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Maxwell's Tensor is to Electromagnetism What Einstein's Tensor is Not to Gravitation

It should have been readily obvious from the exterior and interior solutions to Schwarzschild's problem that the Einstein field equations are completely meaningless. In regard to the exterior problem involving a central mass, the vanishing of the three components of the Ricci tensor were the compatibility conditions for the existence of tidal forces. The internal solution which is a hyperbolic model of the plane of constant negative curvature is a model of centrifugal acceleration. All that needs to be changed on going from the exterior to the interior solution is 2m/r-> w^2r^2, where m is the gravitational parameter and w is the angular speed.

Instead of the vanishing of the Ricci components, as in the exterior solution, they become equal to a constant in the internal solution. That being the case, Einstein equations would have

rest energy + pressure = 0, (1)

which is familiar from the Friedmann model of the early universe. There, Noonan and Robertson remark (Relativity and Cosmology, p. 374) But the nonsense self-propagates itself: In Exact Solutions of the Einstein Equations by Stephani et al., p. 61 it is claimed that (1) is acceptable: the Ricci tensor is of the Lambda-type. They are also known as Einstein spaces. which are defined as

R_(ab)=Lambda g_(ab), (2)

where Lambda is an arbitrary constant, the so-called cosmological constant. Lambda is a constant, either positive or negative, but g_(ab) is an indefinite metric. Thus R_(00) must have an opposite sign to the space components, R_(ij). We shall show that this is not true for a de Sitter universe, alias the inner solution of the Schwarzschild problem.

"There is no physical solution solution to [1] because the only possible negative contribution to the pressure comes from the gravitational energies of the galaxies, and this can hardly be as large as the rest energies of the galaxies. Equation [1] requires both the energy density and pressure to vanish. But then we may ask exactly what it is, if not matter, that follows the fundamental world lines."

Equation (1) follows from the indefinite metric

ds^2= a(r)dt^2- b(r) dr^2 - r^2 d theta^2,

where the coefficients a(r) and b(r) depend only on the radius, r. Defining a new variable, q, we can write the indefinite metric as doubly warped product

ds^2 = F^2 dt^2 - dq^2 - G^2(q) d theta^2.

Then writing r=G(q) and dr/G'=dq we get

ds^2= F^2 dt^2 -dr^2/G'^2 + G^2 d theta^2.

The eigenvalues of the Ricci tensor R_(ij) are then

R_(00) = F"/F +2 F'G'/FG (3)

R_(11)= F"/F + 2G"/G (4)

R_(22)=R_(33)= G"/G +F'G'/FG -(1/G^2)[1-G'^2] (5)

Ricci flat means that all three are zero, while Ricci constant means they are all constant. In respect to Einstein's equations,

R_(ij)-(1/2)g_(ij)(R-2 Lambda)=T_(ij), (6)

where R is the scalar curvature, Lambda is the so-called cosmological constant, and T_(ij) is the energy-stress tensor. According to Besse, (Einstein Manifolds, Sec. 0.13) the condition of Ricci flatness is too strong a condition on most manifolds. " 'No mass' leads to Ricci flat manifolds, R_(ij)=0." However, the eigenvalues of the Ricci tensor are sums of sectional curvatures which are not zero, in general, although their sum may be zero. Why should the energy-stress tensor be zero, if the sectional curvatures are non-zero? And for Ricci constant, why should the energy-stress tensor be constant. The latter, in fact, leads to Eq. (1): the sum of two constants is zero!

For tidal forces, F=G'=(1-2m/G)^(1/2), where m is the gravitational parameter, the sectional curvatures are the tidal forces,

G"/G=m/G^3 ; F"/F= -2m/G^3 ; (1-G'^2)^(1/2)/G^2= 2m/G^3, (7)

Expressions (7) are the compatibility conditions for a path independent scalar potential for they make (3)-(5) to vanish. It makes a statement that all the deformed components of the strain must fit together without lacune. This is the so-called external solution to the Schwarzschild problem.

The internal solution has F=G'(1-w^2r^2)^(1/2), where w is formally an angular speed. Now, the sectional curvatures are

G"/G=-w^2 ; F"/F= - w^2 ; (1-G'^2)/G^2= w^2, (8)

and the eigenvalues of the Ricci tensor are

R_(00)=R_(11)=R_(22) = -3w^2. (9)

This can't even constitute an Einstein space, (2), because all the eigenvalues of the Ricci tensor are negative, and this is incompatible with g_(ab)=(-1,+1,+1,+1). And who says that the metric is Minkowski, when the whole exercise of Einstein equations is to determine the value of the gravitational potentials, g_(ab)? How can we assume that the energy-stress tensor of a perfect fluid is given by

T_(ik) =h u_iu_k +p g_(ik) (10)

where the 4-velocities satisfy u_i u^i=-1, and h is the enthaply density. It is said latin symbol go from 1 to 4 and Greek symbols run over the space indices. So (10) is actually p g_(alpha,beta), [Landau & Lifshitz, Fluid Mechanics, pp. 499-500.] But what gives us the right to assume that the gravitational potentials are Minkowskian in (10) when (10) has to satisfy

R_(ik)-(1/2) R g_(ik)= T_(ik) (11)

the Einstein equations where R is scalar curvature?

In an empty region of space, the rhs of (11) must vanish. This gives

R_(ik) = (1/2)R g_(ik) (12)

which qualifies itself as an Einstein space, (2). The outer solution of Schwarzschild (7) certainly qualifies since R_(ik)=R=0, but the inner solution (8) doesn't since all the g_(ik) must be of the same sign. And we know that the g_(ik) are not Minkowskian since for the outer solution,

g_(00) = 1- 2m/r = 1/g_(11), g_(22) = r^2, (13)

and for the inner solution,

g_(00) = 1- w^2 r^2 = 1/g_(11), g_(22) = r^2. (14)

If, for instance, if we use the eigenvalues of the inner solution, (9), then all gravitational potentials must have the same sign, and the same value. The former excludes the Minkowskian metric, while the latter excludes the gravitational potentials (14).

Even the author of Einstein Manifolds comes clean and writes

"Despite the simplicity of the condition [1] the reader should not imagine that examples are easy to find. If you are not convinced, try to find one yourself which is not in our book. And if you do succeed, please contact us immediately."

The only example I was able to find in the book was R_(ii)=R=0, i.e. the exterior Schwarzschild solution discussed above. Yet, entire books have been written claiming Exact Solutions of Einstein's Field Equations, by Stephani et al. Even the de Sitter solution, which is Schwarzschild's inner solution, is not an Einstein space because the metrical coefficients are not Minkowskian.