Wave Vectors of Light vs. Phonons: A Puzzling Paradox

[Karthiksrao]

In his paper, "Radiative exchange of heat between nanostructures" Pendry makes this statement :

"In general phonons have much smaller wave vectors at a given
frequency than does light "

Isn't wave vector given by 2*Pi /Lambda? So how can wave vectors be
smaller for phonons than for light? What am I missing here ?

Thanks in advance

 

[Cthugha]

I
Isn't wave vector given by 2*Pi /Lambda? So how can wave vectors be
smaller for phonons than for light?

Yes, that is the definition of the wave vector. Nevertheless, the wavelengths are different for photons and phonons at a given energy because their dispersion relations are also different.

 

[andrewr]

In his paper, "Radiative exchange of heat between nanostructures" Pendry makes this statement :

"In general phonons have much smaller wave vectors at a given
frequency than does light "

Isn't wave vector given by 2*Pi /Lambda? So how can wave vectors be
smaller for phonons than for light? What am I missing here ?

Thanks in advance

A phonon is a rippling effect (acoustic/thermal) wave traveling primarily in the atomic areas below the valence band (eg: nucleus). The mass of the nucleus makes it have a different momentum for the same amount of energy that a photon would have; light is even lighter... than the nuclei... What this means, is that the de-Broglie wavelength which is the INVERSE of a wave number (k) -- can be different because the momentum required to achieve a specific frequency is different. Frequency is temporal (time based) --- wave number (and length) are *spatial*. The connection between these two is the mass of the object carrying the wave. Light, too, will travel more slowly in the vicinity of electrons and nucleii it has to travel *partially* through them by virtue of them changing the phase of light as it travels near them. So, I like to remind myself that "Energy" by itself does not equate to speed even knowing the total mass of the thing in motion for not all of it need move in the same way. A ball rolling has more energy than a ball being carried at the same speed. etc.

A wave vector is proportional to frequency and inversely proportional to wavelength.
.

 

[johng23]

Well, I am also confused by that statement. For a given frequency, the wavevector of a phonon should be much larger than that of a photon. The phonon corresponds to vibration of atomic planes so a typical wavelength will be on the order of 10 - 100 A, while a visible photon has wavelength of order 100 nm. So the momentum (or wavevector) of the phonon is much larger.

[PLAIN]http://www.mi.infm.it/manini/images/1D-phonons.gif

The slope of the phonon dispersion near the origin is the speed of sound in the material, and the slope of the photon dispersion is the speed of light, so the photon dispersion relation would be nearly a vertical line on this figure, showing that the wavevector of a photon is much smaller than that of phonons at nearly all frequencies.

Am I missing something?

 

[andrewr]

Well, I am also confused by that statement. For a given frequency, the wavevector of a phonon should be much larger than that of a photon. The phonon corresponds to vibration of atomic planes so a typical wavelength will be on the order of 10 - 100 A, while a visible photon has wavelength of order 100 nm. So the momentum (or wavevector) of the phonon is much larger.

The frequency of the photon is not necessarily in the visible light range; He is arguing that it has the same "frequency"; If E=hf, in both cases... then the energy exciting the nucleus must be the same as that of the photon. According to blackbody radiation laws... there is going to be a peak that corresponds to the chip temperature that has the highest density of light output. From that you can find a good comparison point for a specific frequency to check.

I find his statement perplexing too, but it's out of context and I haven't read the original; Photons have never actually been observed (that I am aware of) even in the VHF band and lower; I am not saying energy hasn't been detected -- or even momentum transfer -- just that the particle like properties that are part of the word "photon" haven't.

So, why is he talking about it? But, assuming they exist and this isn't just a misnomer...
To be fair to him, one needs to do a comparison that is at a statistically likely frequency.
Any takers? (I'm not betting one way or the other...)

 

[johng23]

The frequency of the photon is not necessarily in the visible light range; He is arguing that it has the same "frequency"; If E=hf, in both cases...

So what? You just draw a nearly vertical line on the plot I posted above to represent the photon dispersion. Now pick any point on the y-axis to correspond to the frequency you like, and the wave vector of the phonon is larger. Right? The plot is only for simple model of a 1-D string of atoms but the basic point is the same.

 

[andrewr]

So what? You just draw a nearly vertical line on the plot I posted above to represent the photon dispersion. Now pick any point on the y-axis to correspond to the frequency you like, and the wave vector of the phonon is larger. Right? The plot is only for simple model of a 1-D string of atoms but the basic point is the same.

If your plot is accurate, yes. But he may be using a different model. Like I said -- there is no context.
As I also said, it puzzles me too. Your argument isn't "wrong" in my estimation, and I didn't bother to pull data from your graphs -- but rather took your word that it is true at "nearly" all frequencies to allow that it wasn't in others. I just want to be fair to the author by knowing the context of the statement and doing a calculation that includes the density of states issues that is inevitable in this type of calculation.

 

[andrewr]

So what? You just draw a nearly vertical line on the plot I posted above to represent the photon dispersion. Now pick any point on the y-axis to correspond to the frequency you like, and the wave vector of the phonon is larger. Right? The plot is only for simple model of a 1-D string of atoms but the basic point is the same.

Not that I want to leave everyone in the dark, either. On Hyperphysics, there is a discussion of Debye temperature's and lengths; In computing the heat capacity of a crystal, the main contribution (besides electrons) is acoustic phonons.


I came across one paper with a note on the Debye formula for heat capacity indicating that it is accurate at both low and high temperature extremes but that the acoustic dispersion velocity is non-uniform leading to minor inaccuracies in the middle temperatures. I don't know if that is sufficient to explain what is going on or, if it isn't a large enough difference. If you work the problem out here; perhaps others who know more can correct you/verify your assertions without becoming too egotistically involved.


http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html"
http://http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/debye.html"


Cheers.

 

[SpectraCat]

To the OP, I think the issue is that the phonons travel at the speed of sound (or at least they cannot travel any faster than that), while photons travel at the speed of light. Thus the wavelength (and wavevector) of a phonon are related to its frequency according to the speed of sound, while a photon at the same frequency would have a (much) shorter wavelength, and consequently a (much) larger wavevector.

That is part of what Cthugha meant when he pointed out that their dispersion relations are different. It is also related to what I think andrewr was trying to say in his first post. The reason that the phonons travel so much more slowly is that they must propagate via the collective motion of massive particles.