Schwarzschild Solutions in Respect to the Position of The Sun in a Keplerian Orbit
Updated: Aug 20, 2022
Schwarzschild derived two solutions to his static metric that involve gravitational acceleration and centrifugal acceleration. For the "outer" solution he found a metric of non-constant curvature, while for the "interior" solution he found one with constant, negative curvature. More significantly, the outer solution was found on Einstein's condition of "emptiness"--the vanishing of the eigenvalues of the Ricci tensor---while the interior solution was found with the aid of the Einstein field equations.
Since the exterior contained an finite, central mass, one could question the validity of Einstein's condition of emptiness, while the interior contained a constant mass density which some how transforms the system into a non-empty system for the use of Einstein's equations.
The fact that a metrizable theory is obtained in both cases imply that the Ricci tensor acts as a compatibility criterion, i.e., that the directional eigenvalues vanish in both cases. If we introduce a spatial metric of the form
ds^2 = dR^2 + G^2(R) dS^2, (1)
where r=G(R) is the radial coordinate and dS^2 is an element of a 2-sphere, energy conservation demands
G'^2 = 1-2m/G-w^2G^2, (2)
where m is the gravitational parameter and w is the angular velocity. Expression is what Moller tells us that Schwarzschild found for the exterior solution, but that the last term can be neglected except for large distances, which we are not interested in. In his words, if we neglect this term, (1) "will in the limit of large r go over into the usual line element for a Euclidean space in polar coordinates, and it is interesting that the condition of spherical symmetry is sufficient to secure this result without any explicit use of boundary conditions at infinity."
This is indeed surprising insofar as the metric for the interior solution is (2) with m=0, and the last term is the rotational energy which causes no problems at the origin like the gravitational term does.
In this context, it is interesting to recall that Newton found two power laws of the force share an elegance unlike all others: Onlly for these laws are the orbits conic sections. According to Newton, the inverse square law for the force directed to the sun placed at one of the two focii of the elliptical orbit is a "consequence" of a linear force directed to the centre. It would therefore be an excusable offence of the arts depiction of Newton's law of attraction with a sun placed at the centre of the ellipse in the one pound note.
Rather than a consequence, we argue that the centrifugal force is active at the centre, while the inverse-square law is active at a foci. We maintain that if the Sun were able to traverse these two limits there would be a combination of the two central forces as given in (2).
The simplest case of constant, negative curvature, m=0, in which the radial and tangential sectional curvatures sum to zero,
G"/G + (1-G'^2)/G^2= -w^2 +w^2=0.
Since only two sectional are involved, it is a 2D dimensional system. That is what we would expect for planar motion of a uniformly rotating disc. The Sun is now at the center of the circle.
In terms of the inner Schwarzschild metric, the components of the Ricci tensor do not vanish and Einstein equations are called to intervene. This is surprising for the metric of a uniformly rotating disc of constant curvature.
At the other extreme, w=0, and the Sun is now at the focus. In this case, the difference between twice the radial and tangential sectional curvatures vanish, i.e.,
2G"/G -(1-G'^2)/G^2 = 2m/G^3-2m/G^3+0.
Since two sectional curvatures are involved, we must make reference to an prolate ellipsoid with principal axes, a=2m/G^3, and b=c=m/G^3. This corresponds to Schwarzschild's interpretation of the Ricci tensor as conditions of compatibility.
In the general case of where both terms are present, the Sun is found somewhere between the centre and the foci of the ellipse. Where a single warped metric, G, was sufficient, we now have to introduce a doubly warped metric by setting F=G'. In the conventional intepretation F^2 appears as the coefficient of the dt^2 in the metric. Then taking intersecting planes to form the sectional curvatures, it seems unphysical that a time axis should intersect a spatial axis for form a plane whose intersections would yield sectional curvatures. In other words, "time" does not curve.
The second warped product is needed to get the compatibility conditions
(1-G"2)/G^2 +F"/F =0 (3)
G"/G + F'G'/FG = G"/G + F'/G =0. (4)
In free-fall, each term in (4) vanishes individually.
The radial sectional curvature,
can be expressed in terms of the eccentricity, e^2=w^2R^3/m, whose condition <1, has been linked to a condition of stability that material will not be torn off the planet unless the condition is invalidated.
The vanishing of the Ricci tensors supports the description in terms of stress and strain. The strain is the symmetric gradient of the gravitational acceleration,
E= grad^S g, (5)
where g is the gravitational acceleration. The strain appears as due to tidal forces. The fact that grad div g=0 becomes
div g = (1/r^2) (r^2g)' = 3 w^2= constant, (6)
upon integration where we introduced the divergence in spherical coordinates. Equation (6) gives
g =w^2 r+m/r^2,
where m is constant. Condition (6) is that for constant dilatation.
The principle of equivalence attests to the claim that all forms of accelerations are equivalent. This may be true of radial accelerations but certainly centrifugal acceleration and gravitational acceleration may be distinguished--merely by moving the Sun between the centre and a focus of the ellipse.
As a final comment, it is often stated that the symmetric 2-tensor can be as the spatial components of the 4D Weyl tensor,
E_(ij) =- C_(0i0j), (7)
where "0" is the time coordinate. The Weyl tensor has the same symmetries as the Riemann tensor. In this case, i and j and be switched without change the value of (7) Having uttered in one breath that (7) is symmetric, in the next breath that its curl,
E_[ij]= e_(ikm) E_(mj,k) =-B_(ij,0) (8)
is a contradiction like no other [cf. W B Campbell, T A Morgan, "Maxwell form of the linear theory of gravitation']. If E is defined by (7), (8) is identically zero and there is no Faraday effect in linearized gravitational theory, according to general relativity.
The same can be said about the magnetic-type of field in linear gravitational field. Defining it as the Hodge dual of E,
B_(ij)=(1/2) e_(imn) C_(mn0j), (9)
where e is, again, the Levi-Civita symbol. Since B_(ij)=B_(ji), it should have been evident to these authors and the slew that came before and after that
B_[ij] = e_(ikl) B_(jl,k)= E_(ij,0) (10)
vanishes identifically, i.e., there is no Maxwell displacement current. Since (8) and (10) are both zero, there can be no gravitational waves---at least in linear gravitational relativity.