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Gravitational Diphotons

Don Herbison-Evans ( donherbisonevans@yahoo.com )

SUMMARY

Using simplistic assumptions, the energy of two photons revolving around each other under the influence of each others gravitation is considered. Stationary states were found where the photons have an energy equivalent to approximately 4 x 10-8√n Kilograms, where n = 1,2,3,... However the total energy of such states is zero because the energy of photons exactly equals the loss of potential energy in their interaction. The result is a particle with spin but no mass or charge that travels at the velocity of light.

INTRODUCTION

It is possible to interpret the observed bending of light by a gravitational field as a simple scattering of one heavy body by another, and that photons can behave as if they have a gravitational mass. Then it is also tempting to interpret the observed redshift of light from distant galaxies as loss of energy by the photons as they are scattered slightly many times by interactions with the nuclei of atoms in intergalactic space. This explanation of the redshift would nullify the need for assuming that the universe is expanding after some big bang, and explain the cosmic microwave background as the cast off energy in these scatterings.

THE PHOTONS

This work describes the theoretical investigation into possible bound stationary states of two photons orbiting around each other which appear when gravitation is taken into account. The basic qualitative idea is that the energy of a photon is equivalent to a gravitational mass 'm', which can be approximated using

2.π.h.ν = m.c2      (1) where, in S.I. units ν = the frequency of the light composing the photon, Hz
h = reduced Planck's constant = 1.1 x 10-34 J.s
c = velocity of light = 3.0 x 108 m/s
[Allen, 1964]

Assuming that the light forms a stationary wave around the circumference of the orbit:

2.π.r = n.λ      (2) where r = radius of the orbit, m
λ = the wavelength of the light composing the photon, m
n = 1,2,3,... (any positive non-zero integer)

The gravitational force between the photons will equal their effective mass times their centripetal acceleration:

G.m2/(2.r)2 = m.c2/r      (3) where G = universal constant of gravitation = 6.7 x 10-11 m3kg-1s-2
[Allen, 1964]. Then as c = ν.λ, from (2) ν = n.c/(2.π.r)      (4) and from (1) ν = m.c2/(2.π.h)      (5) so that n.c/(2.π.r) = m.c2/(2.π.h)      (6) or n/r = m.c/h      (7) or r = n.h/(m.c)      (8) Simplifying (3) gives: G.m/(4.r) = c2      (9) or r = G.m/(4.c2)      (10) Taking (8) and (10) together gives: n.h/(m.c) = G.m/(4.c2)      (11) which gives m2 = 4.n.h.c/G ≈ 2.10-15

m ≈ 4.10-8.√n kg
    or 2.1028.√n ev      (12)

The radius of the orbit can then be derived r2 = (n.h/c).√[G/(4.n.h.c)]
r = √[G.h.n/(4.c3)]
≈ 8.10-35.√n m      (13)

TOTAL ENERGY

The total energy E of the these stationary states is the sum of the kinetic energy T and potential energy V:

V = -G.m2/(2.r)      (14)
T = 2.m.c2      (15) Substituting equation 10 into 14 gives V = -Gm2(4.c2)/(2.G.m)
so that
V = -2.m.c2      (16) so that E = 0      (17)

ACKNOWLEDGEMENTS

Thanks are due to my friends for their encouragement, and the staff of the 'Faculty of Engineering and Information Technology' at the 'University of Technology, Sydney' for their assistance and support.

REFERENCE

Allen, C.W., 1964, "Astrophysical Quantities", Athlone Press, London, 2nd Edition.

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(written 10 December 2010, updated 22 January 2012)