1

In his book "Road to Reality" section 19.7 Roger Penrose asks the question:

What is the appropiate analogue of the Maxwell field tensor $F_{ab}$ describing the gravitational degrees of freedom? According to Penrose "... it is more appropiate to choose what is called the Weyl tensor".

I don't really understand this assertion. It might be that Einstein's field equations (EFE) are here only considered as constraint on the Ricci tensor

$$R_{ab} = \kappa \left( T_{ab} - \frac{1}{2}g_{ab} T\right).$$

In Maxwell theory $\nabla \cdot \mathbf{E} = 4\pi \rho$ is sometimes considered as a constraint, but we have in free Maxwell theory at least $\partial_a F^{ab} =0$ whereas in gravitational theory the Weyl-tensor is completely free (there is no field equation for it, at least I don't know any). It is actually this complete freedom which confuses me.

The gravitational physics seems to be independent of the value of the Weyl tensor, but can this be ? The Weyl tensor is known be responsible for tidal forces on matter, without rule for the Weyl tensor they could be arbitrary (large).

Qmechanic
  • 201,751
Frederic Thomas
  • 9,695
  • 1
    see my answer here – KP99 Dec 29 '21 at 13:20
  • While not a strict duplicate, this question is the dual of https://physics.stackexchange.com/questions/162105/why-do-the-einstein-field-equations-efe-involve-the-ricci-curvature-tensor-ins – cesaruliana Dec 29 '21 at 20:03
  • Pinching from my answer above, the Weyl tensor is not free, the field equation involving it are the Bianchi identities, that relate the derivatives of the Weyl tensor with the Ricci tensor – cesaruliana Dec 29 '21 at 20:05

0 Answers0