7
$\begingroup$

In the Einstein field equations, the only tensor that shows up is the Ricci tensor and the metric tensor, together with the Ricci scalar. The Weyl tensor though is a tensor that is a part of the Riemann curvature tensor as well (the off-diagonal part). It represents tidal forces, which are non-local.

Why doesn"t the Weyl tensor show itself in the Einstein field equations? Is it because gravity is based on the equivalence principle, which is a first-order equivalence?

$\endgroup$
7
  • $\begingroup$ Have you learned how the Einstein equations guarantee the local conservation of energy and momentum via the vanishing covariant divergence of the Einstein tensor? $\endgroup$
    – Ghoster
    Commented Sep 2, 2023 at 5:35
  • $\begingroup$ TLDR; the Weyl tensor has a vanishing contraction with the metric i.e., $g^{\mu\nu}W_{\mu\nu\rho\sigma}=0$ and tells us nothing useful (especially with regards to gravitational fields in a vacuum). $\endgroup$
    – joseph h
    Commented Sep 2, 2023 at 6:42
  • $\begingroup$ @Ghoster Is that comparable with the conservation of charge and currents in electromagnetism, as expressed by a vanishing divergence of the charge density and the current? $\endgroup$
    – Il Guercio
    Commented Sep 2, 2023 at 7:24
  • $\begingroup$ @josephh Isn't it in fact the defining tensor for true gravitational fields? $\endgroup$
    – Il Guercio
    Commented Sep 2, 2023 at 7:25
  • $\begingroup$ @josephh looks like that could be an answer (or at least the start of one) $\endgroup$
    – Kyle Kanos
    Commented Sep 2, 2023 at 13:49

3 Answers 3

10
$\begingroup$

Why does the Weyl tensor not show up in the Einstein field equations?

I think that the “moral” reason for that is that the Weyl tensor represents locally free, propagating part of the curvature and is thus not pointwise determined by matter fields.

However, Weyl is not completely independent of matter. By combining differential Bianchi identity, $∇_{[α}R_{βγ]δ} {} ^\epsilon = 0$, with Einstein equations we can arrive at the following constraint on Weyl tensor: $$∇^δC_{αβγδ} = 8πG \left( ∇_{[β}T_{α]γ} −\frac 1 3 g_{γ[α}∇_{β]}T \right).$$ Moreover, the linearized version of this equation can be “repackaged” as a wave equation describing propagation of a massless spin-2 field, thus making clear that these are the propagating degrees of freedom.

$\endgroup$
1
$\begingroup$

The Einstein field equations relate the curvature of spacetime to the distribution of matter and energy. While the Weyl tensor is important for understanding gravitational waves and certain aspects of the gravitational field's geometry, it does not directly involve matter and energy distribution. Therefore, it does not appear explicitly in the Einstein field equations.

$\endgroup$
1
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$ Commented Sep 4, 2023 at 2:02
1
$\begingroup$

My ``incomplete'' answer is that this can be interpreted also by Newtonian analogy. The Poisson equation is the Newtonian analogue to the Einstein field equations.

The Poisson equation in Newtonian gravity fully determines the Newtonian potential, but involves only the trace of the Hessian of the potential.

This Hessian is the Newtonian tidal acceleration matrix/tensor, and it is the Newtonian analogue to the Riemann curvature tensor. The analogy to the Weyl tensor would be the anti-symmetric parts of this matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.