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Is the Ricci Tensor more of a Curse than a Blessing?

General relativity should be a generalization of the special theory where the constant metrical tensor acquires coefficients that may depend on time as well as on space.

Any arbitrary tensor can be split up into a symmetrial and antisymmetrical part. We identify the symmetrical part with the metrical field, and the antisymmetrical part with the electromagnetic field. This assumption was considered by Einstein in Berl. Ber. 1923 and developed more fully by Born and Infeld a decade later.

In vector analysis it is well-known that a vector field is a gradient when its curl vanishes. This guarantees the existence of a scalar potential, like the Newtonian one. Now, in linear elasticity, the strain tensor is one half the symmetric derivative of the displacement field. If it vanishes then there is no deformation caused by the displacement vector field. By construction the strain tensor is always symmetric and a second rank tensor if nonlinear terms can be neglected.

The question now becomes what is the condition that such a strain tensor exists? The result was found in 1864 by the French mathematician Saint-Venant, and it parallels the condition that a vector be a gradient. Instead of the curl vanishing, like in vector theory, he found that it is the curl of the curl of the tensor that must vanish.

Parenthetically, the curl of the curl of a vector field defines a Beltrami flow developed by him in 1889. If we consider Ampere's equation, the curl of the magnetic field is proportional to the current, then the magnetic component of the Lorentz force, the cross-product of the velocity and the magnetic field, vanishes because the vector product of the current with itself is zero. These flows have some special properties, and should be able to be generalized to deal with second-order tensors if general relativity has anything to say.

Now, the results become dimensionally dependent. In 2D, the Rieman tensor can be expressed solely in terms of Gaussian curvature, K,

R_(abcd)= e_(ab)e_(cd) K,

where e is the antisymmetric Levi-Civita tensor, where switching indices introduces a negative sign. The Gaussian curvature K is half the Ricci scalar, K=R/2.

In 3D the Riemann tensor is

R_(abcd)= - e_(abi)e_(cdj)G_(ij)

The computation of the Ricci tensor shows that

R_(ij)= g_(ij)G- G_(ij),

where 2G=R, the Ricci scalar. In general,

R_(ij)= G_(ij) + 1/(2-n) g_(ij) G,

for n other than 2.

Here comes the clincher: If we linearize the metric about the flat metric of Minkowski

g_(ij) = delta_(ij) +h_(ij), by retaining only linear terms in h, the Einstein metric can be expressed as

G_(ij)= (1/2)e_(iac)e(jbd)h_(ab,cd),

where the comma denotes the derivative.

We can now appreciate this as none other than

G_(ij)= (1/2) (curl curl h)_(ij) (1)

whose vanishing is the Saint-Venant criterion that h be metrical! Einstein used this as a criterion of emptiness!

One can lodge the criticism that this pertains to 3D, but the (3+1) decomposition widely used in numerical relativity will testify to its validity in a background Minkowski space. It is, in fact, this decomposition that is used to study gravitational waves.

In terms of the Schwarzschild metric for the outer solution, the vanishing of (1) is nothing else than the sum of the tidal forces, which are given by the sectional curvatures [see Seeing Gravity, formula (2.64) p. 59]. Two of the sectional curvatures are equal, and the third is negative of twice their value making the sum vanish. This has incorrectly been associated with Gauss' law [see Ohanian & Ruffini Gravitation and Spacetime, 3rd ed., equation (1.56)] where the divergence of the tidal force is given in terms of the mass density, instead of the mass. Obviously, this is determined by the spatial part of the metric in 3D space. Yet, the geodesic deviation equation contains the component R^i_(0j0), where i and j refer to the spatial components, which is clear from its representation by a 3 X 3 matrix in their equation (1.53). Such is the confusion that has been generated by trying to make sense out of general relativity.

Equation (1) says that Einstein's equations--at least the linearized versions--do not apply to metrical field, but, rather, to the antisymmetric field--the electromagnetic field. And if sources are included then there is no metrical field at all. Why sources should destroy a metrical field is somewhat surprising, but they are the same sources that appear in Maxwell's theory. There, inductive terms are need for conservation of charge, but such terms have no role in a metrical theory.

Then it is no wonder why the linearized Einstein equations lead to wave equations for the "scalar" and "vector" potentials just like electromagnetic theory. In fact, it is electromagnetic theory pure and simple. And it is no wonder why gravitational waves are, in reality, just electromagnetic waves that propagate, unsurprisingly, at the same speed of light. The displacement of the mirrors in the LIGO interferometer would then be just the action of a radiation pressure.

Having the opposite role to the metrical field theory, the Einstein tensor, or at least its linearized form, has nothing to do with the metrical field for its nonvanishing says that a symmetric strain tensor does not exist. And there will be no scalar potential whose second derivatives are related to the tidal forces, so as to invalidate Newtonian theory.

Thus, it appears that the Ricci tensor--or at least its linearized version--is more of a curse than a blessing as far as a theory of gravitation is concerned.

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