Talk:Debye model

• Does anybody have a good overlay of Debye, Einstein, and experimental data? This was originally in the article proper, but I figure here is a better place for it. Eldereft

I would like a figure showing: The Debye heat capacity, the low temperature limit, and the high temperature limit. Also a formula for the entropy according to Debye, and according to the limit formulae. Bo Jacoby 11:41, 19 October 2005 (UTC)

Hello Eric Kvaalen. You improved the accuracy of the article. I have a question. The factor 3 from polarization assumes that longitudinal and transversal waves move at the same speed. That is known not to be true. Can you please comment on that? Bo Jacoby 13:38, 21 October 2005 (UTC)

Hello again Eric Kvaalen. Your inclusion of the original Debye derivation is welcome, but the placement of it splits the debye formula from the asymptotic formulas, to make the latter unreadable. Find a better place to put it, please. Bo Jacoby 14:09, 31 October 2005 (UTC)

For upstairs

I'm not able to help you. But I know that Einstein's failure of solids' ${\mathcal {C}}_{v}$was he made the assumption of vibration by oscillators is constant. In some sense of physics,it may be yes but in modern common-sense may not. 'Cause as heat inputs into a flat metal(as for instance),energy is enough to thread everywhere on and in it but cannot have to stay the same area always(by macroscopic view). Thus easily gets ${\mathcal {\frac {\partial {U}}{\partial {T}}}}=3R$ which stands only at higher Temperatures but at lower T is in vain. Very simple: The more energy the more T(not absolute) and the fewer the fewer T. So Einstein didn't consider more about physical pictures of oscillators vibrating at lower T. We might image that this kind of a picture. As:

         Money is giving to some people. They are closed in a room.
         When it were lower T so that everyone is lazy.
         This time giving money to them,they just a little happy 
         feeling.
         When it were higher T so that everyone is happy. Giving 
         money to them doesn't "obviously" make them happier.

So Petit's and Einstein's theory of capacity just can show the yes at higher T but cannot at lower T. Something's hide.--HydrogenSu 19:24, 3 February 2006 (UTC)

Answering your question

Quote:

I have a question. The factor 3 from polarization assumes that longitudinal and transversal waves move at the same speed. That is known not to be true. Can you please comment on that? Bo Jacoby 13:38, 21 October 2005 (UTC)

Reply:

The 3 is from the freedom of in-x,y,z. That isn't from a polarization. It's said by my professor the day before yesterday. The main reason he talked in class is that phonon/sound is not like light wave which can have polarization. Just one mode only. Hmm...I might write math formula in physics:

${\mathcal {U}}=f\cdot {\frac {1}{2}}RT=6\cdot {\frac {1}{2}}RT=3RT$

Where f is as freedom,and U is as energy.

Hope I didn't make messy to your original question. However,even if the 3 in your question was supposed to mean 3 of ${\mathcal {,}}3RT,$,it might be still wrong. Because of "The factor 3 from polarization ". The reason was talked about in the early paragraph,you may be back to see. The other freedoms are from 2 of each direction's Kinetic and Potential energy. So $2\cdot 3=6$ is as the show of that formula,respectively. --HydrogenSu 19:46, 3 February 2006 (UTC)

Another reply:

Actually one works with one effective sound velocity $c_{s\,|\,{\rm {eff}}}$, which is a sum of contributions from the longitudinal and transverse sound velocities, respectively. Furthermore, the Debye temperature is proportional to this effective sound velocity, and thus measures the "hardness" of the crystal. More precisely one has

${\frac {1}{T_{D}^{3}}}$$\propto {\frac {1}{(c_{s\,|\,{\rm {eff}}})^{3}}}$$\,:={\frac {1}{c_{s\,|\,{\rm {long.}}}^{3}}}+{\frac {2}{c_{s\,|\,{\rm {trans.}}}^{3}}}$

87.160.85.141 (talk) 08:40, 16 August 2008 (UTC)

This article has many places that is not appropriate

As the title,in which Bose-Einstein's method of statics was put.....--HydrogenSu 15:18, 23 February 2006 (UTC)

Suggestion

[1]'s saying might be more complicated than that can be more easily done originally.

And we shall keep in mind that Debye's original derivation was easier and not yet involved something about Bose-Einstein's. --GyBlop 17:53, 26 February 2006 (UTC)

Beiser's and Blatt's saying might be easier than are here.--GyBlop 17:57, 26 February 2006 (UTC)

Clarification in Derivation section

Regarding the substitution $\nu _{n}=c_{s}/\lambda _{n}$, the article says,"We make the approximation that the frequency is inversely proportional to the wavelength..." Why is this necessarily an approximation? Leif (talk) 14:23, 12 June 2008 (UTC)

For phonons one has (at low frequencies) the approximate relation $\nu \propto {\frac {1}{\lambda }}$ between the frequency $\nu$ and the wavelength $\lambda$ of a wave. Actually this approximation applies only for long wavelengths, precisely for $\lambda \gg a$, where a is the lattice constant. This is in contrast to more complex behaviour, for example $\nu \propto \sin({\vec {k}}\cdot {\vec {a}})\,,$ for shorter wavelengths, e.g. comparable to a. For all wavelengths one has ${\vec {k}}={\frac {2\pi }{\lambda }}{\vec {e}}\,,$ where the vector ${\vec {e}}$ describes the direction of the wave-propagation. The second expression, with the $\sin$ function, corresponds at long wavelength to the expression given in the first-mentioned place.

The Debye model assumes(!), and this was actually at first glance a bold assumption, that the long-wavelength approximation is true (which actually is not the case) throughout the whole $\lambda$-range, up to the absolute high-frequency limit where $\lambda$ becomes as small as, e.g., 2a. Actually, it turned out that the assumption was not only bold, but also very clever; it just met the essential points both at low and at high temperatures. For other quasi-particles, e.g. magnons instead of the phonons, one has totally different relations, e.g. $\nu \propto {\frac {1}{\lambda ^{2}}}\,.$   But the essentials of the Debye model can be transferred even to this seemingly very different problem. - 87.160.123.57 (talk) 19:32, 15 August 2008 (UTC)