Drude Model Permittivity Formula - e^iwt or e^(-iwt)?

[IcedCoffee]

I've been having a sign problem while deriving the permittivity formula using Drude model,

and I found out that the problem came from the fact that complex field vectors are expressed with e^-iwt, not e^iwt, thus producing (-iwt) term when differentiated:

http://photonics101.com/light-matter-interactions/drude-model-metal-permittivity-conductivity (See "Show Solutions")

Now, I guess that when you pick either e^iwt or e^-iwt and derive complex parameters like permittivity, you have to stick with it from then on in order to obtain valid real-part values with correct phase,

but is e^-iwt the conventional one? I've seen the same formula for permittivity in Wikipedia... Is this because i(kx-wt) is more natural for waves propagating to positive x direction?

 

[Charles Link]

In the linear response theory in the time domain, the fundamental equation is $V_{out}(t)=\int\limits_{-\infty}^{t} m(t-t') V_{in}(t') \, dt'$. The Fourier transform $\tilde{F}(\omega)=\int\limits_{-\infty}^{+\infty} F(t) e^{-i \omega t} dt$. The convolution theorem gives $\tilde{V}_{out}(\omega)=\tilde{m}(\omega) \tilde{V}_{in}(\omega)$. Then, using the inverse transform operation, $F(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} \tilde{F}(\omega) e^{+i \omega t} \, d \omega$. The mathematics works equally well if the F.T. is defined with $e^{+i \omega t}$, and the minus sign in $e^{-i \omega t}$ is then used during the inverse F.T. operation. $\\$ Because of the sign convention, the physicist uses impedances $Z_L=j \omega L$ and $Z_C= -\frac{j}{\omega C}$while the EE has the signs reversed in analyzing AC circuits.$\\$ (I think the physicist uses $V(t)=V_o e^{+i \omega t}$ for a sinusoidal voltage in an AC electrical circuit, but when considering traveling waves, they switch to $E(x,t)=E_o e^{i (kx-\omega t)}$. I will need to check this result with some calculations, but I believe that is the case). $\\$ In an AC circuit analysis, the physicist's phasor diagram (because of the sign on $i \omega t$), in a graph of the complex $V$ as a function of time, rotates counterclockwise, while the EE has their phasor diagram rotating clockwise. $\\$ And to conclude: Your assessment is accurate. The complex part of the susceptibility $\tilde{\chi}(\omega)$ will have a different sign dependent on whether $E(t)=E_o e^{+ i \omega t}$ or $E(t)=E_o e^{-i \omega t}$ is assumed. $\\$ The fundamental linear equation here is $P(t)=\epsilon_o \int\limits_{-\infty}^{t} \chi(t-t') E(t') \, dt'$, and $\tilde{P}(\omega)=\epsilon_o \tilde{\chi}(\omega) \tilde{E}(\omega)$ for the expression involving Fourier transforms from the convolution theorem. The susceptibility $\tilde{\chi}(\omega)$ is actually the Fourier transform of the linear response function $\chi(t)$, and it is often written simply as $\chi(\omega)$.

 

[IcedCoffee]

In the linear response theory in the time domain, the fundamental equation is $V_{out}(t)=\int\limits_{-\infty}^{t} m(t-t') V_{in}(t') \, dt'$. The Fourier transform $\tilde{F}(\omega)=\int\limits_{-\infty}^{+\infty} F(t) e^{-i \omega t} dt$. The convolution theorem gives $\tilde{V}_{out}(\omega)=\tilde{m}(\omega) \tilde{V}_{in}(\omega)$. Then, using the inverse transform operation, $F(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} \tilde{F}(\omega) e^{+i \omega t} \, d \omega$. The mathematics works equally well if the F.T. is defined with $e^{+i \omega t}$, and the minus sign in $e^{-i \omega t}$ is then used during the inverse F.T. operation. $\\$ Because of the sign convention, the physicist uses impedances $Z_L=j \omega L$ and $Z_C= -\frac{j}{\omega C}$while the EE has the signs reversed in analyzing AC circuits.$\\$ (I think the physicist uses $V(t)=V_o e^{+i \omega t}$ for a sinusoidal voltage in an AC electrical circuit, but when considering traveling waves, they switch to $E(x,t)=E_o e^{i (kx-\omega t)}$. I will need to check this result with some calculations, but I believe that is the case). $\\$ In an AC circuit analysis, the physicist's phasor diagram (because of the sign on $i \omega t$), in a graph of the complex $V$ as a function of time, rotates counterclockwise, while the EE has their phasor diagram rotating clockwise. $\\$ And to conclude: Your assessment is accurate. The complex part of the susceptibility $\tilde{\chi}(\omega)$ will have a different sign dependent on whether $E(t)=E_o e^{+ i \omega t}$ or $E(t)=E_o e^{-i \omega t}$ is assumed. $\\$ The fundamental linear equation here is $P(t)=\int\limits_{-\infty}^{t} \chi(t-t') E(t') \, dt'$, and $\tilde{P}(\omega)=\tilde{\chi}(\omega) \tilde{E}(\omega)$ for the expression involving Fourier transforms from the convolution theorem. The susceptibility $\tilde{\chi}(\omega)$ is actually the Fourier transform of the linear response function $\chi(t)$, and it is often written simply as $\chi(\omega)$.

Ahh, so it's FT convention. That slipped my mind just thinking about the differential equation. Thank you!

 

[Charles Link]

One additional minor correction to the above: I think in MKS units, they often define the susceptibility so that $\tilde{P}(\omega)=\epsilon_o \tilde{\chi}(\omega) \tilde{E}(\omega)$ with an $\epsilon_o$ in the equation. And yes, I see the "link" you posted uses it with the $\epsilon_o$. Let me make that correction above.

 

[vanhees71]

Yeah, it's convention, and this involves some schizophreny even within the physics community alone (if you switch between physics E&M and electrical-engineering E&M textbooks, which often treat other subjects than the physics textbooks, you get completely confused).

Usually the convention in full Maxwell theory is to choose the time dependence for harmonically evolving fields as $\exp(-\mathrm{i} \omega t)$, while for circuit theory where you deal with integrated quantities like voltages and currents they choose $\exp(+\mathrm{i} \omega t)$, which is of course nuts, because the integrated quantities are just the fields integrated over space after all. Why they choose a different convention, you must not ask me. I don't know; maybe they like to confuse students even more than the subject itself is confusing, and it's confusing enough for the beginner (particularly when the SI units are used, but that's another story).

Then there are the various conventions concerning Fourier transformations of functions. In field theory, at least in the HEP community, the convention usually is like
$$\psi(t,\vec{r}) = \int_{\mathbb{R}} \frac{\mathrm{d} \omega}{2 \pi} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \tilde{\psi}(\omega,\vec{k}) \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{r}).$$
The inverse transformation then follows to be
$$\tilde{\psi}(\omega,\vec{k})=\int_{\mathbb{R}} \mathrm{d} t \int_{\mathbb{R}^3} \mathrm{d}^3 r \psi(t,\vec{r}) \exp(+\mathrm{i} \omega t-\mathrm{i} \vec{k} \cdot \vec{r}).$$
It may well be that in other communities you have to convention used by Charles Link in #2. In the engineering literature it's also not uncommon to write $\mathrm{j}$ for the imaginary unit.