# Electric current dipole moment

Electrical Current Dipole of dipole moment $$\mathbf{p_{EC}}$$ has unit of $$A\cdot m$$. My knowledge of dipole moments unfortunately stopped in high school where I learned that dipole moment $$\mathbf{p}$$ has unit of $$C\cdot m$$. I know also that $$C = A\cdot s$$. From my recent research in the topic I have learned that the $$\mathbf{p_{EC}}$$ is electrodynamic dipole moment and the other one is electrostatic, such that $$\mathbf{p_{EC}}=\frac{d}{dt}\mathbf{p}$$ [link]. The transition between the definitions is clearly about specifying the time.

The question is, when I have a molecule of static dipole moment $$\mathbf{p}$$ illuminated with electromagnetic wave of frequency $$f$$, what time do I have to take into consideration when defining the molecule's dynamic dipole moment? I have an idea, however, please, tell me if it is valid anyhow (if not - what should be changed).

The idea: Because EM wave is like an AC current and the time-average displacement of electrons is zero, I take the half-period ($$2f$$) of the EM wave as in the time the electrons in the molecule move due to electric field direction (before changing the direction for next half-period). This way a molecule having dipole moment of 1 [D] would, in the point-source approximation, have electric current dipole moment of $$1\,[D]\cdot 2f$$.

I think I need to define the dipole moment in terms of point-source approximation because this is what I know is feasible in COMSOL's RF Module in order to simulate plasmonic enhancement of molecule's radiative decay in the vicinity of a plasmonic nanoparticle.

## 1 Answer

I have found that in order to retrieve the time I have to Fourier transform the electromagnetic field.

$$\int_{-\infty}^{\infty} E_0\times\exp(-jkz)\exp(j\omega_0 t)\exp(-j\omega t)dt$$

Then I realised that $$\exp(j\omega_0 t)\exp(-j\omega t)=\exp(-st)$$ , which is like in Laplace transform where $$s=0-j(\omega_0-\omega)$$.

Taking Laplace transform in the ROC:

$$E_0\times\exp(-jkz)\int_{-\infty}^{0} \exp\left(j(\omega_0-\omega) t\right)\,dt$$

I get this: $$E_0\times\exp(-jkz)\times j(\omega_0-\omega)\, ,$$ the magnitude of which is $$E_0\times\exp(-jkz)\times(\omega_0)$$ for monochromatic EM wave (which is the case).

Therefore the only thing that comes out of the transform that is different from the "original" oscillator is its frequency. So $$p_{EC}=p\times\omega_0$$. Which is two times smaller than what came out of my "intuitive" approach.