# Expand Helmholtz energy to fourth order in m (magnetisation)

The energy of a system of spins is given by: $$H=\frac{1}2\sum_{i,j=1}(J\sigma_i\sigma_j-h\sum\sigma_i)$$

I found that we can rewrite in terms of magnetisation, $$m=\sum\sigma_i/N$$: $$E=-N(\frac{Jm^2}2+hm)$$

we can then write the partition function in terms of the magnetisation. To do this, we have to find a new function $$F(h,T)=E-ST=-N(\frac{Jm^2}2+hm)-k_BTlnW$$ where $$W$$ is the number of microstates. This is where I get a bit confused - I carried by saying the number of microstates, to account for degeneracy is given by $$N!$$, and then use stirlings approximation...

Anyway, the next part of the question is expand $$F(h,T)$$ to fourth order in m, and I'm really confused on this because theres no point of m given to expand around..I know i should somehow end up with an expression for $$F$$ that involves an m to the power of 4, $$m^4$$, but I just dont understand how to do that in this context...

• This is the infinite range Ising model. You need to correct the first equation, since the factor $1/2$ does not apply to the term in $h$, and some care is needed with the dependence of the $J$ term on $N$. Often this model is solved exactly, via the Hubbard-Stratonovich transformation, which you can easily look up, but I have sometimes seen it tackled in the way you describe. However, Physics SE is not a homework help site, see physics.stackexchange.com/help/on-topic and I'm afraid that, to be acceptable, you need to narrow this question down to address a specific physics concept. – user197851 Apr 7 at 21:51
• I do not think that you have to pass through $F=E-TS$ in order to find the partition function. It has a direct expression in term of the energy of the states of your system (check for canonical partition function on your preferite Statistical Mechanics textbook). Use it and expand directly. – GiorgioP Apr 8 at 5:40