Einstein's equations, involving the Ricci tensor, are linearized with respect to a Minkowski background under the conditions that the spatial components of the perturbed metric are divergent- and trace free, referred to as the TT-gauge.

In this gauge, the only non-zero components of the Riemann tensor are,

R_{j0k0}=R_{0j0k}=-R_{j00k}=-R_{0jk0},

displaying the fact that the perturbed metric tensor is* time-independent* in the linear approximation that is made [Ohanian & Ruffini, *Gravitation*, p. 255]. It is then incredible, and I repeat incredible, that this component of the Riemann tensor is set equal to the second time derivative of the TT perturbed metric

R_{j0k0}=-(1/2) h_{ij,00}^{TT}, (A)

where the time and zero that follow indicate the second derivative with respect to time! That is a time-independent quantity (left-hand side) is equated with a time-dependent quantity (right-hand side). Then the resounding conclusion follows:

*The use of the TT gauge indicates that a traveling gravitational wave with periodic time behaviour, h~ exp(iwt), can be associated to a local oscillation of the spacetime curvature,*

* R_{j0k0}=-(1/2)w^2h_{jk}^{TT},*

where w is the frequency of the wave. The h^{TT} metric appears in the wave equation that has been obtained from linearizing the Ricci tensor, yet, suddenly it is equated with the time independent component of the Riemann tensor--without contraction of the later! It is then inserted into the geodesic deviation equation for the Riemann tensor, and the concluded that "in the reference frame comoving with particle A, the particle B is seen oscillating with an amplitude proportional to h_{jk}^{TT}.

Now, the linearized Riemann component is

R^k_{0l0}=const.(h_{lk,0,0}-h{0l,k,0}-h_{0k,0,l}+h_{00,k,l}), (B)

where the comma denotes differentiation. Neglecting the cross terms, the first and last terms remain. Neglecting the last term is tantamount to neglecting tidal forces. And all those who claim that gravitational waves are tidal forces (e.g. Ellis) would not agree. If we consider the static case, then it is only the last term that subsists and we have

const. h_{00}=V,

the Newtonian potential. Thus,

R^k_{0l0}=V_{,k,l},

which are the tidal forces.

This point is crucial. For if we consider a static field, the irrotational component of the field is preserved, and we have tidal forces. If we don't, it is tantamount to adding the negative of the time derivative of the vector potential to the electric field, and we no longer have an irrotational component of the field. In any any case there is no justification for the neglect of tidal forces, and this invalidates Thorne's association in (A). It is tantamount to saying that there are no tidal forces, when we know there are. Supposedly, the generation of gravitational waves is due to the contraction and expansion of spacetime, which, in itself, is a tidal effect.

On the other hand, if consider the static case, than the solenoidal component of the field is a Beltrami field. The best we can do is consider standing waves. In the Beltrami field the E and B fields are parallel. They are described by Beltrami's equation,

curl B = k B,

for the magnetic field, where k is a constant. Taking the curl again,

curl (curl B)= k (curl B)=k^2 B,

upon introducing Beltrami's equation. Writing this out explicitly,

grad div B - grad.grad B=k^2 B,

but since div B=0, it reduces to the Helmholtz equation. So, at most we can expect standing waves from the solenoidal component of the field. The question that this poses is: Is there a B component of the gravitational field for masses in motion?

Here, we may contrast Maxwell's versus Beltrami's decomposition of the field. In the static case, Maxwell's field decomposition excludes wave propagation, while in Beltrami's decomposition, standing waves are permitted. Maxwell's decomposition destroys tidal forces since the electric field is not the gradient of a scalar potential, while Beltrami's is compatible with a tidal force description.

Returning to the geodesic deviation equation,

D^2x^j/D^2t = -R^j_{0k0} x^k,

where D stands for the covariant derivative, we have, in a first approximation, flat space surrounding the particle so that the Christoffel connection vanishes. Thus D->d.

the ordinary derivative. Then, the Riemann component is replaced by

R^j_{0k0}=-(1/2) d^2 h^j_k^{TT}/dt^2 (B)

so that the geodesic deviation equations reads

d^2 dx^j/dt^2 = (1/2) d^2h^j_k^{TT}/dt^2 x^k.

But we know already that h^{TT}~exp(iwt), so why introduce the second derivative on the right-hand side of the equation? Notice also that a "d" has been introduced into the x term on the left-hand side [Thorne, "Gravitational radiation", in *300 Years of Gravitation, Hawking, Israel eds.*]. This supposedly makes the "changes dx miniscule compared to the distance of the particle from the origin, x^k can be regarded as essentially constant on the right-hand side, and [it] can be integrated easily to give

dx^j= (1/2) h_{jk}^{TT}x^k.

Thus, aside from a factor of (1/2), h_{jk}^{TT} plays the role of the 'dimensionless strain of space', or the 'time integrated shear of space': it is the ratio of the wave induced displacement of a free particle relative to the origin, to its original displacement from the origin. (In the above equations...it does not matter whether the a spatial index is up or down, since the spatial coordinates are Cartesian.)" [If he takes such care in explaining the difference between co-and contra-variant components, he should also be so careful as to to slip in a "d" before x^j in the geodesic deviation equation. Moreover, Thorne gives a reference to "definition" in Equation (C); yet when we go to consult the reference, he gives another reference--always to himself. Equation (C) is certainly **not **a definition! The question becomes how can the properties gravitational waves be formulated---not to say measured!---on the basis of such discutible mathematics?]

Thorne continues: "Because of their TT nature, gravitational waves produce a quadrupolar, divergence free force field. This force field has two components corresponding to the two polarization states of the wave: the quantity

h_{+}=h_{xx}^{TT}=-h_{yy}^{TT}

produces a force field with the orientation of a '+' sign, while

h_{x}=h_{xy}^{TT}=h_{yx}^{TT}

produces one with the orientation of a 'x' sign."

What is conveniently forgotten is that there is a sum over k in the geodesic deviation equation, (i.e. the Einstein summation convention!) so we cannot write separately

dx=(1/2)h_{+}x and dy=-(1/2)h_{+}y

and

dx=(1/2)h_{x}y and dy=(1/2)h_{x}x.

Another gem can be found in Price & Wang, "The transverse traceless gauge and quadrupole sources" [*American Journal of Physics* **76** (2008) 930]. They call the fact that the trace of the xx and yy components of \bar{h}_{jk}=h_{kj}-g_{jk}h doesn't vanish. Rather, what is required to vanish is

h_{xx} + h_{yy} = \bar{h}_{xx} + \bar{h}_{yy} + \bar{h}_{00}. (*)

This, the trace of the space components is required to vanish. The paradox to which these authors allude to is that while the first two components on the right hand side depend on angles, the third, which is the time integral of the density,

\bar{h}_{00}(x,t) = (4G/c^2r) \int T_{00}(x,t')/|x-x'| DV, (**)

where the integrand is evaluated at the past time, t', doesn't appear to be. Introducing the Lienard-Wiechert potentials, they show that it does and is given by

\bar{h}_{00} =âˆ’ 2A sin2\theta cos 2[w(t âˆ’ r/c)âˆ’phase factor] , (***)

where A is the amplitude. Although this is precisely what is necessary to cancel the sum of the first two terms in (*), it does not explain how a positive definite term like (**) could become sign indefinite in (***). They might claim that this is due to the neglect of the non-radiative terms, yet, if these terms were to be included, the trace (*) would still not vanish!

Now we come to the polarization of the gravitational wave. Not surprisingly, it is possible to decompose the gravitational wave into two linearly polarized waves and two circularly polarized ones. The linearly polarized waves are propagating in the z-direction with polarization tensors + and x, e_{+} and e_{-}, while the two tensors describing the states of circular polarization are their linear combinations: e_{+}+ i e_{x} and e_{+}- i e_{-}.

But these are just the eigenvalues of the curl operator [Moses "Eigenfunctions of the curl operator, rotationally invariant Helmholtz theorem, and applications to electromagnetic theory and fluid mechanics" 1971]. The zero eigenvalue corresponds to laminar or irrotational flow. So what is attributed to gravitational waves is the solenoidal component of the tensor--which must be time-independent--and is completely independent of the geodesic deviation equation which contains the symmetric, time-independent Riemann tensor. If the Riemann tensor were to acquire a time-component, just like the derivative of the vector potential in the expression for the electric field, it would lose its symmetry and no longer would be a Riemann tensor. So we may say that the symmetry the time-independent Riemann tensor stands in the defence of the decomposition of the total tensor into irrotational and solenoidal fields. What LIGO has measured is undoubtedly the magnetostatic component of a stationary field that is somehow generated in the cavity--if, at all, there is a gravitational B field which is solenoidal.

The upshot of the matter is this: With a constant vector potential the gravitational analog of the E-field splits off from that of the B-field. This is necessary to retain the symmetry of the Riemann tensor. So whatever the gravitational analog of the B-field is, it cannot be described by general relativity. In other words, tidal forces are restricted to the E-field. This result pertains to the Riemann tensor, and not to its contracted form, the Ricci tensor. So linearizing the Ricci tensor and saying that, after applying all the gauge conditions to get rid of unwanted terms, it results in a wave equation is meaningless. The gravitational analog, if you can say that, satisfies the Helmholtz equation. But this equation is not governed by the Riemann tensor, but, rather, in the criterion of a complete decomposition a stress tensor. Recall that Schaefer added on the dyadic (the E-field) to the Beltrami solution (the B-field) whose completeness was proved by Gurtin.

That is, to Beltrami's solution of div **S**=0,

**S**= curl curl **A**, (1)

where **A** is an arbitrary second-order symmetric tensor, was added the symmetric tensor 2E=grad E+grad E^+, where "+" is the transpose to get the complete decomposition,

**S**= curl curl **A** + 2**E. (2) **

It differs from Maxwell's decomposition insofar as there is no bridging term involving the time rate of change of vector potential, that would appear in the second term and destroy its symmetry. That would be tantamount to invalidating the Riemann tensor, just as a non-symmetric connection would do.

But there is another possibility here: Elasticity considers both components of (2) to be spatial. Thus, the decomposition is mutually exclusive. Since in GR, the metric is indefinite, (2) could be related to the time and space components of the metric tensor. That is,

**E**_{ik}=R_{i0k0} (3)

and

e_{ijr}e_{kln}(curl curl **A**)_{rnr}=R_{ijkl}, (4)

where the subscripts in the Riemann tensor would vary over the spatial components, and e is the permutation tensor. It is evident that the space components of the Riemann tensor vanish in (3) in the static case in the linear approximation where (B) applies. The tidal force,

f^k=-R^k_{0l0}x^l,

will then be able to be represented graphically by field lines since

f^k_k=R^k_{0k0}=-R_{00}=0.

This can be considered as a dynamic condition of equilibrium, when referenced to an ellipsoid, instead of considering it the condition of 'emptiness" as in GR [cf. *Seeing Gravity*].

Since (3) is referred to as the "tidal force tensor," this is supportive of the fact that the Riemann tensor splits into irrotational and rotational components of the stress tensor, or to the time and space components of the perturbed metric, h. This alludes to the fact that the cross terms, should there be any, are without physical significance. In the case of the (stationary) Schwarzschild metric, (3) consists of a 3 X 3 matrix with diagonal entries that sum to zero. This is easily referred to an ellipsoid so that (3) can be extracted from the spatial part of the tensor, (4).

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