Mechanical Proof of the Maxwell Speed Distribution

This article derives the probability density function ψ(ξ; x, x) of the resulting speed ξ from the collision of two particles with speeds x and x′. This function had been left unsolved for about 150 years. Then uses two approaches to obtain the Maxwell speed distribution: (1) Numerical iteration: using the equation Pnew(ξ) = ∫ ∫ ψ(ξ; x, x ′) ∙ Pold(x) ∙ Pold(x ′) dxdx ∞ 0 ∞ 0 to get the new speed distribution from the old speed distribution. Also, after 9 iterations, the distribution converges to the Maxwell speed distribution. (2) Analytical integration: using the Maxwell speed distribution as the Pold(x), and then getting Pnew(ξ) from the above integration. The result of Pnew(ξ) from analytical integration is proved to be exactly the Maxwell speed distribution.

Maxwell first provided the Maxwell speed distribution in 1860 on statistical heuristic bases (Maxwell, 1860a,b).Maxwell in 1867 (Maxwell) and Boltzmann in 1872 (Boltzmann) carried out some more investigations into the physical meaning of the distribution.The simplest way to prove the Maxwell speed distribution is from the statistical view: beginning from the Boltzmann distribution of energy state which is proportional to the square of velocity, and extending to three velocities in three directions and summing the same speed distribution in all three directions to get the Maxwell speed distribution (Brush, 1966, Landau et al., 1969, McQuarrie, 1976, Garrod, 1995, Maudlin, 2013).Therefore, the distribution is also known as the Maxwell-Boltzmann distribution.The standard speed distribution function is listed as follows along with a more compacted parameter ℎ which is the inverse of the most probable speed   , i.e.,   = ℎ −1 .
where  is the Boltzmann constant,  is the equilibrium temperature,  is the particle mass, and ℎ = √  2 .
In 1872, Boltzmann gave the following equation: (2) where f(x, t)dx is the number of particles with speed between x and x+dx, and similarly for f(x′, t)dx′, dn is the number of particles with speed between ξ and ξ + dξ.If we let f(ξ, t + dt) = dn/dξ, and rewrite Eq.(1) as As  → ∞, f(x, t) → (), the correct distribution, (), should satisfy the following new integral equation Boltzmann said that "Since this calculation ((ξ; x, x ′ ), add by authors), although tedious, is not at all difficult, …".However, until now, this calculation is still missing in the literature.As shown in Section 2, the function can be derived based on Newton's laws of motion, and therefore it is also a mechanical proof of the Maxwell speed distribution.
After we get the function (ξ; x, x ′ ), we use two approaches to get the Maxwell speed distribution: (1) Numerical iteration: using the following equation to get the new distribution from the old one.Also, found that the final distribution after 9 iterations converges to the Maxwell speed distribution as shown in Section 3.
(2) Analytical integration: using the Maxwell speed distribution as   to get   from integration.And the   from analytical integration is exactly the Maxwell speed distribution as shown in Section 4.
2. Derivation of (; ,  ′ ) Before processing to derive the function (ξ; x, x ′ ), we change the variables ξ to , x to   and x′ to   and rewrite the function as (;   ,   ).For ease of reference, the resulting function is listed as follows.Since   and   are exchangeable, only the functions for   ≥   are listed.(5)

𝜓(𝑣
2.1 For Special Case of   = 0 Let v 0 =   be the speed of particle 1 with mass M 1 which will hit particle 2 with mass M 2 at rest.After a collision, the new particle speeds are v 1 and v 2 as shown in Fig. 1.
Figure 1.The collision of two particles where particle 2 is at rest Based on Newton's laws of motion, the total momenta before and after the collision are the same (Eqs.(6-7)).Also, for elastic collision, the total energies before and after the collision are also the same (Eq.(8)).
8) For M 1 = M 2 , we get the solutions as (Note 1) The solutions can be represented as Fig. 2, where v 1 =  ̅̅̅̅ , and v 2 =  ̅̅̅̅ .Note that, after the collision, P and Q are always located on the sphere surface and the probability is uniform on this surface.Since the probability of the point inside the circle in Fig. Figure 3. Solution P, Q located on a sphere surface Then for fixed magnitudes of   and   , if   is fixed in the horizontal direction but changed the direction of   , then point A will be located on a spherical surface as shown in Fig. 4.And the probability of point A on the surface is uniformly distributed since   has equal opportunity in any direction.The center of the surface A is at point O and its radius is   .S is the middle point between B and A and is the center of the sphere surface P (also in Fig. 2 and 3).The point S will be located on a smaller sphere surface center at C (middle point of O and B) with radius   /2.Since  ̅̅̅̅ is always parallel to  ̅̅̅̅ , the point S on the sphere surface S is also uniformly distributed similarly to point A on the sphere surface A. Although the sphere surfaces S and A are fixed, the surface P is variable in center S and radius r 2 .Next, the surface S and the surface P are used to find the probability of   .The location of S is defined by α (representing the relative moving direction before collision), and the location of P is defined by β (representing the particle moving direction after collision).
1) The probability density of point S located on surface S at angle is  ( ) = 1 2 sin .
2) The probability density of point P located on surface P at angle is The  1 and  2 as shown in Fig. 4 can be computed from as where  = 0 for   ≤  ≤   , else  are where  =  ̅̅̅̅ or  = ′ ̅̅̅̅̅ (Fig. 4) as follows Both equations have identical solutions for cos  as (Note 4) Substitution of Eq.( 20) into Eq.(18) yields the probability density function as (Regions are shown in Fig. 6)

Numerical Iteration
It is easy to do the iteration by 65 equal spaced discrete speeds beginning from v 1 = 0.5 in increments of 1.0 and end up to v 65 = 64.5,where speeds over 64.5 are truncated, and therefore the probabilities are assumed to be zero.For discrete speeds, the integration changes to summation as follows (∆  ∆  = 1) P  (  ) = ∑ ∑ (  ;   ,   )P  (  )P  (  ) If we assume the Root-Mean-Square speed is 16.5, and the initial speeds of all particles are 16.5, that is P  ( 17 ) = 1 and all others P  (  ) = 0, for  ≠ 17.Use the equation above to get P  (  ), and set P  (  ) = P  (  ) for next iteration.After nine iterations, the distribution curves converge to the Maxwell speed distribution as shown in Figure 5.As shown in Fig. 5, the horizontal axis for the speed, , has been normalized by the most probable speed,   .Therefore the peak dirtribution density is just at    ⁄ = ℎ = 1 as we would expect.Where   = √2/3  = 13.47, he peak distribution density is (4ℎ √ ⁄ ) −1 = 0.0616.Any initial distribution may be assumed, as long as the initial RMS speed less than 25% of the maximum speed used, i.e., 64.5 in the presented case, the distribution curve always converges to the Maxwell speed distribution.

Analytical Integration
Let   be the Maxwell speed distribution () = 4ℎ 3 √  2  −ℎ 2  2 and compute   from the following equation with four regions as shown in Fig. 6.
The analytical integration result of   () is just the Maxwell speed distribution as we would expect.This concludes the proof that the Maxwell speed distribution is correct from the random collisions of the particles.

Conclusions and Further Studies
It is not only interesting but also very important to get the function (;   ,   ) since this is from which the Maxwell speed distribution can be proved.From the derivation of the function (;   ,   ), we can reveal the basic mechanism behind the macroscopic phenomenon.The mechanics of the collision of particles is a bridge between microscopic behavior and macroscopic phenomenon.
This paper only investigates the collisions of equal mass particles.Further study may be on the collisions of unequal mass particles and may be used to give a mechanical proof of Avogadro's law.The procedures of this paper may also be used for the collisions of charged particles.
2) The area of the ring inside the circular plane disk (Fig. 2(b)) is 2( sin )( sin ) = 2 2 sin  cos 3) The ratio of the two areas is v 0 2 / 2 , it is not dependent on .
4) When the center of the particle hits inside the ring of the disk, the Q point must locate inside the ring on the sphere surface.
5) It has equal opportunity to hit on any point inside the circle.

Figure 2 .
Figure 2. The relation between collision point as shown in (b) and new velocity BQ ̅̅̅̅ in (a) 2.2 For General Case of   ≥   > 0 Let   = |v ̅ 1 | and   = |v ̅ 2 | be the speeds of two particles before a collision.After the collision, the new particle speeds are  1 = |v ̃1| and  2 = |v ̃2|.Let v 0 = v ̅ 1 − v ̅ 2 , and follow the same procedures of Section 2.1 to get v 1 and v 2 , and therefore to get v ̃1 = v ̅ 2 + v 1 and v ̃2 = v ̅ 2 + v 2 as shown in Fig. 3.

Figure 4 .
Figure 4. Three sphere surfaces A, S, P with centers at O, C, S with radius   ,   /2, r 2