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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.

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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.

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