# Computing the average energy $\overline{E}$ in thermodynamic equilibrium for a paramagnet

I am having trouble with the following physics exercise:

In the case of a paramagnet in a magnetic field $$H$$, show that from the requirement that $$F(E)$$ (the Helmholtz free energy) be minimal in thermodynamic equilibrium, one can produce the known expression for the average energy $$\overline{E}$$ of this system.

I am really not sure about how to proceed. I have an example in my book labeled "minimum property of the free energy of a paramagnetic system," in which it they assert that the entropy of the paramagnetic system with $$N$$ magnets and spin excess $$2s$$ is approximately equal to

$$-\left(\frac{1}{2}N + s\right)\log\left(\frac{1}{2} + \frac{s}{N}\right) - \left(\frac{1}{2}N - s\right) \log\left(\frac{1}{2} - \frac{s}{N}\right).$$

It then asserts that the energy in a magnetic field $$B$$ is $$-2smb$$, where $$m$$ is the magnetic moment of an elementary magnet. Through some computation, it is then shown that $$\partial F/\partial s$$ (the derivative of the Helmholtz free energy with respect to spin excess) occurs when $$2s = N\tanh(mB/t)$$. I'm not sure if this is relevant though, because, from my understanding, the energy $$-2smb$$ is the total internal energy rather than the average energy?

Although this question arises from an exercise, you have already done almost all of the work to solve it. Your question is basically "should I be using the energy $$E$$ or the average energy $$\overline{E}$$ here?" or "am I justified in solving the problem this way?". This is a conceptual question, connected with the transformation of ensembles between microcanonical and canonical, which I believe this exercise is trying to help illustrate. So this answer will concentrate on that context.
You are correct that the energy $$E$$ of your paramagnet system is a simple linear function of the spin excess $$2s$$. So, go ahead, as you imply in your question: use that equation to substitute for $$s$$ in your entropy equation, to give you an entropy function $$S(E)$$. Use that to calculate a Helmholtz function, at a given $$T$$, which for the moment I'll call $$\mathcal{F}(E)=E-TS(E)$$. Do as the exercise asks: differentiate $$\mathcal{F}(E)$$ to find the energy $$E=\hat{E}$$ which minimizes $$\mathcal{F}(E)$$ and compare with the average energy $$\overline{E}$$ which comes from analyzing this model in the canonical ensemble (at temperature $$T$$). I'm leaving all of that to you.
The conversion between microcanonical and canonical ensembles is found in many statistical mechanics books. In the canonical ensemble at temperature $$T$$, the probability distribution function for energy may be written $$\mathcal{P}(E) \propto \Omega(E)\exp(-E/k_BT)$$ where $$\Omega(E)$$ is the density of states (number of states per unit energy), and $$\exp(-E/k_BT)$$ is the Boltzmann factor for energy $$E$$. NB, here I am glossing over the fact that $$E$$ only takes discrete values in this spin model: I am treating it as a continuous variable (as the exercise expects you to do, when you differentiate with respect to $$E$$).
$$\Omega(E)$$ is a very rapidly increasing function of $$E$$, for large $$N$$. The Boltzmann factor is a very rapidly decreasing function of $$E$$ (which is, after all, an extensive variable). The consequence is that $$\mathcal{P}(E)$$ has a very sharp peak at some value $$E=\hat{E}$$. Moreover, because of the sharpness, this value will be very close to the average energy $$\overline{E}$$. Before differentiating to find the maximum of $$\mathcal{P}(E)$$, it is convenient to take logs: $$-k_BT \ln \mathcal{P}(E) = E- k_BT\ln\Omega(E) + \text{const} = E-TS(E) + \text{const} = \mathcal{F}(E) + \text{const},$$ where we recognize $$S(E)=k_B\ln\Omega(E)$$, the formula which was used to obtain your entropy expression. Differentiating $$\mathcal{F}(E)$$ with respect to $$E$$, setting to zero, and hence finding $$\hat{E}$$, is the key step. In the process of doing that, one gets to match up the energy $$E$$ of the microcanonical ensemble with the temperature of the canonical ensemble, by requiring $$\left . \frac{\partial S(E)}{\partial E} \right|_{E=\hat{E}}=\frac{1}{T}$$
One can go further and discuss the width of the $$\mathcal{P}(E)$$ distribution, the link between $$\mathcal{F}(\hat{E})$$ and the thermodynamic free energy $$F$$, and so on, but that would go beyond the scope of the question. The main point is that, provided $$N$$ is large, we get the desired result: the temperature $$T$$ in the canonical ensemble is chosen to make $$\overline{E}\approx \hat{E}=E$$ in the microcanonical ensemble.