Speculation on the Maxwell-Boltzmann exp(-ei/T) From the View of Conservation of Momentum Part 2

Speculation on the Maxwell-Boltzmann exp(-ei/T) From the View of Conservation of Momentum Part 2 Francesco R. Ruggeri Hanwell, N.B. Oct. 3, 2025 It is possible to obtain the Maxwell-Boltzmann (MB) distribution through reaction balance applied to a 2-body elastic collision, or by maximizing ln of a number of arrangements (entropy) subject to constraints. In Part 1, we tried to obtain the MB distribution by using momentum considerations (with no knowledge of kinetic energy). Here, we try to provide an alternative argument, using nonrelativistic momentum alone, and the idea of a Galton board, introduced in (1). Basically, we argue that given conservation of momentum, there is equal probability to have changes dv and -dv to velocity. We suggest creating the velocity k dv - (n-k) dv using k dv’s and (n-k) dv’s, where n is a very large number. This leads to a binomial distribution describing the arrangements n!/ (k! (n-k)!)) which (1) shows is equivalent to a Gaussian, exp(- C v dot v). This result is obtained with no maximization or constraints and no notion of a reaction in the system. In other words, one is able to assign a probability to each v (one dimension) by considering it to be made of various different dv and -dv units for a fixed n which tends to infinite. Physically, however, there are reactions which must conserve momentum. This is already implied above by giving -dv and dv equal probability weights. Thus, if one has v1 and v2 as initial momenta, then one has: exp(-C v1 dot v1) exp(-C v2 dot v2) as the probability for a v1, v2 collision to occur in a simplified scenario. By time reversal balance, exp(-C v3 dot v3) exp(-C v4 dot v4) must hold for v3 and v4 outcome velocities suggesting that there exists a quantity C v dot v which is conserved in these collisions. If one had known about this conserved quantity a priori, one could have obtained the MB distribution directly from p(e1)p(e2) = p(e3)p(e4) and e1+e2 = e3+e4, where e= C v dot v. Energy Based Approaches to Finding the Maxwell-Boltzmann Distribution For some reason, the MB distribution is usually obtained through energy (kinetic energy in the case of no potential) considerations. One common method is to use reaction balance (time reversal balance) for elastic 2-body collisions: E1+e2 = e3+e4 and p(e1)p(e2)=p(e3)p(e4) —> p(ei) = C exp(-ei/T) ((1)) An alternative approach is to consider the number of arrangements existing when a total energy E is distributed among N particles such that n(ei) = N p(ei) is the number of particles with kinetic energy ei. Then, ln of the number of arrangements is: ln(N!/ Product over i n(eiI)!) ((2)) Using Stirling’s approximation and maximizing ((2)) with respect to: Sum over i p(ei) =1 and Sum over i ei p(ei) = eva ((3)) yields the MB distribution. Given that momentum is also conserved in 2-body collisions, one may ask: Is it possible to obtain the MB distribution through momentum considerations alone, without even knowing the concept of kinetic energy? In Part 1, we tried to present one approach, and here we present a second. Obtaining the MB Distribution Through Momentum Conservation Our basic argument is that given momentum conservation, the probability to have a velocity change dv should equal the probability linked with -dv. We then argue that any given nonrelativistic momentum p= mo v (one dimension) may be constructed from various dv and -dv components. In particular, we consider a fixed n tending to infinite and k dv units together with (n-k ) -dv units, to create: p = mo ( k dv - (n-k) dv) = mo (2kdv - ndv) ((4)) The number of arrangements is: n!/ { k! (n-k)!) ((5)) Thus, ((5)) gives different weights to different v values. V may be a component vx in one dimension or the magnitude of v. Now, a given p may be created using k=0, k=1 … We note that the above math describes the results of a Galton board. Such a board consists of a peg in the center of the first row. On the second row, two pegs are placed an equal distance delta and -delta to the right and left of any above peg. This rule is extended to subsequent rows to create a triangle. If one releases particles at the top, they have equal probabilities to go to the right or left of each peg and this is equivalent to the equal probability for dv and -dv suggested above. Experimentally, it may be shown by repeated trials that one obtains a Gaussian distribution. (1) show mathematically, how one may use the binomial distribution to obtain a Gaussian. ((5)) implies a probability for v being made of k dv units of: p(k) = n!/ { k! (n-k)!) .5 power k .5 power (n-k) ((6)) According to (1): p(k) = Integral (xk-e, xk+e) p(x) dx proportional to exp(- (x-u)(x-u)/(2ss)) ((6)) Here x-u= (2k-n) e (equivalent to ((4)) and s = 2e sqrt(n .5.5) ((7)) The point is that one obtains a Gaussian in v dot v (or vv for one dimension). As a result, one may suggest that ((6)) represents the distribution of velocity vectors in an ideal gas. The above argument makes no use of the notion of two body reactions or the existence of kinetic energy. It does, however, use the factorial counting scheme (also used for entropy) , but involves no maximization subject to constraints as explicitly pointed out in (1). In other words, the above scenario uses the same probability scheme as a coin toss, i.e. a probability of .5 to obtain dv and .5, to obtain -dv, i.e. a uniform distribution to ultimately obtain a Gaussian. The Notion of Two-Body Collisions and Kinetic Energy The above argument makes no use of the notion of kinetic energy or 2-body collisions. Nevertheless, the assumption that dv and -dv are linked with equal probabilities suggests that one has conservation of momentum, we argue. Physically, there are two body collisions. The simplest argument for the combined probability of a v1 velocity vector with v2 is: exp(-C v1 dot v1) exp(-C v2 dot v2) ((8)) using the Gaussian obtained above. The outcome is a set of v3 and v4 velocity vectors. By time reversal, however, one must have: ((8)) = exp(-C v3 dot v3) exp(-C v4 dot v4) ((9)) Thus, there exists a quantity C v dot v which is conserved together with momentum in these collisions, and this quantity arises directly from statistical arguments for nonrelativistic momentum mo v. This quantity C v dot v is the usual kinetic energy of classical mechanics. If one knows a priori about the existence of kinetic energy (i.e. through Newtonian mechanics), one may write ((9)) a priori and obtain the Maxwell-Boltzmann distribution in a simpler manner than the one given above for momentum. Nevertheless, the momentum approach seems to hold and gives rise to the notion of kinetic energy, just as this quantity arose in Part 1. Conclusion In conclusion, the Maxwell-Boltzmann distribution is usually obtained using notions of kinetic energy for a system with no potential V(x). One may use either reaction balance for elastic two body collisions or create ln of the number of arrangements of Etotal distributed among N particles (n(ei)=Np(ei)) and then maximize this subject to the constraints Sum over i p(ei) = 1 and Sum over i ei p(ei) = eave. The ln factorial function is proportional to what is called entropy. We argue here that one does not need any notion of kinetic energy to obtain the MB distribution. One may simply use the notion of nonrelativistic momentum mov and the idea of conservation of momentum, i.e. assign equal probabilities for a dv and -dv to occur. One may then consider a large n with k dv’s and (n-k) (-dv)s as creating v. The probability for such a v= k dv - (n-k)dv = (2k-n)dv is: n!/ ( k! (n-k)!) .5 power k .5 power (n-k). This is the same math form that arises from a Galton board, as discussed in (1). In (1), it is shown that this probability is equivalent to a Gaussian in the large n limit, i.e. exp(-C (2k-n)dv (2k-n)dv). As a result, one obtains the MB distribution using the probability ideas of a coin toss and factorials (describing arrangements). Given that physical two body interactions occur, the simplest joint probability for v1 and v2 is exp(-Cv1 dot v1) exp(-C v2 dot v2) and by time reversal this must equal exp(-C v3 dot v3) exp(-C v4 dot v4), for v3, and v4 as outcome velocities. Thus, in addition to conservation of momentum, there is a conserved quantity C v dot v. This is the kinetic energy of classical physics which arises strictly from statistics. References 1. Keppens, A. and Lambert, J-C. Thermodynamics of Observations Entropy 2025, 27, 968 https://pdfs.semanticscholar.org/38b4/2b6a342ccc910f81f85b6716d6a823ee7545.pdf