# Gauss' law in differential form

1. Sep 15, 2015

### Quicksilvr

I'm trying to understand how the integral form is derived from the differential form of Gauss' law.
I have several issues:
1) The law states that $\nabla\cdot E=\frac{1}{\epsilon 0}\rho$, but when I calculate it directly I get that $\nabla\cdot E=0$ (at least for $r\neq0$).
2) Now $\iiint\limits_\nu \nabla\cdot E d\tau$ should be zero no matter what the value of the divergence is at 0, since the divergence is zero everywhere but 0 (in contrast to the law which states it is non-zero).
3) a. The proof itself goes on to use the divergence theorem to state that for any volume $\nu$, $\iiint\limits_\nu \nabla\cdot E d\tau = \iint\limits_{\partial\nu} E d a$, however the divergence theorem requires E to be continuously differentiable everywhere in $\nu$ (it is not differentiable at 0, let alone continuously differentiable there).
b. The function cannot be corrected at any way at 0 since the derivative goes to infinity around 0.
c. The point 0 cannot be removed from the integrated volume because the divergence theorem requires that the volume of integration be compact.
d. In light of the former I don't see how the divergence theorem can be used here.

2. Sep 15, 2015

### Orodruin

Staff Emeritus
1) Are you talking about a point charge here? Obviously the charge density for a point charge is zero away from the charge.

2) No, this is not true. For the point charge, the divergence at the charge is a delta function.

3) If you just look at the version of the theorem for smooth functions yes. The theorem can be extended to distributions which do have well defined derivatives even if they are not continuous functions.

Edit: If you are uncomfortable with applying the law for the point charge, just see it as a limit of a continuous charge distribution which is getting more and more peaked.

3. Sep 15, 2015

### Quicksilvr

1) I am treating the divergence as a function at each point (is there any other way to define it? the wikipedia definition states that the definition of the divergence as an integral around the point is equivalent to this definition when the function is continuously differentiable), defined as $\nabla \cdot E := \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}$.
2) What is a delta function? Assuming the formula $\nabla \cdot v = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r) + \frac{1}{r\sin\theta}\frac{\partial v_\theta}{\partial \theta}(\sin\theta v_\theta) + \frac{1}{r\sin\theta}\frac{\partial v_\phi}{\partial \phi}$, my calculation was: (latter two terms vanish) $\frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r) = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{1}{r^2}) = 0$.
3) Could you write the definitions here?

Edit: I am really talking about a point charge when saying that the divergence is vanishes everywhere but 0. However, when talking about a collection of point charges, while the divergence is not zero, the electric field is still not differentiable around any contained charge which means that the divergence theorem can't be used. Additionally, how does Gauss' law hold for a point charge?

Thanks

Last edited: Sep 15, 2015
4. Sep 15, 2015

### Orodruin

Staff Emeritus
Yes, I am treating it like that too. But my comment was directed at the fact that you seemed to find it surprising that the divergence was zero away from the point charge, but this is just as it should be since there is no charge away from the point charge. The charge density is a scalar field and is equal to zero where there are no charges, nothing strange here.

Your computation is only valid for $r\neq 0$. At $r = 0$, your coordinate system is singular and your computation does not apply. The definition of the delta function (or rather, delta distribution, if you want to be formal) is that
$$\iiint_V \delta(\vec x - \vec a) f(\vec x) dV = f(\vec a).$$

No, distribution theory is an entire subject of its own. You may think of it as a relatively straight forward extension of derivatives of functions, where (for example) the derivative of the Heaviside step function ($\theta(x) = 1$ for $x > 0$ and 0 otherwise) is a delta function. That this makes sense is based on the relation
$$\int f'(x) g(x) dx = -\int f(x) g'(x) dx$$
assuming that the boundary terms disappear. For the derivative of the Heaviside function, you would find
$$\int \theta'(x) g(x) = -\int \theta(x) g'(x) = -\int_0^\infty g'(x) dx = g(0) - g(\infty) = g(0) = \int \delta(x) g(x) dx$$
(again assuming that $g(x) \to 0$ as $x\to \infty$ -- distributions are defined by their action on sufficiently nice test functions).

The bottom line (and all you really need to keep in mind until you study the mathematical details more) is that the divergence of the point charge field is a delta function. As I said, this is not strange, the delta function is zero away from the point charge and so the divergence of the field there should be zero.

5. Sep 15, 2015

### Quicksilvr

I'm trying to figure out the mathematical details right now. I'm trying to avoid usage of delta functions because there was no rigorous definition for them in my textbook (and the distribution definition in wikipedia seems complicated). In any case, while true that the divergence does not vanish at 0 (since it is undefined), changing a function at a single point does not affect the value of the integral of it. So even if the divergence is non-zero (any other finite value there) at 0, the integral should still be 0 if the divergence vanishes everywhere but 0.

In any case, Gauss' law states that the divergence is non-zero everywhere, not just at 0.

Additionally, you're saying there is a different version of the divergence theorem which accepts non-smooth functions. Can you give any reference to this?

Thanks

6. Sep 15, 2015

### Orodruin

Staff Emeritus
But this is the entire point, the delta function is not finite at 0. As I said earlier, the easiest way of thinking of the delta function is as a limit such as
$$\delta(x) = \lim_{a\to 0^+} \frac{1}{2\sqrt{\pi a}} e^{-x^2/4a}.$$
You will notice that this goes to zero for all $x \neq 0$ and infinity for $x = 0$. The three-dimensional delta can be seen as a product of the deltas in each direction.

7. Sep 15, 2015

### Quicksilvr

But the divergence is not defined as a delta function. Can you show me why you think it is a delta function using the wikipedia definition?

Additionally as I said before, Gauss' law states that the divergence is non-zero everywhere.

Last edited: Sep 15, 2015
8. Sep 15, 2015

### Orodruin

Staff Emeritus
It is not a matter of thinking, it is a delta function. All you have to do is to compute
$$\iiint \varphi(\vec x) \nabla\cdot \vec E \, dV.$$
You will need to define this as the limit of the integral over all of space except for a small ball around $\vec x = 0$ of radius $\varepsilon$ as $\varepsilon \to 0$. You should find that the integral evaluates to $\varphi(0)$ (up to overall constants) and therefore $\nabla\cdot\vec E$ is a delta function.

What do you mean by this? Of course it is not defined as a delta function - but the divergence of the field from a point charge is a delta function because this follows directly from Gauss law on differential form -- the charge distribution is a delta function. Again, this is absolutely consistent with what you expect. You expect a point charge to give zero charge density everywhere but at the charge itself. Since the volume of the charge is zero, you expect the charge density to be formally infinite at $\vec x = 0$ and that the integral of the charge density should give you the charge whenever $\vec x = 0$ is included in the integration domain, i.e.,
$$\iiint_V \rho dV = Q$$
if $0 \in V$. The delta function (multiplied by $Q$) is the distribution which satisfies this requirement.

9. Sep 15, 2015

### Quicksilvr

And what about the fact that the law states that the divergence is non-zero everywhere?

10. Sep 15, 2015

### Orodruin

Staff Emeritus
It does not.

11. Sep 15, 2015

### Quicksilvr

12. Sep 15, 2015

### Orodruin

Staff Emeritus
The charge density is generally a function of space.

Of course not, the law is more general than the case to which you are attempting to apply it. You are applying it to the field of a point charge (this is the field which has the 1/r^2 form) - a general field does not need to be of this form.

13. Sep 15, 2015