What is the meaning of an imaginary part of position in Lorentz-Drude model?

Question

Some time ago I got to know about Lorentz-Drude model, which predicts dependence of complex permittivity $\varepsilon = \varepsilon_1 + i \varepsilon_2$ from classical assumptions.

In the model, firstly, we derive position-time dependence of position of momenta by treating them like oscillators: $$ m \ddot{x} + m \gamma \dot{x} + m \omega_0^2 x = -e E_0 \exp(i \omega t) $$ Here, we have damping ($\gamma$), restoring force ($m \omega_0^2 \equiv k$) and external electric force ($e$ is an elementary charge). The equation has solution: $$ x(t) = \frac{e E_0 \exp(i \omega t)}{m} \frac{1}{\omega^2 - \omega_0^2 + i \gamma \omega} $$ which, as we see is a complex function.

Next, we use definitions of dipole and polarization: $$ P = Np = - N e x(t) \quad \land \quad P = \varepsilon_0 (\varepsilon - 1) E_0 \exp(-i \omega t) $$ where N is number of oscillators. Using these, we get: $$ \varepsilon (\omega) = 1 - \frac{N e^2}{m \varepsilon_0} \frac{1}{\omega^2 - \omega_0^2 + i \gamma \omega} $$ and this is complex too. Now, we can calculate real and imaginary part of permittivity: $$ \varepsilon_1 (\omega) = 1 - \frac{N e^2}{m \varepsilon_0} \frac{\omega^2 - \omega_0^2}{(\omega^2 - \omega_0^2)^2 + \gamma^2 \omega^2} $$ $$ \varepsilon_2 (\omega) = \frac{N e^2}{m \varepsilon_0} \frac{\gamma \omega}{(\omega^2 - \omega_0^2)^2 + \gamma^2 \omega^2} $$

I was told that imaginary part of the position $x(t)$ has no physical meaning. I also know, that $\varepsilon_2$ is responsible for energy losses.

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How can we get the meaningful results ($\varepsilon_2$) from something that has got no meaning ($\mathfrak{Im}~x(t)$)? We could perform derivation using trigonometric functions, only with real part of electric force and then getting only real position and real permittivity. Is there any interpretation we can give to the imaginary part of position here?

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Edit: Suggested in comments answer to another question is explanation, why imaginary part of the position $\mathfrak{Im}~x(t)$ carries no new information. In case of permittivity we have a difference - real part tells us about dispersion and imaginary about absorption. Those are two different quantities we only for convenience hold together in one complex value. We are be able to do all calculations from above only in real space ($F_e=- e E_0 \sin (\omega t)$), but then we are losing information about absorption ($\varepsilon_2$).

Answer 1

In the model, firstly, we derive position-time dependence of position of momenta by treating them like oscillators: $$ m \ddot{x} + m \gamma \dot{x} + m \omega_0^2 x = -e E_0 \exp(i \omega t) $$

By writing the above equation you have implicitly agreed to solving for a complex function $x(t)$, only the real part of which is the physical position. You can see this is the case, because you are already using a complex force on the RHS of the equation. If this was an equation for the physical position, you would write: $$ m \ddot{x}_R + m \gamma \dot{x}_R + m \omega_0^2 x_R = -e E_0 \cos(\omega t)\;, $$ where $\Re(x)=x_R$.

The equation has solution: $$ x(t) = \frac{e E_0 \exp(i \omega t)}{m} \frac{1}{\omega^2 - \omega_0^2 + i \gamma \omega} $$ which, as we see is a complex function.

This is a solution for the complex $x(t)$, not the physical position, which is real.

Using these, we get: $$ \varepsilon (\omega) = 1 - \frac{N e^2}{m \varepsilon_0} \frac{1}{\omega^2 - \omega_0^2 + i \gamma \omega} $$ and this is complex too.

Yes.

I was told that imaginary part of the position $x(t)$ has no physical meaning.

Presumably whoever told you this meant that the real part of $x(t)$ is the physical position. This is true, but not quite the same thing as saying the imaginary part has "no physical meaning."

How can we get the meaningful results ($\varepsilon_2$) from something that has got no meaning ($\mathfrak{Im}~x(t)$)?

As mentioned above, whoever told you "no meaning" presumably meant that the real part of x is the physical position and the imaginary part is not. The imaginary part is usually associated with losses and shows up in a real part of some other calculation when, say, you multiply two complex quantities together. Or, say, when you are interested in an odd number of time derivatives.

We could perform derivation using trigonometric functions, only with real part of electric force and then getting only real position and real permittivity. Is there any interpretation we can give to the imaginary part of position here?

Yes. For example, consider the effect of a single time derivative on your solution: $$ v_R = \frac{d x_R}{dt} = \Re(\frac{dx}{dt}) = \Re(i\omega x) = -\omega x_I\;. $$

So, in the above case, we see the the imaginary part of the position is proportional to the physical velocity.