- #1
kde2520
- 16
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I'm just reposting this question that someone else asked a long time ago but got no responses. Any help would be great.
Hi,
Studying for my thermal physics final and want to make sure I'm doing this right as our book doesn't have answers. Pretty simple question so I need to make sure I'm doing the basics right.
Find the equilibrium value at temperature \tau of the fractional magnetization
\frac{M}{Nm} = \frac{2<s>}{N}
of the system of N spins each of magnetic moment m in a magnetic field B. The spin excess is 2s. Take the ntropy as the logarithm of the multiplicty g(N,S) where
{\sigma}(s) \approx {\sigma}_o - \frac{2s^2}{N}
where {\sigma}_o = \ln{g(N,0)} and g is the multiplicity function. Using the fact that U = 2smB, we can say
{\sigma}(U) = {\sigma}_o - \frac{U^2}{2m^2B^2N}
Now, the equilibrium of the system will occur when the entropy is maximized, so I take the derivative with respect the U, set it equal to zero, solve for U and then use that to find s. This gives me that <s> = 0 which makes sense as the temperature isn't going to affect whether you have spin up or spin down, right?
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Hi,
Studying for my thermal physics final and want to make sure I'm doing this right as our book doesn't have answers. Pretty simple question so I need to make sure I'm doing the basics right.
Find the equilibrium value at temperature \tau of the fractional magnetization
\frac{M}{Nm} = \frac{2<s>}{N}
of the system of N spins each of magnetic moment m in a magnetic field B. The spin excess is 2s. Take the ntropy as the logarithm of the multiplicty g(N,S) where
{\sigma}(s) \approx {\sigma}_o - \frac{2s^2}{N}
where {\sigma}_o = \ln{g(N,0)} and g is the multiplicity function. Using the fact that U = 2smB, we can say
{\sigma}(U) = {\sigma}_o - \frac{U^2}{2m^2B^2N}
Now, the equilibrium of the system will occur when the entropy is maximized, so I take the derivative with respect the U, set it equal to zero, solve for U and then use that to find s. This gives me that <s> = 0 which makes sense as the temperature isn't going to affect whether you have spin up or spin down, right?
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