $G$ is conserved ($\dot G(t)=0$). Yet, it could be that some quantity $G_0$ vanished instantaneously in a small area, and appeared instantaneously in another area : $G$ would still be conserved: $G(t+\mathrm{d}t)=G(t)+G_0-G_0=G(t)$. Now, if continuity (of the density of $G$?) is imposed, this is not possible. Commented Sep 5, 2014 at 20:46

Continuity equations are an embodiment of local conservation laws, and they both reflect the fact that there is no 'quantity teleportation'. That said, the local transport of a quantity is perfectly possible within local conservation laws and it is precisely this that the continuity equation models.

Your distinction between global and local conservation laws could use some refinement, though. Consider a quantity $G$ whose local density is $g(\mathbf r)$, and whose flow density (i.e. flux) is $\mathbf j(\mathbf r)$. With this notation, a global conservation law establishes only that the total amount of $G$, i.e. $$G=\int g(\mathbf r)\text d\mathbf r,$$ where the integral is over all space, is constant over time.

Saying that $G$ additionally obeys a local conservation law is a stronger statement, and it is exactly the statement that $g$ and $\mathbf j$ obey the continuity equation. This one comes in two flavours:

• differential, $$\frac{\partial g}{\partial t}+\nabla\cdot\mathbf j=0,$$

• and integral, $$\frac{d}{dt}\int_Vg(\mathbf r)\text d\mathbf r +\bigcirc \!\!\!\!\!\!\!\!\!\iint_{\partial V}\mathbf j(\mathbf r)\cdot \text d\mathbf a.$$

It is important to note that both of these forms are completely equivalent (modulo technical assumptions on point, line and surface charges). The differential form holds at every point $\mathbf r$, and the integral form holds for every volume $V$, and one can use appropriate calculations to translate between both forms and therefore between both freedoms.

The reason that we say continuity equations embody local conservation laws is that they make precise the intuition that all the $g$ that "comes out" of some region can be "seen crossing the boundary", which is measured by the surface integral / the divergence term.

This is as opposed to, for example, a quantity with a density of the form $$g(\mathbf r,t)=g_0 \cos^2(\omega t)e^{-(\mathbf r-\mathbf r_1)^2/\sigma^2} + g_0 \sin^2(\omega t)e^{-(\mathbf r-\mathbf r_2)^2/\sigma^2}$$ where $\mathbf r_1$ and $\mathbf r_2$ are in principle far apart. Here $G$ stays constant, but between $t=0$ and $\pi/2\omega$, all of the $G$ near $\mathbf r_1$ has disappeared without there being any flux through the plane between it and $\mathbf r_2$. Here $G$ obeys a global conservation law, but not a local one.

For clarity, I should note that your statement that "A local conservation law is the result of writing that the integrand of the global law are equal" is incorrect, and depending on exactly what you mean by it, there may be exceptionally few systems that obey that.

• That's what I was asking for, TY. Two remarks though: 1) when you say "the integral form holds for every volume", do you mean for any volume with any velocity, or for any volume attached to matter (material velocity field)? 2) Answer to your last paragraph: I meant $\forall\mathcal{D},\ \int_\mathcal{D}f=0\Longrightarrow f=0$. I now understand that in order to apply this, all the terms must be gathered in an integral over a volume (integrated on $\mathcal{D}$, which requires continuity. Still have to assimilate all your answer though. Commented Sep 5, 2014 at 21:05
• Your definition is not that useful unless you specify what D and f are. The volumes in the integral continuity equation are fixed in space (i.e. in an inertial frame) and do not move, either with the fluid or on their own. Commented Sep 6, 2014 at 8:23
• "all of the G near r1 has disappeared without there being any flux through the plane between it and r2." Can you explain why? Assuming $\partial_t g + \nabla \cdot \mathbf j = 0$ and some boundary conditions on $\mathbf j$, appropriate $\mathbf j$ might be found. Commented Sep 6, 2014 at 8:43
• @Ján yes, that's true, particularly if $g$ is a signed measure. Nevertheless, the lack of $g$ anywhere in between does rule out many physical situations. Commented Sep 8, 2014 at 9:37
• I do not follow. Why is sign of any importance? What do you mean by lack of g in between? Commented Sep 8, 2014 at 18:42

In the equation (i.e. mathematically), where do you see the differences between continuity equations and conservation laws?

The continuity equation is not sufficient to derive conservation of something. For example, continuity equation for fluid flow in non-relativistic theory is

$$\partial_t \rho + \nabla \cdot (\rho \mathbf v) = 0$$

wherer $\rho$ is density of fluid and $\mathbf v$ is its velocity. Integrating this equation over some region $V$ with boundary surface $\Sigma$, switching the order of integration and differentiation and using the Gauss-Ostrogradskii theorem we obtain $$\frac{d}{dt}\int_V \rho \,dV = -\oint_\Sigma \rho \mathbf v \cdot d\boldsymbol{\Sigma}.$$ As the region $V$ is expanded to contain all space in a limit, the left-hand side gives rate of range of mass in the whole space. This is not necessarily zero, for the right hand side may be generally have non-zero limit.

To get conservation, we have to impose additional condition that there is a surface $\Sigma$ such that value of the surface integral is zero or at least goes to zero as the surface is expanded to infinity. This may not be always possible.

For example, the Poynting theorem for current-free region

$$\partial_t \left(\frac{1}{2}\epsilon_0E^2 + \frac{1}{2\mu_0}B^2 \right) + \nabla \cdot (\mathbf E\times \mathbf B/\mu_0) = 0$$ has the same form as the above equation of continuity, but the integral version $$\frac{d}{dt}\int_V \left(\frac{1}{2}\epsilon_0E^2 + \frac{1}{2\mu_0}B^2 \right)\,dV = -\oint_\Sigma \mathbf E\times \mathbf B/\mu_0 \cdot d\boldsymbol \Sigma$$ may have nonvanishing surface integral on the right-hand side, if there are EM waves present at infinity. From this it follows that the rate on the left-hand side is not zero and there is no conservation of value of the integral there (Poynting energy).

So continuity does not automatically mean conservation.

Similarly, conservation does not imply continuity - mass or energy can jump from place to place suddenly while the net value remains the same.

• "So continuity does not automatically mean conservation." It seems to me you're talking about global conservation, because you take the limit of $\Sigma$ to infinity. Do you agree with "So continuity does not automatically mean global conservation" but "continuity does automatically mean local conservation"? Helpful answer also by the way, thank you very much. Commented Sep 5, 2014 at 21:23
• I think "continuity does automatically mean local conservation" is wrong, sorry. What is true (see Emilio's answer) is that global conservation + continuity implies local conservation. Additionally, you showed that continuity does not imply global conservation and that global conservation does not imply continuity. Agree? Commented Sep 5, 2014 at 21:28
• "Do you agree with <So continuity does not automatically mean global conservation> but <continuity does automatically mean local conservation>?" Yes, but I think it is just a definition - local conservation seems to be defined by the validity of continuity equation. Commented Sep 6, 2014 at 8:45
• for context ,that poyenting theorem ,means energy transferd is connected to how much momentum flows one surface to another.left hand side is energy term and right hand side is telling me the momentum of the electromagmetic field.this is also called a conservation ,does not? Commented Mar 26, 2023 at 23:45
• The Poynting theorem is interpreted in terms of energy, not momentum. For momentum, there is another similar set of equations, involving the Maxwell stress tensor. Commented Mar 26, 2023 at 23:59