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 1


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.

  • $\begingroup$ What does this Fourier transform procedure mean in an intuitive way? $\endgroup$ Sep 24, 2020 at 10:12
  • $\begingroup$ Fourier transform for "future" EM wave has no sense due to causality, switching to Laplace transform allows for "past" (infinite past, t=-inf, to now, t=0) EM wave frequency retrieval, when the wave has known states or frequencies. This transform allows to retrieve all frequencies from time representation of EM wave in the "past". Since this is monochromatic wave, there is only omega_0 frequency. Does that anyhow answer your question? $\endgroup$
    – Radek D
    Sep 25, 2020 at 9:13

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