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Tidal Forces Do Not 'Bend' Spacetime

Updated: May 21, 2022

Ever since the 0i0j0 component of the Riemann tensor, where 0=time and i,j are space coordinates, gave the tidal force components, varying inversely as the cube of the distance between the bodies, it was believed that the Riemann tensor could account for tidal forces. Since the Newtonian potential appears in the time component, and the Riemann tensor attests to the non-commutability of the second derivatives of the metric coefficients, the tidal forces naturally appear. Yet the bending of spacetime exists---even in the absence of matter--- as the Schwarzschild metric clearly shows. So how can tidal forces be the culprit of twhy the geometry of spacetime is non-Euclidean?

It clearly can't. If the moon wasn't there, there would be no tides. (The tide generating force of the sun is reduced by about 59 million times that compared to the moon.) So once the external perturbation is removed, the system relaxes to its unperturbed state. We say that the perturbation is exogenic, in comparison with the endogenic nature of the putative warping of spacetime.

Tidal forces have radial and tractive, or normal and tangential, components. The radial component stretches or compresses radially, which the tractive component acts laterally. For instance, at 54.7 degrees from the earth-moon line, the vector difference in the forces happens to be parallel to the earth's surface. At this point the tidal forces act tangentially, with no radial component that would increase or decrease the radial component of the stress. The radius of the earth coincides with the radius of the unstressed earth.

In spherical coordinates, the condition of equilibrium is

{(l+2m)u' + 2 l u}' +2(l+2 m)(u/r)' =0, (A)

where l is the first Lame coefficient, m the shear modulus and the prime denotes differentiation with respect to r. The stress is determined uniquely by the strain, and the strain is the gradient of the generalized displacement, u. The solution to (A) is

u = A r + B/r^2, (B)

where A and B are constants. This shows that the generalized displacement, u, is the sum of harmonic oscillator and inverse square forces. In electromagnetism, u would be the electric field. An evential magnetic field component would be given by the non-gradient part of the tensor, or the Riemann component in linearized theory the neglects the bilinear sum of Christoffel coefficients that make up Einstein's pseudo tensor. And if Einstein's tensor is second-order how can it contribute to gravitational energy?

Since the strain is the symmetric gradient of the displacement. there is no tine derivative of a vector potential to the field that would destroy its gradient characeter. The existence of any such component would destroy the very nature of tidal forces!

Since tidal forces can be related to the axes of an ellipsoid, cylindrical coordinates are more appropriate. If the z-axis being the axis of symmetry, the equilibrium condition is

(l+2m)u" +2(l+2m) (u/r)' =0,

which is independent of (l+2m). Obviously, the inverse-square law is a solution.

If we represent the tidal stress as the difference betwen radial and tangential stresses,

(2/r) (S_rr-S_theta theta)= 4 m (u/r)', (C)

it is independent of the dilatation, l. With (C) as the body force, the equilibrium equation becomes

(l+2m)u" +2(l+3m)(u/r)' =0. (D)

This has a first integral,

{(l+2m)/2(1+3m)} u'+ (u/r) = const. (E)

In the case l=0, giving pure traction, the solution is

u = A r + C/r^3. (F)

where C is a constant. Unlike the inverse square law in (B), (F) now has an inverse-cubic law implying the presence of tidal forces.

In the general case,

ln u = -2(l+3m)/(1+2m) ln r. (G)

In the limit m=0, (G) becomes an inverse square law, while for l=0, it become an inverse cubic law.

The vanishing of the traction on the shell requires S_rr to vanish. This is independent of whether tidal forces (C) are present or not. The condition that u=Ar is that the Young modulus, E=n(3 l+2m)/(l+m), vanishes. Solution (B) has no condition on the inverse square term to vanish since m>0, while (F) would require l>-6 m, which would contradict the strong ellipticity condion l+2m>0.

Therefore, everything evolves in Euclidean space. The vanishing of the tidal stress (C) gives back the unperturbed condition. The strain is the gradient of the displacement, and is not related to the Riemann tensor. Rather, the latter is related to the incompatibility condition: Curl(Curl e) does not vanish. This corrects the presentation of the gravitational equations of Costa and Heidero in terms of the Riemann tensor and its Hodge dual. Furthermore, there is nothing in gravity to make it evolve in time: i.e., no time derivative of the vector potential to destroy the tidal forces. Yet standing waves can occur, and these are equivalent to the orbital equations.

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