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 }v^{2} e^{ }^{-(mv^2)/(2kT)}.
If we assume that the velocity distribution is isotropic, it follows that
^{ }ò_{4}_{π }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(v_{x},v_{y},v_{z}) dv_{x}dv_{y}dv_{z}
(transformation from spherical to Cartesian velocity space) and since dΩ
= sin θ dθ dφ and
dv_{x}dv_{y}dv_{z} = v^{2} sin θ
dθ dφ dv, it follows that
N(v_{x},v_{y},v_{z}) = (1/v^{2})N(v,θ,φ) = N(v)/(4πv^{2}) = N [m/(2πkT)]^{3/2 } e^{ }^{-(mv^2)/(2kT) } = N [m/(2πkT)]^{3/2 }e^{ }^{-[m/(2kT)](v}_{x}^{2+v}_{y}^{2+v}_{z}^{2)} .
But
N(v_{x},v_{y},v_{z})/N = N(v_{x})/N ´ N(v_{y})/N ´ N(v_{z})/N.
It follows from isotropy that
N(v_{x}) = N [m/(2πkT)]^{1/2 }e^{ }^{-mvx^2/(2kT)}, and so on for N(v_{y}) and N(v_{z}).
From these one-dimensional velocity distributions is follows immediately that
since
N(v_{x}) peaks at v_{x}= 0_{
}[(¶N(v_{x})/¶v_{x})_{vx=0}
= 0], the most probable mean one-dimensional speeds are
v_{xP} = v_{xP}
= v_{xP} = 0,
the mean one dimensional speeds are
<v_{x}> = <v_{y}>
= <v_{z}> = 0 (because the speed distributions are
symmetric about the v = 0 peaks),
the mean magnitudes of the one
dimensional speeds are <|v_{x}|>
= <|v_{y}|> =<|v_{z}|> = [2kT/(πm)]^{½},
and the root
mean square speeds are v_{x
rms} = v_{y rms }=_{ }v_{z rms}
= (kT/m)^{½}.
Finally we note that
v_{rms}_{ }=
(v_{x
rms}^{2} + v_{y rms}^{2} + v_{z
rms}^{2})^{½} =
(3kT/m)^{½}.