top of page

Central Force Duals Derived from the Compatibility and Equilibrium Elasticity Conditions

Updated: Oct 17, 2022

The fact that the strain is a gradient follows from the vanishing of the Riemann tensor in linearized elasticity theory where it is incompatibility, or non-vanishing, tensor Curl(Curl e)_ij . The strain is the linear part of the Saint-Venant tensor. Then the radial dependencies of the displacement field follow from vanishing of the divergence of the stress, i.e., the equilibrium condition in the absence of body forces.

The gravito-electromagnetic analogy based on tidal tensors, although flawed, gives the correct conclusion that there can be no time-dependent inductive terms in the gravitational equations analgous to Maxwell's equations [Costa & Herdeiro]. In the absence of the circuit equations there can be no waves; in other words, the absence of inductive terms outlaws the existence of gravitational waves.

The tidal tensor analogy replaces the electric and magnetic fields by their covariant derivatives. Mistakenly, the electric field tensor is taken as the Riemann tensor in its symmetric indices and the magnetic field tensor is that of its Hodge dual: what is symmetric for the electric field tensor becomes antisymmetric for the magnetic field one. The symmetry of the former implies the vanishing of Faraday induction (i.e., the time rate of change of the magnetic field), while the non-vanishing of the latter makes it proportional to the current, and the current being proportional to the magnetic field makes it a Beltrami flow. But this would make the curl of the curl of the magnetic field nonvanishing and this violates the imcompatibility condition which guarantees that the electric field tensor be symmetric. The upshot is that the Riemann tensor and its Hodge dual cannot both be used to represent the field tensors.

The generalized displacement vector, u, corresponds to the electric field, and the electric field tensor is the covariant derivative of the electric field tensor, e_kl=u_k,l. And the compatibility condition, Curl(Curl e)_kl=0 guarantess that the electric field tensor, or strain, be symmetric. This guarantess that it is a tidal tensor. For linearized elasticity, this is ensured by the vanishing of the Riemann tensor, which is thecommutivity of the derivatives of the Christoffel symbols, or connections. Yes, tidal tensors, enshired in a Hessian metric, occur in flat Euclidean space. So tidal forces do not warp space- or space-time which is meaningless.

Once the strain is gradient then the central forces follow from the condition of equilibrium, or the vanishing of the divergence of the stress tensor,

(s_ij),j= (C_ijkl u_l,k),j=0,

where C is the elasticity tensor, and the comma means differentiation. Implied in this equation is that the strain is a gradient so that the compatibility condition is automatically satisfied, the curl of a gradient is zero. The elasticity tensor is given in terms of the Lame, l, and shear moduli, m,

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

where the prime indicates differentiation with respect ot r.

In a system which manifest spherical symmetry, the displacement, or electric, field is purely radial, i.e., u=u(r). The condition of


{s_(rr)-s_(theta theta)-s_(phi phi)},r = 0,

in the absence of body forces. This would be analogous to the Laplace equation of the electric field in the absence of matter.

Provided the Lame and shear moduli are constant, this condition reduces to

u" + (2/r)u' -2u/r^2 =0.

The general solution is

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

where A and B are constants. The displacement, or electric field, is a sum of an harmonic oscillator force and and inverse square one. These are dual laws: In the Kepler problem, when the sun is placed at the center of the ellipse, B=0, while when it is placed at one of the foci, A=0. Remarkably, they come out of the equilibrium, elasticity, condition when the moduli are assumed to be constant.

Two things should be pointed out: (1) the nonlinear term in the expression for the strain, the square of the gradient of the displacement is omitted as well as the bilinear product of the Christoffel symbols in the Riemann tensor R_ijkl, Gamma_isk Gamma^s_jl =0, and (2) the tidal forces are destroyed by a non-vanishing Riemann tensor. So to say that the tidal forces are in the 00 component R^i_0j0, where 0 is time, of the Riemann tensor is meaningless. A 4X4 matrix is suddenly replaced by a 3X3 one, and the omission of the bilinear product of Christoffel symbols is tantamount to setting the Einstein pseudo-tensor equal to zero. This is, indeed, strange since that pseudo-tensor has been identified as gravitational energy. So even if linearized gravitational waves would exist, they would not carry energy! So General Relativity with its non-vanishing Riemann tensor destroys the gravitational phenomenom that it seeks to predict.

8 views0 comments

Recent Posts

See All

What Is Time in Spacetime?

It is well-known that the hyperbolic plane is "too big" to be embedded in Euclidean 3-space, but it is not "too big" to be embedded in Minkowski 3-space. Whereas the metric of the former is positive d


bottom of page