Dr. Huerta
Speed Distributions
Phy 210
Say we have a collection of N particles that have many different speeds. We want to
know things like what is the average speed, or the most probable speed. We define a
function P (v) such that P (v)dv equals fraction of particles with speeds in the
range (v − dv/2, v + dv/2). We can also think of P (v)dv as the probability that a particle
will have speed in the range above. So, for example the fraction of particles with speeds
in the range (14.95, 15.05) would be P (15) × 0.1.
Rv
The fraction of particles with speeds in the interval v1 ≤ v ≤ v2 is f rac = v12 P (v)dv.
The actual umber of particles with speed in the above range would be dN = N P (v)dv.
If we add up all the particles that have all the possible speeds, we should get the total
number N of particles, so
R
R∞
R∞
N = dN = N 0 P (v)dv. Therefore we must have that always 0 P (v)dv = 1,
or because the sum of all the fractions must be equal to 1. We say that the function P (v)
must be normalized.
The most probable speed vp is the one that maximizes P (v), or 0 = dP
dv |vp .
Different kinds of averages are computed using P (v), so
R∞
R∞
R∞
< v >avg = 0 vP (v)dv, < v 2 >avg = 0 v 2 P (v)dv, or f (v) avg = 0 h(v)P (v)dv, any h(v)
PM
vM
v for 0 ≤ v ≤ vM , and P (v) = 0 for v > vM . The
R∞
Rv
2
vM
M v
M
normalization condition requires that 1 = 0 P (v)dv = 0 M PvM
vdv = PvM
= PM2vM ,
2 |0
and therefore we must have that PM = v2M , and the function P (v) = v2v
2 . Obviously the
M
most probable speed is vp = vM because P (vM ) is the maximum value, even though the
derivative is never zero.
Rv
R vM 2
2
1 2
2
Then vavg = 0 M v × v2v
v × v2v
2 dv = 3 vM , and (v )avg =
2 dv = 2 vM
0
Example Say the P (v) =
M
M
Derivation of the Maxwellian speed distribution (difficult): We begin by considering
the velocity (not the speed) distribution. Let f (vx )dvx be the fraction of particles with
vx in the range (vx , vx + dvx ). We assume that vx , vy and vz behave the same way, so
f (vx )f (vy )f (vz )dvx dvy dvz is actually the velocity distribution that gives the fraction of
particles with velocity in the range (~v , ~v + d~v ). We assume there is no preferred direction,
so this should only depend on the speed, so
q
f (vx )f (vy )f (vz ) = F (v), where v = vx2 + vy2 + vz2 .
(2)
Differentiating Eq.(2) with respect to vx we get
dF (v) ∂v
dF (v) vx
df (vx )
f (vy )f (vz ) =
=
.
dvx
dv vx
dv v
(3)
Dividing Eq.(3) by F (v)vx we get
1
df (vx )
1 dF (v)
=
.
vx f (vx ) dvx
v dv
(4)
Dr. Huerta
Speed Distributions
Phy 210
But v and vx are independent variables because we can change v by changing vy or vz
without changing vx . Therefore the only way that Eq.(3) can be valid is if each side of the
equation is equal to the same constant K, so
1 1 df (vx )
1 dF (v)
d ln F (v)
=K=
, and then
= Kv, or
(5)
vx f (vx ) dvx
v dv
dv
ln F (v) = K
2
v2
− B or F (v) = Ae−Bv where A and K are constants to be determined.
2
(6)
−Bv 2
Therefore F (v) = Ae
= f (vx )f (vy )f (vz ).
(7)
We note that dvx dvy dvz is an element of volume in velocity (vx , vy , vz ) space. To find P (v)
we need to count the particles with speed v which are found in the volume
dvx dvy dvz → 4πv 2 dv of velocity space inside a shell of radius v and thickness dv
Z
2
P (v)dv =
F (v)dvx dvy dvz = F (v)4πv 2 dv = Ae−Bv 4πv 2 dv,
(8)
shell
2
Therefore P (v) = 4πv 2 Ae−Bv , and we need to find A and B
(9)
R∞
We know the P (v) should satisfy the normalization condition 0 P (v)dv = 1, and that
3
1
2
2 m < v >avg = 2 kT. To do this we need two ”advanced” integrals calculated at the end
of this paper.
√
√
Z ∞
Z ∞
Z ∞
π −1/2
d
π −3/2
2 −av 2
−av 2
−av 2
a
,
v e
dv = −
e
dv =
a
e
dv =
2
da 0
4
0
0
(10)
√
Z ∞
Z ∞
π −5/2
d
4 −av 2
2 −av 2
and
v e
dv = −
v e
a
dv = 3
da 0
8
0
3/2
√ 3/2
Z ∞
Z ∞
π 1
B
−Bv 2
2
Therefore
= 1, or A =
, and
P (v)dv =
Ae
4πv dv = 1 so 4πA
4 B
π
0
0
√
Z ∞
3kT
3
m
π −5/2
2
−Bv 2
4
=
B
=
, so B =
. (11)
Ae
< v >avg =
4πv dv = 12Aπ
m
8
2B
2kT
0
So, finally
3/2
mv 2
m
P (v)dv =
e− 2kT 4πv 2 dv.
(12)
2πkT
From this we can get that
3/2
q
R∞
R∞
mv 2
m
8kT
< v >avg = 0 P (v)vdv = 0 2πkT
e− 2kT 4πv 2 vdv =
πm . Finally the most
q
(vP )
= 0, or vP = 2kT
probable speed is the one that makes dPdv
m .
R ∞ −av2
R∞
2
To evaluate the integral I = 0 e
dv in Eq.(10) consider J = −∞ e−av dv = 2I. We
R∞ R∞
2
2
write J 2 = −∞ −∞ e−a(v +w ) dvdw. Change variables to ”cylindrical coordinates”
R 2π R ∞
R∞
2
2
v = r cos θ, w = r sin θ, and then dvdw → rdrdθ, so J 2 = 0 0 e−ar rdrdθ = 2π 0 e−ar rdr
∞
p
p
−ar 2 so J 2 = 2π e 2a . So, J 2 = πa , so J = πa , and finally I = J2 = 12 πa as in Eq. (10).
0