# General formula for the variation of the chemical potential with temperature

For small temperatures $$T$$, such that $$k_\mathrm{B}T\ll \mu(T=0)\equiv \mu(0)$$, the variation of chemical potential with temperature is given by $$\mu(T)=\mu(0)\left[1-\frac{\pi^2}{12}\left(\frac{k_\mathrm{B}T}{\mu(0)}\right)^2\right].\label{eqn:1}\tag{1}$$ Is there an exact analytical solution for the chemical potential at any temperature $$T$$ in a way that $$\eqref{eqn:1}$$ and its high temperature limit $$(k_\mathrm{B}T\gg\mu(0))$$ can be deduced from that? I am interested in such a situation because one often has to deal with Fermi gases at very high temperatures.

## 1 Answer

The short answer is that there is no explicit formula for $$\mu(T)$$ valid at any $$T$$. In Peter Young's notes you can see his attempt to connect the low-$$T$$ and high-$$T$$ limits, and the basic equations can be found in most books on condensed matter physics. I'll just give the starting point here. One can equate the standard expression for $$N$$ at $$T=0$$ with the expression at higher $$T$$, given the 3D density of states for free electrons and the known occupation number formula at chemical potential $$\mu(T)$$. This gives $$\frac{2}{3}\epsilon_F^{3/2} = \int_0^\infty d\epsilon \, \frac{\epsilon^{1/2}}{\exp[\beta(\epsilon-\mu)]+1}$$ where $$\epsilon_F=\mu(0)$$ and $$\beta=1/k_BT$$. The chemical potential as a function of $$T$$ may be obtained by solving this equation.

As those notes point out, it is convenient to express it in dimensionless form by defining $$x=\epsilon/\epsilon_F$$, $$\tilde{\mu}=\mu/\epsilon_F$$, and $$\tilde{\beta}=\beta\epsilon_F$$ or equivalently $$\tilde{T}=k_BT/\epsilon_F$$: $$\frac{2}{3} = \int_0^\infty dx \, \frac{x^{1/2}}{\exp[\tilde{\beta}(x-\tilde{\mu})]+1} = \int_0^\infty dx \, \frac{x^{1/2}}{\exp[(x-\tilde{\mu})/\tilde{T}]+1} .$$ For $$\tilde{T}=0$$ the Fermi function is zero for $$x>\tilde{\mu}$$ and the equation is satisfied for $$\tilde{\mu}=1$$. You are after the solution $$\tilde{\mu}(\tilde{T})$$ for nonzero $$\tilde{T}$$.

Evidently, this solution cannot be obtained analytically. It can be done numerically (see Fig 1 of the above-cited notes) and, of course, one can get low-$$T$$ and high-$$T$$ asymptotic forms for $$\mu(T)$$. The low-$$T$$ form you have already. The high-temperature limit follows from the fact that $$\mu$$ becomes large and negative at high $$T$$, and so the $$+1$$ in the denominator of the integrand may be dropped in this limit. Then the $$\mu$$-dependence may be taken out of the integral, and the integral itself can be done; the resulting equation can be rearranged to give $$\mu$$. See this answer for details.