# Is stimulated emission emitted by a rotating classical dipole subject to a resonant oscillating wave beamed?

Consider a dipole, $$\pm q$$ connected by a rigid rod of length $$2L$$, spinning around its center in the $$x-y$$ plane with angular frequency $$\omega$$, such that the charges follow $$\vec{r}_{\pm q}(t) = \pm (-L\sin\omega t,L\cos\omega t, 0)$$. An external electric plane wave $$\vec{E}_{\rm ext} = E_{\rm ext} (\cos \omega t,\sin \omega t, 0)$$ is incident on the dipole with the same $$\omega$$. The dipole changes its energy at a rate $$P_{\rm ext} = -2q\vec{E}_{\rm ext}\cdot \vec{v} = -2qE_{\rm ext} L \omega$$ directly due to the field and $$P_{\rm Larmor}=q^2 a^2 /(3\pi\varepsilon_0)$$ due to Larmor radiation where $$a^2=L^4\omega^2 + (q/m)^2 E_{\rm ext}^2$$. The energy lost by the dipole must increase the energy density in the radiation field due to Poynting's theorem, i.e. the dipole emits stimulated emission. Let us assume that the values of $$(q,m,E_{\rm ext},r,\omega)$$ are such that this dominates $$|P_{\rm ext}|\gg |P_{\rm Larmor}|$$.

What is the amplitude of the vector potential at $$r\rightarrow \infty$$ as a function of direction and the corresponding Poynting flux? Is the radiation pattern close to that of classical dipole radiation field or is it collimated and nonzero only along $$(0,0,1)$$?

Notes: I am interested in a nonrelativistic ($$v\ll c$$) first-principles derivation of this problem in classical (i.e. non-quantum) electromagnetism. In Lorentz gauge, the vector potential satisfies the wave equation $$\square A=\mu_0 j = \mu_0 q L \omega \delta(\vec{r}-\vec{r}_{+q})-\mu_0 q L \omega \delta(\vec{r}-\vec{r}_{-q})$$ which can be solved naively using the Green's function $$A=\frac{\mu_0}{4\pi}\int d^3 x' j(\vec{r}',t_{\rm ret})/|\vec{r}'-\vec{r}|$$ which leads to the Lienard-Wiechert potential, but this does not satisfy the boundary condition posed in the problem with $$\vec{E}_{\rm ext}$$ and seems to only describe $$P_{\rm Larmor}$$ but not the stimulated emission part. It seems like the simulated emission part may come from an interference between the field corresponding to the spinning dipole's Lienard-Wiechert potential and the external wave, but only if the emission is beamed. But given that the dipole is nonrelativistic $$\omega L \ll c$$ its size is much less than the wavelength $$L\ll c/\omega = \lambda/(2\pi)$$, I am surprised if this produces beamed emission. If so, setting the relative phase of the dipole oppositely to cause absorption of the external wave's energy, this would manifest in the wave as a shadow, which is unexpected given $$L\ll \lambda$$.

• All you do is superpose the usual rotating dipole field with the plane wave you put in. "Stimulated emission" in QM corresponds to the fact that in classical field theory energy is quadratic in fields, and fields superpose, leading to interference terms. Nov 12, 2022 at 21:56
• You should find that at large radii the interference terms cancel themselves out in any direction besides the direction the plane wave is going, which is why you get a large enhancement of energy flux only in that direction. Nov 12, 2022 at 21:56
• Thanks @knzhou, what I am confused by is that the rotating dipole radiation field has an infinitesimal total power across an infinitesimally narrow angle around the axis, so I don't see how the interference term of this with the plane wave may be finite. Also, the analogous stimulated absorbtion implies that you can cast a narrow shadow with an object much smaller than a wavelength? Nov 13, 2022 at 0:09
• Yes, I agree that taking the limit $r \to \infty$ is subtle; it's a very singular limit and you should find rapid oscillations of the radiated flux as a function of $\theta$, because the phase relation between the radiated wave and the plane wave changes quickly. You also can't easily extract the "size" of the shadow in such a limit. Perhaps the best thing to do would be to compute the Poynting vector exactly at finite $r$ (which is not hard in this case) and then carefully take the limit of that. Nov 13, 2022 at 20:25

The dipole changes its energy at a rate $$P_{\rm ext} = -2q\vec{E}_{\rm ext}\cdot \vec{v} = -2qE_{\rm ext} L \omega$$ directly due to the field

Yes, except there should not be a minus sign in the power formula. The correct expression for rate of change of dipole energy due to external field is

$$P_{\rm ext} = 2q\vec{E}_{\rm ext}\cdot \vec{v} .$$

and $$P_{\rm Larmor}=q^2 a^2 /(3\pi\varepsilon_0)$$ due to [Larmor radiation][1] where $$a^2=L^4\omega^2 + (q/m)^2 E_{\rm ext}^2$$. The energy lost by the dipole must increase the energy density in the radiation field due to [Poynting's theorem][2], i.e. the dipole emits stimulated emission.

Larmor formula gives Poynting energy passing through a sphere per unit time, solely due to retarded field of a single compactly moving charged body in accelerated motion in side the sphere.

This number has obvious energy interpretation as energy put into EM field by the body only under some conditions: the charged body has to have non-singular distribution of charge (no point/line charge) and there must not be other EM fields on the sphere far from the charge where the Poynting flux is calculated. It makes no sense to calculate Poynting energy of the dipole field when there is also the external field present. One must always calculate Poynting energy based on the total field if one wants to obtain a number that has energetic interpretation in the sense of work-energy theorem or local conservation of EM energy.

Here we have two charged bodies that may interact with each other via EM forces, the positive and negative particle of the dipole, and the external field. So we have three correlated fields. Larmor's formula does not describe EM energy in such complicated situations, unless we can show all the interactions have zero impact and the system behaves as two isolated accelerating charged bodies.

What is the amplitude of the vector potential at $$r\rightarrow \infty$$ as a function of direction and the corresponding Poynting flux?

given $$L\ll \lambda$$.

This depends on the gauge, but it is not very interesting. Electric field will have two components, constant with distance (due to external field plane wave) and a one decaying with distance as $$1/r$$ (due to retarded radiation field of the dipole).

If the charges are so close, their radiation fields will be similar to the case where they are at the same point of space, moving with opposite accelerations. Thus the radiation field will be twice that of single particle, and will have the torus-like angular pattern in space, with torus plane perpendicular to acceleration, radiating zero intensity in direction of acceleration, and radiating the most in the perpendicular directions. Since the acceleration rotates in plane, so will the torus pattern. In the plane of rotation, from which the apparent motion of the charged bodies is oscillation in line, radiation will be oscillating in intensity, going through zero when real acceleration points towards the observer. From points on the z axis, where the apparent motion of the particles is orbiting in circles, the radiation field will be constant in intensity, only changing in direction.

but this does not satisfy the boundary condition posed in the problem with $$\vec{E}_{\rm ext}$$ and seems to only describe $$P_{\rm Larmor}$$ but not the stimulated emission part.

The retarded field of the charged particles manifests both the "spontaneous" radiation due to rotation of the dipole, and the "stimulated" radiation due to external field slowing the rotation down. But indeed this does not obey the "boundary condition" where the external field is present. It is not supposed to, external field is an additional field. Only total field will obey that "boundary condition".

It seems like the simulated emission part may come from an interference between the field corresponding to the spinning dipole's Lienard-Wiechert potential and the external wave,

In the total field in space, stimulated emission will be seen indeed as due to interference of these two fields. But since the external field is not changed by anything, the pattern is already in the radiation field of the particles.

but only if the emission is beamed.

Not sure why you think this is important. There is no reason there would be unidirectional beam created here. There will be broad directionality to the interference pattern, appropriate to combination of the plane circularly polarized external field and dipole radiation field. The shadow will be there, but it won't have "sharp boundaries".

• Thanks. I'd like to see explicitly the total outgoing energy in radiation as a function of direction at distance $r\rightarrow\infty$ and see that when integrating over the sphere this is increased by $-q\vec{E}_{\rm ext}\cdot\vec{v}$, relative to the case when $\vec{E}_{\rm ext}=0$. Nov 13, 2022 at 4:05
• All fields are known, so this seems to be a straightforward, albeit quite a laborius task. Nov 14, 2022 at 3:42

The incorrect way: just superpose the solutions, compute Poynting flux

Far from the dipole we have the leading dipole solution $$\vec{A}_d = -\frac{\mu_0 \omega}{4 \pi r} e^{i \omega(t-r)} \vec{d} +c.c., \phi_d = - \frac{\mu_0 \omega}{4\pi r} e^{i \omega(t-r)}\vec{d}\cdot\hat{r}+c.c.$$ Here I am using $$c=1$$ units, $$c.c.$$ stands for complex conjugate, $$\hat{r} = (x,y,z)/r$$ is the unit distance vector, and the formal complex dipole vector reads $$\vec{d} = 2Lq(1,i,0)$$. The wave potential is then most conveniently expressed in the Gibbs gauge $$\phi=0$$: $$\vec{A}_W = \frac{\vec{a}}{\omega} e^{i\omega(t-z)}+c.c., \phi_W = 0$$ Here $$\vec{a} = E_{ext} (1,i,0)$$.

By adding these potentials, it is then easy to obtain the total $$\vec{E}$$ and $$\vec{B}$$. The Poynting flux vector can then be computed simply by "turning the crank". I am only showing the final result for the radial flux through the angle element $$d \theta$$, integrated over $$\varphi$$ ($$r,\theta,\varphi$$ standard polar coordinates)

$$\int_0^{2\pi}\frac{r^2 \sin\theta}{\mu_0} \hat{r}\cdot \left( \vec{E}\times\vec{B}\right) d\varphi = F_{dip.}+F_{cross.}+F_{wave}$$ $$F_{dip.}$$ and $$F_{wav.}$$ are the same terms as with the dipole on its own and the wave on its own. The $$F_{cross}$$ term is new and reads: $$F_{cross} = \frac{1}{2} E_{ext.}L q r \omega ^2 \sin \theta \left(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right)^4 \cos (r \omega (\sin \theta -1))$$ It has a weird behavior that I find hard to understand. As $$r\omega \gg 1$$, the cross-flux wildly oscillates, here it is plotted at $$r\omega=150$$ as a function of $$\theta$$:

The total flux $$\int F_{cross} d\theta$$ changes sign depending on the value of $$r\omega$$, and it seems to converge to zero. Here it is plotted as a function of $$r\omega$$:

So what is the meaning of this? After some thought, I believe that $$F_{cross}$$ balances out the energy needed to keep the system in a steady state for an indefinite time, which follows from implicit assumptions. The rotating dipole is not really moving in the external field, no equations of motion are being solved - so one cannot get consistent momentum-energy balances.

We could then instead choose to solve equations of motion. This would require allowing for a dynamical and independent $$\omega(t)$$ of the dipole (or rather phase $$\varphi(t)$$). The evolution would depend (nonlinearly) on the masses of the charges, and the solution would be nonstationary. As a result, one should get a self-consistent radiative field and also the correct balance reflecting $$P_{ext.}$$ in the fluxes. I think that verifying this would amount to a neat Bachellor's or even Master's thesis.

• I'm usually a fan of your answers but I disagree with lots of this. Yes, $F_{\mathrm{cross}}$ oscillates in $\theta$, but that's a sensible result. That always happens as energy sloshes in and out of the near field, and that behavior survives to large $r$ in this problem because the plane wave is infinite. Furthermore, there's nothing wrong with forcing the charges to move in a circle. That requires an external source of energy and momentum, but away from the charges themselves energy and momentum are locally conserved. Nov 13, 2022 at 20:29
• Finally, this really is the exact classical analogue of stimulated emission, and there is net momentum absorbed from the incoming plane wave. It comes from the magnetic force, and it's compensated by whatever external force keeps the charges moving in a circle in the $xy$ plane. Nov 13, 2022 at 20:30
• Actually, as a final note, I think you already completely solved the OP's question, even though you don't seem to believe you did! At large but finite $r$, your solution implies there's a large positive contribution to the forward energy flux at angles $\theta \lesssim 1 / (r \omega)$. That implies a "spike" of linear size $r \theta \sim 1 / \omega \sim \lambda$, on the scale of the wavelength, which is precisely the physical behavior OP expects! Nov 13, 2022 at 20:34
• @knzhou I adjusted the post to reflect your comments. I took another numerical look and it seems that the total flux of the cross term through infinity actually goes to zero.
– Void
Nov 13, 2022 at 21:30
• @bkocsis Yes, it seems weird given that the far-field of the isolated dipole predicts correct radiation reaction. Then again, infinite energy is pumped into the system and leaves it at every point through the plane wave, so perhaps this is some sort of breakdown of the theory.
– Void
Nov 20, 2022 at 18:44