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Gravitational Energy Is Negligible to the Gravitational Waves They Produce in General Relativity

Einstein dissected gravity from matter by constructing "energy components" components of the gravitational field in a matter-free universe. He realized that his construct was not a tensor, thereby violating general covariance. Since his expression for gravitational energy is second-order, and belongs to the non-derivative components of the Ricci tensor, it is neglected for small deformations which are described by stationary Helmholtz equation. The same must be true for gravitational waves.

According to Levi-Civita, Einstein

"had already felt some uneasiness, in particular when after having outline with genial simplicity the theory of gravitational waves, he was led to the unacceptable result that also spontaneous waves should as a rule give rise to the dispersion in energy through irradiation."

Einstein adds his gravitational energy to the energy-stress tensor; Levi-Civita equates the gravitational energy with the energy-stress tensor: "when the energy tensor T_{ik} vanishes, the same occurrence must happen with the gravitational tensor A_{ik}. This fact entails a total lack of stresses, of energy flow, and also a simple localization of energy."

Levi-Civita sets the gravitational tensor equal to the Einstein tensor. Thus, the sum of the stress-tensor and the gravitational energy vanishes, and his conclusion follows. But you can't add oranges to apples. When Levi-Civita comes to writing down the Einstein expression, he realizes that it is not a differential invariant: coefficients of the metric and their derivatives (which make up the Christoffel coefficients) are "not admissible". They are not tensors, and, consequently, cannot be equated with tensors.

The general decomposition of a tensor field consist of the sum of an irrotational part and a solenoidal part. If g are the coefficients of the metric then this composition reads

T_{ij}=E_{ij} + (Curl(Curl g))_{ij}.

In mechanics, the first, irrotational, term is the symmetric gradient of the displacement. Since the cross derivatives are equal, it implies the existence of a scalar potential. For gravity it would be the tidal potential--a completely static quantity. The second, solenoidal, component is related to the Riemann curvature tensor,

R_{klmn}=e_{ikl}e_{jmn}(Curl(Curl g))_{ij}

where e_{ijk} is the Levi-Civita symbol. Now, if we consider finite deformations in 3D Euclidean space, the Riemann tensor, based on g, must vanish. If this results in a set of 6 non-linear partial differential equations, called the equations of compatibility, for the strain tensor which is the difference between the deformed and un-deformed metric coefficients.

Now, the Riemann tensor can be operated upon by the Levi-Civita symbols in the reverse direction, viz.,

e_{ikl}e_{jmn}R_{klmn}=2R_{ij}=(Curl(Curl g))_{ij} (A)

only if the quadratic terms in the Christoffel connection coefficients

{ ih_l}{jl_h}-{ij_l}{lh_h} (B)

where the subscript mean write under the pair of indices, can be neglected since they are of higher order than the linear terms in the derivatives. But, this bilinear difference is precisely what Levi-Civita has written for the gravitational potential! By neglecting it, means that gravity does not make itself in the linear wave equation which results from the solenoidal component of the tensor field. If one insists on calling these gravitational waves, then they have no energy, for, otherwise, they would appear as source terms in the wave equation. One could argue that the free-wave equation would be valid far from the sources. However, the association of the the Ricci tensor, R_{ij}, in Eq (A) demands that the Einstein gravitational potential (B) be of higher order than the linear terms in (A). In other words, the second equality holds in (A) only if (B) can be neglected everywhere. [If my memory doesn't fail me, this was noted by Hermann Weyl.]

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