The Maxwellian One-Dimensional Velocity Distribution Functions

As a starting point we again restate the speed distribution function equation we derived earlier:

N(v) = N (2/π)½ (m/kT)3/2 v2 e -(mv^2)/(2kT).

If we assume that the velocity distribution is isotropic, it follows that

ò4π N(v,θ,φ) dΩ = 4πN(v,θ,φ) = N(v), and therefore that N(v,θ,φ) = N(v)/(4π).

But N(v,θ,φ) dΩ dv = N(vx,vy,vz) dvxdvydvz (transformation from spherical to Cartesian velocity space) and since = sin θ and
dvxdvydvz
= v2 sin θ dv, it follows that

N(vx,vy,vz) = (1/v2)N(v,θ,φ) = N(v)/(4πv2) = N [m/(2πkT)]3/2 e -(mv^2)/(2kT) = N [m/(2πkT)]3/2 e -[m/(2kT)](vx2+vy2+vz2) .

But

N(vx,vy,vz)/N = N(vx)/N ´ N(vy)/N ´ N(vz)/N.

It follows from isotropy that

N(vx) = N [m/(2πkT)]1/2 e -mvx^2/(2kT), and so on for N(vy) and N(vz).

From these one-dimensional velocity distributions is follows immediately that

• since N(vx) peaks at vx= 0 [(N(vx)vx)vx=0 = 0], the most probable mean one-dimensional speeds are vxP = vxP = vxP = 0,

• the mean one dimensional speeds are <vx> = <vy> = <vz> = 0 (because the speed distributions are symmetric about the v = 0 peaks),

• the mean magnitudes of the one dimensional speeds are  <|vx|> = <|vy|> =<|vz|> =  [2kT/(πm)]½,

• and the root mean square speeds are vx rms = vy rms = vz rms = (kT/m)½.

• Finally we note that vrms = (vx rms2 + vy rms2 + vz rms2)½ = (3kT/m)½.