Conservation of charge with Dirac delta

Frostman
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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? o_O
 
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Frostman said:
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))]
 
Last edited:
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.
 
So that factor ##3## really isn't there, because there's ##\sum_n## in front of everything, right?
 
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##.
 
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.
 
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