# Continuity equation for charge and current densities of an accelerated point charge

For a point charge that moves with the trajectory $\vec r(t)$, we know that it has the singular charge and current densities

$$\rho (\vec x, t) = q \delta^3(\vec x - \vec r(t))$$ $$\vec J(\vec x, t) = q \frac{d\vec r}{dt} \delta ^3 (\vec x - \vec r (t))$$

How do you prove that these densities satisfy the continuity equation?

$$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0$$

I considered using the property of the dirac delta

$$\delta (f(x)) = \sum_i^N \frac{\delta (x - x_i)}{|\frac{df}{dx}|_{x = x_i}}$$

where $x_i$s are the roots of $f(x)$. In the problem at hand, we see that the dummy variable x is t, and $f(x) = \vec x - \vec r(t)$ so that the only root is $t_0$ which makes $\vec r(t_0) = \vec x$.

However I just don't know how to apply the derivative to the dirac delta. I know that

$$\delta '[f] = f'(0)$$

Any help please? Thank you very much!

• The continuity equation is just the time derivative of a density plus the divergence of a flux equal to sources and sinks. In the absence of sources or loss terms, the right-hand side should be zero. Since a flux is just a density multiplied by a velocity (from dimensional analysis), you can see that $\mathbf{J}$ is the flux related to $\rho$. By the way, the easiest way to get rid of a delta function is to integrate. – honeste_vivere Apr 24 '15 at 12:13

We have, for a point charge $q$ at position $\vec r(t)$: $$\rho(\vec x, t) = q\delta^3(\vec x - \vec r(t))$$ $$\vec J(\vec x, t) = q \frac{d\vec r}{dt}\delta^3(\vec x - \vec r(t))$$ Let us for now work without worrying about what the derivative (more precisely, gradient) of the delta function actually is. We will also enforce the convention that $\vec\nabla$ denotes differentiation with respect to $\vec x$ unless specified otherwise.
We will first evaluate the rate of change of the charge density: $$\frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t}\left[q\delta^3(\vec x - \vec r(t))\right]$$ $$= q\frac{d\vec{r}}{dt}\cdot\vec\nabla_r\delta^3(\vec x - \vec r(t))$$ as all the time dependence is contained (through $\vec r(t)$) in the delta function. Now, for $\delta(y-z)$, we have the result that $$\frac{\partial}{\partial y} \delta(y-z) = -\frac{\partial}{\partial z} \delta(y-z)$$ We can use this here to get: $$\frac{\partial \rho}{\partial t} = -q\frac{d\vec{r}}{dt}\cdot\vec\nabla\delta^3(\vec x - \vec r(t))$$
The divergence of the current density is: $$\vec\nabla\cdot\vec J = \vec\nabla\cdot\left[q\frac{d\vec{r}}{dt}\delta^3(\vec x - \vec r(t))\right]$$ $$= q\frac{d\vec{r}}{dt}\cdot\vec\nabla\delta^3(\vec x - \vec r(t))$$ where the field-like dependence on position is once again contained only in the delta function.
Now, whenever $\vec x \ne \vec r(t)$, we know that any derivative of the delta function (which is uniformly zero in this region) must be zero and the continuity equation trivially holds for this case. However, at $\vec x = \vec r(t)$, it still holds because of the coefficients of the derivative of the delta function in the above expressions (Note: we can treat the delta function as the limiting case of a well-behaved, but sharply peaking function, and it is this consideration that underlies this argument).
Therefore, the equation of continuity holds for all $\vec x$ (and $t$): $$\frac{\partial \rho}{\partial t} + \vec\nabla\cdot\vec J= 0$$
• Thank you so much! How do I prove that $\frac{\partial}{\partial y} \delta (y-z) = -\frac{\partial}{\partial z} \delta (y-z)$ – quarkleptonboson Apr 24 '15 at 18:11