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Bias Removal and a Momentum Treatment of the Maxwell-Boltzmann Distribution Part 2 Francesco R. Ruggeri Hanwell, N.B. Oct. 8, 2025 In Part 1, we argued that if one wishes to describe the distribution of ideal gas particles in terms of momentum, P1(p) = P1(-p), where P1 is the probability distribution for momentum p. This forces a functional form of P1 to be quadratic in p (e.g p dot p or something else). The point we made is that one may also consider the problem in terms of energy and elastic collisions ei+ej = ek+el and p(ei)p(ej) =p(ek)p(el) ((1)). The P1(p) distribution cannot contradict p(ei) and the ei one is already quadratic in p, i.e. e = pp/2m and so P1(p) should equal p(ei) according to the arguments made in Part 1. That, however, does not explain why one may consider a velocity v as being constructed from k (dv) units and (n-k) (-dv) units with a probability of: n!/ (k! (n-k)!) .5 power n. The approach of ((1)) deals with a two body elastic collision which also conserves momentum and this is a physical process in an ideal gas. At the same time p(ei)p(ej) = p(ei+ej) for p unnormalized means that one may use the bias removal of ((1)) to actually create a probability for p(ei+ej). We suggest then that the approach of ((1)), linked to energy conservation, should allow one to construct v in a probabilistic manner which removes bias. This approach, however, should be linked to physical reality as well. In the case of momentum which is linear in velocity (nonrelativistic case), we consider that from the point of view of a particle, one could imagine a somewhat strange system in which dv and -dv hits are given with constant probability. This means that energy delivery changes, but here we are only concerned with momentum, i.e. one has a random walk as an unbiased creation of v in v space. Thus, the bias removal of an AND situation (ei,ej) and (ek,el) having the same joint probability if ei+ej=ek+el is now changed to bias removal of building blocks of v. These seem like two very different kinds of bias removals, but the goal is to remove as much bias as possible. The construction of v from k (dv) units and (n-k) (-dv) units does not seem to violate any constraints in the system and so a particle in a Maxwell-Boltzmann gas seems to change motion at random (nonrelativistic case). In other words, the notion of bias removal through complete randomness of momentum hits seems to be identical to the condition ((1)). It seems reasonable that the factorial probability P1(p) = n!/(k! (n-k)!) .5 power n (for p = 2k-n) should yield the MB distribution value p(e) = C exp(-ei/T). It does seem that there is a link between a Maxwell-Boltzmann gas, time reversal balance to remove bias, and the idea of bias removal through randomness represented by a Gaussian. The issue is that this argument only holds in the nonrelativistic case, whereas ((1)) holds in both the relativistic and nonrelativistic cases, with ei following eiei = pcpc + momocccc in the relativistic scenario. Nevertheless, the randomness (bias removal) of a Gaussian (random walk in v space) seems to be fine for a nonrelativistic picture in which p=mov and so one can construct p in essentially the same way as v.
Bias Removal In Part 1, we argued that a Maxwell-Boltzmann distribution must remove bias. The question is: How does one mathematically describe bias? We argued that one kind of bias that must be removed is: n(ei)n(ej) not= n(ek)n(el) if ei+ej = ek+el ((1)) This is done by imposing: n(ei)n(ek) = n(el)n(ek) for ei+ej = ek+el ((2)) This is a very specific bias linked to physical two body elastic scattering and it leads to the solution of the MB distribution: p(ei) = C exp(-ei/T) ((3)) The question we ask is: Is there another seemingly different bias removal which is equivalent? We try to argue here that there is, in the nonrelativistic situation. We suggest that there is a hint of this second bias contained in ((2)). In particular, for unnormalized p(ei): p(ei+ej) = p(ei)p(ej) ((4)) In other words, one may actually consider creating an ei+ej probability by using various ei, ej values. One may extend this to p(ei+ej) = p(ek)p(el)p(em) for ei+e = ek+el+em etc. As a result, one may consider creating ei+ej in many different ways and the product of probabilities must be the same for bias to be removed. The key idea here, we suggest, is the notion of creating ei+ej in an unbiased manner and its equivalence to the two body elastic collision ((2)). Momentum Considerations We suggested in the previous section that one may consider an unbiased creation of a key variable (say energy or momentum) without any bias as leading to the MB distribution. In the above section, we considered energy. We now focus on momentum, which in the nonrelativistic case is mov. Thus, this argument only holds in the nonrelativistic case. We suggest that an unbiased approach to creating v is to have equal probability of adding dv or (-dv) to the existing momentum, i.e a random walk in v space. This does not mean that kinetic energy is being added in constant units, but we only focus on momentum here. Bias removal in the nonrelativistic momentum case seems to be linked to a random walk type of building of velocity and in one dimension this means dv and (-dv) steps with equal probability. In Part 1, we argued that this different form of bias removal (random velocity walk) must be equivalent to the bias removal of ((2)) as ultimately all possible bias must be removed. Thus, one may use the binomial (Galton board) probability: P1(p)= P1( dv(2k-n)) = n!/ ( k! (n-k)!) .5 power n ((5))
One sees that interchanging k and (n-k) leaves ((5)) unchanged, but that: p= dv k + (n-k) (-dv) becomes -p ((6)) P1(p) = P1(-p). ((5)) is the random walk scenario and one already knows that for complete randomness in the large k, n limit, a Gaussian will result. One may, however, formally show that that ((5)) becomes a Gaussian as done in (1). Here we use the simplest form of Stirling’s approximation to show that this is the case, using: K = n/2+m/2 from (1) such that ((5)) becomes: n!/ ( (n/2 + m/2)! (n/2-m/2)! ) ((7)) Then using b! = b power b for b very large, one has: ln(P1) = (n/2+m/2) ln(n/2+m/2) + (n/2-m/2) ln(n/2-m/2) For n>>m ln(P1) = constant mm so p = C1 exp(C2 mm) , but m= 2k-n which is momentum ((8)) One obtains the random walk Gaussian in the nonrelativistic case which removes bias from v, and p (because p=mov), but this bias removal cannot contradict the bias removal of time reversal balance ((2)). Thus, two seemingly different kinds of bias removal actually lead to the same result in the nonrelativistic case. Relativistic Case In the relativistic case, one has a problem. The approach of bias removal of ((2)) still holds, but eiei = ppcc + momocccc. On the other hand, momentum p is no longer mo v and so if one wishes to construct p in an unbiased way, one can no longer use the k (dv) and (n-k) (-dv) approach. The approach of the random walk and Gaussian as being equivalent to reaction balance only holds in the nonrelativistic case. Nevertheless, we argue that it is interesting that it holds at all. Conclusion In conclusion, in Part 1, we argued that the Maxwell-Boltzmann distribution is driven by removing all possible bias from the ideal gas problem. We suggested that concretely, this bias is represented by n(ei)n(ej) not= n(ek)n(el) if ei+ej = ek+el. The solution is to force n(ei)n(ej) =n(ek)n(el) for ei+ej=ek+el which yields the MB distribution. We then showed that a binomial (Galton board) factorial probability for creating momentum yields a distribution which is consistent with the exp(-ei/T) obtained from reaction balance. Here we try to explain why this is the case. We suggest that reaction balance is also linked to the notion of constructing an ei+ej, i.e. p(ei+ej) = p(ei)p(ej) for an unnormalized p(ei). We suggest that one might try to extend this idea of constructing a variable to the momentum case. In the nonrelativistic scenario (and not the relativistic one), p=mov, so one may consider creating v in an unbiased manner. We suggest that a random walk (i.e. equal probability for dv
and (-dv) steps in velocity space) is an unbiased construction approach (nonrelativistically). This leads to: P1(p) = n!/ (k! (n-k)!) .5 power n, where p = (dv)k + (n-k) (-dv). Thus, P1(p) = P1(-p). This removal of bias should lead to the same result as the reaction balance one because ultimately one wants all bias removed. This happens because the binomial factor becomes a Gaussian for large n,k as shown above. Thus, two seemingly different bias removals (reaction balance) and random walk construction of v yield the same result, but only in the nonrelativistic case. References 1. https://people.bath.ac.uk/pam28/Paul_Milewski,_Professor_of_Mathematics,_University _of_Bath/Past_Teaching_files/stirling.pdf
