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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!

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  • $\begingroup$ 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. $\endgroup$ – honeste_vivere Apr 24 '15 at 12:13
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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 $t$): $$\frac{\partial \rho}{\partial t} + \vec\nabla\cdot\vec J= 0$$

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  • $\begingroup$ Thank you so much! How do I prove that $ \frac{\partial}{\partial y} \delta (y-z) = -\frac{\partial}{\partial z} \delta (y-z)$ $\endgroup$ – quarkleptonboson Apr 24 '15 at 18:11
  • $\begingroup$ The delta function (treated as a limit of a suitable peaking function) is symmetric about zero, and a small displacement in any direction makes the delta function go from the peak to zero. Thus, the "incremental change" in the delta function for a small change in the argument, is of the same magnitude for a given change, and is always of a fixed sign (i.e. negative). So the sign of the derivative changes with the sign of the increment in the argument. While on the negative side of the argument, the function has positive slope, it has negative slope on the positive side. $\endgroup$ – AV23 Apr 25 '15 at 15:02

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