Electric field due to a point dipole

[Hypochondriac]

Let's say I have a point dipole (as an approximation for an atom) at the origin and it oscillates in the $$z$$ axis. The (theta component of the) electric field due to this dipole in the far field will be

$$E = \frac{d}{4\pi\epsilon_0}\frac{k^2\sin\theta}{r}\exp i\left(kr-\omega t\right)$$

where $$d$$ is the dipole moment, $$k=2\pi/\lambda$$, $$\theta$$ is the angle made with the $$z$$ axis, $$r$$ is the radial distance, $$\omega$$ is the angular frequency of the oscillation and $$t$$ is the time.

Due to the zero $$\phi$$ dependance, i.e. the angle in the equatorial plane, there is a cylindrical symmetry. Instinct tells me that I should I have two lobes of electric field, one in the $$+z$$ direction and one in $$-z$$, which will oscillate, alternatively between positive and negative. However the equation I quoted implies a doughnut shaped electric field. For a given $$r$$, $$E$$ increases as $$\theta$$ goes from 0 to $$\pi/2$$, then decreases from $$\pi/2$$ to $$\pi$$. I.e. I think the equation should have a $$\cos\theta$$ in it instead of a $$\sin\theta$$

Where have I gone wrong? I think it's my definition of $$\theta$$, however it is always defined from the $$z$$ axis.

PS. I'm pretty sure the equation is in Jackson.

 

[G01]

This is the correct form for a point dipole oscillating along z. The intensity profile for a point dipole source is indeed like a donut.

Since the dipole is oscillating along z, there is 0 intensity at $\theta=0$, and the intensity is maximum $\theta=\pi /2$ in the x-y plane. I think if you take a cross section of the donut on the z-y or z-x planes you'll find the lobes you are picturing. However, the lobes are centered in the x-y plane, not along z.

For reference, see Griffith's Eq. 11.18 and Figure 11.4