3
$\begingroup$

I need help understanding the physics behind this insightful comment below the question Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?:

Please note: This is a relaxation spectrum. The transitions do not come from a resonance (absorption, like in NMR, IR, etc.), but an applied electric field has a lossy (at low frequencies) or elastic (at higher f.) effect on some mode of molecular orientation of electrical dipoles in your sample.

The following comment links to https://en.wikipedia.org/wiki/Dielectric_spectroscopy which confirms this.

Both mechanisms can to a peak in the imaginary part of the dielectric permittivity at a certain frequency or energy, and at this point the slope of real part is steepest.

But is there a simple yet intuitive way to understand the physical difference between a spectral absorption feature due to a resonance process versus a relaxation process?

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ Two things. Concerning the actual example, the link and comments should suffice. It is the nature or level of interaction that differs. The interaction between the applied em field and the sample results in a different orientation of the dipoles in the sample and it is not quantised at molecular level. Each same dipole would equally interact with the field as in a merely electrostatic interaction. The spectroscopic character is due to scanning the frequency at which the field is reversed, so that different samples adapt to it differently, promptly or not, or even not at all. $\endgroup$
    – Alchimista
    Jul 15, 2019 at 9:07
  • 2
    $\begingroup$ Concerning terminology, relaxation can be encountered also in a more molecular spectroscopy context, when indeed transitions due to resonance between the applied em field and quantised energetic levels (electronic, vibrational, rot. and their mixes) take places. Except that you won't scan for them, but parameters like the time needed for the sample to return back. A simple example: you excite a molecule at an energy you know it suffices to give luminescence emissions, than you follows the luminescence decay in time. Here you have a resonance process to pump a relaxation phenomenon. $\endgroup$
    – Alchimista
    Jul 15, 2019 at 9:19
  • 1
    $\begingroup$ For me nothing. The overall result is well discussed in the linked answer. For you: you are the only one knowing if my comments were useful or not. I did compress anything. Just unable to formulate a nicely concise answer. If the comments weren't useful I can delete them. Let me know. $\endgroup$
    – Alchimista
    Jul 15, 2019 at 9:33
  • 1
    $\begingroup$ Didn't notice you weren't polemic so perhaps I have been so in last comment. But it answers well. I do have an intuitive answer, the problem is to convey that. I am confident I went the right direction. I can add one more thing: in the original Q you mention absorption peak. But while this might be done in a colloquial scenario, the peaks are in Permittivity, at the end. Not Absorbance. I think this note helps, too. $\endgroup$
    – Alchimista
    Jul 15, 2019 at 9:49
  • 1
    $\begingroup$ @Karl distinction without a difference. If you have a sample of ordinary matter and you have a non-zero imaginary part of the permittivity you have absorption, and vice-versa. Use $(n+ik)^2=\epsilon'+i\epsilon"$ to get $k$ then Beer–Lambert to get absorption of power per unit depth. $\endgroup$
    – uhoh
    Jul 17, 2019 at 12:46

1 Answer 1

Reset to default
3
$\begingroup$

The difference can be viewed in two ways, I guess.

Firstly, for relaxation spectroscopy, you apply a forced oscillation on your sample. That can be mechanical (dynamic rheology, mHz to .1kHz, several kHz with very special, not commerically available equipment) or electrical (dielectric spectroscopy, low mHz to GHz range). The mathematical concepts behind both methods are very similar. For vibrational, MR, UV, etc. spectroscopy, the sample does just not interact with frequencies outside of the linewidth of some QM transition. The width of a resonant peak is given by the energy needed for the transition, and the lifetime of the excited state. The width of a feature in a relaxation spectrum is typically at least half an order of magnitude, from "sample can follow the external stimulus and stay close to a (dynamic) equillibrium" to "external stimulus is much too fast for the sample to follow at all".

The other point is that in relaxation spectroscopy, energy is transferred and dissipated into the sample below a specific frequency (invariably as heat), and reflected (elastically) at higher frequencies. The energy is not quantised, because the induced change in the sample is translatory. The peak frequency itself has no real significance, it's inverse is called a "relaxation time" of the sample. A statistical quantity, mostly. For QM transitions, energy is only transmitted at the given frequency/energy, and nothing happens otherwise, except Rayleigh scattering.

Basically, resonant spectroscopy deals with individual quantum mechanical species, and relaxation spectroscopy deals with the classical, unquantised newtonian properties of the whole sample. NMR is a bit of a hybrid: The induced transitions are purest QM, but you can only observe the classic induction of the ensemble, never individual photons.

Technically, relaxation spectra are usually taken by applying the stimulus at discrete frequencies, stepwise, and you want to have a phase-sensitive detector (voltage - current, deformation - torque).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for your post, I will now give this some thought. It would be great if you could add a supporting link or two to provide an opportunity for further reading for me and future readers. In both cases you apply an electromagnetic stimulus using an antenna or pair of plates or something similar; why would it be called a forced oscillation only for relaxation spectroscopy and not in the case of resonant spectroscopy? $\endgroup$
    – uhoh
    Jul 15, 2019 at 23:41
  • $\begingroup$ Well, outside of resonance, oscillation can only be forced. The difference is that you are actually interested in the nonresonant reaction of the sample. In e.g. optical spectroscopy, the sample also interacts with off frequencies, by elastic (Raleigh) scattering. $\endgroup$
    – Karl
    Jul 16, 2019 at 6:35

Your Answer

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

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