Minimum wavelength of phonons under the Debye aproximation

  • Thread starter AngelFis93
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  • #1
AngelFis93
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Homework Statement:
It's stated on an example problem that under Debye aproximation on a monoatomic cubic lattice of lattice constant a= 3.7 Å, sound speed v=3000 m/s (in both longitudinal and transverse directions) and Debye frequency ω=3.2·10^(13) rad/s, to find the minimum phonon wavelength. They give you the solution λ=4.27 Å .
Relevant Equations:
λ=v/f
Since in Debye aproximation Debye's frecuency is defined as the maximum frecueny , the corresponding wavelenght should be the minimum one, due to the inverse relation among those

λ=v/f=v·2π/ω=5.9 Å , which is higher than the given result.

I believe I should be using the information 'cubic lattice' somehow ,but can't see it.


Thanks.
 
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Answers and Replies

  • #2
trurle
508
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It a bit strange.
Problem is over-defined. You do not need Debye frequency here because it is calculable from lattice parameter and sound speed. Actually as i calculate Debye approximation using equation
ω/(2*pi)=(Cs/2a)*[(9/(4*pi))^(1/3)]
, ω for 3.7 Å lattice should be 2.277*10^13
Higher value of 3.2*10^13 will actually give wavelength 4.19 Å.
 
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  • #3
Dr_Nate
Science Advisor
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Homework Statement:: It's stated on an example problem that under Debye aproximation on a monoatomic cubic lattice of lattice constant a= 3.7 Å, sound speed v=3000 m/s (in both longitudinal and transverse directions) and Debye frequency ω=3.2·10^(13) rad/s, to find the minimum phonon wavelength. They give you the solution λ=4.27 Å .
Homework Equations:: λ=v/f

Since in Debye aproximation Debye's frecuency is defined as the maximum frecueny , the corresponding wavelenght should be the minimum one, due to the inverse relation among those

λ=v/f=v·2π/ω=5.9 Å , which is higher than the given result.

I believe I should be using the information 'cubic lattice' somehow ,but can't see it.


Thanks.
Is the intention of the example to show you where the Debye approximation breaks down? Because, the minimum wavelength is actually ##2a## because of the Nyquist theorem (https://en.wikipedia.org/wiki/Phonon#Lattice_waves).

You should try to take the value of ##2a## and divide by the factor you get from taking the body diagonal of a cube. You will get the answer you posted, which to me seems wrong.
 
  • #4
AngelFis93
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Is the intention of the example to show you where the Debye approximation breaks down?
I don't think so, at least there isn't anything in the statement that makes me think that way.
You should try to take the value of 2a2a and divide by the factor you get from taking the body diagonal of a cube. You will get the answer you posted, which to me seems wrong.
I came to the same conclusion, a·2/√3 gives the exact solution, I just can't find in the theory were it's justified why it is calculated this way. A mistake in the solution could be posible aswell, so i probably ask the proffesor directly.

Thanks for the reply

It a bit strange.
Problem is over-defined. You do not need Debye frequency here because it is calculable from lattice parameter and sound speed. Actually as i calculate Debye approximation using equation
ω/(2*pi)=(Cs/2a)*[(9/(4*pi))^(1/3)]
, ω for 3.7 Å lattice should be 2.277*10^13
Higher value of 3.2*10^13 will actually give wavelength 4.19 Å.

I get 3.2 ·10^(13) rad/s using ω =v·k=v·a^(-1)(6π^2)^(1/3) for ω,k on Debyes aproximation (a^(-1) for being a cubic lattice, and using the definition of Debye's k).

Thanks for the reply.
 

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