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?


4 Answers 4


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 $V$

  • $\begingroup$ So the divergence depends on the choice of V. $\endgroup$
    – Subhra
    Commented Jul 13, 2014 at 19:56
  • $\begingroup$ The divergence is a function of position. I haven't been terribly clear and have used $V$ to mean both the set of points being integrated over and the volume of that set of points. I hope its clear from the context which is meant. (Hey this is Physics SE not Math SE) Clearly if I consider $\nabla\cdot\vec{E}$ if two different regions of space it will, in general, be different. $\endgroup$ Commented Jul 13, 2014 at 20:04
  • $\begingroup$ +1 because the key point is that what we measure is the integral form of Maxwell's laws. You should probably remove the part that says $\rho \rightarrow \inf$. Even though it links back with the divergences he brought up I think it's just confusing Subhra. What he needs to understand is that is that if we ever measure a non zero integral of $d^2 \vec{S} . \vec{B}$ over a closed surface, we will admit the existence of magnetic monopoles, so there is no double standard. $\endgroup$
    – ticster
    Commented Jul 13, 2014 at 22:48
  • $\begingroup$ @ticster: I fully agree with your last sentence and I realized it at the very beginning of my study of electricity and magnetism. If you are allowed to generate electric field only from magnetic flux can a non zero divergence of electric field be found anyway? $\endgroup$
    – Subhra
    Commented Jul 14, 2014 at 5:59
  • $\begingroup$ @Subhra No you could not, nor could you find the integral equivalent. While an $\vec{E}$ field would be generated, any closed surface integral of it would be null. $\endgroup$
    – ticster
    Commented Jul 14, 2014 at 9:27

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.

  • $\begingroup$ This is a very smart approach to bypass the problem. But what if the closed surface includes equal amount of positive and negative magnetic charges? $\endgroup$
    – Subhra
    Commented Jul 13, 2014 at 19:15
  • 1
    $\begingroup$ The Gaussian surface is arbitrary. $\endgroup$
    – Qmechanic
    Commented Jul 13, 2014 at 19:17
  • $\begingroup$ Why should $g$ be necessarily non-negative? $\endgroup$
    – Ruslan
    Commented Jul 14, 2014 at 5:44
  • $\begingroup$ It can be generalized to complex-valued $g$. I just didn't want to discuss issues like, say, $\infty-\infty$, to keep the answer short. $\endgroup$
    – Qmechanic
    Commented Jul 14, 2014 at 7:06
  • $\begingroup$ For more on distributions, see e.g. math.stackexchange.com/q/285642/11127 $\endgroup$
    – Qmechanic
    Commented Jul 14, 2014 at 15:19

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.

  • $\begingroup$ For a distribution of charges, there is no problem with the first equation. Can you express the first equation for a single point charge? $\endgroup$
    – Subhra
    Commented Jul 13, 2014 at 19:07
  • $\begingroup$ @Subhra: Read the wiki link - distribution does not mean what you think it means. And yes, for a charged point particle and its Coulomb electric field, these equations are prefectly valid. $\endgroup$
    – ACuriousMind
    Commented Jul 13, 2014 at 19:08
  • $\begingroup$ This is another fantastic trick. For a point charge r=0 so the definition of the delta function is justified. What if the charge be an electron whose dimension is not zero? $\endgroup$
    – Subhra
    Commented Jul 13, 2014 at 19:49
  • $\begingroup$ @Subhra The electron (as far as we know) is a point, the distribution of charge in a volume around it is a Dirac delta fuction. But for a finite (non point-like) particle the distribution is just a normal function, possibly similar to a 3D bell curve (the density of charge in 3 dimensions). So the 2 Maxwell equations still hold $-$ in their differential form for a particular point in space, and in their integral form for some surface enclosing particular amount of charge. $\endgroup$
    – Lurco
    Commented Jul 13, 2014 at 21:05

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.

  • $\begingroup$ The B field has neither any source/sink (div(B)=0) nor it can do any work(Lorentz force Law), although it has a finite energy. If it doesn't have any source of energy, can't do any work, how can I accept a finite non zero energy stored in the field? Furthermore, work done by a magnetic field is dependent on the applied B field and velocity of the charged particle while the energy stored in the field is dependent only on the B field. Can you please explain what's going wrong? $\endgroup$
    – Subhra
    Commented Jul 14, 2014 at 5:46
  • $\begingroup$ @Subhra You are using terminology too informally. This, I think, is masking some deeper misunderstanding. How can you say "B does no work" and then start a sentence with "work done by B"?! I know what you mean, "a direction changing force caused by B", but still. You should also stop using the word "source" if you don't mean it! If you ask someone to name a source of a magnetic field, they might answer a wire with a current going down it. In some sense, the wire is a source of the magnetic field. We still have $\nabla \cdot B=0$, I just mean to demonstrate that "source" is ambiguous. $\endgroup$
    – user12029
    Commented Jul 14, 2014 at 6:19
  • $\begingroup$ @Subhra There are further ambiguities in what you're writing. For example, a constant magnetic field does no work on a moving charge, that much is true. But "B does no work" is less true. Through Ampere/Maxwells's equation linked, $\nabla \times B$ is proportional to $dE/dt$ (when there is zero current density). Since $E$ stores energy, $B$ must be doing work! It isn't doing work on a charged particle, however. It is doing work on a field. $\endgroup$
    – user12029
    Commented Jul 14, 2014 at 6:23
  • $\begingroup$ @Subhra and the magnetic field does have a source of energy. In the same way that a block resting on a table has a "source of energy" - someone lifted it up there in the first place. Some current density or changing electric field lifted the magnetic field up, and to do that it took energy. This is an important principle of the inductor. $\endgroup$
    – user12029
    Commented Jul 14, 2014 at 6:25
  • $\begingroup$ I am not saying "B does no work", Lorentz force law says it. I would rather be extremely happy if this statement be false and consequently the Lorentz force law. Div(B)=0 still. So I don't find any harm saying "B field has neither any source/sink". "E stores energy, B must be doing work" - if this is so, is E doing any work? How the work is distributed between E and B? "It isn't doing work on a charged particle, however. It is doing work on a field." - So E is doing work on charged particle and B is doing work on E (I guess). $\endgroup$
    – Subhra
    Commented Jul 14, 2014 at 6:57

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