6
$\begingroup$

So my question is quite simple I suppose, and perhaps trivial. It is known that the frequency domain susceptbility $\chi(\omega)$ is complex, and that the two parts can be related with the Kramers-Kronig relations. But the time domain susceptibility, $\chi(t)$, is said to be real, according to my textbook.

Now, I know that in a linear response type of framework we often write that the polarization density (lets suppress the spatial dependence and only talk about a particular point in space) as $$P(t) = \int_{-\infty}^{\infty}{\epsilon_0\chi(t-t')}E(t')\mathrm{d}t'$$

So in this case the susceptibility is an impulse response function for a time invariant system.

Moreover, I suppose it would only make sense for $P(t)$ to be real itself, as it is the density of electric dipole moments, which itself is just a measure of the separation of positive and negative electrical charges in a system. That has to be real, surely.

But then I get a little confused. Don't we often take $E(t)$ to be complex in our calculations? So then why can $\chi(t)$ not be complex as well? I'm probably missing some very simple ingredient, but I can't seem to figure it out.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ No, E(t) is never complex. It is taken to be complex after we get the complex frequency-dependent susceptibility, because it just makes the math easier. However when we take E complex, we're just omitting the fact, that we have a complex E(f) and the complex conjugate E(-f) which cancel out the imaginary part. $\endgroup$
    – LLlAMnYP
    Commented Jan 22, 2016 at 18:17
  • $\begingroup$ @LLlAMnYP Hm, yes that is very true of course. So then it is a very logical consequence, the time domain susceptibility being real. Doesn't really warrant a seperate answer I suppose; what does one typically do here, close the question? $\endgroup$
    – user129412
    Commented Jan 25, 2016 at 17:47
  • $\begingroup$ Nah, let it be. It's a matter of formalism, really. I'd add further, that we don't even take $E(t)$ to be complex after we got the frequency-dependent part, but rather we take the amplitude of the $\omega$-th Fourier component of the Fourier transformed $E(t)$ to be non-zero. If we only consider that part without the $-\omega$ part, then we get a complex $E(t)$. $\endgroup$
    – LLlAMnYP
    Commented Jan 25, 2016 at 17:53
  • $\begingroup$ Which, in hindsight seems like the same thing I wrote in the first comment, just worder differently. Huh. $\endgroup$
    – LLlAMnYP
    Commented Jan 25, 2016 at 17:54

1 Answer 1

Reset to default
2
$\begingroup$

Everything classical is real in the time domain. However, you'll notice that response functions involve convolutions with the "input" field. This makes them non-local, which is the only way to achieve causality. In the Fourier domain this is equivalent to having a complex part. In fact, both the real and complex part MUST be nonzero or else you violate causality. That's KK. Many textbooks fake you out by writing the Maxwell equations in the time domain and then showing the constitute relations (e.g., $D = \epsilon E$) in the frequency domain to avoid the convolution. This causes endless confusion. See: Linear response laws and causality in electrodynamics AJ Yuffa, JA Scales European Journal of Physics 33 (6), 1635. Promoting E to the Fourier domain is a mathematical convenience. If you do linear calculations, then everything is fine.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.