I am trying to check if the classical electromagnetic sources from a point electric/magnetic dipole do form a true four-current. In this SE post, it is shown that a point electric charge do transform covariantly between inertial reference frames (IRF). As a point dipole is just a case where two opposite point charges are placed very close to each other, and also as Lorentz transformations (LT) are linear, my first guess to the question would be that such four-current is indeed valid.

Mathematically, the charge density $\rho$ and current density $\mathbf{J}$ due to a static point dipole at rest are known to be $\rho = -(\mathbf{p}\cdot\nabla)\delta^3(\mathbf r)$ and $\mathbf J=-(\mathbf m \times \nabla)\delta^3(\mathbf r)$, where $\mathbf p = q \mathbf d $ is the electric dipole moment, with $q$ being the charge and $\mathbf d$ the charges' distance, $\mathbf m = (1/2)\int(\mathbf r \times \mathbf J) \mathrm{d}^3\mathbf r$ is magnetic dipole moment, $\delta^3(\mathbf r)$ is the three-dimensional Dirac delta distribution and $\mathbf r$ is the position vector. These equations are obtained from the electromagnetic potentials -- see J. D. Jackson's book, for example.

For simplicity, let us assume another IRF moving with velocity $\mathbf u = (u,0,0)$ relative to the dipole's rest frame, which shall be denoted with primed variables. Applying the LT, a four-current $(c\rho,\mathbf J)$ transforms as $\rho'=\gamma \rho-\gamma u J_x/c^2$ and $J_x' = -\gamma u \rho + \gamma J_x $, $J_y'=J_y$, $J_z'=J_z$.

For $t=0$, $\mathbf r$ and $\mathrm{d}^3\mathbf r$ transform as $\mathbf r'=(\gamma x,y,z)$ and $\mathrm{d}^3\mathbf r'=\gamma^{-1}\mathrm{d}^3\mathbf r$, respectively. Thus, the electric dipole moment must transform as \begin{eqnarray} \mathbf p' &=& \int \rho'\mathbf r'\mathrm{d}^3\mathbf r'\nonumber\\ &=& \int (\gamma \rho-\gamma J_x u/c^2)(\gamma x,y,z) \gamma^{-1}\mathrm{d}^3\mathbf r\nonumber\\ &=& \gamma \mathbf p_{\parallel} + \mathbf p_{\perp}-\frac{1}{c^2}\int (\mathbf u \cdot\mathbf J)(\gamma x,y,z)\mathrm{d}^3\mathbf r. \end{eqnarray} To solve the last integral, we use the result $\int r_i J_j \mathrm{d}^3\mathbf r=\epsilon_{ijk}m_k$, which is valid for static configurations. Therefore, \begin{eqnarray} \mathbf p' &=& \gamma \mathbf p_{\parallel} + \mathbf p_{\perp}-\frac{u}{c^2}(0-m_z+m_y)\nonumber\\ \mathbf p' &=& \gamma \mathbf p_{\parallel} + \mathbf p_{\perp}-\frac{1}{c^2}\mathbf u \times \mathbf m. \end{eqnarray} Using this result and recalling that $\nabla'=(\gamma^{-1}\partial_x,\partial_y,\partial_z)$ and $\delta'^3(\mathbf r')=\gamma \delta^3(\mathbf r)$, it is possible to show that $\rho'$ has the expected form. However, as the volume transforms as $V'=V/\gamma$, this result for $\mathbf p'$ also leads to an incorrect transformation equation of the polarization field $\mathbf P$, which is known to be $\mathbf P= \mathbf{P}_{\parallel}+\gamma(\mathbf P_{\perp}-\mathbf u \times \mathbf M/c^2 )$.

The magnetic dipole moment $\mathbf m$, on its turn, transforms as \begin{eqnarray} \mathbf m' &=& (1/2)\int \mathbf r'\times\mathbf J'\mathrm{d}^3 \mathbf r'\nonumber\\ &=& \gamma^{-1}\mathbf m_{\parallel}+ \mathbf m_{\perp} +\mathbf u \times \mathbf p, \end{eqnarray} which now leads to a correct transformation of the magnetization field $\mathbf M$, but an incorrect transformation of $\mathbf J'$. This result for $\mathbf m'$ can be found, for example, in this old paper and, to first order in $u/c$ (i.e., $\gamma \approx 1$), in this paper.

Alternatively, one could look directly into the transformations of the four-potential of static dipoles. I won't discuss this here for brevity, but this procedure also shows a covariant transformation for $\mathbf p'$ and $\mathbf m'$ only if $\gamma \approx 1$.

The steps developed here show that, contrary to my initial intuition, the dipolar sources generally do not transform convariantly between IRFs, and therefore can not be associated to a four-current. I would like to understand this result better. As any physical, electromagnetic source can be described as a multipole expansion, a reasonable explanation is that transforming the dipolar sources between IRFs actually mixes them with higher other terms in the expansions, making the interpretation of the new sources impossible in the terms of dipoles only. In this scenario, the transformation of point charges would just be a special case where there is no such mixture with higher order terms. Is this reasoning correct?

  • $\begingroup$ Where in Jackson did you see this? \rho = -(\mathbf{p}\cdot\nabla)\delta^3(\mathbf r) $\endgroup$ Jun 12, 2023 at 18:31
  • $\begingroup$ Problem 6.21 of 3rd edition. Alternatively, you can also find in Andrew Zangwill's book "Modern Electrodynamics". Also, you can quick check for yourself that $\int \rho \mathbf{r} \mathrm{d}^3\mathbf{r} = \mathbf p$, as required for self-consistency. $\endgroup$
    – Woe
    Jun 12, 2023 at 18:51

2 Answers 2


I think that you made a mistake in your calculations. Essentially, I think that it boils down to the fact that you forgot to change time as well. For relativistic calculations, it is always best to set $c=1$ as it only clutters the equations.

The Lorentz transformation is given by: $$ \begin{align} t' &= \gamma(t-ux) & x' &= \gamma(x-ut) & x_\perp' &= x_\perp \\ t &= \gamma(t'+ux') & x &= \gamma(x'+ut') & x_\perp &= x_\perp' \\ \end{align} $$ with $\gamma = \frac{1}{\sqrt{1-u^2}}$. You need to be more careful with the delta and the partial derivatives: \begin{align} \delta(\vec x) &= \delta(x)\delta(x_\perp)\\ &= \frac{1}{\gamma}\delta(x'+ut')\delta(x_\perp')\\ \partial_x &= \gamma(\partial_x'-u\partial_t') \\ \nabla_\perp &= \nabla_\perp' \\ \partial_x\delta(x) &= \gamma(\partial_x'-u\partial_t') \frac{1}{\gamma}\delta(x'+ut') \\ &= (1-u^2)\delta'(x'+ut') \\ &= \frac{1}{\gamma^2}\partial_x'\delta(x'+ut')\\ \vec p \cdot \nabla \delta(\vec x) &= \gamma^{-1}(\gamma^{-1}p_x\partial_x'+p_\perp\cdot \nabla_\perp)\delta(x'+ut')\delta(x_\perp') \\ (\vec m \times \nabla \delta(\vec x))_x &= \gamma^{-1}m_\perp\times \nabla_\perp'\delta(x'+ut')\delta(x_\perp') \end{align} Note that already the $\gamma$ factors are different than yours. You don't need to integrate, it's easier to use only the deltas (one less source of mistake when converting the volume element): $$ \begin{align} \rho'(t',\vec x')&= \gamma(\rho(t,x)-uj_x(t,x)) \\ &= -(\gamma^{-1}p_x\partial_x'+p_\perp\cdot \nabla_\perp'-u m_\perp\times \nabla_\perp')\delta(x'+ut')\delta(x_\perp') \\ p_x' &= \gamma^{-1}p_x\\ p_\perp' &= p_\perp-\vec u\times \vec m \end{align} $$ so you got the wrong $\gamma$ factor for the parallel component. Btw, the advantage of actually calculating the new charge density is also to check that it is the result of an electric dipole and that there are no new charge distributions that intervene.

Similarly for the magnetic dipole: $$ (\vec m \times \nabla \delta(\vec x))_\perp = \gamma^{-1}(m_xe_x\times \nabla_\perp'+\gamma^{-1}m_\perp\times e_x \partial_x')\delta(x'+ut')\delta(x_\perp') $$ so: $$ \begin{align} j_x'(t',\vec x')&= \gamma(j_x(t,x)-u\rho(t,x)) \\ &= -(m_\perp\times \nabla_\perp'-u(\gamma^{-1}p_x\partial_x'+p_\perp\cdot \nabla_\perp'))\delta(x'+ut')\delta(x_\perp') \\ m_\perp' &= m_\perp+\vec u\times\vec p \\ j_\perp'(t',\vec x') &= j_\perp(\vec x)\\ &= -\gamma^{-1}(m_xe_x\times \nabla_\perp'+\gamma^{-1}m_\perp\times e_x \partial_x')\delta(x'+ut')\delta(x_\perp')\\ m_x' &= \gamma^{-1}m_x \end{align} $$ So I recover your transformation law for the magnetic moment. There is a small catch though. The point of looking at the Dirac deltas is to check whether the current distribution is still a magnetic dipole. You can see, there are some extra terms that cannot arise from a magnetic dipole: $$ \begin{align} \vec j' &= \vec j'_s+\vec j'_d \\ \vec j'_s &= -\vec m'\times\nabla\delta(x'+ut')\delta(x_\perp') &&\text{(static)}\\ \end{align} $$ This is because the charge distribution is not stationary (moving dipole). The dynamic component can be explained by the continuity equation: $$ \partial_t'\rho'+\nabla'\cdot\vec j_d' =0 $$ Due to the dynamic component, you'll need to take into account the velocity of the dipoles as well in the LR (given by the usual velocity addition formula).

Things should now be consistent.

Hope this helps.

  • $\begingroup$ You wrote $\mathbf{p}\cdot\nabla = \gamma^{-1}p_x\partial_x'+\mathbf{p}_{\perp}\cdot\nabla_{\perp}$. Should not it be $\mathbf{p}\cdot\nabla = \gamma p_x(\partial_x'-u\partial_t')+\mathbf{p}_{\perp}\cdot\nabla_{\perp}$? Even if the time derivative term is zero, your version has $\gamma^{-1} p_x$ while mine has $\gamma p_x$. I think this is the only difference in the $p_x$ transformation, even though I obtained it by integration. $\endgroup$
    – Woe
    Jun 14, 2023 at 23:59
  • $\begingroup$ I don't think that I made a mistake. I've added intermediate steps for clarity. $\endgroup$
    – LPZ
    Jun 15, 2023 at 7:59
  • $\begingroup$ In these new added steps, you wrote $(\partial_x'-u \partial_t')\delta(x'+u t') = (1-u^2)\partial_x' \delta(x'+u t')$. Taking one extra intermediate step for this calculation (the chain rule), one would have $(\partial_x'-u \partial_t')\delta(x'+u t') = \delta'(x'+u t')-u^2\delta'(x'+u t')$. These $\delta'$ are derivatives w.r.t. different variables ($x'$ and $t'$), no? Why in your result you took them to be equal? $\endgroup$
    – Woe
    Jun 15, 2023 at 22:52
  • $\begingroup$ Actually I did put the intermediate step with the chain rule. They are equal simply due to the general fact: $$\partial_tf(x+ut) = u\partial_xf(x+ut)$$ $\endgroup$
    – LPZ
    Jun 15, 2023 at 22:57
  • $\begingroup$ Correct, thank you. I was troubled by these deltas. On the other hand, generally, I see from your answer that if $\mathbf{p}$ and $\mathbf{m}$ transform the way you showed, the electromagnetic sources are indeed covariant (provided we consider the extra term from the continuity equation). Therefore, it would remain only to show these transformations explicitly. It should, of course, be possible to do that from the integral definitions of $\mathbf{p}$ and $\mathbf{m}$ as I tried, but my results are incorrect for $\mathbf{p}$, and this is not due to the deltas. Do you have any clues on that? $\endgroup$
    – Woe
    Jun 15, 2023 at 23:40

Thank you for the references. I guess I had not noticed them because I do not look at problems in textbooks. As a minor point, Eq.(4.20) in Jackson has a $1/(3\epsilon_0)$, which I think should propagate to the later equation.

More importantly, the equations for E of a dipole in Jackson and Zangwill are averaged over solid angle. This is because they guess from an integral over $\Omega$ that the function inside has no angular dependence. I show in my textbook that that guess is wrong. A direct derivation, without averaging, leads to my equation (2.66): ${\bf E(r)}= \frac{3({\bf p\cdot{\hat r}){\hat r}}-{\bf p}}{r^3}-{\bf{\hat r}({\hat r}\cdot p)}\delta({\bf r})$, in Gaussian units. Then, the equation for $\rho$ becomes $4\pi\rho=-\left[\frac{3{\bf(r\cdot p)\delta(r)}}{r^2}\right] -{\bf({\hat r}\cdot p)}\delta({\bf r})$.

If your equations are not averaged over angle, these new equations may resolve things, although with considerable complication.


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