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For such a simple question I'm finding it remarkably hard to get a definitive answer. Googling has not helped me. Consider an ideal electric dipole that is constant i.e. neither its magnitude nor direction change with time. If we apply an acceleration (possibly time dependent) to this dipole then does it produce electromagnetic radiation in an analogous way that an accelerating charge radiates?

Bonus question: does the same apply to a magnetic dipole?

I am interested in the purely classical theory as described by Maxwell's equations, so not in effects arising from quantum electrodynamics.

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    $\begingroup$ Is the dipole orientation fixed in space? If not, then you can have a trivial trajectory and a rotating dipole which obviously radiates. If yes, then you can get the full fields by differentiating the Liénard-Wiechert fields in the direction of the dipole, since the charge distribution of a dipole is $\delta'(\mathbf r) $. $\endgroup$ – Emilio Pisanty Mar 11 '18 at 13:07
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    $\begingroup$ For any charge configuration, if the lowest non-vanishing multipole moment is the $n$th multipole moment, then all higher multipole moments are dependent on position. So the configuration you're describing would have a quadrupole moment with a non-zero acceleration, and so I would expect an accelerating dipole to emit quadrupole radiation. $\endgroup$ – Michael Seifert Mar 11 '18 at 13:30
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    $\begingroup$ I believe the answer is no, as can be seen from Feynman's delayed field expression derived from the Liénard-Weichert potentials; its form is here. Only the third term has a $1/R$ dependence, and thus a nonzero contribution to the power flux through a sphere of radius $R$ as $R\to\infty$. Therefore, when you put two identical but opposite charges near to one another, the sum of the two third terms in the Feynman formula has a $1/R^2$ dependence, analogously to the static dipole, and so the power flux through the big sphere is nought ..... $\endgroup$ – Selene Routley Mar 12 '18 at 5:26
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    $\begingroup$ @WetSavannaAnimalakaRodVance: Your argument doesn't account for the fact that the retarded time is slightly different for the two charges. If there is in fact radiation in this case (and I'm not 100% sure that there is now), I suspect that it's due to this fact. $\endgroup$ – Michael Seifert Mar 12 '18 at 19:25
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    $\begingroup$ @MichaelSeifert That sounds like a good point, and if true it would be borne out in a full calculation of from the LW potentials. The clincher is then the effect of the delay on the dependence. I also suspect relativistic effects are also highly significant, and one would also need to specify the rigidity of the dipole in the face of accelerations (i.e. Born rigidity or otherwise). $\endgroup$ – Selene Routley Mar 12 '18 at 21:05
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Summary: A dipole moving through space radiates. Specifically, the power radiated depends both on the dipole's acceleration and its jerk.

Finding the potentials:

Consider an idealized dipole $\vec{p}$ moving along a trajectory $\vec{w}(t)$. We assume that this dipole has a constant magnitude and direction in an inertial frame. This is somewhat artificial, particularly for motions along the direction of $\vec{p}$ (one would expect the dipole moment to Lorentz-contract); but the calculations below are complicated enough as it is, and so long as the speed of the dipole is non-relativistic, this assumptions should be valid.

The charge distribution for this dipole will be $$ \rho(\vec{r},t) = \vec{p} \cdot \vec{\nabla} \delta^{(3)}(\vec{r} - \vec{w}(t)) $$ and the current distribution will be $$ \vec{J}(\vec{r},t) = \dot{\vec{w}} \rho(\vec{r},t) = \dot{\vec{w}} \left[ \vec{p} \cdot \vec{\nabla} \delta^{(3)}(\vec{r} - \vec{w}(t)) \right]. $$

To find the potentials, we use the retarded Green's functions for the wave operator. We have $$ V(\vec{r}, t) = - \frac{1}{\epsilon_0} \iiiint G(\vec{r} - \vec{r}',t - t') \rho(\vec{r}',t') \, d\tau' dt' $$ $$ \vec{A}(\vec{r}, t) = - \mu_0 \iiiint G(\vec{r} - \vec{r}',t - t') \vec{J}(\vec{r}',t') \, d\tau' dt', $$ where $$G(\vec{r} - \vec{r}',t - t') = - \frac{c}{4\pi} \begin{cases} \delta(\mathcal{R} - ct)/\mathcal{R} & t > t' \\ 0 & t < t' \end{cases} $$ with $\vec{\mathcal{R}} \equiv \vec{r} - \vec{r}'$.

Let's consider the integral for $V$ first. We have $$ V(\vec{r}, t) = \frac{c}{4 \pi \epsilon_0} \iiiint \frac{\delta( \mathcal{R} - (t - t'))}{\mathcal{R}} \vec{p} \cdot \vec{\nabla}' \delta^{(3)}(\vec{r}' - \vec{w}(t')) \, d\tau' dt' $$ where $\vec{\nabla}'$ denotes the gradient with respect to $\vec{r}'$. Integrating by parts, the boundary term vanishes, and this becomes $$ V(\vec{r}, t) = - \frac{c}{4 \pi \epsilon_0} \iiiint \vec{p} \cdot \vec{\nabla}' \left[ \frac{\delta( \mathcal{R} - (t - t'))}{\mathcal{R}} \right] \delta^{(3)}(\vec{r}' - \vec{w}(t')) \, d\tau' dt' $$ The term being acted on by the gradient only depends on $\vec{r}'$ through $\mathcal{R}$, and for any such function, $\vec{\nabla} f(\mathcal{R}) = - \vec{\nabla}' f(\mathcal{R})$ (where $\vec{\nabla}$ is the "normal" gradient.) Thus, $$ V(\vec{r}, t) = \frac{c}{4 \pi \epsilon_0} \iiiint \vec{p} \cdot \vec{\nabla} \left[ \frac{\delta( \mathcal{R} - (t - t'))}{\mathcal{R}} \right] \delta^{(3)}(\vec{r}' - \vec{w}(t')) \, d\tau' dt' $$ Since the final delta function is independent of $\vec{r}$, we can pull the gradient operator out of the integral to obtain $$ V(\vec{r}, t) = \vec{p} \cdot \vec{\nabla} \left[ \frac{c}{4 \pi \epsilon_0} \iiiint \frac{\delta( \mathcal{R} - (t - t'))}{\mathcal{R}} \delta^{(3)}(\vec{r}' - \vec{w}(t')) \, d\tau' dt'\right] $$ But the quantity in square brackets is precisely what we would get for a point charge (with unit magnitude) moving along the trajectory $\vec{w}(t)$; in other words, this is the standard Liénard-Wiechert potential! If we denote the Liénard-Wiechert potential for such a point charge as $V_1 (\vec{r},t)$, we can conclude that $$ V(\vec{r}, t) = \vec{p} \cdot \vec{\nabla} \left[ V_1 (\vec{r},t) \right]. $$ A similar line of argument involving the vector potential results in a very similar-looking result: $$ \vec{A}(\vec{r}, t) = \vec{p} \cdot \vec{\nabla} \left[ \vec{A}_1 (\vec{r},t) \right] = \vec{p} \cdot \vec{\nabla} \left[ \frac{\dot{\vec{w}}}{c^2} V_1 (\vec{r},t) \right]. $$ Note that $\dot{\vec{w}}$ (as well as all the other distances and positions involved in the Liénard-Wiechert potentials) are evaluated at the retarded time.

Finding the radiation fields:

To obtain the electric and magnetic fields, we would now want to take the time derivatives, gradients, and curls of these expressions. But since the potentials of an accelerated dipole are related to those of an accelerated unit point charge by taking the directional derivative $\vec{p} \cdot \vec{\nabla}$, and since this operator commutes with all spatial and time derivatives, it follows that the electric and magnetic fields of an accelerated dipole are similarly related to those of an accelerated unit point charge: $$ \vec{E}(\vec{r},t) = (\vec{p} \cdot \vec{\nabla}) \vec{E}_1(\vec{r},t) \qquad \vec{B}(\vec{r},t) = (\vec{p} \cdot \vec{\nabla}) \vec{B}_1(\vec{r},t) $$

For a unit point charge in arbitrary motion, the piece of the electric field responsible for radiation is the acceleration field: $$ \vec{E}_1 \sim \frac{\mathcal{R}}{4 \pi \epsilon_0} \frac{ \vec{\mathcal{R}} \times (\vec{u} \times \vec{a})}{(\vec{\mathcal{R}} \cdot \vec{u})^3}, $$ where $\vec{u} = c \hat{\mathcal{R}} - \vec{v}$, $\vec{v} = \dot{\vec{w}}$, $\vec{a} = \ddot{\vec{w}}$, and all quantities are evaluated at the retarded time. Note that this quantity scales as $\mathcal{R}^{-1}$ at large distances; we use the symbol $\sim$ to denote "equality up to highest order in $\mathcal{R}$."

We now want to take the directional derivative of this quantity. This quantity depends on position in two ways: first, via the explicit dependence on $\vec{\mathcal{R}}$; and second, via the implicit dependence of $\vec{w}$, $\vec{v}$ and $\vec{a}$ on the retarded time $t_r$. (Note that $\vec{\mathcal{R}} = \vec{r} - \vec{w}(t_r)$ also depends implicitly on $t_r$.) Taking this derivative in the most general case is left (out of self-preservation) to the reader; instead, I will focus on the case where the charge is instantaneously at rest at time $t_r$ (cf. the usual simplification made in the calculation of the Larmor formula.) In such a case, we have $$ (\vec{p} \cdot \vec{\nabla}) \vec{\mathcal{R}} = (\vec{p} \cdot \vec{\nabla}) (\vec{r} - \vec{w}(t_r)) = \vec{p} - \dot{\vec{w}} (\vec{p} \cdot \vec{\nabla} t_r) = \vec{p}. $$ This means that any time the operator $\vec{p} \cdot \vec{\nabla}$ acts on a function of $\vec{\mathcal{R}}$, the resulting expression will scale by one power of $\mathcal{R}$ less than the original function. Since the radiation fields are those which scale as $\mathcal{R}^{-1}$, and $\vec{E}_1$ already scales as $\mathcal{R}^{-1}$, we can effectively treat $\vec{\mathcal{R}}$ as constant when taking the derivative of $\vec{E}_1$: all the terms arising from $\vec{p} \cdot \vec{\nabla}$ acting on $\vec{\mathcal{R}}$ will fall off faster than $\mathcal{R}^{-1}$.

We will still need to take the directional derivatives of $\vec{u}$ and $\vec{a}$, though. The first one works out to be $$ (\vec{p} \cdot \vec{\nabla}) \vec{u} = (\vec{p} \cdot \vec{\nabla}) (c \hat{\mathcal{R}} - \vec{v}(t_r)) \sim - (\vec{p} \cdot \vec{\nabla})\vec{v}(t_r) = - \dot{\vec{v}} (\vec{p} \cdot \vec{\nabla} t_r) = \vec{a} \left(\frac{\vec{p} \cdot \hat{\mathcal{R}}}{c} \right), $$ where we have used the fact that for a charge at rest at time $t_r$, $\vec{\nabla} t_r = - \hat{\mathcal{R}}/c$. Similarly, the second one becomes $$ (\vec{p} \cdot \vec{\nabla}) \vec{a} = - \vec{\jmath} \left(\frac{\vec{p} \cdot \hat{\mathcal{R}}}{c} \right), $$ where $\vec{\jmath}$ is the jerk of the dipole. Finally, if the dipole is at rest at time $t_r$, then $\vec{u} = c \hat{\mathcal{R}}$.

Thus, the overall radiation field is $$ \vec{E}(\vec{r}, t) = (\vec{p} \cdot \vec{\nabla}) \vec{E}_1 \\ \sim \frac{\mathcal{R}}{4 \pi \epsilon_0} \left\{ \frac{ \vec{\mathcal{R}} \times [ ((\vec{p} \cdot \vec{\nabla})\vec{u} ) \times \vec{a}]}{(\vec{\mathcal{R}} \cdot \vec{u})^3} + \frac{ \vec{\mathcal{R}} \times [ \vec{u} \times ((\vec{p} \cdot \vec{\nabla}) \vec{a})]}{(\vec{\mathcal{R}} \cdot \vec{u})^3} - 3 \frac{\vec{\mathcal{R}} \times [ \vec{u} \times \vec{a}]}{(\vec{\mathcal{R}} \cdot \vec{u})^4} \vec{\mathcal{R}} \cdot [ (\vec{p} \cdot \vec{\nabla}) \vec{u} ] \right\} $$ The first term vanishes, since $(\vec{p} \cdot \vec{\nabla}) \vec{u}$ is parallel to $\vec{a}$; and the result is that $$ \vec{E}(\vec{r}, t) \sim - \frac{\vec{p} \cdot \vec{\mathcal{R}}}{4 \pi \epsilon_0 c} \left\{ c \frac{\vec{\mathcal{R}} \times [ \hat{\mathcal{R}} \times \vec{\jmath}]}{(c \mathcal{R})^3} + 3 c \frac{\vec{\mathcal{R}} \times [ \hat{\mathcal{R}} \times \vec{a}]}{(c \mathcal{R})^4} \vec{\mathcal{R}} \cdot \vec{a} \right\} \\ = - \frac{\vec{p} \cdot \hat{\mathcal{R}}}{4 \pi \epsilon_0 c^3 \mathcal{R}} \left\{ \hat{\mathcal{R}} \times [ \hat{\mathcal{R}} \times \vec{\jmath}] + \frac{3}{c} (\hat{\mathcal{R}} \cdot \vec{a}) \hat{\mathcal{R}} \times [ \hat{\mathcal{R}} \times \vec{a}] \right\} $$ It is easy to see that we can construct trajectories $\vec{w}(t)$ for which this quantity does not vanish.

Thankfully, we don't have to go through this all again with the magnetic field. We will have $$ \vec{B}(\vec{r},t) = (\vec{p} \cdot \vec{\nabla}) \vec{B}_1 (\vec{r},t) = (\vec{p} \cdot \vec{\nabla}) \left[ \frac{1}{c} \hat{\mathcal{R}} \times \vec{E}_1 \right] \sim \frac{1}{c} \hat{\mathcal{R}} \times \left[ (\vec{p} \cdot \vec{\nabla}) \vec{E}_1 \right] = \frac{1}{c} \hat{\mathcal{R}} \times \vec{E}(\vec{r},t). $$ In the third step, we have again used the fact that any derivatives acting on $\hat{\mathcal{R}}$ will lead to terms that do not contribute to the radiation fields. The Poynting vector will then be $$ \vec{S} \propto \vec{E} \times \vec{B} \propto \hat{\mathcal{R}} E^2, $$ since $\vec{E}$ is perpendicular to $\hat{\mathcal{R}}$ and $\vec{B}$ is perpendicular to both of these. Thus, for a general motion of the dipole, there will be a finite amount of power radiated to infinity.

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  • $\begingroup$ Your expression $\hat{R}\times(\hat{R}\times j)+3(\hat{R}\cdot a)\hat{R}\times(\hat{R}\times a)$ sort of smells right when you compare it to expressions like $P=-(2/3) e^2 v\cdot j$ for a charge (see Feynman Lectures on Gravitation, p. 124, or mathpages.com/home/kmath528/kmath528.htm ). It would be interesting to see if the power resulting from this expression has similar behavior, such as vanishing when $j=0$. I'm still very skeptical about any calculation of this type because of the lack of a self-consistent theory of radiation from pointlike particles and of any universally [...] $\endgroup$ – user4552 May 22 '18 at 15:33
  • $\begingroup$ [...] acceptable definition of what constitutes radiation, as well as the problem of the apparent contradiction with standard formulations of the equivalence principle. $\endgroup$ – user4552 May 22 '18 at 15:36
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Assuming the field is purely retarded as usual, so the Lienard-Wiechert formula applies, and assuming the dipole is a real dipole made of two point charges of magnitude $q$ with finite distance $d$.

Also assuming the dipole electric moment

$$ \mathbf p = q(\mathbf r_+ - \mathbf r_-) $$

is constant in its rest frame, which means it may change in the observer's frame due to Lorentz contraction of lengths (this was not specified in the question but is more natural than $\mathbf p$ being constant in the observer frame).

Then the presence of $1/r$ electric field depends on details of how the dipole is accelerated.

If the acceleration $\mathbf A$ of the dipole as a whole is parallel to $\mathbf p$, the two point charges won't have exactly the same acceleration, due to Lorentz contraction of the distance between the two charges. The radiation fields will combine as (ignoring the angular dependence a neglecting the difference between distances $r_+$ and $r_-$):

$$ q\frac{A-\epsilon}{r} - q\frac{A+\epsilon}{r} = \frac{-2q\epsilon~~~}{r} $$ where $\epsilon$ is the Lorentz contraction correction to acceleration and $r$ is distance between the dipole and the observer. So there will be some weak residual field decaying with distance as $1/r$, which is the usual condition of radiation.

Since $\epsilon$ is proportional to $d$, we can express the acceleration field as a function of electric moment $p$: $$ \frac{-p\alpha~~~}{r} $$ where $\alpha$ is a factor depending on speed of the dipole and its time derivative.

If the acceleration $\mathbf A$ is perpendicular to electric moment vector $\mathbf p$, there is no Lorentz contraction.

EDIT the following ignores the fact that the acceleration field term depends on the particle velocity which has to be evaluated at different times for the two particles. This probably invalidates the conclusion. Thanks to Michael Seifert for pointing this out.

... the terms combine as (again, ignoring angles, but taking into account the difference between $r_+$ and $r_-$):

$$ \frac{qA}{r - \delta} - \frac{qA}{r+\delta} = A\frac{-2q\delta~~~}{r^2-\delta^2} $$ where $\delta$ is the correction to distance due to finite length of dipole.

When we expand this into Taylor series in $1/r$, there are only even powers of $1/r$, there is no $1/r$ term. Hence there is no radiation in this case.

The residual field is however proportional to acceleration and decays one order slower than the static field, similarly to radiation field of accelerated charge, so the acceleration definitely "does something".

Since $\delta$ is half the projection of dipole size $d$ onto line of observation, we can express the residual acceleration field as $$ A\frac{-p f~~~}{r^2-\delta^2} $$ where $f$ is some angular factor.

For the case of ideal dipole, we should take the limit of these results where $d\rightarrow 0,q\rightarrow \infty$ while $qd = p=\text{const}$. It is easy to see that neither of the two residual fields vanishes in this limit; there is always a field proportional to electric moment $p$.

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  • $\begingroup$ I'm not sure what you mean when you define $\epsilon$ as "the Lorentz contraction correction to acceleration." If $\epsilon$ is nonzero for constant acceleration, then I think your result contradicts the well-known fact that a charge with constant acceleration doesn't radiate. $\endgroup$ – user4552 May 22 '18 at 4:13
  • $\begingroup$ If two particles accelerate as a whole and their distance is constant in their rest frame, in the observer frame, their distance is Lorentz contracted. Thus the particles do not have the same acceleration (nor is it constant in time). Regarding the radiation during constant acceleration: charged particle with constant acceleration definitely does radiate in the sense there is $1/r$ field around it. This is easily seen from the expression of electric field based on the Lienard-Wiechert formula. Perhaps you mean some other definition of radiation. $\endgroup$ – Ján Lalinský May 22 '18 at 4:17
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    $\begingroup$ Interesting argument — and a very different approach from my own answer. However, I'm not sure that I agree with your analysis for the perpendicular case. The radiation fields also depend on the velocity of the particle at the retarded time; and since the charges are accelerating, this will be different for the two charges (unless you happen to be equidistant from both of them.) So it's not accurate to simply say that the radiation fields are proportional to $qA/r_\pm$. $\endgroup$ – Michael Seifert May 22 '18 at 14:24
  • $\begingroup$ @JánLalinský: I'm asking for a mathematical definition of $\epsilon$, which you haven't provided. Re constant acceleration, there is a discussion of this in the Feynman Lectures on Gravitation, p. 123, and a longer discussion of the same ideas here: mathpages.com/home/kmath528/kmath528.htm . It can be fun to play around with this stuff, but there is ultimately no self-consistent physical theory of radiation from pointlike objects, nor has anyone come up with a universally satisfactory definition of what constitutes a radiation field. If we believed that these foundational [...] $\endgroup$ – user4552 May 22 '18 at 15:17
  • $\begingroup$ [...] issues were unimportant, then you would deserve the Nobel prize for this calculation, because you would have shown that electrically neutral bodies can have different free-fall trajectories in the same gravitational field, contradicting one of the standard ways of stating the equivalence principle. $\endgroup$ – user4552 May 22 '18 at 15:20

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