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).
dv = N(vx,vy,vz) dvxdvydvz
(transformation from spherical to Cartesian velocity space) and since dΩ
= sin θ dθ dφ and
dvxdvydvz = v2 sin θ dθ dφ 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) .
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
N(vx) peaks at 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
Finally we note that
rms2 + vy rms2 + vz