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I'd like to know the how the number density of longitudinal optical (LO) and longitudinal acoustic (LA) phonons varies as a function of temperature of the material. Is there a simple expression for these two cases?

I'm guessing that this would work,

$N_{LO} = \int g_{LO}(E) f(E, T) dE$

$N_{LA} = \int g_{LA}(E) f(E, T) dE$

where $g(E)$ is the density of states for LO and LA phonons and $f$ is the Bose-Einstein distribution. What would be appropriate limits for the integrals. Does anybody know a reference where the density of states for these two modes is given?

EDIT

To improve the question, I'm interested in semiconductor 3D crystals. But maybe I left this too long, sorry.

Best regards,

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Sure seems right to me. Of course thee Bose-Einstein distribution allows occupation numbers greater than one. –  Carl Brannen Jun 9 '11 at 20:05
    
Question number one: do you want experimental or theoretical results? Question number two: what kind of materials? Pure crystals, crystals with defects, arbitrary solids? –  Marek Jun 10 '11 at 8:56
    
@Marek; It seems to me that the equations should work irrespective of the material. Can you correct me on this? I'm thinking that they would simply have different $g(E)$. –  Carl Brannen Jun 12 '11 at 4:33
    
@Carl: are you saying the temperature dependence for any material is only present in the Bose-Einstein part of the formula? This statement seems quite unobvious and is probably wrong too. –  Marek Jun 12 '11 at 8:08
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It's never too late to try for an answer ;-) –  David Z Feb 1 '12 at 4:51
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1 Answer

Different types of phonons can not be considered as separate systems. They are oscillations of the same crystal and interact with each-other.

For example, LO phonon lifetime is about $10^{-12}$ - $10^{-11}$ seconds while the period of the oscillations is about $10^{-13}$ seconds (GaAs). At the end it turns into two LA phonons that run opposite directions.

Bose-Einstein distribution describes thermodynamical equilibrium of the whole phonon system. You should integrate over all the modes.

The density of states can be estimated numerically or measured experimentally. Both usually give similar results that can be found e.g. in chapter 3 of "Fundamentals of Semiconductors" by Peter Y. Yu and Manuel Cardona.

The main experimental techniques are

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This answer is technically correct, but if the interaction is weak enough (as I believe it will be for this case), you can describe the optical phonons and the acoustic phonons both with separate Bose-Einstein distributions for free fields and not be too far off. The question is not whether they interact, but whether the average energy of an optical phonon in the thermal background of acoustic phonons is going to be changed significantly (and vice versa). My impulse is that opticals are absent at reasonable temperature, while long-wavelength acoustic phonons have a near-perfect Debye spectrum. –  Ron Maimon Feb 2 '12 at 12:24
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@Maksim, Would suggest that you include neutron scattering with appropriate application of the incoherent approximation rather than solely Raman spectroscopy, since not all optical phonon modes are Raman-active. –  Jen Feb 2 '12 at 16:58
    
@RonMaimon, I have never heard anything about quasi-equilibrium state of optical phonons. In order to let this equilibrium exist there should be some mechanism of LO phonon - LO phonon interaction with characteristic time much shorter then the decay time. This is an interesting question. –  Maksim Zholudev Feb 2 '12 at 17:38
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