Search

Could Kepler have discovered Quantum Mechanics?

Updated: Jun 29

In his derivation of the nonrelativistic wave function for the hydrogen atom, Schrodinger refers to it as the "Kepler" problem because the Coulomb potential is analogous to the Newtonian potential.


But the Kepler problem treats elliptical orbits, while the Bohr atom the orbit is circular. Even not relying on the Bohr model, Schrodinger's radial wave function is spherically symmetric. It has been proposed that the fractional time probability distribution is the bridge linking classica and quantum theory in the form of the old quantization conditions [Curtis & Ellis, Eur J Phys 27 (2008) 485]. But how can, what is essentially, the beta probability distribution have a quantum mechanical overtone?


The answer is that it doesn't. But, the derivation of the beta distribution through the process of subordination does [Thermodynamics of Extremes, Sec II 4.2.] In fact. it is fundamentally related to the uncertainty relation. Recall that a Gaussian wave packet in space is a Gaussian wave packet in momentum space. The dispersion of the two distributions swap places under a Fourier transform.


The hydrogen ground state probability density has to be multiplied by an element of area of the surface. This transforms the exponential distribution into a Gamma density which has a maximum at the Bohr radius. Feymnan in his lectures doesn't due this an treats the exponential decay radial distributions by claiming that the most probable place to find an electron is at the center where the nucleus is located. But, again, this assumes a spherically symmetric distribution which is incompatible with the classical Kepler orbit, which has a finite eccentricity. Moreover, the beta distribution is an extreme value distribution making it more probable to find the system at the limits rather than at the center. So one can rightly question is there any connection between the classical Kepler problem and the quantum mechanics of a single electron orbiting about its proton? Quantum mechanically, we would say that the more problem place to be found is in the nucleus corresponding to pendulum orbits. However, this would have provided no connection with the old quantum theory and its quantization prescriptions.


Just as the wave packets for increments in distance and wavenumber are Fourier transforms of one another, the gamma density for the radial coordinate

p(r|k')= k' (k'r)^(m-1)/G(m) exp(-k'r) (1)

which is parameterized by the wavenumber, k_o, in units of Planck's constant, and G(m) is the Gamma function. Randomizing the wave number, and setting k'r=kr', emphasizing the inverse relationship between k and r, we obtain the Bayes't distribution for k, i.e.,

p(k|r')= r'(kr')^(m-1)/G(m) exp(-r'k), (2)

where the radial coordinate, r', is now the parameter. This probability density is to be looked upon as one in likelihood of a given value of k, rather than a frequency one in terms of a value of r.


Imposing that the two values of the wavenumber coincide, k=k', and integrating over all values of k result in

p(r|r')=r^(m-1)r'^m /B(m,m)(r+r')^(2m),

where B is the beta function, B(m,m)= G^2(m)/G(2m). Now, in terms of the fraction,

t=r/(r+r')=A/r-1, and (1-t)=r'/(r+r')=1-P/r,

where A and P are the apehelion and perihelion of the orbits, i.e, the apsides, In rwema of the fraction t, the Beta distribution is the arcsine law

p(t)=[t(1-t)]^m/B(m,m). (3)

We want this to be an extreme value distribution so m<1; in fact in this case m=1/2. For as m increases from 1 to 3 the proability density transforms from a U to an inverted U probability density. The index, m, is half-degrees of freedom.


Now, in terms of the Kepler problem the velocity is

v= 2 pi a/Tr^2 n(r),

where a is the semi-major axis, T, the period, proportional to a^(3/2), through Kepler's III law, and n(r) is the refractive index of the Eaton lens,

n^2(r)=2/r-1/a.

This is due to the Newtonian gravitational potential. The radial velocity is

dr/dt=2 pi/T(a/r){(A-r)(r-P)}^1/2.

If we identify the time-fraction dt/T with the probability density then

p(r)dr=dt/T=1(2 pi a) dr/{(A/r-1)(1-P/r)}^(1/2). (4)

This is the same as the arcsine law (3).where t=A/r-1=P/r.


Thus, the index in the Gamma density is m=1/2. In quantum mechanics, m=1, and the Gamma density (1) is multiplied by r^2 to get the unimodal density whose most probable value is 1/k', the Bohr radius. But a circular Bohr orbit is not allowed since it would imply A=P, and the probability distribution would collapse. The Gamma density is only allowed for elliptical orbits.


By introducing the equation of the Kepler orbit,

(L/r)=1+e cos phi,

where L is the semi-latus rectum and e is the eccentricity, (4) is converted into

d phi/2pi, and the averages of the inverse radii are given by

<1/r^n>=(1/2pi) int_0^(2pi)(1+ecos phi)^n d phi=(b/a)^n P_n(a/b),

where b is the semi-minor axis and P_n are the Legendre polynomials. So clearly everything depends on the ellipticity of the orbit.





2 views0 comments

Recent Posts

See All

From the generalized Helmholtz's representation theorem for tensors, E + curl S = (grad g + g grad) + curl curl g, (1) where E and S are symmetric 2-tensors, g, will