How Maxwell derived his velocity distribution law

Peter Haggstromgotohaggstrom.com

mathsatbondibeach@gmail.com

April 24, 2019

Maxwell’s velocity distribution law is one of the foundational building blocks of statistical me-chanics and was derived by Maxwell in 1859 in the context of his work on gases: “Illustrationsof the Dynamical Theory of Gases”, read at a meeting of the British Association at Aberdeen,21 September 1859 [ 1, pp 380-383 ]

Maxwell’s derivation is short and I will set it out in full and then flesh out intermediate steps.

1 Proposition IV

“To find the average number of particles whose velocities lie between given limits, after a greatnumber of collisions among a great number of equal particles.

Let N be the whole number of particles. Let x, y, z be the components of the velocity of eachparticle in three rectangular directions, and let the number of particles for which x lies betweenx and x+ dx be Nf(x)dx , where f(x) is a function of x to be determined.

The number of particles for which y lie between y and y + dy will be Nf(y)dy; and the numberfor which z lies between z and z + dz will be Nf(z)dz, where f always stands for the samefunction.

Now the existence of the velocity x does not in any way affect that of the velocities y or z, sincethese are all at right angles to each other and independent, so that the number of particles whosevelocity lies between x and x+ dx, and also between y and y+ dy and also between z and z+ dzis:

Nf(x)f(y)f(z)dx dy dz (1)

If we suppose the N particles to start from the origin at the same instant, then this will be thenumber in the element of the volume (dx dy dz) after unit of time, and the number referred tounit of time will be

Nf(x)f(y)f(z) (2)

1

1.1 Comments

The resolution of a velocity vector onto the 3 rectangular coordinate axes ensures they areindependent.

Now Maxwell moves to the critical functional relationship that leads to the Gaussian. He saysthe following:

”But the directions of the coordinates are perfectly aribitrary, and therefore this number mustdepend on the distance from the origin alone, that is:

f(x) f(y) f(z) = φ(x2 + y2 + z2) (3)

Solving this functional equation, we find:

f(x) = CeAx2

, φ(r2) = C3eAr2

” (4)

Maxwell notes that if A is positive the number of particles will increase with velocity and thiswould result in an infinite number of particles. Thus physical considerations ensure that A isnegative ie A = − 1

a2 .

He then integrates from −∞ to +∞ to find the number of particles to be NC√πa = N so that

C = 1a√π

. The details are not difficult and are set out below:

N =

∫ ∞−∞

NCe−x2

a2 dx = NCa√π =⇒ C =

1

a√π

(5)

Note here that:

∫ ∞−∞

e−u2

du =√π (6)

and we let u = xa .

Thus the required functional form is:

f(x) =1

a√πe−

x2

a2 (7)

Maxwell does does not elaborate on the proof of the assertions in (3)-(4) but (3) indicates afunctional relationship which maps sums of reals to products. Equation (3) basically says thatthe function is radial in nature since there is no angular dependence. This is reasonable sinceMaxwell proves earlier in his papers ( see [1], pages 378-9) that the probability of the direction ofthe velocity after impact of two perfectly elastic spheres lies between given limits is independentof angle of rebound. I have set out how he did this in the Appendix.

2

We know that the exponential function establishes an isomorphism from the additive group ofreal numbers (R,+, 0) and the multiplicative group of positive reals (R+, ·, 1). The isomorphismis effected by the map x → ex. This is bijective with inverse y → ln y and gives rise to thefunctional equation ex+y = ex ey.

It is worth noting here that the use of a functional relationship with symmetry was used by Gaussin his investigations of the distribution of errors in which he was seeking the form of a functionwhich gave the maxiumum probabiity of the mean of the observed values. That function hadthe form of a Gaussian. Laplace’s original proof of the central limit theorem ultimately involveda limiting argument which relied upon Stirling’s formula (see [3] )

To solve (3) there are two essential preliminary results both associated with Cauchy.

1.2 Cauchy’s functional equation

If f : R→ R is a continuous function which satisfies f(x+ y) = f(x) + f(y) for all real x, y, thenthere exists an a ∈ R such that f(x) = ax for all real x.

1.3 Cauchy’s exponential functional equation

If f : R→ R is a continuous function (not identically zero) which satisfies f(x+ y) = f(x) f(y)for all real x, y, then there is a real b > 0 such that f(x) = bx for all real x.

Applying these principles to the assumptions Maxwell made in positing the functional relation-ship (3) we can conclude that φ(x2 + y2 + z2) must be non-zero at the origin since we assumethe system starts there with N particles. By its very construction it must be positive and (3)implies that the relationship holds for negative values. If we take a neighbourhood around theorigin we see that φ(x2 + y2 + z2) must be positive (recalling that we are assuming continuity)and hence this holds everywhere by extension to the whole of the real line.

If we let x2 = X, y2 = Y and z2 = Z then (3) becomes:

f(√X) f(

√Y ) f(

√Z) = φ(X + Y + Z) (8)

Now let g(X) = f(√X) = φ(X). Therefore:

g(X + Y ) = f(√X + Y ) = φ(X + Y ) = f(

√X) f(

√Y ) = g(X) g(Y ) (9)

Hence g satisfies Cauchy’s exponential equation (see 1.2). Thus g(X) = g(X2 + X2 ) = g(X2 )2 > 0

for all X. As noted above (3) holds for negative values of x,y and x ie φ(x2 + y2 + z2) =f(−x) f(−y) f(−z) so f(x) = f(|x|).

Now because g(X) = g(x2) = f(x) > 0 for all x we can let g(X) = ln(f(√X)). We then have:

3

g(X + Y ) = ln(f(√X + Y ))

= ln(φ(X + Y ))

= ln(f(√X) f(

√Y ))

= ln(f(√X)) + ln(f(

√Y ))

= g(X) + g(Y )

(10)

Hence g satisfies Cauchy’s functional equation (1.2). Thus g(X) = AX for some real A > 0. Itfollows that:

ln(f(√X)) = AX

f(√X) = eAX

f(x) = eAx2

(11)

Since f(0) = 1 and there are N particles at the origin we must have f(x) = C eAx2

where C > 0.

Going back to (3) we must have f(x) f(y) f(z) = C3 eA(x2+y2+z2) = C3 eAr2

as Maxwell stated.

Maxwell goes on to deduce four things from the formula for f given in (7). First, he says thatthe number of particles whose velocity, resolved in a certain direction, lies between x and x+dx is:

N1

a√πe−

x2

a2 dx (12)

That simply follows by weighting (3) by the number of initial particles.

The second implication of (7) Maxwell states is that the number of particles whose actual velocitylies between v and v + dv is:

N4

a3√πv2 e−

v2

a2 (13)

Maxwell does not explain how (13) was derived. If we think like a physicist the approach mightbe to visualise the actual velocity v like a radius vector of a sphere - the magnitude of the radiusis the magnitude of the velocity (where v2 = x2 + y2 + z2 recalling that the x, y and z are therelevant velocity components) and the radius can be distributed anywhere on the sphere. Whatwe are after is the distribution of particles with a velocity in the range v to v+dv and this can bevisualised as the infinitesimal volume dV = d( 4

3πv3) = 4πv2 dv. Now because the f’s are multi-

plicative and there are N particles to start with, the number of particles in the range v to v+dv is:

4

N1

a√πe−

x2

a21

a√πe−

y2

a21

a√πe−

z2

a2 4πv2 dv = N4

a3√πv2 e−

(x2+y2+z2)

a2 = N4

a3√πv2 e−

v2

a2 (14)

The third thing Maxwell deduces from the basic functional equation is the mean value of v whichhe finds by adding the velocities of all the particles and dividing by the number of particles giving:

mean velocity =2a√π

(15)

Although Maxwell does not set out the calculation what he did was this integral:

mean velocity =

∫∞0v N 4

a3√πv2 e−

v2

a2 dv

N=

∫ ∞0

4

a3√πv3 e−

v2

a2 dv (16)

To evaluate (16) let u = v2

a2 so that du = 2v dva2 hence the integral becomes:

∫ ∞0

4

a3√πv3 e−

v2

a2 dv =1

2

∫ ∞0

4

a3√πa4 ue−udu =

2a√π

(17)

where∫∞0ue−udu =

[− u e−u

]∞0

+∫∞0e−u du = 1 since limu→∞ u e−u = 0

The fourth thing that Maxwell deduces is that the mean value of v2 = 32a

2. This is greater thanthe square of the mean velocity (as you would expect). To derive the mean of v2 involves thisintegral: ∫∞

0N 4a3√πv2 v2e−

v2

a2 dv

N=

∫ ∞0

4

a3√πv4 e−

v2

a2 dv (18)

To integrate (18) involves some tedious integration by parts but you quickly get to 6a√π

∫∞0v2 e−

v2

a2 dv

and a further integration by parts establishes that (18) finally evaluates to 3a2

2 . Given that kineticenergy is

∑i12mv

2i and there are 3 spatial dimensions, if you had to guess what the constant might

be for the mean of v2, 32 would be a reasonable guess. The Appendix contains all the gory details.

Maxwell notes that ”the velocities are distributed among the particles according to the samelaw as the errors are distributed among the observations in the theory of the method of leastsquares. The velocities range from 0 to ∞, but the number of those having great velocities iscomparatively small.” This is a reference to Gauss’s earlier work on the distribution of errors.

5

2 The details of Maxwellian velocity distribution integrals

Recall that in (18) we had to integtrate∫∞0

4a3√πv4 e−

v2

a2 dv and the details were left out. The

details are now set out. We first calculate this integral since we use it a couple of times:

∫v e−v2

a2 dv =a2

2

∫e−u du = −a

2

2e−v2

a2 (19)

where u = v2

a2

Our first integration by parts yields:

I =∫∞0

4a3√πv4 e−

v2

a2 dv =∫∞0

4a3√πv3 v e−

v2

a2 dv = 4a3√π

{[−v3a2 e−

v2

a2

2

]∞0

+∫∞0

3a2

2 v2 e−v2

a2 dv}

=

6a√π

∫∞0v2 e−

v2

a2 dv. Noting here that[−v3a2 e−

v2

a2

2

]∞0

= 0 since limv→∞ vk e−v2

a2 = 0 for any power

of k.

Hence I = 6a√π

∫∞0v2 e−

v2

a2 dv and to integrate this we have to do another integration by parts

(I did say it was tedious).

6a√π

∫∞0v2 e−

v2

a2 dv = 6a√π

∫∞0v v e−

v2

a2 dv = 6a√π

{[−a2ve−

v2

a2

2

]∞0

+∫∞0

a2

2 e− v2

a2 dv}

= 3a√π

∫∞0e−

v2

a2 dv =

3a√π

√πa2 = 3a2

2

6

3 Appendix

7

Maxwell proves Proposition 1 by geometric methods and his conclusion, given his assumptionthat the balls have velocities inverse to their masses, is that the velocity of each ball is the samebefore and after impact and line in the same plane with the line of centres and make equalangles with it. The assumption of perfect elasticity allows Maxwell to draw this conclusion - thecoefficient of restitution is 1. To obtain an expression for the coefficient of restitution we need tosolve the kinetic energy and linear momentum equations for horizontal and vertical components.In Maxwell’s set up of the diagram ∠APN = ∠BQN

′= α where NN

′is the line through the

centres. Similarly ∠aPN = ∠bQN′

= β. Thus if the masses are m1,m2 and the before collisionvelocities are u1, u2 and the post collision veloctities are v1, v2 we have the following where itis understood that these velocities are placeholders for the horizontal and vertical componentsresolved along NN

′and will thus reflect relevant sine and cosine factors. Note here that the

magnitude of the velocities pre and post collision are simply the inverses of the masses ie:

8

| ~u1| = |~v1| =1

m1

| ~u2| = |~v2| =1

m2

(20)

Thus the conservation of kinetic energy and linear momentum equations are respectively (forconvenience I will save keystrokes by writing uk = | ~uk| etc in the kinetic energy and subsequentformulas) :

1

2m1u

21 +

1

2m2u

22 =

1

2m1v

21 +

1

2m2v

22

m1(u21 − v21) =m2(v22 − u22)

m1(u1 − v1)(u1 + v1) =m2(v2 − u2)(v2 + u2)

(21)

m1 ~u1 +m2 ~u2 =m1 ~v1 +m2 ~v2

m1( ~u1 − ~v1) =m2(~v2 − ~u2)(22)

Taking the horizontal components in (22) we have:

m1(u1 cosα− v1 cosα) =m2(v2 cosβ − u2 cosβ)

m1 cosα(u1 − v1) =m2 cosβ(v2 − u2)(23)

Taking the perpendicular components in (22) we have:

m1(u1 sinα− v1 sinα) =m2(v2 sinβ − u2 sinβ)

m1 sinα(u1 − v1) =m2 sinβ(v2 − u2)(24)

Squaring and adding (23) and (24) we get:

m21(u1 − v1)2(cos2 α+ sin2 α) = m2

2(v2 − u2)2(cos2 β + sin2 β) (25)

which implies that:

m1(u1 − v1) = m2(v2 − u2) (26)

Dividing (21) by (26) we have:

u1 + v1 =v2 + u2

−( v1 − v2u1 − u2

)=1

(27)

9

Equation (27) reflects the requirement for perfect elasticity and this must hold for horizontal andvertical components. The initial horizontal components are u1 cosα, u2 cosα and the final hori-zontal components are v1 cosβ, v2 cosβ. The vertical components are similarly u1 sinα, u2 sinαand v1 sinβ, v2 sinβ . Thus from (27) we have for horizontal components:

v2 cosβ − v1 cosα =u1 cosα− u2 cosβ

(u2 + v2) cosβ =(u1 + v1) cosα

(u1 + v1) cosβ =(u1 + v1) cosα

cosβ = cosα

(28)

and for vertical components we have:

v2 sinβ − v1 sinα =u1 sinα− u2 sinβ

(u2 + v2) sinβ =(u1 + v1) sinα

(u1 + v1) sinβ =(u1 − u2) sinα

sinβ = sinα

(29)

Multiplying (28) by sinα and (29) by cosα and subtracting we have:

sinα cosβ − cosα sinβ =0

sin(α− β) =0(30)

This implies that α = β so that Maxwell’s statement about equal angles is correct (one canignore the possibility that α− β = nπ where n is a non-zero integer as being unphysical).

To prove Proposition 2 Maxwell notes that in order for a collision to take place, ”the line ofmotion of one of the balls must pass the centre of the other at a distance less than the sum oftheir radii; that is, it must pass through a circle whose centre is that of the other ball, and radius(s) the sum of the radii of the balls. Within this circle every position is equally probable, andtherefore the probability of the distance from the centre being between r and r + dr is”:

2r dr

s2(31)

This is seen as follows. The area of the annulus between r and r + dr is 2πrdr and the area ofthe circle of radius s is πs2, hence the probability is:

2πr dr

πs2=

2r dr

s2(32)

The next step is to note that if phi is the angle APa between the original direction and thedirections after impact then ∠APN = 1

2φ and r = s sin 12φ. The following diagram sets out the

geometry for the relationship r = s sin 12φ:

10

The required probability is then:

2s sin 12φ×

s2 cos 1

2φdφ

s2=

1

2sinφdφ (33)

Maxwell then notes that ”the area of a spherical zone between the angles φ and φ+dφ is:”

2π sinφdφ (34)

Maxwell is assuming a sphere of unit radius so that the area of the angular strip is 2π×sinφ×dφ.

11

The final step is the observation that if ω is any small area on the surface of a unit sphere theprobability of a direction of rebound passing through this area is:

ω

4π(35)

since the surface area of the unit sphere is 4π. Thus the probability is independent of φ so thatall angles of rebound are equally likely.

4 References

[1] James Clerk Maxwell, ”The Scientific Papers of James Clerk Maxwell”, Volume 1, Dover,2003.

[2] Christopher G Small,”Functional Equations: How to solve them”, Springer, 2007.

[3] Peter Haggstrom, ”The central limit theorem - how Laplace actually proved it”, https://

gotohaggstrom.com/The%20central%20limit%20theorem%20-%20how%20Laplace%20actually%

20proved%20it.pdf

12

5 History

Created 21 April 2019 24 April 2019 - typo corrected: ”elastic” for ”inelastic”. I doubt anyonewas fooled!

13