# Difference between relaxation and resonance leading to an absorption spectral feature?

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?

• 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. – Alchimista Jul 15 '19 at 9:07
• 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. – Alchimista Jul 15 '19 at 9:19
• 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. – Alchimista Jul 15 '19 at 9:33
• 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. – Alchimista Jul 15 '19 at 9:49
• @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. – uhoh Jul 17 '19 at 12:46