Tidal Forces do not cause Spacetime Curvature
- bhlavenda
- Apr 20, 2022
- 5 min read
Updated: Apr 22, 2022
General Relativity affirms that the tidal forces are the causes of the curvature of spacetime. However, tidal forces are exogenic in origin; they are cause by external forces and once the forces are removed the system returns to its normal state. Tidal forces are described by Hessians which imply the existence of a scalar potential. Rather, curvature is endogenic, caused by intrinsic deformations. More generally we will show that the compatibility conditions for Beltrami flows are stationary points of Ricci flows!
In the decomposition of a general symmetric tensor field into irrotational and solenoidal components, the presence of tidal forces alone are described by the irrotational component and the vanishing of the solenoidal component is given by the compatibility condition; in Saint Venant's case, it is the vanishing of the Riemann tensor. Thus, the presence of tidal forces has nothing to do with the curvature of the manifold.
In the case of tidal forces, the vanishing of the Ricci tensor is the compatibility condition! This means that the manifold in question is Euclidean, and the conditions under which it applies are called the equations of compatibility. One compares two states, one before the deformation has been applied and the other after it has. The difference in metric coefficients is twice the strain tensor, and the Riemann tensor, based on either of the metrics, undeformed or deformed, must vanish.
The Riemann tensor is fully determined by the Ricci tensor. In 3D one may dualize the pairs a,b and c, d in the Riemann tensor, R_(abcd) into pairs e, f through the antisymmetric Levi-Civita tensor e_(eab) and e_(fcd) so that the new tensor, R_(ef) is symmetric in the consolidated indices. It has 6 components, just like the original Riemann tensor.
The vanishing of the Riemann tensor, or the consolidated Ricci tensor, is the integrability condition for the existence of a symmetric rank 2 tensor, F_(ij), in nD where n greater or equal to 2. The integrability condition
R_(ijkl)=F_(ij,kl)+F_(kl,ij)-F_(il),jk -F_(jk),il=0
which is precisely the vanishing of the Riemann tensor. That the Riemann tensor can be expressed in terms of the incompatibility condition
R_(ijkl)=e_(fij)e_(gkl) [(Curl(Curl g)]_(fg) =0,
means that in a simply connect domain the stress is the symmetric derivative of some vector field.
In gravity, it is the existence of the tidal potential It always seemed strange that the R^i_(0k0) components of the Riemann tensor give the tidal forces, as expressed in a 3X3 matrix instead of a 4X4 one, which is dictated by the indefinite metric. The radial tidal forces are given by the diagonal metric with entries, V_(rr), (1/r)V_r, (1/r)V_r as components along the diagonal where V is the gravitational potential. The metric coefficients are g_(11)=2GM/r^3, g_(22)=g_(33)=-GM/r^3. The gravitational metric coefficients form a Hessian matrix of the scalar potential, V=-GM/r; the positive sign represents a tensile force, while the negative sign a compressive force. This is why the association of an oblate ellipsoid makes the components become alive.
Because the radial tidal force matrix is diagonal, the compatibility condition is just that the sum of all diagonal terms vanish. In fact, we find (1=r, 2=theta, and 3=phi)
{11,1}=3/2r, {22,1}={33,1}=-3/4r, {12,2}={13,3}=-3/2r,
and all other connection coefficients are zero. It is then an easy matter to show that
R= R_(11)+ R_(22) + R_(33) =0
implying the scalar curvature, R, vanishes. The manifold is Euclidean, not by Einstein's condition of emptiness which would require all the eigenvalues of the Ricci tensor to vanish, [Petersen, Riemann Geometry, p. 69 and p. 72 for the Schwarzschild solution treated as a doubly warped product] but by the very fact that there exists a scalar potential which is ensured by the integrability condition (A), or as it commonly referred to as the equation of compatibility.
This says that tidal forces are exogenic. Endogenic forces must be sought after in the interior of the earth, say, by dislocations and disclinations. causing curvature and torsion, respectively. The former is accounted for by Riemann geometry, the latter not. Rather it is accounted for by Riemann-Cartan theory. Tidal forces do not produce spacetime curvature.
In fact, tidal forces only operate in Euclidean space. The Beltrami decomposition (1886) states that any symmetric tensor field can be split into a divergence-free part and an incompatibility-free part.
In this optic, Einstein's equations take on a whole new meaning. We can write them as
e_(ik)e_(jl)G,_(ij)= T_(kl),
where the comma denotes the ordinary derivative. The stress field, T_(kl) must satisfy the condition of equilibrium, div T=T_(kl),l=0. But if A=e_(ij), the symmetric strain tensor, then
e_(ik)e_(jl)e_(ij,kl)=0,
and the stress field vanishes. The stresses T_(kl) are endogenic, arising from dislocations or defects in the material, while e_{ij) represent plastic deformations, which, upon their removal, the body returns to normal. It has nothing whatsoever to do with stresses applied from the outside.
In comparison with electromagnetism, the irrotational field would correspond to the electric field, and the solenoidal one with the magnetic field. However, the presence of a time varying vector potential in the electric field destroys the fact that it can be derived from a scalar potential, the Coulomb potential. The total field would be
T_(ij) = E_(ij) +(Curl(Curl A))_(ij) (B)
where A is analogous to the vector potential. If it is divergence-free and is a Beltrami flow,
Curl A = k A, (C)
where k is a constant, then the second term in (B) becomes the Helmholtz equation
grad^2 A +k^2 A=0,
since div A=0. So in the static case, where E_(ij) is potential, the nonvanishing of the incompatibility condition gives rise to stationary waves if it is a Beltrami flow (C). In fact, in the last part of Beltrami's 1889 paper, he remarked the applicability of (C) to electromagnetism. He was probably thinking about Maxwell's reference to vortex atoms. But, even more important is Fresnel's earlier relation to circularly polarized waves. Plane waves with circular vibration ellipses satisfy Curl A=k A and div A =0, at a give circular frequency and wave number, k. So what Fresnel envisioned was a time-periodic Beltrami flow {A. Lakhtakia, "V Trkal, Beltrami fields, and Trkalian flows"]
It may seem coincidental but the condition that the metric be Einsteinian is
R_(ij)=c g_(ij), (D)
where c is uniform on the whole manifold. k is referred to as the "cosmological" constant. Rather than referring to cosmology one should refer to a Beltrami flow. Since
R_(ij)(g) = (Curl(Curl g))_(ij),
in the linear approximation, condition (D) is none other than that the flow be Beltrami (C), i.e.,
Curl( k g)= k^2 g,
where k^2=c is the putative cosmological constant in (D). From (D) it follows that tidal forces have zero scalar curvature, being the sum of the diagonal terms g_(ii).
Again we refer to electromagnetism. The magnetic component of the the Lorentz force is
v x B. Maxwell's equation without the displacement current is Curl B = v, and if it is a Beltrami flow then v is proportional to B, and the magnetic component of the Lorentz force vanishes, i. e. B x B=0. So Einstein metrics are zero force laws in the electromagnetic analogy.
It is also not surprising that the Einstein metrics are the stationary points for Ricci flows. As we have shown that Beltrami flows are characterized by a periodic stationary Helmholtz equation. Another interesting property is that for dimensions up to three, the Einstein metrics are ones of constant curvature. The zero value corresponds the compatibility condition for a gravitational potential is that the Ricci tensor vanishes . This does not mean at all that the tidal forces are zero. Thus, tidal forces do not curve space--or spacetime, and live in Euclidean space! But, more generally, the constant curvature Einstein metrics refer to Beltrami flows of the form (C).
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