Does Energy Discreteness Follow Directly From the Maxwell-Boltzmann Probability?

  In Newtonian mechanics there exists a continuous x i.e. for a conservative force: .5mv(x)v(x) + V(x) = E with x taking any value. This is associated with the idea of dx→0. As a result one may choose an x and x+dx with dx→0 and v(x) and v(x+dx) become arbitrarily close in value. It is almost as if their identities merge.

      In (1) entropy and average energy are calculated using the Maxwell-Boltzmann MB factor  exp(-e/T) and a continuous energy. The results  E= NT  (k=1)  and S = -ln(T/To) with To being an integration constant show that one must have T>To because T→0 does not lead to S→0. Thus there is a problem if one uses continuous energy. In other words continuous energy is rejected.. (1) then considers an equally spaced energy i.e. e=bn, n=0,1,2,3 … and calculates energy and entropy again. In this case S→0 as T→0. Thus (1) concludes that it is entropy which leads to the discretization of energy

  In this note we try to argue that discretization of energy is already implied by the MB factor.

The MB factor exp(-energy/T) where T is a math parameter called temperature represents a probability. In other words, different energies have different probabilities i.e. one must distinguish between them. Having two energy values merge is a problem now because one has a second parameter T which may also go to zero i.e. e2-e1→0 and T→0 leads to trouble because exp(-e2/T) does not tend to zero as it should if e1=0.

  We argue that the underlying reason for e1 and e2 being discretely different is that one cannot use the probability/T formalism i.e. MB factor unless interactions occur. We argue that such two body interactions cannot ultimately transfer or remove a de→0 and keep e1 and e2 distinct. In other words the interaction/collision is ultimately linked to a finite energy transfer even though T→0.

   As a result energy should be discretized in an equilibrium scenario. This argument does not make use of entropy explicitly although implicitly having exp(-e2/T)-->0 as T→0 (say for a 2 level system) means that the particle is in state e1 which may be associated with an entropy of 0 if one ignores internal features of this state. We also note that the continuous energy result E=NT, and not just the entropy, leads to trouble because as T becomes extremely tiny E which is an average of say e1 and e2 with e1=0 as an example approaches T in magnitude. The smaller T becomes the smaller e2 becomes which e2 and e1 no longer have separate identities, but in such a case one does not have an equilibrium scenario either. In other words exp(-e/T) implies interactions between distinguishable states with respect to energy.