I came across the problem of calculating the interaction energy of two point charges separated by some distance a in Griffith's Introduction to Electrodynamics. Here and everywhere else that I look, I find that the calculation is done by using the expression for interaction energy: $$\epsilon_{0}\int\vec{E}_{1}\cdot\vec{E}_{2}\,d\tau$$ where the integration is done on all space.

Why does this give correct result when $\vec{E}_{1}$ and $\vec{E}_{2}$ are both undefined at the points where the two charges are situated? How come does the integral comes out as finite when the integrand goes to infinity at two places in the domain of integration?

  • $\begingroup$ You need to learn about improper integrals, particularly those of the second kind. $\endgroup$
    – Ruslan
    Commented Dec 7, 2021 at 20:45

3 Answers 3


This converges because the integral can be thought of as finite. To put it more rigorously, let's look at the integral of two vector functions $\vec{v}_1(\vec{r})$ and $\vec{v}_2(\vec{r})$ near the origin, where $\vec{v}_1$ is bounded and $$ \vec{v}_2 = \frac{1}{r^2} \vec{w}(\vec{r}), $$ where $\vec{w}(\vec{r})$ is bounded.

Now, the integral we want to perform is $$ I = \iiint_\text{all space} \vec{v}_1 \cdot \vec{v}_2 \, d\tau $$ but we are worried that the integrand diverges as $r \to 0$. We can "isolate" the problematic region by splitting up the integral: $$ I = \iiint_{|\vec{r}| \geq \epsilon} \vec{v}_1 \cdot \vec{v}_2 \, d\tau + \iiint_{|\vec{r}| \leq \epsilon} \vec{v}_1 \cdot \vec{v}_2 \, d\tau. $$ We'll assume that the first integral is well-defined and look more closely at the second. If we look at this carefully, we can put a bound on the integral into a ball of radius $\epsilon$ around the origin and the rest of space: \begin{align*} \left| \iiint_{|\vec{r}| \leq \epsilon} \vec{v}_1 \cdot \vec{v}_2 \, d\tau \right| &= \left| \iiint_{|\vec{r}| \leq \epsilon} \frac{\vec{v}_1 \cdot \vec{w}}{r^2} \, d\tau \right| \\ &\leq 4 \pi \max_{|\vec{r}| \leq \epsilon} |\vec{v}_1 \cdot \vec{w}| \int_0^{\epsilon} \frac{1}{r^2} r^2 \, dr \\ &= 4 \pi \max_{|\vec{r}| \leq \epsilon} |\vec{v}_1 \cdot \vec{w}| \epsilon. \end{align*} So we can see that in the limit as $\epsilon \to 0$, the contribution from this portion of the integral becomes negligible. This means that it is well-defined to talk about the "integral over all space" as the limit as $\epsilon \to 0$ of "the integral over all space except for small spheres of radius $\epsilon$ around points where $\vec{v}_2$ diverges."1

In a real sense, this works because the singularity of $\vec{v}_2$ is sufficiently "weak" that we can still get a sensible limit out of this. This happened because the singularity was proportional to $1/r^2$, which was cancelled out by the volume factor of $r^2$ in the integral. If you applied the same logic to the quantity $\vec{v}_2 \cdot \vec{v}_2$ near the origin, the integrand would be proportional to $1/r^4$ near the origin, which is "too strong" of a singularity to be cancelled out by the volume factor. This is why Griffiths has to discuss the charge's "self-energy" separately from the "interaction energy."

1 To really be rigorous about this (and avoid doing a suspect $0/0$ cancellation in the demonstration above), we would instead want to fix some radius $r_0$ and take a sequence of $\epsilon$ values $\epsilon_1, \epsilon_2, \dots$ such that $\lim_{n \to \infty} \epsilon_n = 0$. We can then show, using similar logic, that the sequence of integrals $$ I_n \equiv \iiint_{\epsilon_n \leq |\vec{r}| \leq r_0} \vec{v}_1 \cdot \vec{v}_2 \, d\tau $$ also converge to some finite limit. This limit would then be what we mean by "the integral over the sphere of radius $r_0$ centered at the origin." The logic is much the same, but somewhat more opaque a first pass, which is why I phrased my main answer the way I did.


$|E|$ blows up as $r^{-2}$ but the measure in spherical coordinates centered about one of the charges the integration measure is $d^3x= r^2 \sin\theta\,dr\, d\theta\, d\phi$, so the $r^2$'s cancel to give a finite integrand.

  • $\begingroup$ But what if we consider rectangular coordinates? if one particle is situated at the origin and the other at $(0,0,a)$, the integral looks like: $(...)\,dx\,dy\,dz\,[(x^2+y^2+z^2)^{(3/2)}]$, and there are no terms to cancel. $\endgroup$ Commented Dec 7, 2021 at 18:41
  • $\begingroup$ Do it iteratively. First the inegral over $x$ clearly converges absolutely, as does the integral over $y$. The same must be true for the ilast integral over integral over $z$, but I have not done it explicitly. (Absolute convergence allows the use of Fubini's theorem). It can't matter what coordinates you use. $\endgroup$
    – mike stone
    Commented Dec 7, 2021 at 20:54
  • $\begingroup$ The convergence of generalized integrals in dimension greater than 1 is simpler than in one dimension. There is no more difference between simple convergence and absolute convergence. And in the case of a positive function, either the integral converges or it tends to infinity. Here, in spherical coordinates, we show that the integral converges absolutely and this result is independent of the chosen division of space (cartesian, spherical....). $\endgroup$ Commented Dec 8, 2021 at 7:01

This is a way to prove convergence and also to understand why this integral represents potential energy between two charge distributions. Another way to look at this is

$$\iiint \frac{1}{2} \epsilon_0(\vec{E}_1+\vec{E}_2)^2d\tau $$ : $$\iiint \frac{1}{2} \epsilon_0(\vec{E}_1)^2d\tau +$$

$$\iiint \frac{1}{2} \epsilon_0(\vec{E}_2)^2d\tau +$$

$$\iiint \epsilon_0\vec{E}_1\cdot\vec{E}_2\,d\tau$$ The last component represents PD between charge distributions, $$=\iiint \epsilon_0\vec{E}_1\cdot(-\nabla V_{2})\,d\tau.$$ Rearranging the vector field identity $$\nabla \cdot (V_{2}\vec{E}_1) = V_{2} \nabla \cdot \vec{E}_1 + \vec{E}_{1} \cdot \nabla V_{2},$$ gives $$\vec{E}_{1} \cdot (-\nabla V_{2})= -\nabla\cdot(V_{2}\vec{E}_1) + V_{2} \nabla \cdot\vec{E}_1.$$ Substituting this in the integral gives $$\iiint\epsilon_0\left[-\nabla \cdot (V_{2}\vec{E}_1) + V_{2} \nabla \cdot \vec{E}_1\right]d\tau$$ $$=-\iiint\epsilon_0 \nabla \cdot (V_{2}\vec{E}_1)\,d\tau+\iiint \epsilon_0V_{2} \nabla \cdot \vec{E}_{1}\,d\tau.$$

Invoking Stokes' Theorem on the first integral gives $$-\iint\epsilon_0 (V_{2}\vec{E}_1) \cdot d\vec{S}+ \iiint\epsilon_0 V_{2} \nabla \cdot \vec{E}_1\,d\tau.$$ With the integration volume as all of space, the surface integral evaluates to zero for localized sources, since at large distances $(r\rightarrow\infty)$, the integrand falls off as $V_{2}\vec{E}_{1}\sim r^{-3}$. This leaves $$\iiint\epsilon_0V_{2} \nabla \cdot \vec{E}_1\,d\tau.$$ Invoking Gauss' Law, $$\iiint\epsilon_0 V_{2} \frac{\rho_1}{\epsilon_0}\,d\tau$$ $$=\iiint V_{2} \rho_1\,d\tau$$

In this form, it is clear to see why it represents potential energy, as you are building up charge distribution 1 in the presence of potential 2.

In this form you can also understand why it converges, as the $V\rho$ is always a finite number, because $V$ is never evaluated at any point that causes it to be infinity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.