UNISA
Opens: Monday, 10 August 2026, 8:00 AM
Due: Thursday, 20 August 2026, 11:00 PM
All the problems are clearly answered fully bellow
, Problem 5.16: Behaviour of the Specific Heat Near 𝑻𝒄
The problem states:
Use
𝐸 1 𝐽𝑞𝑚 2
= − 𝐽𝑞 [tanh ( )]
𝑁 2 𝑘𝑇
and the fact that
3(𝑇𝑐 − 𝑇)
𝑚2 ≈ (𝑇 ≲ 𝑇𝑐 )
𝑇𝑐
to show that
3𝑘
𝐶(𝑇 → 𝑇𝑐− ) = .
2
Simplify the energy near 𝑻𝒄
Close to the critical temperature, the magnetisation 𝑚is very small.
Therefore
tanh 𝑥 ≈ 𝑥
for 𝑥 ≪ 1, where
𝐽𝑞𝑚
𝑥= .
𝑘𝑇
Hence
𝐸 1 𝐽𝑞𝑚 2
≈ − 𝐽𝑞 ( ) .
𝑁 2 𝑘𝑇
Since
𝑘𝑇𝑐 = 𝐽𝑞,
we obtain
𝐸 1 𝑇𝑐 2 2
≈ − 𝐽𝑞 ( ) 𝑚 .
𝑁 2 𝑇