Conservation of charge with Dirac delta

[Frostman]

Homework Statement

Demonstrate conservation of charge

Relevant Equations

Dirac's property

Hello, I was reviewing a part related to electromagnetism in which the charge and current densities are defined by the Dirac delta:

$\rho(\underline{x}, t)=\sum_n e_n \delta^3(\underline{x} - \underline{x}_n(t))$
$\underline{J}(\underline{x}, t)=\sum_n e_n \delta^3(\underline{x} - \underline{x}_n(t))\frac{d\underline{x}_n}{dt}$

At this point, when I want to evaluate the current density divergence, I find a $3$ coefficient that shouldn't appear when I apply a Dirac delta property. I show you here:

$\nabla \cdot \underline{J}(\underline{x}, t)=\sum_n e_n \frac{\partial}{\partial x^i}\delta^3(\underline{x} - \underline{x}_n(t))\frac{dx_n^i}{dt}$

I apply the property for which

$\frac{\partial}{\partial x}\delta(x-y(t))\frac{dy}{dt}=-\frac{\partial}{\partial t}\delta(x-y(t))$

In the three-dimensional case, shouldn't I work like that?

$\frac{\partial}{\partial x^i}\delta^3(\underline{x} - \underline{x}_n(t))\frac{dx_n^i}{dt}=$
$\frac{\partial }{\partial x}\delta(x-x_n(t))\delta(y-y_n(t))\delta(z-z_n(t))\frac{dx_n}{dt}+ \delta(x-x_n(t))\frac{\partial }{\partial y}\delta(y-y_n(t))\delta(z-z_n(t))\frac{dy_n}{dt}+ \delta(x-x_n(t))\delta(y-y_n(t))\frac{\partial }{\partial z}\delta(z-z_n(t))\frac{dz_n}{dt}$

Applying the above property on each of the three addends should produce the same result three times, so:

$-\frac{\partial }{\partial t}\delta(x-x_n(t))\delta(y-y_n(t))\delta(z-z_n(t))- \delta(x-x_n(t))\frac{\partial }{\partial t}\delta(y-y_n(t))\delta(z-z_n(t))- \delta(x-x_n(t))\delta(y-y_n(t))\frac{\partial }{\partial t}\delta(z-z_n(t))$

Which is therefore equal to:

$-3\frac{\partial }{\partial t}\delta(x-x_n(t))\delta(y-y_n(t))\delta(z-z_n(t))=-3\frac{\partial }{\partial t}\delta^3(\underline{x} - \underline{x}_n(t))$

So when putting it into the current density divergence, I would have a $3$ factor:

$\nabla \cdot \underline{J}(\underline{x}, t)=-3\sum_n e_n \frac{\partial}{\partial t}\delta^3(\underline{x} - \underline{x}_n(t))=-3\frac{\partial \rho(\underline{x}, t)}{\partial t}$

Wrong, I would have:

$\nabla \cdot \underline{J}(\underline{x}, t) + 3\frac{\partial \rho(\underline{x}, t)}{\partial t}=0$

I can't find the point where I'm wrong, can you tell me?

 

[anuttarasammyak]

Homework Statement:: Demonstrate conservation of charge
Relevant Equations:: Dirac's property

Which is therefore equal to:

−3∂∂tδ(x−xn(t))δ(y−yn(t))δ(z−zn(t))=−3∂∂tδ3(x―−x―n(t))

So when putting it into the current density divergence, I would have a 3 factor:

$$-[\frac{\partial}{\partial t}\delta(x-x_n(t))]\delta(y-y_n(t))\delta(z-z_n(t))+similar\ for \ y\ and \ z$$
$$=-\frac{\partial}{\partial t}[\delta(x-x_n(t))\delta(y-y_n(t))\delta(z-z_n(t))]$$

 

[stevendaryl]

You don't get a factor of 3. Since $J = \sum_n \text{ something} \dfrac{d \vec{x}_n}{dt}$, then

$\nabla \cdot J = \sum_n \dfrac{\partial \text{ something}}{\partial x} \dfrac{d x_n}{dt}$
$+ \sum_n \dfrac{\partial \text{ something}}{\partial y} \dfrac{d y_n}{dt}$
$+ \sum_n \dfrac{\partial \text{ something}}{\partial z} \dfrac{d z_n}{dt}$

Those are three different terms, not 3 copies of the same term. There is no factor of 3.

 

[Frostman]

So that factor $3$ really isn't there, because there's $\sum_n$ in front of everything, right?

 

[stevendaryl]

Well, yes, there's a sum over $n$, but that's not really the point.

You're computing $\nabla \cdot J$ which involves derivatives of $J$ with respect to $x_n^i$. There is no derivative with respect to $t$.

 

[stevendaryl]

I think your mistake was in generalizing from the one-dimensional case.

It is true that
$\frac{d}{dt} f(x-y(t)) = - \frac{d}{dx} f(x-y(t)) \frac{dy}{dt}$

But in 3 dimensions, you have:

$\frac{d}{dt} f(\vec{x}-\vec{y}(t))$
$\ \ \ = - \frac{d}{dx^1} f(\vec{x}-\vec{y}(t)) \frac{dy^1}{dt}$
$\ \ \ - \frac{d}{dx^2} f(\vec{x}-\vec{y}(t)) \frac{dy^2}{dt}$
$\ \ \ - \frac{d}{dx^3} f(\vec{x}-\vec{y}(t)) \frac{dy^3}{dt}$

That is not three copies of the same thing.