top of page

General Relativity vs Ricci Flow: Propagation or Diffusion?

Whereas General Relativity operates in 4D, Ricci flows occur in 3D. But we can always contract the Riemann tensor, not with the metric tensor but, with its projection onto 3D space, which is a subspace normal to the unit time vector.


The vector identity

curl(curl A) =grad(div A) - div(grad A) (1)

also holds for second order symmetric tensors, g_(ij). If h_(ij) is its linearized version with respect to a flat metric then in 3D we have

G_(ij)=R_(ij)-(1/2)R =(1/2)(curl curl h)_(ij) (2)

where the linear Ricci tensor is

R_(ij) = (1/2){h_(ik,kj) + h_(kj,ik)-h_(kk,ij)-h_(ij,kk)} (3a)

and the scalar curvature is

R = h_(jk,jk)-h_(jj,kk). (3b)

Equations (3a,b) hold in any dimension. In 4D, specializing to the time component (ij)=(00),

and considering the time-independent case, only the last term subsists in (3a) showing that the Ricci tensor is -1/2 the laplacian, and the scalar curvature vanishes.


In 3D, the interpretation is that the Ricci tensor is the "laplacian of the metric", and by an analogy with heat, or diffusion, it is set equal to the time-rate-of-change of the metric. Whereas in 4D, with the application of the Lorentz gauge, or traceless condition:

h_(ij,i)-(1/2)h_(ii,j) =0 (4)

and the condition that the fields be time-independent, the linearized Einstein equations become

(d_t^2 - div grad)h_(ij)=-16 pi G T_(ij), (5)

where T_(ij) is the symmetric energy-stress tensor of Einstein. In the absence of sources, (5) becomes the free wave equation. So, it would appear on going from from 3D to 4D we go from diffusion to wave propagation!


In general, if we take the trace of the Einstein equations

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

where k is the Einstein gravitational constant, then we get

R=-trace T,

and when this vanishes in empty space,

R_(ij) =0.

This is none other than Einstein's condition of emptiness. Yet, Einstein introduced his cosmological constant, C, so that his equations would read

R_(ij)- (1/2)R g_(ij) + C g_(ij) = k T_(ij).

Now in empty space, the trace condition reads R=-2C so that the condition of emptiness is

R_(ij) -R g_(ij) =0, (7)

The Riemann metric is proportional to the metric itself in all cases where there is constant curvature. Equation (7) is said to define an Einstein manifold [e.g., Besse, Einstein Manifolds].







2 views0 comments

Recent Posts

See All

Never has so much confusion, and conflicting results, have been made that a theory can be made compatible with just about anything. Einstein built into his general theory of relativity that particles

The Schwarzschild metric is unique in general relativity, and its shows. The contradictions that it provides casts doubts on the validity of the whole theory. Depending on whether we reason from the

The idea that plane waves propagating away from electromagnetic sources are gravitational waves because the coefficients of the indefinite metric are called "gravitational" potentials is nonsense. The

bottom of page