Do lattice vibrations emit radiation?

[Master J]

I know that accelerating electrons in a solid are responsible for the emission, absorption and reflection of light.

However, what role do the lattice vibrations play? These oscillating atoms are charge distributions and should emit radiation too, right?

As far as I can remember, the maximum lattice vibration frequency (Debye frequency) is in the infrared range typically. Does this mean that the lattice DOES emit radiation, but it is of such low frequency and intensity that it is negligible?

 

[the_emi_guy]

Ever seen someone heat up a metal rod in a hot fire? When you pull the rod out and witness it glowing visibly red and emitting lots of heat, these are electromagnetic radiation (visible and infared) due to thermal motion within the metal.
I think that the Debye model is just one of several models that attempt to predict this behavior. For example there is the Einstein model which treats the solid as lots of non-interacting harmonic oscillators.

 

[nonequilibrium]

the_emi_guy, are you sure that is due to *lattice* vibration? And not due to internal atomic vibration? E.g. the electron cloud vibrating around the nucleus? After all, the OP is correct in stating that the maximum lattice vibration frequency is usually below optical frequencies. You're right that the Debye model is just one model, but the fact that the frequencies lie below the optical region is model-independent (i.e. one doesn't necessarily need the concept of Debye frequency to make this claim).

To the OP: well theoretically of course we know that there must be at least *some* emission, since indeed we're dealing with vibrating charges. Also it is experimentally detectable, quick google search: http://adsabs.harvard.edu/abs/1968JPSJ...25.1091H . But I'm not sure how important the contribution is and how to argue it. If we're dealing with spherically symmetric electron clouds, the dipole/quadrupole/... etc moments are zero, so no emission, but in the more general case I don't have a real feel for the matter. Hopefully others can give more insight.

 

[Master J]

Thanks for the input guys!

I was also thinking, considering an atomic mass is typically thousands of times that of a single electron, the frequencies of the vibrating lattice atoms must be far less. What kind of radiation would they emit, if say the electrons were emitting IR. Would it be radio, microwave?

 

[nonequilibrium]

I think you're confusing yourself. None of the above applied to electrons, we were talking about atoms. E.g. the maximum lattice vibration frequency you quoted *is* for the lattice vibration (being the atoms).

 

[Master J]

I understand that, I'm not confused. Perhaps we have digressed too far.

My original question was, since accelerating electrons in a solid are responsible for reflection, emission and absorption of light, and since the lattice (the atoms, which themselves are charge distributions) oscillates, does the LATTICE also emit radiation?

 

[vanhees71]

Of course, any accelerated charges radiate electromagnetic waves.

On a fundamental level there is only the electromagnetic field written as two three-dimensional vector fields, \vec{E} and \vec{B} or, more elegantly, as a antisymmetric Minkowski-space tensor field, the Faraday or field-strength tensor F_{\mu \nu}. On a microscopic level, a solid is described quantum mechanically (or in classical approximation) as a lattice of atoms bound together through electromagnetic interactions.

Under usual circomstances you can treat the problem in the Born-Oppenheimer approximation, i.e., (for a body at rest) as a lattice of atomic nuclei, to which the electrons are bound. In metals you also have conducting electrons that are (quasi-)freely moving.

Now, if you perturb this static system in thermal equilibrium by shining in an electromagnetic wave, this wave acts via the Lorentz force,
\vec{F}=q \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right )
on all the charges. Approximately the bound electrons start to vibrate around their equilibrium locations, and the conducting electrons start to drift and thus make an induced AC current. This in turn radiates em. waves which superpose with the external em. wave you shine in.

Of course, you cannot resolve the details of this highly complicated macroscopic problem of the self-consistent motion of particles and fields in all detail. Fortunately this is often not necessary, but you can coarse-grain over distances large compared to the spacing of the atoms or molecules in the lattice, provided the typical wavelength of the incoming radiation is large compared to the lattice spacing. Further, if the magnitude of the external em. field is small compared to the em. field of the charges making up the body, you can calculate the reaction of the matter in linear approximation ("linear response theory"). What you get from this is nothing else than "macroscopic" electrodynamics with the coarse grained electromagnetic field (\vec{E},\vec{B}) and the auxiliary fields (\vec{D},\vec{H}) that are connected with each other through the dielectric tensor and magnetic permeability tensor which for isotropic media even simplifies to scalars. The auxiliary fields turn out to include the reaction of the medium in terms of polarization and magnetization, and the charge and current distributions in the macroscopic equations are only those external sources added to the medium in equilibrium.

You find an excellent treatment of macroscopic electromagnetism from this modern perspective, providing a much better physical understanding of the meaning of the various quantities than in the usually given purely phenomenological treatment, in the Feynman Lectures, vol. 2.

Of course, if the external electromagnetic wave is chosen in the infrared frequency range, despite the reaction of the bound and conducting electrons you also excite lattice vibrations. Here also the same concepts hold, and the macroscopic em. field can be understood as the superposition of the imposed external fields and the fields created by the motion of the lattice ions.