# Ferromagnet $\leftrightarrow$ paramagnet at Curie temperature

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)?

• What are $\beta$ and $\mu$? – jacob1729 Jun 15 '19 at 15:38
• 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)? – jacob1729 Jun 15 '19 at 18:16
• $\mu$ is the magnetic moment – Mari3 Jun 16 '19 at 5:33
• But $m$ too seems to indicate magnetic moment. You should check your notation. Anyway, the critical point is at zero external field. – GiorgioP Jun 16 '19 at 8:13
• m is the magnetisation, $m= \mu * 1/N * (\sigma_1 + .... + \sigma_N)$ – Mari3 Jun 16 '19 at 9:40