Conservation of charge with Dirac delta

AI Thread Summary
The discussion centers on the application of the Dirac delta function in evaluating the divergence of current density in electromagnetism. The original poster encounters a factor of three when attempting to apply a property of the Dirac delta, leading to confusion regarding charge conservation. It is clarified that the divergence involves separate derivatives for each spatial dimension rather than three identical terms, which eliminates the erroneous factor. The key point is that the computation of the divergence in three dimensions requires careful consideration of the individual contributions from each coordinate, rather than assuming a simple multiplication by three. This distinction is crucial for correctly demonstrating the conservation of charge.
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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