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I think it's like this: $\, m=\tanh\left(\frac{Bμ}{k_bT}\right)$. If now the temperature decreases, then $\mu$ increases, until it flattens out ($\tanh$ function). Is the a point where $m$ flats out, the critical point, because then all spins are aligned, and thus the paramagnet becomes a ferromagnet? If the temperature increases, $\mu$ goes to zero, and thus the spins are all random (=paramagnet)?

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  • $\begingroup$ What are $\beta$ and $\mu$? $\endgroup$ – jacob1729 Jun 15 at 15:38
  • $\begingroup$ The argument of your tanh doesn't make sense. Is it meant to be something like $\mu_B B / kT$ (with $\mu_B$ the Bohr magneton)? $\endgroup$ – jacob1729 Jun 15 at 18:16
  • $\begingroup$ $\mu$ is the magnetic moment $\endgroup$ – Mari3 Jun 16 at 5:33
  • $\begingroup$ But $m$ too seems to indicate magnetic moment. You should check your notation. Anyway, the critical point is at zero external field. $\endgroup$ – GiorgioP Jun 16 at 8:13
  • $\begingroup$ m is the magnetisation, $m= \mu * 1/N * (\sigma_1 + .... + \sigma_N)$ $\endgroup$ – Mari3 Jun 16 at 9:40

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