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unscientific
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Homework Statement
(a) Show the curie-weiss behaviour.
(b) Estimate ##\lambda## and ##B_e## and exchange energy.[/B]
Homework Equations
The Attempt at a Solution
Part(a)
Since even when applied field is zero, ##B_{total} \neq 0## which gives rise to ##M\neq 0##. This is a fundamental property of ferromagnetism.
Consider the spin-1/2 ising model:
H = \sum\limits_{\langle i,j \rangle} J_{ij} \sigma_i \sigma_j - \sum\limits_i g \mu_B \sigma_i \cdot \vec B
H_i = \left( g \mu_B B_{applied} - J \sum\limits_j \sigma_j \right)\sigma_i
We represent the term in the brackets by an effective field ##B_e##:
H_i = \left( g \mu_B B_e \right)\sigma_i
Average spin at site ##i## is given by
\langle \sigma \rangle = -\frac{1}{2} \tanh \left( \frac{\beta g \mu_B B_e}{2}\right)
By the mean-field approach, we assume that ##\langle \sigma \rangle## is the same at all sites, giving:
g \mu_B B_{app} - J z \sigma_j = g \mu_B B_e
where ##z## is the number of nearest neighbours, or coordination number.
This leads to a 'self-consistency' equation
\frac{1}{2}\tanh \left[ \frac{\beta}{2} \left( Jz\langle \sigma\rangle - g\mu_B B \right) \right] = \langle \sigma \rangle
At zero appleid field, Curie temperature is given when gradient = 1, so
\frac{1}{2} \left( \frac{\beta}{2}Jz \right) = 1
T_c = \frac{Jz}{4 k_B}
For the curie-weiss behaviour, we expand for small ##\langle \sigma \rangle##:
\langle \sigma\rangle = \frac{ \frac{\beta}{4}g \mu_B B_{app} }{frac{\beta}{4} Jz - 1}
\langle \sigma\rangle = \frac{\frac{1}{4} \frac{g \mu_B B_{app}}{k_B}}{T-T_c}
M = g n \mu_B \langle \sigma \rangle = \frac{\frac{1}{4} \frac{(g \mu_B)^2 n B_{app}}{k_B}}{T-T_c}
\chi = \mu_0 \frac{\partial M}{\partial B_{app}} = \frac{1}{4} (g\mu_B)^2 \frac{n}{k_B} \frac{1}{T-T_C}
Part(b)
I found the coordination number at ##T = 1024 K## and ##J=1## using
T_c = \frac{Jz}{4 k_B}
It gave ##z = 5.7 \times 10^{-20}## which seems wrong as the number of neighbours should be a whole number..