Electric field and Fourier transform

Question

I am having problems understanding the following equations due to difficulties with Fourier transforms.

The frequency dependent electric field is related to polarization, vacuum permittivity and susceptibility by

$$ \boldsymbol P(\omega)= \varepsilon_0 \boldsymbol \chi(\omega)\boldsymbol E(\omega) $$

I am considering a field $\boldsymbol E(t)=\boldsymbol E_0 e^{i\omega t}$.

In this article, the author does

$$ \frac{\boldsymbol P(\omega)}{\boldsymbol E(\omega) } = \frac{\boldsymbol P_0}{\boldsymbol E_0} \frac{1}{\omega_0+i\omega} $$

while considers for some systems that $\boldsymbol P(\omega)=\boldsymbol P_0 \frac{1}{\omega_0+i\omega}$

Intuitively, I found that reasonable but this imply that the Fourier transform of the electric field is $\boldsymbol E_0$. But, the Fourier transform should be $2\pi \boldsymbol E \delta(\omega^\prime - \omega)$, due to

$$\mathcal{F}^{-1}\{\delta(\omega^\prime - \omega)\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega^\prime-\omega)e^{i\omega^\prime t}d\omega^\prime=e^{i\omega t}$$

So I am confused with $2\pi\delta(\omega^\prime - \omega)$. How to get the expression obtained by the author?

Answer 1

There are a few technical issues with the mathematics in that article, but I think the overall logic/presentation is great.

To begin, the relation:

$$\mathbf{P}(\omega)=\frac{\mathbf{P}_0}{\omega_0+i\omega}\tag{1}$$

was derived using the model of a step-function electric field and an exponentially-decaying response $\mathbf{P}(t)$ with characteristic time $\tau\equiv 1/\omega_0$. If you notice though, this was derived using an odd definition for the Fourier transform:

$$\mathbf{P}(\omega) \overset{?}{=}\int_{\color{red}{0}}^{\infty}\mathbf{P}(t)e^{-i\omega t}\,dt\tag{2}$$

The lower limit should be $-\infty$, but that would lead to a $\delta$ function for both $\mathbf{P}(\omega)$ and $\mathbf{E}(\omega)$. What the author has effectively assumed then is a model in which $\mathbf{E}(t)=\mathbf{E}_0\delta(t)$ (nevermind units) and:

$$\mathbf{P}(t)=\begin{cases}0 &,t<0\\ \mathbf{P}_0 e^{-\omega_0 t} &, t>0\end{cases}\tag{3}$$

This is a little odd, so let me clear the air a little.

The relation $\mathbf{P}(\omega)=\chi(\omega)\mathbf{E}(\omega)$ is an abstract/general, but very physical, relation between the polarization of an object and the applied electric field (assuming sufficient uniformity of the field, or sufficient smallness of the object, or both). You can easily get it by assuming that in the time-domain, $\mathbf{P}(t)$ is related to the applied field $\mathbf{E}(t)$ through a linear integral operator:

$$\mathbf{P}(t)=\int_{-\infty}^{t}\epsilon_0\chi_e(t-t')\mathbf{E}(t')dt'\tag{4}$$

where I have limited the upper integration bound to $t$ under the assumption of causality: $\mathbf{P}$ can't possibly respond to the future values of $\mathbf{E}$. Notice that in this formulation, if $\mathbf{P}$ responds only to the instantaneous value of the applied field $\mathbf{E}$ at, say, the same instant $\mathbf{P}$ is measured (i.e. infinitely fast information travel + no hysteresis (Markovian)), then the kernel $\chi_e(t-t')$ will now be a $\delta$ function:

$$\mathbf{P}(t)=\int_{-\infty}^{t}\epsilon_0\chi_e\delta (t-t') \mathbf{E}(t')dt'=\epsilon_0\chi_e \mathbf{E}(t)\tag{5}$$

but in general this will not be the case - the response (which we are assuming to be linear!), will generally depend on the past values of the applied field. Notice, if we now take the Fourier transform of eq. (4), we get:

$$\mathbf{P}(\omega)=\epsilon_0\chi_e(\omega)\mathbf{E}(\omega)\tag{6}$$

where every function with the argument $\omega$ is understood to be the Fourier transform of its respective time-domain counterpart, i.e. $\chi_e(\omega)\equiv \mathcal{F}\{\chi_e(t)\}$.

This frequency-domain relationship is something you literally measure by applying an electric field $\mathbf{E}=\mathbf{E}_0\sin\omega t$, and then measuring the resulting $\mathbf{P}(t)$ which we physically expect to also be a sinusoid (with a possible phase shift). The relation between the strengths (amplitudes) and phase-shift between $\mathbf{E}(t)$ and $\mathbf{P}(t)$ are meant to be captured by that general relationship.

But as you notice, we cannot just say that $\mathbf{E}(\omega)$ is the Fourier-transform of $\mathbf{E}(t)$. The Fourier transform of this is a $\delta$-function (well two, technically). This is a problem, of course, because we cannot "divide" $\delta$-functions, because they are not really functions. They are distributions, which are not defined point-wise.

This problem is an artifact of the assumed continuous nature, and infinite extent, of $\omega$. If we assume $\omega$ can take values from a continuum, of course you're going to get a distribution in the end. This problem wouldn't exist if we had used a Fourier series. You can kind of see these by noticing the mere units of a Fourier transform are not the same as the Fourier-series coefficients.

In reality, we do not sample the material for an infinite amount of time, and thus we do not have a $\delta$-function Fourier transform. We have a finite sample time $T\gg 1/\omega$. If we assume that over that sample time our applied field was a single frequency $\mathbf{E}(t)=\mathbf{E}_0e^{i\omega_0t}$, you get a $\text{sinc}$ function for the Fourier transform, which of course converges to a $\delta$-function in the limit $T\rightarrow\infty$. My point is, if you do all your work before/without taking the limit $T\rightarrow\infty$, you will have no problems. Similarly, if you work in the real world (with finite and discrete sampling), you will work with Fourier coefficients (discrete Fourier transform), and you will never have any problems of this kind (arising from $\delta$-functions).

So when we apply an Electric field of the form $\mathbf{E}(t)=\mathbf{E}_0e^{i\omega_0 t}$, it must be understood that $\mathbf{E}(\omega)=\delta_{\omega,\omega_0}\mathbf{E}_0$, where that is a Kronecker-delta, which is not the Fourier-transform.