# Intepretation of the Fermi Dirac distribution graph [closed]

Is there a significance or an important interpretation of the fact that the plots in the Fermi-Dirac distribtion graph, all intersect at the point with coordinates ($$\epsilon_F$$,$$\frac {1}{2}$$)?

The curve shown in the image quoted in OP correspond to different values of temperature, but the same value of the Fermi level. It is a simple mathematical fact that the Fermi function $$f_{\epsilon_F}(\epsilon)=\frac{1}{\exp\left(\frac{\epsilon-\epsilon_F}{k_BT}\right)+1}$$ takes value $$1/2$$ when $$\epsilon=\epsilon_F$$. What changes with temperature is the spread of the function about the Fermi level, which characterizes how many electrons from below the Fermi level have been excited to above the Fermi level.

• But the formula I have for the distribution is: $\overline n_i=\frac{1}{e^{\beta(\epsilon_i-\mu)}+1}$. But $\mu$ is dependent on Temperature. In your formula, you are making the assumption that $\mu=\epsilon_F$ why? Commented Mar 27, 2023 at 21:11
• $\mu\rightarrow\epsilon_F$ is just a change of notation - Fermi level and chemical potential are the same thing (but perhaps you are right, that using $\mu$ would avoid a confusion with Fermi energy.) Your question does not say anything about temperature dependence - in fact, if $\mu$ were a function of temperature, the curves of course would not cross at the same point... but what is more important here is that we are working in the grand canonical ensemble where $\mu$ is among the independent variables (while the number of particles $N$ is a function of temperature.) Commented Mar 28, 2023 at 8:10
• In our class we said that $\mu(T=0,V,N)=\epsilon_F$. Meaning that the value that the chemical potential has at absolute zero, is taken as the fermi energy, which is the highest energy level at T=0. But I see we keep the $\mu$ constant Commented Mar 28, 2023 at 8:28