Relativistic Gravitational Atoms

Don Herbison-Evans ,   donherbisonevans@yahoo.com
revised 8 June 2017, but beware: this work is still in progress

SUMMARY

Schrödinger's equation has beeen used to investigate theoretically the bound states in a system consisting of two object matter waves held together by their mutual gravitational attraction, so constituting a Gravitational Atom. As in much of physics, the component objects are anticipated to be complex systems in themselves, but which are small enough and separated by large enough distances for their mutual actions to be described well by their gravitational attraction, and for which other natural forces are only a minor consideration. Examples are

  1. two Xenon atoms orbiting each other m ~ 10-25 Kg
  2. two grains of sand orbiting each other, m ~ 10-6
  3. two meteorites orbiting each other, m ~ 10+2
  4. two asteroids orbiting each other, m ~ 10+14
  5. a planet orbiting a star, m ~ 10+27
  6. a star orbiting the central galactic black hole, m ~ 10+31
  7. a pair of galaxies circling each other, m ~ 10+40
  8. a pair of galactic clusters circling each other, m ~ 10+50.
In seeking the stationary stable states of such systems: we include the normal gravitational attraction, but also include an approximation to the gravitational effects of the relativistic increase in mass due to the velocities of the orbiting objects.

In solving the equations, we find a series of solutions. These are of states corresponding to stationary matter waves that one might infer are stable, and would emit no gravitational radiation except in transitions to other states. Gravitational Atoms in these states may exist and be observable.

INTRODUCTION

In the study reported here, we consider theoretically the bound states of a gravitational atom consisting of the matter waves of two objects, one light and one heavy, moving only under the influence of their mutual gravitation. The basic qualitative idea is that the gravitational mass of the rapidly orbiting lighter object is increased as suggested in Special Relativity by its orbital kinetic energy.

We shall use the following variables and constants in MKS units [Allen, 1964] :

r = separation of the two objects, metres
M = rest mass of the heavier object, kg
m = rest mass of the lighter object, kg
u = travelling mass of the lighter object, kg
T = kinetic energy of the lighter object, Joules,
V = potential energy, J,
c = velocity of light ≈ 3.0 x 108 m/s
G = universal constant of gravitation ≈ 6.7 x 10-11 m3kg-1s-2
h = reduced Planck's constant ≈ 1.1 x 10-34 J.s

MATHEMATICAL DETAILS

Quantitatively we may set up the Schrödinger equation to find the characteristic energies of the stationary states of this Gravitational Atom. In this simple exploratory study, we take the approximations that the objects are point masses, that the lighter object moves around the stationary heavier object which is centred on the coordinate origin, and use the simple model of Special Relativity [eg. Schwartz, 2007] so that the mass of the lighter object can be written:

u ≈ m + T/c2 = m.[ 1 + T/(mc2) ]      (1) then we may write : V ≈ - GMu/r = - (GMm/r).[ 1 + T/(mc2) ]      (2) Then with E = total energy of the atom, we have E = T + V      (3) or E = T - (GMm/r).[ 1 + T/(mc2) ] = T.[1 - GM/(rc2)] - (GMm/r)      (4) or T.[1 - GM/(rc2)] - (GMm/r) - E = 0      (5)

Following the usual development to find the stationary states (e.g [Houston, 1959]) we transform this using de Broglie's relationship:

T → -(h2/2m).Δ      (6) where Δ is the Laplacian operator Thus we obtain: { (h2/2m).[1 - GM/(rc2)]Δ + GMm/r + E }.ψ = 0      (7) where ψ = wave function of the smaller object

Thus

{ Δ + 2m.(GMm + E.r)/[ h2.(r- GM/c2) ] }.ψ = 0      (8) We can rewrite this as { Δ + [ B + C.r ]/[ r-A ] }.ψ= 0      (9) with A = GM/c2 (half the Schwartzchild Radius of the heavier object)      (10)
B = 2GMm2 / h2      (11)
C = 2mE/h2      (12)
Writing: [ B + C.r ]/[ r-A ] = F/[ r-A ] + C.[ r-A ]/[ r-A ]      (13) we have F = AC + B      (14) Changing the dependent variable to s: s = r - A      (15) we obtain the usual form of the hydrogen atom energy equation: { Δ + F/s + C }.ψ = 0      (16) Writing ψ in spherical coordinates as the product of a radial function R and an angular function W: ψ = R(s).W(θ,φ)      (17) Assuming an angular momentum k about the point s = 0 (i.e. r = A), we can remove the angular dependence to obtain: (d2R/ds2) + (2/s).(dR/ds) + [ C + F/s - k(k+1)/s2 ].R = 0      (18)
k = 1,2,3,...
We seek solutions for R that decay exponentially to zero as s → ∞, so let: R = Q(s).e-s.v      (19) where: v2 = -C = -2mE/h2      (20) Then: (d2Q/ds2)- 2(v - 1.s).(dQ/ds) + [ (F-2v)/s - k(k+1)/s2 ].Q = 0      (21) This equation is known to have two solutions, which at the origin behave as Q ≈ sk or s-(k+1)      (22)

The normalisation condition requires a finite value of

  ∞   π   2π
  ∫     ∫     ∫ ψ.ψ*.r2.sin(θ).dφ.dθ.dr     (23)
  0    0    0
or a finite value of   ∞   π   2π
  ∫     ∫     ∫ ψ.ψ*.(s+A)2.sin(θ).dφ.dθ.ds      (24)
-A    0    0
which requires a finite value of   ∞
  ∫ Q2.e(-2s.v).(s+A)2.ds      (25)
-A
In order to examine the solutions near the origin of s: let Q(s) = sk.P(s)      (26) where P(s) is a polynomial in s.
The solutions of the form sk with k positive are those of conventionally thought to correspond to the physical situation, but the solutions with k negative are usually rejected because the singularity at r = 0 impedes normalisation of ψ. In the current case, the singularity is not at the lower limit of the integral, and these solutions may have some physical meaning, even though the integral encompasses the singularity.

Changing the variable by the substitution

y = 2s.v      (27) and assuming v ≠ 0 and C ≠ 0, we obtain a form of Kummer's Equation [Abramowits & Stegun, 1972]: y.d2P/dy2 - (k+y).dP/dy + {(2k-F/v)/4}.P = 0      (28) Writing j = (2k-F/v)/4,      (29) we get y.d2P/dy2 - (k+y).dP/dy + j.P = 0      (30) which has as a solution the Confluent Hypergeometric Function: Pj,k(y) = H[ j, k, y ]      (31) which reduces to a finite length Laguerre polynomial if j is a positive integer. So let j = (2k-F/v)/4,    where    j = 0,1,2,..., k = 1,2,...,      (32) Then for example, for the first 3 values of j have P0,k(y) = 1      (33)
P1,k(y) = 1 - y      (34)
P2,k(y) = 1 - y + y2/2      (35)
Rearranging equation (32) we obtain F/v = 2k - 4j      (36) or ( AC + B )/v = 2n,    where    n = k-2j = ...,-2,-1,0,1,2,...      (37) or substituting C from equation (20): A.v2 + 2.n.v - B = 0      (38) We shall refer to n as the state number, and j as the level number.

GROUND STATES

From equation (37) a set of Ground States exist when n = 0 (i.e. when k = 2j ):

AC + B = 0      (39) or ( G.M/c2 )( 2m.E0/h2 ) + 2GMm2/h2 = 0      (40) or (aha) : E0 = -mc2      (41) So in these ground states: the smaller object has its mass energy cancelled by the energy of its state, and it will appear to have vanished.

GENERAL STATES

Using equation (38) the general states are found from the solutions of this quadratic equation for v, namely

v = (1/A) . [ -n ± ( n2 + A.B )1/2 ] = - (n/A) . [ 1 ± ( 1 + A.B/n2 )1/2 ]      (42) The value of AB from equations (10) and (11) is A.B = [G.M / c2] . [2G.M.m2 / h2]   = 2.( G.M.m/hc )2 ≈ 1031.(M.m)2      (43) In equation (42) if A.B/n2 << 1, the square root may be expanded as a Taylor Series in A.B/n2, and the higher powers of this quantity truncated. Then 1031.(M.m)2 << n2,      (44) and M.m << 10-16.n      (45) We shall call this the case of light objects.

If instead, we rewrite equation (42) as

v = (1/A) . [ - n ± {A.B.[1 + n2/(A.B)]}1/2 ]      (46) and if n2/(AB) << 1, then the square root may be expanded as a Taylor Series in n2/(AB), and the higher powers of this quantity truncated. This will be the case if n2.10-31/(M.m)2 << 1,      (47) or M.m >> 10-15.n      (48) We shall call this the case of heavy objects.

GENERAL STATES FOR LIGHT OBJECTS

From equation (43) with A.B/n2 << 1, the general states for light objects can be found from

v ≈ -(n/A){ 1 ± [ 1 + A.B/(2.n2) ] }      (49) so that v+ ≈ -(2n/A)      (50)
v- ≈ B/(2n)      (51)
From equation (20) the corresponding energies of these states are given by E = -(h.v)2/(2m)      (52) giving E+n ≈ - [ 2/m ] . ( h.n/A )2      (53)
E-n ≈ - [ 1/(8m) ] . ( h.B/n )2      (54)
Expanding the quantities A and B from equations (10) and (11) gives E+n ≈ - (2/m) . [ (h.n.c2)/(GM) ]2 ≈ - [ (2.h2.c4)/G2) . n2 / (m.M2) ] ≈ - 10-7n2/(m.M2)      (55)
E-n ≈ - [ 1/(8m) ] . (h/n)2 . (2.G.M.m2/h2)2 ≈ - [G2/(2.h2)] . (m3.M2) / n2 ≈ - 10+47(m3.M2)/n2      (56)

GENERAL STATES FOR HEAVY OBJECTS

If, in equation (42) we have n2/(AB) << 1, and the square root be expanded as a Taylor Series in n2/(AB), and the higher powers of this quantity truncated, then for small values of n :

v ≈ (1/A) . { - n ± (A.B)1/2 . [ 1 - n2/(2AB) ] }
    = ( (A.B)1/2/A ) . { n/(A.B)1/2 ± (1 - n2/(2AB) ]
    ≈ ±[ (A.B)1/2 / A ]
    = ± (B/A)1/2      (57)
From equations (10), (11), and (20) the corresponding energies of these states are given by E±n ≈ ± [ h2/(2.m) ] . (B/A)
     = [ ± h2/(2.m) ] . [ 2.G.M.m2/h2 ] . [ c2/(G.M) ]
     = ± m.c2      (58)

SIZE of LEVEL 1, STATE -1

For the next level above the ground state :

j = 1, k = 1, n = -1      (59) Then P1,1(y) is the Laguerre polynomial: P1,1(y) = L1 = 1 - y      (60) so Q1,1(s) = s2.(1-2v.s)      (61)
R1,1(s) = s2.(1-2v.s).e-v.s      (62)
and v = (-C)1/2 = (-2.m.E)1/2/h      (63) This radial function R1,1(s) of ψ has two zeroes, at s = 0, and s = 1/(2v), after which it decays exponentially to zero at s = ∞ . So the peak value of R will be somewhere around halfway between the two zeroes, at approximately s1,1 ≈ 1/(4v) = (h/4)/(-2.m.E)1/2 ≈ h/(-5.m.E)1/2      (64)
using 4x21/2 ≈ 5
so that by equation (15) r1,1 = s1,1 + A
≈ h/(-5.m.E)1/2 + G.M/c2      (65)

Noting that G.M/c2 is half the Schwartzchild Radius of the heavier object, we need

h/(-5.m.E)1/2 >> 2.G.M/c2      (66) or m1/2.M << h.c/[10.G.(-E)1/2]
or, assuming that only E is negative :
mM2 << 10-35/(-E)1/2.      (67)
Noting that this radius is imaginary for E > 0, we may now dismiss the solutions with positive energy in equation (58).

THE EXAMPLES

  • two Xenon atoms orbiting each other : m,M ≈ 10-25 , so that (m.M) ≈ 10-50, and the Light Objects approximations apply.
    Thus using equation (55) :
    E+1 ≈ - 10-7/(m.M2) ≈ - 10-7/(10-75) ≈ - 10+68 Joules
    E+n ≈ - 10+68n2 J

    Using equation (56) :
    E-1 ≈ - 10+47(m3.M2)/n2 ≈ - 10+47.10-125 ≈ - 10-72 Joules E-n ≈ - 10-72/n2 J


  • two grains of sand orbiting each other : say m ≈ 10-6 , M ≈ 10-6
    so that (m.M) ≈ 10-12, and the Heavy Objects approximations apply.
    Thus using expression (48) for this and the heavier examples :
  • two meteorites orbiting each other,
  • two asteroids orbiting each other,
  • a planet orbiting a star,
  • a star orbiting the central galactic black hole,
  • a pair of galaxies circling each other,
  • a pair of galactic clusters circling each other : M.m >> 10-15 so the Heavy Masses approximations apply, and the states all have the same approximate energy formula :
    E = -Gm/c2
    in which the lighter object disappears.

    ACKNOWLEDGEMENTS

    Many thanks are due to many friends and colleagues, particularly Michael Partridge, who examined early drafts of this work and explained to me some of errors therein.

    REFERENCES

    Abramowitz, M., and Stegun, I.A. (eds.), 1972,
    "Handbook of Mathematical Functions", Dover, New York, 9th Printing, p. 504.

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

    Behr, B.B., 2007.
    "Universe", in Volume 19, "Encyclopedia of Science and Technology", McGraw-Hill, New York, 10th Edition, pp. 80-89.

    De Volp, A., 2007.
    "Hydrogen Bomb", in Volume 8, "Encyclopedia of Science and Technology", McGraw-Hill, New York, 10th Edition, pp. 712-713.

    Fry, E.S., 2001.
    "Neutron", in Volume 14, "World Book", World Book, Chicago, 2001 Edition, p. 155.

    Houston, W.V, 1959,
    "Principles of Quantum Mechanics", Dover, New York, pp. 58-82.

    Schwartz, H.M., 2007,
    "Relativistic Mechanics", in Volume 15, "Encyclopedia of Science and Technology", McGraw-Hill, New York, 10th Edition, pp. 330-332.

    APPENDIX 1

    Equations defining Intermediate Variables

    A equation (10)
    B equation (11)
    C equation (12)
    F equation (14)
    H equation (31)
    j equation (29)
    k equation (18)
    n equation (32)
    P equation (26)
    R equation (17)
    Q equation (19)
    s equation (15)
    v equation (20)
    W equation (17)
    y equation (27)
    ψ equation (7)
    ~~~~~~~~~~~~~~~~~~~~~~

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