Intuition for continuity equation

Question

Suppose, for example, we have an equation,

$$\frac{\partial \rho}{\partial t}=-\nabla \cdot\vec J$$

which we can apply to show conservation of electric charge. How can we interpret this intuitively?

Answer 1

The intuition is rather straightforward with the use of vector calculus theorems. Suppose we have a conserved current $j^\mu = (\rho, \vec j)$ with the continuity equation,

$$\partial_\mu j^\mu = \frac{\partial \rho}{\partial t} + \nabla \cdot \vec j = 0.$$

Consider now defining a conserved charge,

$$Q = \int_{\mathbb R^3} d^3x \, j^0 = \int_{\mathbb R^3} d^3x \, \rho.$$

If it is conserved, then we expect its time derivative to vanish. Let us verify this by taking the time derivative and using the continuity equation:

$$\frac{\partial Q}{\partial t} = \int_{\mathbb R^3} d^3x \, \frac{\partial \rho}{\partial t} = - \int_{\mathbb R^3} d^3x \, \nabla \cdot \vec j = 0.$$

This is providing that $\vec j$ vanishes sufficiently quickly as $|\vec x| \to 0$. The continuity equation implies a strong statement, namely that $Q$ can only change if there is some flow, since,

$$\frac{\partial Q_V}{\partial t} = \int_V d^3x \, \frac{\partial \rho}{\partial t} = -\int_V d^3x \, \nabla \cdot \vec j = - \int_S \vec j \cdot dS$$

using the divergence theorem to relate the surface (flux) integral to the integral of the divergence of the vector over the volume bounded.

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Response to comment

If $\nabla \cdot \vec j > 0$, it implies that the flux at a point is outward, since from the divergence theorem, we can interpret $\nabla \cdot \vec j$ as the volume density of flux, for an infinitesimal volume around a point.

The fact that $\dot \rho$ and $\nabla \cdot \vec j$ have opposite signs in the continuity equation is to ensure that if $Q$ decreases, then the flux is indeed negative indicating charge escaping the volume, rather than entering it.

Answer 2

The first thing we need to note about this equation in order to gain intuition (in my opinion of course) is that it is a differential equation. We can rewrite the flux density to show this: $$\vec{J}=\rho \vec{v}$$ where $v$ is the mean velocity of the charge carriers (e.g. electrons).

Therefore the equation becomes: $$\frac{\partial\rho}{\partial t}=-\vec{\nabla}\cdot (\rho\vec{v}) $$

Writing the equation this way, we see that the solution is the function: $\rho(\vec{r})$.
In other words, the equation describes the physical constrains on the charge distribution (as a function of $\vec{r}$.)

The second thing we need to note, is that the equation is not an ODE, but a PDE: there is a relation between the time derivative and the space derivative. In other words, in order to change the charge distribution in time (LHS), you must create a current: changing the charge distribution is only possible by displacing charges, at a certain rate, which is exactly the definition of current. The current (or flux density) is described by the RHS.

In conclusion, the equation is a differential equation, i.e. we are looking for the a function that describes the charge distribution while satisfying the following physical constraints: a change in charge density must result in a charge flux.

Answer 3

The equation basically says charge does not get destroyed or created. To show this, consider a field of charges which move only in the x-direction with speed $u(x)$. Now consider a small cube with dimensions $\Delta x$, $\Delta y$ and $\Delta z$ in the x-, y- and z directions. A cross section is shown in the diagram below. Every second a volume of $u(x)\cdot \Delta y\Delta z$ passes through the left boundary (as shown in $\color{green}{\text{green}}$); since that volume travels a distance of $u(x)$ every second it will be just enough to move completely inside the boundary. This means every second an amount of charge equal to $u(x)\rho(x)\ \Delta y\Delta z$ enters the cube through the left boundary. Speed times charge density is the current so this becomes $J(x) \Delta y\Delta z$.

At the same time an amount of $J(x+\Delta x)\ \Delta y\Delta z$ is leaving the cube at the right boundary (as shown in $\color{red}{\text{red}}$). So the change in charge per second of the cube becomes $\frac{\partial q}{\partial t}=\Delta y\Delta z\ (J(x)-J(x+\Delta x))$. If we take the cube infinitesimally small we get $$J(x)-J(x+\Delta x)=-\frac{J(x+\Delta x)-J(x)}{\Delta x}\cdot \Delta x\rightarrow-\frac{\partial J}{\partial x}\Delta x$$ Plugging that in and using that $q=\rho \Delta x\Delta y\Delta z$: \begin{align}\frac{\partial(\rho \Delta x\Delta y\Delta z)}{\partial t}&=-\Delta x\Delta y\Delta z\ \frac{\partial J}{\partial x} \\\frac{\partial\rho}{\partial t}&=-\frac{\partial J}{\partial x}\end{align} By now allowing the other directions to have speed, it can be proven using similar reasoning that $$\frac{\partial\rho}{\partial t}=-\nabla \cdot\vec J$$

This diagram now gives intuition for the divergence: the divergence in this case is positive, since $\frac{\partial J}{\partial x} > 0$. So a positive divergence means charge is leaving the cube, which means locally the charge density is descreasing.