Consider a 1d array of \(N\) particles, \(\sigma_i\), each in some state \(\in\{-1,+1\}\). The state can be thought of as the spin of the particle, and each particle can flip between the two states.
Each particle \(\sigma_i\) interacts with each of its two next-door neighbors, \(\sigma_{i-1}\) and \(\sigma_{i+1}\), and each interaction has an energy \[E(\sigma_a,\sigma_b) = -\sigma_a \cdot \sigma_b \mbox{ Joules}\]
This interaction encourages adjacent particles to have like state, and discourages adjacent particles from having unlike states.
Another energetic contribution comes from entropy. Energy states that have many possible arrangements are favored over those with few. A state with \(\Omega\) arrangements has entropy \(S = k_B\log\Omega\), and has a free energy contribution \(-TS\).
In general, at very high temperatures, entropy dominates energy and we expect each particle state to be \(-1\) or \(+1\) with probability \(\frac12\). At absolute zero, energy dominates entropy, and we expect the system to be in the lowest possible energy states, i.e. \[\ldots,-1,-1,-1,\ldots\] or \[\ldots,+1,+1,+1,\ldots\]
As we lower the system from high temperatures to absolute zero, we expect a transition from disorder to order at some critical temperature, \(T_c\): a battle between energy and entropy.
Suppose we have such a 1d lattice in the limit \(N\rightarrow\infty\). As we lower \(T\), at what temperature \(T_c\) does the system first become ordered, i.e. trapped in one of the lowest energy states?
by
Josh Silverman
