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I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.

$$ \nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$

That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.

I know the math looks better in spherical coordinates, but I will be using Cartesian.

So when I calculate the divergence I obtain:

$$ \nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

This can further be simplified:

$$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

Qmechanic
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Ben
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1 Answers1

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What you want to compute is essentially $$\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3}$$ at the origin. Of course, that doesn't exist as a function since the field is singular. On the other hand, you have already shown that it vanishes everywhere else.

So you need to interpret the expression in a weak sense, i.e. as a distribution, and consider the integral $$\int_{B_\epsilon}\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3} \,\text{d}^3x$$ over some volume containing the origin, conveniently chosen as a ball of radius $\epsilon$, convert it to a surface integral which does not include the singularity and see that the result is finite.

Toffomat
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  • You used a phrase here that I would like to ask you about. You said "that doesn't exist as a function since the field is singular."

    I have not studied (much to my dismay) abstract algebra. When mathematicians use the phrase vector FIELD, are they actually talking about a type of field in the AA sense?

     – Ben Jan 11 '18 at 10:15
  • @Ben: No, I'm using it in the simple sense of "function defined at every point in space": The electric field assigns a vector to every point $x$, i.e. it is a function of a variable $x\in \mathbb{R}^3$, and in this case, the function is sigular and not differentiable at $x=0$. – Toffomat Jan 11 '18 at 12:04