Maxwell Boltzmann distribution

[Rzbs]

TL;DR Summary

Difference between Maxwell Boltzmann and fermi dirac distribution

In the Aschcroft & Mermin solid state book there is a curve to compare F.D and M.B distribution. I can't understand the concept of M.B curve; what does mean exactlly when x =0? It means the probability of zero energy for particles is most or ...?

 

[mjc123]

Figure 2.1 seems odd; it appears to be plotting the distribution of velocity (note, velocity, not speed) versus kinetic energy. The energy distribution does not have a maximum at zero. Look up the MB distribution; observe its forms for the distribution of velocity, speed and energy.

Equation 2.1 is the distribution of vectorial velocity, i.e. with a particular direction as well as magnitude (or alternatively, a particular set of components {v_x, v_y, v_z}). It has a maximum at zero velocity. The distribution of the speed is found by summing Eq. 2.1 over all directions (or all sets of {v_x, v_y, v_z} that give the same magnitude v), and is given by
f(v) = 4πv²f(v)
This has a value of zero at v=0, and a maximum at a finite v. The energy distribution goes as E^1/2e^-E/kT

 

[Rzbs]

Thanks so much for your simple description.
Now I can understand the difference between graph for speed, energy and velocity.

But I can't understand "what fig2.1 try to say", it shows if x decreases then the probability increases. The plot is Ae^(-x). The meaning of x=0 is have zero velocity.
Does it mean if T increases then x decreases and the probability for have less velocity increases?

 

[mjc123]

Figure 2.1 is a very bad graph. It is plotting the probability of one variable (velocity) against a different variable (energy/kT). All it really shows, qualitatively, is that MB and FD are very different.
If T increases, the distribution broadens, and the probability at low velocities decreases. The x-axis in Fig 2.1 is E/kT, and the distribution of (E/kT) is independent of temperature.

 

[Rzbs]

Thanks. I thought so but I wasn't sure. You really help me to understand this. Thanks so much.

 

[Charles Link]

Usually there is a $v^2$ factor on the Maxwell-Boltzmann distribution, and the peak speed ($v_{mp}=$most probable speed) occurs at $v_{mp}=(2kT/m)^{1/2}$.