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Speculation on the Maxwell-Boltzmann exp(-ei/T) From the View of Conservation of Momentum Part 3 Francesco R. Ruggeri Hanwell, N.B. Oct. 3, 2025 In Part 2, we noted that one may consider, in the nonrelativistic case, the creation of v through k dv units and (n-k) (-dv) units, where p =mov. We then noted that if the probability for a dv and (-dv) is the same, i.e. .5, then p(k) = n!/ (k! (n-k)!) .5 power n. One may convert a binomial to a Gaussian in the n→ infinite limit, yielding p(v) = C exp(- C1 vv) or C exp(- C1 v dot v) in three dimensions. We noted that in an equilibrium system, the probability of v1 colliding with v2 to yield v3 and v4 should be time reversal invariant, suggesting that in a simple case, exp(-C1 v1v1) exp(-C1 v2v2) = exp(-C1 v3v3) exp(-C1 v4v4). This suggests that in equilibrium one has a conservation law v1v2 + v2v2 = v3v3 + v4v4. If this, however, is a real conservation condition, it holds whether or not there is equilibrium (as long as there are no physical processes which cause this conserved quantity to change, i.e. sound production etc.) One may also note that if one considers special relativity, one may see the term .5mvv emerge from mocc/sqrt(1-vv/cc) in the v<<c limit. We note that in the above scheme, the idea was to consider dv and (-dv) changes, which in the nonrelativistic case is analogous to dp changes. Thus, regardless of the velocity of a particle, one has a physical system which yields changes of dv or -dv. Now, changes in velocity are due to a push or pull, called a force by Newton. If one imagines that force is microscopically equivalent to a certain number of mo dv/N (N large) deliveries per sec, then F dt = dv m. This scheme then yields, upon multiplying by dx and dividing the LHS by dx, Integral Fdx = .5 mvv if the initial v is 0. In other words, one sees how .5mvv appears in a nonrelativistic scheme by trying to interpret a force as the delivery of dp/N per sec. This delivery of momentum is consistent with the scheme of Part 2 in which a v is created by k dv units and (n-k) (-dv) units. In particular, one has the idea of action-reaction because each dv delivered by F may be linked to a -dv and -F. Thus, one may create v and -v and both should have the same probability. Thus, we argue that Newtonian mechanics is implicitly contained in the statistical argument of Part 2. Speed is not a quantity, like the number of particles in a gas, but rather is changed dynamically through mechanics. . The Gaussian resulting from the binomial factor is associated with a conserved quantity C1 vv, but this in turn is linked to a physical scheme which delivers mo dv hits. The probability then becomes linked to a conserved quantity and this quantity is associated with work done creating v, i.e. .5movv (nonrelativistic case). In other words, stating that one may create v from k (dv)s and (n-k) (-dv)s really means that each dv or -dv process is linked with work (and a change in kinetic energy and this is directly linked to probability in an equilibrium situation. Thus, one should be able to obtain the C1 v v term (C1 v dot v in 3D) from “work” considerations which are linked to the notion of adding dv or mo dv units, which seems to be the case.
Part 2 In Part 2, we argued that one may obtain the Maxwell-Boltzmann distribution without any knowledge of the concept of kinetic energy. One only needs to consider velocity and consider creating a particular v through k dv and (n-k) (-dv) additions, where n→inifnite. If dv and -dv have the same probability, .5, then the probability for v is: P(v) = n!/ ( k! (n-k)!) .5 power n ((1)) As noted in Part 2, ((1)) becomes a Gaussian when n→ infinite, i.e. N→ infinite P(v) = C exp(- C1 vv ) (or vv→ v dot v in 3D) ((2)) If one considers v1 and v2 combining in a collision, the probability should be: exp(-C1 v1v1) exp(-Cv2v2) ((3)) (in a simple analysis) If the outcome is v3, v4, then to have time reversal balance: ((3)) = exp(-C1v3v3) exp(-C1v4v4) ((4)) This implies that there is conservation in a collision, namely: V1v1 + v2v2 = v3v3 + v4v4 ((5)) ((5)) should hold whether or not there is equilibrium. In other words, equilibrium yields probabilities which reflect the conserved quantity in the problem. The question is: If one does not know about the existence of kinetic energy, why does C1 vv emerge as a conserved quantity? How does it arise aside from statistics and not physical properties of interactions? Connection to Special Relativity We first note that the form C1 vv appears in special relativity for v<<c, where c is the speed of light in a vacuum, i.e. mocc/ sqrt(1-vv/cc) → mocc + .5movv ((6)) Thus, .5movv is actually a property of a particle for v<<c and not simply an abstract math result as in the above section. This suggests that one should examine the ideas of the previous section for clues to the physical origin of C1 vv.
The Emergence of Force In the first paragraph, we argued that v may be created by k (dv) units and (n-k) (-dv) units. In a physical situation, one must examine what this really means. Newton argued that an object changes its velocity when acted upon by a force (push or pull). If the C1 vv result is physical, it should somehow be linked to force which creates or changes v. The question then is: How can one interpret force in a way which is consistent with C1 vv? We suggest that force physically means that: Mo dv/N tiny momentum units are delivered per sec ((7)) Thus, F dt = the change in momentum or dp. ((8)) (nonrelativistic result) ((8)) is in fact Newton’s second law: F = dp/dt ((9)) One may then multiply and divide the LHS by dx yielding: F dx = mo v dv which integrates to: Integral F dx = .5 mvv (if v=0 initially). Thus, one has a mechanical (nonstatistical) approach to obtaining the conserved quantity and sees that the statistical approach implicitly includes this mechanical physics through the notion of creating v using dv and -dv units regardless of the original speed. For a constant force in a nonrelativistic scenario: dv = F/mo dt ((10)) ((10)) shows that the constant force F is “not interested” in the initial velocity, but adds mo dv if it acts over a time dt. This idea of always adding mo dv if Fdt remains constant, or -modv for -Fdt seems to be the picture behind creating v from k dv units and (n-k) (-dv) ones. Velocity is not a quantity like the number of particles in a box. Rather it is created dynamically and so one must examine the statistical argument closely to see that it is consistent with a physical approach to changing v. We suggest that the dv (-dv) approach does seem to be consistent with Newtonian mechanics and so yields C1 vv as a conserved quantity which is equivalent to the conserved quantity which would appear in an ideal gas. One may note that if one has F and -F, i.e. action-reaction from Newton, then creating k dv and (n-k) (-dv)s for one particle leads to the reverse for a second particle, hence creating -v. Then, the probability for both must be the same which is consistent with C1 vv (or C1 v dot v in 3D). Conclusion In conclusion, in Part 2, we argued that one may obtain the Maxwell-Boltzmann distribution C exp(-C1 vv) directly from velocity/momentum arguments without having any notion or definition of kinetic energy. We considered a speed v (1D) as being created by k dv units and (n-k) (-dv)
ones, with n approaching infinite. For dv and -dv having equal probability (i.e .5), the overall probability for a v is n!/ (k! (n-k)!) .5 power n. In Part 2 (see references therein) it was shown that this is equivalent to a Gaussian C exp(-Cvv). If one considers a reaction between v1,v2 yielding v3, v4, then the combined initial probability should equal the combined final for time reversal balance. This leads to the notion of a conserved quantity C1vv associated with a particle. Here we try to explain why such a quantity should arise. We argue that one must examine closely the statistical scenario of adding dv and (-dv) units. Speed is not a quantity like the number of particles in a box. Rather, it is changed dynamically through force (Newton). Thus, the physics of a force must be consistent with the dv (-dv) addition for the statistical approach to make sense. We argue that if one considers force as delivering mo dv/N (N large) momentum pieces per sec, then F dt is the overall change in momentum, i.e. Fdt = dp. This is Newton’s second law and multiplying and dividing by dx and then integrating (using dp/dt = mo dv/dt and v= dx/dt): Integral Fdx = .5mvv (v=0 initially). This leads to .5mvv as a conserved quantity. The point seems to be that dp= mo dv = Fdt and so as long as Fdt is constant, units of mo dv may be added. Given that dv and -dv have the same probability, it seems as if this is like actionreaction with v and - v being created with the same probability. Thus, the probabilistic scheme introduced in Part 2 seems to be consistent with Newtonian mechanics and this seems to be why C1 vv appears. It is not the case that statistics yields the concept of kinetic energy, rather mechanics seems to be contained in the statistical approach used, albeit in an implicit manner.
