Divergence of a field and its interpretation

Question

The divergence of an electric field due to a point charge (according to Coulomb's law) is zero. In literature the divergence of a field indicates presence/absence of a sink/source for the field.

However, clearly a charge is there. So there was no escape route.

To resolve this, Dirac applied the concept of a deltafunction and defined it in an unrealistic way (the function value is zero everywhere except at the origin where the value is infinity). However the concept was accepted and we became able to show that

$\nabla \cdot \vec E=0$, everywhere except at the origin.

Conclusion: The source of the electric field exists although its divergence is zero everywhere except at the source point.

In the case of the magnetic field we are yet to observe its source or sink. However, the zero divergence of this field implies that no magnetic charge exists and since we don't have any real magnetic monopole at hand, there is no question of finding the field at the source point.

Isn't this a double standard? Do we really need to find a non-zero divergence of a field for its source to exist?

Answer 1

This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law \begin{equation} \nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0} \end{equation} If we integrate this over some volume $V$ and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} \int_V\mathrm{d}^3\vec{x}\;\nabla\cdot\vec{E}= \int_{\partial V}\mathrm{d}^2\vec{S}\cdot \vec{E}\end{align} where $\partial V$ is the boundary of $V$. While the right hand side gives \begin{equation} \int_V\mathrm{d}^3\vec{x}\;\frac{\rho}{\epsilon_0}= \frac{Q}{\epsilon_0}\end{equation} Where $Q$ is the total charge enclosed in $V$. Combining the two gives \begin{equation}\int_{\partial V}\mathrm{d}^2\vec{S}\cdot \vec{E} = \frac{Q}{\epsilon_0}\end{equation} I words the electric flux entering any closed region is equal to the charge contained in that region, i.e. electric field lines only start and stop on charges.

Conversely we can apply this equation over an arbitrary volume, $V$. In particular we can choose a volume so small that $\nabla\cdot\vec{E}$ and $\rho$ are approximately constant, so so we can recover the differential form of Gauss' Law.

Now let's see what these equation's look like for a point charge, $q$, at the origin. For any volume $V$ that does not include the origin, $Q = 0$, so by taking $V$ small we find that $\nabla\cdot\vec{E} = 0$. If however we consider a volume which does include the origin then $Q = q$ and the integral of $\nabla\cdot\vec{E}$ is non-zero. If we let the volume of $V\rightarrow 0$ we find that $Q$ remains constant as long as the origin is still contained, so \begin{equation}\frac{Q}{V}\rightarrow\rho\rightarrow \infty\end{equation} So $\rho$ must diverge for a point charge! Further more this behaviour where the value of an integral is given by the value of the integrand at a point is the definition of the Dirac delta. If you find this unsatisfying you can push back to the question of whether point charges actually exist, but this is an empirical, rather than theoretical question. (we currently have little reason to think that fundamental particles are not pointlike.)

A similar analysis can be done with magnetic fields, where we find that \begin{equation} \int_{\partial V}\mathrm{d}^2\vec{S}\cdot \vec{B} = 0\end{equation} for any volume $V$

Answer 2

I) Right, the differential form of Gauss's law

$$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0} $$

uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges

$$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$

Note in particular, that it is technically wrong to claim (as OP seems to do) that the Dirac delta distribution $\delta^3({\bf r})$ is merely a function $f:\mathbb{R}^3\to [0,\infty]$ that takes the value zero everywhere except at the origin where the value is infinity:

$$\tag{3} f({\bf r})~:=~\left\{ \begin{array}{rcl} \infty& {\rm for}& {\bf r}={\bf 0}, \cr 0& {\rm for}& {\bf r}\neq {\bf 0}.\end{array}\right. $$

For starters, for an arbitrary test function $g:\mathbb{R}^3\to [0,\infty[$, the Lebesgue integral$^1$

$$\tag{4} \int_{\mathbb{R}^3} \! d^3r~f({\bf r})g({\bf r}) ~=~0 $$

vanishes, in contrast to the defining property of the Dirac delta distribution

$$\tag{5} \int_{\mathbb{R}^3} \! d^3r~\delta^3({\bf r})g({\bf r}) ~=~g({\bf 0}). $$

The Dirac delta distribution $\delta^3({\bf r})$ is not a function. It is instead a generalized function. It is possible to give a mathematically consistent treatment of the Dirac delta distribution. However, it should be stressed that the analysis does not reduce to the investigation of two separate cases ${\bf r}= {\bf 0}$ and ${\bf r}\neq {\bf 0}$, but instead (typically) involves (smeared) test functions. To get a flavor of the various intricacies that can arise with distributions, the reader might find this Phys.SE post interesting.

II) To avoid the notion of distributions, it is more safe (and probably more intuitive) to work with the equivalent integral form of Gauss's law

$$\tag{6} \Phi_{\bf E}~=~ \frac{Q_e}{\varepsilon_0}. $$

The corresponding Gauss's law for magnetism

$$\tag{7} \Phi_{\bf{B}}~=~ 0 $$

expresses (without employing double standards) the fact that there is no magnetic charge $Q_m$.

--

$^1$ Eq. (4) relies crucially on the fact that in integration theory for non-negative functions, one defines multiplication $\cdot: [0,\infty]\times[0,\infty]\to[0,\infty]$ on the extended real halfline $[0,\infty]$ so that $0\cdot\infty:=0$. Eq. (4) is essentially caused by the fact that $f$ is zero almost everywhere. Also we should mention the well-known fact that integration theory can be appropriately generalized from non-negative functions to complex-valued functions.

Answer 3

Maxwell's equations state

$$ \nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$$

$$ \nabla \cdot \vec B = 0 $$

If we accept Maxwell's equations as true, there is no source/sink of the magnetic field, since the divergence of the magnetic field is zero no matter what. Yet, no matter how you feel about the Dirac delta, where there is charge, there is non-zero divergence of the electric field. And, conversely, where there is non-zero divergence, there is charge.

Now, it is not the Dirac delta that is "unrealistic" (it is a perfectly well defined distribution), it is the concept of a "point charge". Every charged thing we know has this charge distributed over a - however small - area of space, and the Dirac delta is a way to model that this area is so small that we don't care that its not point-like. And if there truly was a point-like charge, the Dirac delta would exactly describe its charge density - because the volume of a point is clearly zero, and whatever charge the thing has divided by zero is infinite. (Do not take this as a rigorous statement, this is as handwavy as it gets)

The more serious thing to learn here is that densities are distributions - they do not make sense unless integrated over, and if we integrate over a point charge with $\rho(r) = q \delta(r)$, we obtain the perfectly finite charge $q$. There is nothing wrong with the Dirac delta as a charge (or other) density.

Answer 4

You write, "In the case of the magnetic field we are yet to observe its source or sink."

If you mean "we are yet to observe a source or sink", you're correct.

However, consider the magnetic vector field (ignoring units/speaking qualitatively):

$$\vec{B}=(0, \frac{z}{(1+r^2)^2},\frac{y}{(1 + r^2)^2})$$

This is a valid field because it's the curl of the vector potential $(\frac{1}{1+r^2},0,0)$.

It's a valid instantaneous magnetic field. You could use Maxwell's equations to find a current density or changing electric field, but that's beyond the point. The point is:

1. There is no source or sink of $\vec{B}$,
2. Since $\vec{B}$ goes to zero at infinity, no source or sink has been "pushed off to infinity".
3. The energy stored in the field is finite.

This is really a finite magnetic field with no source or sink. It's not a matter of observing its source or sink. There's no source or sink to observe!

I've been using the term "source or sink" to mean $\nabla \cdot \vec{V}\neq 0$. But you could also use the term "source" to mean "cause of", in which case "source" is not synonymous with $\nabla \cdot \vec{V}\neq 0$. You can look at Ampère's circuital law and say that the $\vec{B}$ field is caused by a current or a changing electric field. So it's not as if $\nabla \cdot \vec{B}=0$ implies that the B-field has no "source", in the general meaning of the word.

If $\vec{B}$ represented the velocity field of a liquid filling up space, then zero divergence implies no water being injected/removed anywhere. But $\vec{B}$ does not represent the velocity field of a liquid filling up space.