Statistical Physics, Second Revised and Enlarged Edition

The Maxwell–Boltzmann distribution ofspeeds 67

1 2 3

max rms



n()

///T

Fig. 6. 1 The Maxwell–Boltzmanndistribution ofspeedsin a gas. The representative speedsvmax,v ̄and

vrmsaredefinedinthe text.

Also since thegasisisotropicandhas no net velocitythethree component velocities
must satisfy

Mvvx^2 / 2 =Mv^2 y/ 2 =Mv^2 z/ 2 =M(vv^2 x+v^2 y+v^2 z)/ 6

=Mv^2 rms/ 6 =kkkBT/ 2

This is another example of the classicallaw of equipartition of energyreferred to

earlierin section 3.2. Eachdegree offreedom ofthesystem,i.e. the translational

motion of each molecule in each of the three dimensions of space, contributes^12 kkkBT

to the internal energy of the system. We shall see this result again in the following

section andinChapter 7.

Finally, before leavingthe topic, we maynote that other more detailed statistical

information can be compiled and used in a similar way. Again the key isni=gi×fffi,

i.e. number ofparticlesinagroup = number ofstatesinthegroup×fillingfactor.

Suppose for example one wants the number of particles with a velocity close to

a particular valuev.We maydefine the usualdensityfunctionsfornandgsuch

thatn(vvx,vy,vz)dvvxdvvydvzisthe number ofparticles andg(vvx,vy,vz)dvvxdvydvzthe

number of states with velocity in the element dvvxdvydvzat velocityv=(vvx,vy,vz).

Fitting wavesintoboxesinthis case gives thegfunction tobe a constant; the states

are evenlyspreadink-space, andhenceinv-space also. AndtheBoltzmannfactorin

fffiis simply exp(−Mv^2 / 2 kkkBT)withv^2 =vvx^2 +v^2 y+v^2 z.Therefore the result is

n(vvx,vy,vz)dvvxdvvydvz

=const×exp(−Mv^2 / 2 kkkBT)dvvxdvydvz (6.10)

with the value of the normalization constant left as aproblem (Appendix E).