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A New Take on Gravitational Interactions

Updated: Jun 18, 2021

Ever since Einstein wrote down his field equations, the vanishing of the eigenvalues of the Ricci tensor has been the condition that the universe be empty. Balancing the Einstein tensor, which has a term appended onto the Ricci tensor so that its divergence vanishes, is the energy-stress tensor. This tensor is problematic since it contains all forms of energy except gravitational energy. The gravitational interactions supposedly required a pseudo-tensor, which is a bilinear product of Christoffel symbols. It does have the attribute that it can be made to vanish at a mere waiving of the magic wand through a coordinate transformation. Supposedly, this is related to Einstein's equivalence principle whereby a mere change in the frame of reference, gravity can be made to disappear or appear at will.


But, what if the vanishing of the eigenvalues of the Ricci tensor had a completely different meaning---rather, that of a force balance laws expressed in terms of sectional curvatures? A good example is the Brachistochrone problem that Johann Bernoulli challenged all the great physicists of Europe to solve in 1696.


Bernoulli imagined a steel bead with a hole bored through it so that it could be threaded with a metal wire. The bead is able to move along the wire freely without any friction. The problem is to choose the shape of the wire so that the time of descent of the bead under gravity is least when it begins from a state of rest.


This turned out to be a classic example in the use of the calculus of variations. The kinetic energy is balanced by the gravitational potential

(1/2)(ds/dt)^2=gy,

with the y-axis directed downwards, and g is the constant gravitational acceleration at the surface of the earth. Rearranging the above we have

dt=[(dx^2+dy^2)/2gy]^1/2 =[1+y'^2]^1/2 dx/(2gy)^1/2,

where y'=dy/dx. Calling the integrand L, a first integral exists

L-y'(dL/dy')=1/[(2g)y]^1/2 [(1+y'^2)^1/2-y'^2/(1+y'^2)^1/2]=1/C,

where C is a constant. This can be rearranged to read

y'^2=C^2/(2g y)-1.

Differentiating with respect to x yields and dividing through by y yield

y"/y= - C^2/(4g y^3).

This is the value of the radial curvature, while the tangential curvature is

[1+y'^2]/y^2=C^2/2gy^3,

which is twice the negative of the radial curvature.


Thus, the path of quickest descent is determined by the vanishing of the of the sum of twice the radial and tangential curvatures:


2y"/y + [1+y'^2]/y^2 = 0,


This is, in fact, the path followed is a cycloid.


As a second example (which will turn out to be an eye catcher!) consider the motion of a spring which is extended in the horizontal x-direction. Beginning from a state of rest, energy conservation demands

(ds/dt)^2=kx^2,

where k is the spring constant. The extremal trajectory we now want to determine is x=x(y).

Requiring the integrand of

dt=ds/k^1/2 x=(1+x'^2)^1/2 dy /(k^1/2 x)

to be an extremum, where the prime now stands for differentiation with respect to y. This requires the balance of the radial sectional curvature

x"/x = - C^2/kx^4

by the tangential section curvature

[1+x'^2]/x^2= C^2/k x^4,

or

x" /x + [1+x'^2]/x^2 = 0,


whose solution is a half-circle. All the particulars of the model have disappeared: the spring constant k, and the constant of integration C. This is, in fact, the Poincare' half-plane model of hyperbolic geometry!


Why the Poincare model is exemplificatory of the hyperbolic plane is because it has negative, constant curvature, -1. The Poincare model is likewise independent of the constants used in its derivation, namely, k and C.


The Brachistochrone, on the other hand, has Gaussian curvature,

K=-g/y.

If the gravitational acceleration were not constant, but, rather proportional to GM/y^2, the sectional curvature would again be negative but now proportional to a constant density, M/y^3, equal to the square of the inverse Newtonian free-fall time.


It is well-known that the bead falling along the wire can be made to conform to Snell's law: Light entering a different medium will arrive at its destination in the shortest possible time. The integral of the motion,

{(2gy) (1+y'^2)^(-1/2)=1/C,

is nothing other than an expression for Snell's law which relates the sine of the angle \theta to the index of refraction, where

sin\theta=dx/[dx^2+dy^2]^1/2=1/[1+y'^2]^1/2.

The speed v=(2gy)^1/2, (the first equation, v^2/2=gy) which associates (2gy)^(-1/2) with the index of refraction n so that Snell's law:

n sin\theta= constant

is obeyed. The only thing that changes in the Poincare' half-plane model is the expression for the velocity, which is now v^2=ky^2, which is the first equation of this paragraph!


It may come as a surprise, but the Schwarzschild metric is just free-fall. Energy conservation would dictate

v^2 =(d r/dt)^2=2E-2GM/r

where E is the total energy, and the prime now stands for differentiating with respect to time.

The conservation of energy should read

v^2=2E+2GM/r,

where E is the total energy per unit mass.


Unlike general relativity there is only one time variable, not two. Differentiating the first equation again gives


d^2 r/dt^2= GM/r^2,


which would be the equation for free-fall --- if the sign were reversed. This is why repulsive gravitational energy enters into the discussion of the Schwarzschild metric.


General relativity doctors this up to look like the vanishing of one of the eigenvalues of the Ricci tensor. The metric is completely static; i.e. all coefficients depend on the radial coordinate and not on time. The radial sectional curvature is

GM/r^3


which would be constant for a constant mass density as in the inner Schwarzschild problem, and a tangential curvature,


2 GM/r^3.


Whereas the latter is twice the former, the vanishing of the eigenvalue of the Ricci tensor the difference between twice the radial curvature and the tangential curvature to vanish.

It has nothing whatsoever to do with the condition that the universe be "empty."


We have only used one condition for emptiness. There are another two conditions so that it would seem we have not exhausted all the information on the exterior solution of the Schwarzschild problem. This, in fact, is not the case. The metric is usually expressed in terms of a warped product, with F^2 the coefficient of the time increment squared and G^2, the coefficient of the 2-sphere increment. It can always be arranged that the coefficient of the square of the radial increment is unity. It turns out that F=G' so all three conditions reduce to the one we have employed. [For greater detail see: A Besse, Einstein Manifolds, p. 103.]


For instance, the condition

F"/F+2G"/G=0

is really

G"'/G'+ 2G"/G =0. (*)

Multiplying through by GG' makes it exact and leads to the first integral

GG"+ (1/2)G'^2=C,

where C is an arbitrary constant of integration which we shall set equal to one. Dividing

the last equation by G^2 shows that the sectional curvatures satisfy

G"/G + (1/2)(G'^2-1)/G^2=0,

which is remarkably similar to the equation of a cycloid that we derived for the Brachistochrone. It would be an exact match if the last term were +1/G^2. This is a consequence of the negative curvature. That is to say, the tangential sectional curvature can be written as

(2E+G'^2)/G^2,

where E is the total energy. For positive sectional curvature, E<0, while for negative curvature, E>0.


Finally, the third condition,

F"/F+2 F'G'/FG=0,

is identical to the condition given above in Eq (*). So there is only one condition that is afforded by the vanishing of the Ricci tensor.


Consequently, the Schwarzschild metric,

g=F^2(r)dt^2-dr^2-G^2(r)d\theta^2,

is actually

g=G'^2 dt^2-dr^2 -G^2(r)d\theta^2

We can always consider sectional curvatures given by Gauss's law. In terms of the (r,\theta) pair, the Gaussian curvature is

K=-G"/G,

while for the pair (t,\theta) it is identically zero since G is not a function of t. This calls into question the entire significance of the metric, g.


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