- #1
WWCY
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Hi everyone, I need a little help understanding how periodic reciprocal space applies to the Debye model for solids. Many thanks in advance!
If we start with the general derivation of a dispersion relation for a 1D system, with atoms coupled by springs, one gets the following relation
$$\omega = 2\sqrt{\frac{k}{m}} |\sin \big( \frac{ka}{2} \big)|$$
whereby it is possible to show that the function is periodic with ##\omega (k + 2\pi/a) = \omega (k)##.
However if we consider Debye's model, which comes as a result of taking the long wavelength (and small k) limit, we get
$$\omega = v_s k$$
where ##v_s## is the speed of sound.
##\textbf{Question}##: Does the concept of a periodic ##k## still apply here? Also, is the number of allowed modes for a chain of N masses still equal to N?
For example if I were to calculate energy contributions by individual frequencies ##\omega (k)## using the following equation
$$\langle E \rangle _k= \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
clearly energy for the ##k=0## mode differs from that of the ##k = 2\pi /a## mode. Should I pick ##k = 0## for my calculations or ##k = 2 \pi /a##?
If we start with the general derivation of a dispersion relation for a 1D system, with atoms coupled by springs, one gets the following relation
$$\omega = 2\sqrt{\frac{k}{m}} |\sin \big( \frac{ka}{2} \big)|$$
whereby it is possible to show that the function is periodic with ##\omega (k + 2\pi/a) = \omega (k)##.
However if we consider Debye's model, which comes as a result of taking the long wavelength (and small k) limit, we get
$$\omega = v_s k$$
where ##v_s## is the speed of sound.
##\textbf{Question}##: Does the concept of a periodic ##k## still apply here? Also, is the number of allowed modes for a chain of N masses still equal to N?
For example if I were to calculate energy contributions by individual frequencies ##\omega (k)## using the following equation
$$\langle E \rangle _k= \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
clearly energy for the ##k=0## mode differs from that of the ##k = 2\pi /a## mode. Should I pick ##k = 0## for my calculations or ##k = 2 \pi /a##?