The analysis of the distribution of spins for a paramagnetic solid in a B field shows that the probability of a dipole being aligned/anti-aligned with the B field ##\to 0.5## as ##T \to \infty##.(adsbygoogle = window.adsbygoogle || []).push({});

The intuitive justifications that I've read say that this is "expected" as thermal motion tends to cause disorder (e.g. this is from Mandl). However, what's wrong with this intuitive argument:

1. It costs energy E, say, to align a dipole with the B field.

2. A dipole will align if it receives energy E from the thermal motion of its heat bath.

3. As ##T \to 0##, the probability that a dipole-heat bath interaction exchanges energy E ## \to 0##, since the proportion of particles in the heat bath with energy at least E ##\to 0## as ##T \to 0## (so we have agreement with the Boltzmann distribution prediction here)

4. As ##T \to \infty##, the probability that a dipole-heat bath interaction exchanges energy E ## \to \infty##, since the proportion of particles in the heat bath with energy at least E ##\to 1## as ##T \to \infty##

5. Hence, we would expect that, on average, in the high T limit, that each dipole spends an increasingly large proportion of its time in the aligned state, so p(alignment) ## \to \infty## as ##T \to \infty##

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# I Boltzmann distribution for spin-1/2 dipole: high T limit

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