# What is the origin of the Dirac delta term in the dipole electric field?

I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The Wikipedia article contains a delta function in the formula, but this paper gives only the formula without the delta function,

$$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) )$$

Is it intentionally left out?

My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.

I'll give you the derivation from my book which includes a nice way to see how the delta functions arise: ..............................................................................................................................................................

We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and $\vec{B}$ of a vector point dipole and an axial point dipole in the same way as we derive these in the case of a the point monopole. We start with a static point charge $\delta(\vec{r})$ and derive the dipole charge/current densities with the help of differential operators.

We apply the same differential operators for $\vec{A}$, $\vec{E}$ and $\vec{B}$ to obtain the dipole fields from the monopole fields. First we recall the fields of the monopole. The point charge obtains (over time) a potential field given by.

$\mbox{field}\Big\{\,\delta(\vec{r})\,\Big\} ~~=~~ \frac{1}{4\pi r}$

The reversed operator which derives the source from the field is just the Laplacian operator. $-\nabla^2\Big\{\,\frac{1}{4\pi r}\,\Big\} ~~=~~ \delta(\vec{r})$

Integrating the delta function over space shows equal contributions from the three spatial components.

$\int \delta(\vec{r})~d\vec{r} ~=~ -\int\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right] \frac{d\vec{r}}{4\pi r} ~=~ \frac13+\frac13+\frac13 ~=~ 1$

We can define vector dipole and axial dipole sources by using differential operators on the monopole $\delta(\vec{r})$ and derive their potential and electromagnetic fields.

The vector dipole is obtained by differentiating the monopole along the direction that we want the dipole to have. The result is a combination of a positive and a negative delta function. The axial dipole is defined by the curl of the monopole so that it gives a circular point current in the direction of the dipole.

charge/current densities, potentials and fields of dipoles:

$\begin{array}{|lcll|} \hline &&& \\ j^o &=& ~~~\mbox{div}\,\left(~\vec{\mu} \,\delta(\vec{r})~\right) & \mbox{vector dipole charge density} \\ \vec{j} &=& ~~~\mbox{curl}\left(~\vec{\mu}\, \delta(\vec{r})~\right) & \mbox{axial dipole current density} \\ &&& \\ A^o &=& ~~~\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{vector dipole electric potential} \\ \vec{A} &=& ~~~\mbox{curl}\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{axial dipole magnetic potential} \\ &&& \\ \mathsf{E}&=& -~\mbox{grad}\left(\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right)\right)& \mbox{vector dipole electric field} \\ \mathsf{B}&=& +~\,\mbox{curl}\left(\mbox{curl} \left(~\vec{\mu}\,\frac{1}{4\pi r}\,\right)\right)& \mbox{axial dipole magnetic field} \\ &&& \\ \hline \end{array}$

The potential fields are obtained by applying the same differential operators on the field $1/4\pi r$ of the monopole rather than on the charge distribution. For a dipole moment in the $z$-direction we obtain in SI-units.

Potential fields of the electric and magnetic dipoles:

$\Phi ~=~ \frac{z}{4\pi\epsilon_o r^3}\,\mu~~~~~~~~ \vec{A} ~=~ \Big( -y ~,~ x~ ~,~ 0 \Big)~ \frac{\mu_o}{4\pi r^3}\,\mu$

The expressions for the electromagnetic fields do implicitly contain delta functions at the center with the right magnitude. These delta functions are easily lost if the derivation isn't careful enough. The E and B fields are related to each other by the standard vector identity.

$\mbox{curl}(\mbox{curl}\vec{X}) ~=~ \mbox{grad}(\mbox{div}\vec{X})-\nabla^2\vec{X}$

We have seen that the last term (the Laplacian) yields $\delta(\vec{r})$. The two therefor differ only at the center by a delta function in the $\vec{\mu}$ direction. We can see this explicitly if we align the dipoles with the z-axis, so that $\vec{\mu}$ has only a z-component, and write out the fields.

The only difference is in the z-components. The total difference between the two is the Laplacian and thus $\delta(\vec{r})$. The vector dipole gets -1/3 while the axial dipole gets +2/3.

\begin{aligned} \mathsf{E} &= &\frac{1}{\epsilon_o}\bigg[~ \mathbf{\hat{x}}\, \frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\, \frac{\partial}{\partial y}\frac{\partial}{\partial z} ~+~~~ \mathbf{\hat{z}}\, \frac{\partial^2}{\partial z^2} ~~~~~~~~~~~~~ &\bigg] ~ \frac{\mu}{4\pi r}\\ \\ \mathsf{B} &= &\mu_o\bigg[~ \mathbf{\hat{x}}\,\frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\,\frac{\partial}{\partial y}\frac{\partial}{\partial z} ~-~ \mathbf{\hat{z}}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)~ &\bigg] ~ \frac{\mu}{4\pi r}\\ \end{aligned}

Which we can see from the third equation with the three 1/3 parts. In the general case, with arbitrary dipole direction $\vec{\mu}$ we get for the electromagnetic fields in vector form.

Electromagnetic dipole fields:

\begin{aligned} \mathsf{E} &= &\frac{1}{\epsilon_o}\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~-~\frac13\vec{\mu}\,\delta(\vec{r})~\right) \\ \\ \mathsf{B} &= &\mu_o\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~+~\frac23\vec{\mu}\,\delta(\vec{r})~\right) \end{aligned}

Alternatively we can look at the fields explicitly expressed in the individual $x$, $y$ and $z$ components which gives. (with the dipole moment in the $z$-direction)

\begin{aligned} \mathsf{E} &= &\frac{\mu}{4\pi\epsilon_o}\bigg[~ \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~ \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~ \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} - \frac{4\pi}{3}\delta(r)\right)~ &\bigg] \\ \\ \mathsf{B} &= &\frac{\mu_o\mu}{~4\pi~}\bigg[~ \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~ \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~ \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} + \frac{8\pi}{3}\delta(r)\right)~ &\bigg] \\ \end{aligned}

In general the fields decrease with the third order of $r$ compared to $1/r^2$ for the charge which makes it possible not many effects at larger scale. At the other hand, when we go to smaller scales, the magnetic dipole fields become more powerful relative to the charge.

• which book and which page? – hhh Aug 30 '11 at 15:39
• @hhh: The book is "a work in progress" of 900+ pages now but you can find selected chapters here: physics-quest.org. (The text above in the post is not yet available). – Hans de Vries Aug 31 '11 at 7:15
• could you tell how you deduced the factor $\frac{1}{4\pi r}$? – hhh Sep 2 '11 at 19:49

There are two common types of dipole vector fields in physics:

The "divergence-free" dipole field:

$\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} + \frac23\vec{\mu}\,\delta(\vec{r})$

The "curl-free" dipole field:

$\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} - \frac13\vec{\mu}\,\delta(\vec{r})$

The divergence of the first one is zero everywhere, even at the origin; the curl of the second one is zero everywhere, even at the origin. (Proof outline in the appendix.)

The electric field of an ideal electric dipole has zero curl, so it has to be the second one. The magnetic B field of an ideal magnetic dipole has zero divergence, so it has to be the first one. The magnetic H field of an ideal magnetic dipole has zero curl, so it has to be the second one. And so on!

It's certainly important to include the delta-functions: If you leave it out, Maxwell's equations are violated at the origin! For example, if you leave out the delta-function, then the electric field has nonzero curl, which violates Faraday's law.

A nice example is the field inside a sphere which is uniformly filled with dipoles. It's just proportional to the delta-function term. So the E-field inside a uniformly polarized sphere points antiparallel to the dipole moments, the B-field inside a uniformly magnetized sphere points parallel, the H-field points antiparallel.

Appendix: Proof outline that the "divergence-free dipole field" is really divergence-free everywhere (including the origin), and the "curl-free dipole field" is really curl-free everywhere (including the origin).

This is only tricky because of the origin. Away from the origin, the fields are the same, and they have both zero divergence and zero curl. You can prove this by straightforward calculus and algebra. To include the origin requires a different approach.

One approach -- a kind of physical approach -- is to write the divergence-free dipole field as a limit of finite, smooth, divergence-free fields, and to write the curl-free dipole field as a limit of finite, smooth, curl-free fields. For the first one, you can use the magnetic field of a circular wire loop carrying a current that gets larger and larger while the loop simultaneously gets smaller and smaller, keeping the product of current and area constant. For the second one, you can use the electric field of a pair of equal and opposite charges that have larger and larger magnitude as their separation gets smaller and smaller, keeping the product of charge and separation constant. This approach requires considerable work and care to make sure you're keeping track of exactly what's happening at and around the origin, to find out what multiple of the delta-function it's approaching there.

An alternative approach -- more mathematical -- is to write the divergence-free one as the curl of something, and the curl-free one as the gradient of something. (You can look up the "something" easily: It's the electric dipole potential in the first case, the magnetic dipole vector potential in the second case.) To prove the equalities away from the origin, it's just straightforward calculus. To prove the equalities at the origin, you do an integral transform (e.g., Kelvin-Stokes theorem) to relate what's happening at the origin to the well-behaved stuff that's happening away from the origin.

A dipole is formed of two opposite charges. By bringing them ever closer together, all the while increasing their charge but keeping the product of the charge and the separation $p = q \times d$ constant, we can form an ideal, elementary dipole.

The field on a line in between the charges goes as $~ 1/d^4$ and thus goes to infinity as $d \to 0$. This forms the Dirac delta. As this dipole is an ideal model, it's only necessary to keep the theory consistent. The only two examples I know for its applications are

1. To get a correct result for the average electric field over a sphere, cfr. Griffiths' "Introduction to Electrodynamics", pg 157.

2. Hyperfine splitting. Although I doubt the classical expression is meaningful at that level, it apparently gives rise to the Fermi contact interaction

• the argument of the delta is not $d$, but $x$. The magnitude of the fied on a line between the two charges in the dipole approximation goes as $p/r^3$. Without this approximation on that line the magnitute of the field id $qd / (r^2+d^2)^{3/2}$ – wiso Aug 29 '11 at 22:02

The origin of the problem is the special point $$\mathbf{r}=0$$. In the usual derivation (when using easily-derived formula for the potential $$\varphi = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$$): $$E_{\alpha} = -\frac{1}{4 \pi \epsilon_0} \nabla_{\alpha} \left( \frac{p_{\beta} x_{\beta}}{r^3}\right) = -\frac{1}{4 \pi \epsilon_0} p_{\beta} \left[ \frac{\nabla_{\alpha} x_{\beta}}{r^3} + x_{\beta}\nabla_{\alpha}\left(\frac{1}{r^3}\right)\right] = - \frac{1}{4 \pi \epsilon_0} p_{\beta} \left[ \frac{\delta_{\alpha \beta}}{r^3} - 3 \frac{x_{\alpha} x_{\beta}}{r^5}\right] =$$ $$=-\frac{1}{4 \pi \epsilon_0} \left[ \frac{p_{\alpha}}{r^3} - 3 \frac{x_{\alpha} (p_{\beta} x_{\beta})}{r^5} \right]$$ the delta-term is missing, because we do not treat the special point $$\mathbf{r}=0$$ in this derivation. To account for this term, we need to look closely on $$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right)$$ and rewrite it: $$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right) = - \nabla_{\alpha} \nabla_{\beta} \frac{1}{r}~~~~(1)$$ The trace of this term must satisfy the well-known $$\Delta \frac{1}{r} = -4 \pi \delta^3(\mathbf{r}).$$ The above relation (1) caries two indexes, it is symmetrical and invariant with respect to rotations (when $$r=0$$). Thus, the missing term must be proportional to identity matrix (that is, $$\delta_{\alpha \beta}$$) and we finally have: $$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right) = \left( \frac{\delta_{\alpha \beta}}{r^3} - 3 \frac{x_{\alpha} x_{\beta}}{r^5} \right) + \frac{4 \pi}{3} \delta_{\alpha \beta} \delta^3(\mathbf{r})$$ from which it is easy to get the formula from Wikipedia article.

This is, if you wish, a mathematical explanation of the problem. To get a physical explanation, read D. J. Griffiths, Am. J. Phys. 50, 698 (1982). Note also that this term (the delta-term) arises only in quantum theory and in classical theory can be neglected.

P.S. I'm using the following notation: $$(\text{grad}\,\varphi)_{\alpha} = \nabla_{\alpha} \varphi$$ where $$\alpha \in \{1,2,3\}$$. Thus, for example, $$\nabla_{\alpha} x_{\beta} = \frac{\partial x_{\beta}}{\partial x_{\alpha}} = \delta_{\alpha \beta}$$ $$\nabla_{\alpha} r = \nabla_{\alpha} \left(\sqrt{x_{\beta} x_{\beta}}\right) = \frac{1}{2 \sqrt{x_{\beta} x_{\beta}}} \nabla_{\alpha} \left(x_{\beta} x_{\beta}\right)= \frac{x_{\beta} \nabla_{\alpha} x_{\beta}}{ \sqrt{x_{\beta} x_{\beta}}} = \frac{x_{\alpha}}{r}$$ $$(\mathbf{a},\mathbf{b}) = a_{\alpha} b_{\alpha}$$ and so forth.

One difference is that this formula is for a dipole with $p$ in the $z$ direction only.

For the delta function I think that the derivate (the gradient components) is done in a distributional sense. You can safetely ignore it if you don't want to evaluate the field in the origin.

I) The Dirac delta distribution (and derivative thereof) in the dipole field

$$\Phi ~=~\frac{1}{4\pi\varepsilon}\frac{\vec{p}\cdot \vec{r}}{r^3} \tag{1}$$ $$\Downarrow$$ $$\vec{E}~=~-\vec{\nabla}\Phi ~=~ \frac{1}{4\pi\varepsilon}\frac{3(\vec{p}\cdot \vec{r})\vec{r}-r^2\vec{p} }{r^5} -\frac{\vec{p}}{3\varepsilon}\delta^3(\vec{r}) \tag{2}$$ $$\Downarrow$$ $$\rho~=~\varepsilon\vec{\nabla}\cdot \vec{E} ~=~ - (\vec{p}\cdot\vec{\nabla})\delta^3(\vec{r}) \tag{3}$$

is crucial. In particular, the charge distribution $$\rho$$ of the dipole is the derivative of a Dirac delta distribution. Else we couldn't use eq. (3) to confirm the formula for the dipole moment

$$\vec{p}~=~\iiint_{\mathbb{R}^3} \! d^3r~\vec{r}\rho(\vec{r})\tag{4}$$

Here the arrow $$\Downarrow$$ denotes differentiation. Note that the $$\frac{2}{3}$$ $$\left(\frac{1}{3}\right)$$ of the rhs. of eq. (3) comes from the first (second) term on the rhs. of eq. (2), respectively! To understand how the above differentiations are performed, see next Section II.

II) Mathematically, distributions can be given a rigorous meaning with the help of test functions, see e.g. my Phys.SE answer here. A more physical approach is via regularization

$$\Phi_{\delta} ~=~\frac{1}{4\pi\varepsilon}\frac{\vec{p}\cdot \vec{r}}{\left(r^2+\delta\right)^\frac{3}{2}} \tag{1'}$$ $$\Downarrow$$ \begin{align} \vec{E}_{\delta}~=~-\vec{\nabla}\Phi_{\delta} ~=~& \frac{1}{4\pi\varepsilon}\left[\frac{3(\vec{p}\cdot \vec{r})\vec{r} }{\left(r^2+\delta\right)^\frac{5}{2}} -\frac{\vec{p}}{\left(r^2+\delta\right)^\frac{3}{2}} \right] \cr ~=~& \frac{1}{4\pi\varepsilon}\left[\frac{3(\vec{p}\cdot \vec{r})\vec{r}-r^2\vec{p} }{\left(r^2+\delta\right)^\frac{5}{2}} -\frac{ \vec{p}\delta}{\left(r^2+\delta\right)^\frac{5}{2}} \right]\end{align} \tag{2'} $$\Downarrow$$ $$\rho_{\delta}~=~\varepsilon\vec{\nabla}\cdot \vec{E}_{\delta} ~=~ - \frac{15(\vec{p}\cdot \vec{r})\delta}{4\pi\left(r^2+\delta\right)^\frac{7}{2}} ~=~ - (\vec{p}\cdot\vec{\nabla}) \frac{3\delta}{4\pi\left(r^2+\delta\right)^\frac{5}{2}} \tag{3'}$$

by introducing a regularization parameter $$\delta>0$$. Note that the regularized potential $$\Phi_{\delta}\in C^{\infty}(\mathbb{R}^3)$$ is a smooth function everywhere, so differentiation of eq. (1') is well-defined, and it results in eqs. (2') and (3'), as the reader can easily verify.

Finally recall that the 3D Dirac delta distribution can be represented as

$$\delta^3(\vec{r})~=~\lim_{\delta \searrow 0^+} \frac{3\delta}{4\pi\left(r^2+\delta\right)^\frac{5}{2}} \tag{5}$$

to derive eqs. (2) and (3) from eqs. (2') and (3'), respectively.

To prove formula (5), smear with a test function $$f\in C^{\infty}_c(\mathbb{R}^3)$$:

$$\iiint_{\mathbb{R}^3} \! d^3r~ f( \vec{r})\frac{3\delta}{4\pi\left(r^2+\delta\right)^\frac{5}{2}} ~\stackrel{\begin{matrix}\text{substitution}\\ \vec{r}~\to~\sqrt{\delta}~\vec{r}\end{matrix}}{=}~ \iiint_{\mathbb{R}^3} \! d^3r~ f( \sqrt{\delta}\vec{r})\frac{3}{4\pi\left(r^2+1\right)^\frac{5}{2}}$$ $$~\stackrel{\begin{matrix}\text{Leb. dom.}\\ \text{conv. thm.}\end{matrix}}{\longrightarrow}~ f( \vec{0})\iiint_{\mathbb{R}^3} \! d^3r~\frac{3}{4\pi\left(r^2+1\right)^\frac{5}{2}}~=~f( \vec{0})\int_0^{\infty} \! dr~\frac{3r^2}{\left(r^2+1\right)^\frac{5}{2}}$$ $$~=~f( \vec{0})\left[ \! \frac{r^3}{\left(r^2+1\right)^\frac{3}{2}}\right]_{r=0}^{r=\infty}~=~f( \vec{0}) \quad\text{for}\quad \delta ~\searrow~ 0^+.\tag{6}$$