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Help with Stat Mech density of states problem?

  • #1

Homework Statement



It's easier to post a picture of the problem:

NKGqG.png


Homework Equations



In picture, and boson occupation number:

[itex]\left\langle n_k \right\rangle = \frac{1}{e^{\beta E(k)} - 1}[/itex]

Where E is the energy of the state with k and [itex]\beta = 1/k_B T[/itex]

The Attempt at a Solution



Goal: Find M(T), where M is the magnetization. Now my usual understanding of magnetization is that it's the net number of spins in a chosen direction. I'm most familiar with it from the Ising Model, where to find it, it's (number of spins up) - (number of spins down). I'm not really sure what they mean here though, so I assumed they simply mean the total number of magnons.

If I call the number of modes of wavevector less than k per unit volume

[itex]N(k) = (1/2\pi)^3 (4\pi k^3/3)[/itex]

Then my best guess on how to get the answer is:

[itex]M(T) = V \int_0 ^\infty N(k) \left\langle n_k \right\rangle D(\omega)d\omega[/itex]

Where I use the given dispersion relation to find [itex]dk/d\omega[/itex] and convert [itex]k[/itex] to [itex]\omega[/itex] and vice versa.

But I have a few problems... First off is, what is E in my boson occupation number? In the past we've always used [itex] E = \hbar^2 k^2/2m[/itex], but there is no mention of mass here because the magnons are waves, not particles. Also, the dispersion relation given is in the form of an energy (one that's familiar from waves, [itex]\hbar \omega[/itex]), so maybe that could be E(k)? So I don't know what to do there.

Additionally, I'm not even sure I'm putting these equations together correctly. I'm also a little worried that we don't seem to be using the first equation given (U) at all, though I guess it's just supposed to be an example, and not to be actually used?

Another thing I'm not sure about is the limits of integration of whatever we're integrating. But, the hint shows an integral with limits of (0,inf). If I had to guess, I'd say that we're going to use that trick where we say, because the denominator is so large in this problem (because T << 1, β >> 1), the only non negligible pieces of the integrand are going to be for low k anyway, so we can extend the upper limit to inf.

If anyone could point me in the right direction, that would be great! Thanks!



Edit: So I noticed that [itex]D(\omega) = \frac{dN(k)}{d\omega}[/itex], which leads me to think that they give us N(k) just for illustration porpoises. Also, if I just use [itex]E = \hbar \omega[/itex] and say

[itex]M(T) = \int_0 ^\infty \left\langle n_k \right\rangle D(\omega)d\omega[/itex]

It leaves me with the form of the integral in the hint (power of 1/2 in the numerator). Does this possibly seem right?

thanks!
 
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Answers and Replies

  • #2
938
9
Edit: So I noticed that [itex]D(\omega) = \frac{dN(k)}{d\omega}[/itex], which leads me to think that they give us N(k) just for illustration porpoises. Also, if I just use [itex]E = \hbar \omega[/itex] and say

[itex]M(T) = \int_0 ^\infty \left\langle n_k \right\rangle D(\omega)d\omega[/itex]

It leaves me with the form of the integral in the hint (power of 1/2 in the numerator). Does this possibly seem right?

thanks!
Everything seems to be in order here now. Except, of course, at low temperatures the background magnetization is N (all spins point to the same direction), and the excited waves decrease that number by 1 (or 2?) each.
 
  • #3
Everything seems to be in order here now. Except, of course, at low temperatures the background magnetization is N (all spins point to the same direction), and the excited waves decrease that number by 1 (or 2?) each.
Right, because it's ferromagnetic obviously they all have to be pointing in the same direction at T = 0, and for huge T they'll probably have a third of them pointing in each direction because the thermal noise overpowers any interaction energy terms.

But the physics still doesn't make sense to me. I can't really say why we're using D(ω) rather than N. Is the volume of the magnet already included in D? I'm also not really sure about the form of the energy to be used... the dispersion relation is convenient, but it says these magnons are "spin waves" and as far a I know, E = [itex]\hbar[/itex]ω is just for EM waves.

Thanks!
 
  • #4
938
9
Right, because it's ferromagnetic obviously they all have to be pointing in the same direction at T = 0, and for huge T they'll probably have a third of them pointing in each direction because the thermal noise overpowers any interaction energy terms.

But the physics still doesn't make sense to me. I can't really say why we're using D(ω) rather than N. Is the volume of the magnet already included in D? I'm also not really sure about the form of the energy to be used... the dispersion relation is convenient, but it says these magnons are "spin waves" and as far a I know, E = [itex]\hbar[/itex]ω is just for EM waves.

Thanks!
N(ω) is the number of modes whose energy is less than ω while D(ω) dω is the number of modes whose energy is in the interval [ω,ω+dω]. The magnons are bosons, so you have nontrivial statistics, and therefore you need to integrate n(ω) D(ω) dω. If every value of ω would be equally likely, then the number of magnons would simply be N(ω) = ∫ D(ω) dω.

It is said in the problem that N(k) is the number of magnons in a "unit volume", so you should multiply it with the number of spins (N) to get the total number of magnons.

Magnons are just quantized spin waves, so if you remember how quantum harmonic oscillators work, that's basically where [itex]E = \hbar \omega [/itex] comes from. The waves are excitations in the oscillators, each corresponding to a jump of one energy level.

EDIT: I forgot, you should also probably calculate the first order correction to the ground state magnetization due to the usual thermal fluctuations.
 
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