1
$\begingroup$

Any action expressed as the integral of a 4-form in 4-dimensional spacetime is diffeomorphism invariant. For example the following 4-form topological (Pontryagin) action $$ S = \int F\wedge F $$ is diffeomorphism invariant, where $F$ is the electromagnetic field strength 2-form. This action has nothing to do general relativity or gravity.

General relativity is usually alleged as the result of diffeomorphism invariance. But should we say that general relativity is resulted from local Lorentz gauge invariance in stead of diffeomorphism invariance?

Diffeomorphism invariance is rather a general condition satisfied by all actions pertaining spinors/fermions, Standard Model gauge fields, gravity spin connection (local Lorentz group) gauge field $\omega$, and Lorentz-covariant tetrad/frame field $e$.

For example, the massless Dirac spinor fermion action is a 4-form (thus diffeomorphism invariant) of exterior product of 3 frame field $e$ 1-forms and the co-derivative $(d+\omega)$ 1-form (for abbreviation, Lorentz indices are not shown here and only the Lorentz gauge interaction $\omega$ is included), $$ S_{fermion} \sim \int{i\bar{\psi}e\wedge e \wedge e \wedge (d+\omega)\psi}. $$

The local Lorentz gauge action for gravity is a 4-form (thus diffeomorphism invariant) of exterior product of 2 frame field $e$ 1-forms and the Lorentz curvature 2-form $F_{\text{Lorentz}} = d\omega + \omega\wedge\omega$, $$ S_{gravity} \sim \int e \wedge e \wedge F_{\text{Lorentz}}. $$

$\endgroup$
  • $\begingroup$ Related/possible duplicate: physics.stackexchange.com/q/12461/50583, physics.stackexchange.com/q/346793/50583 and their linked questions. Also, this is the fourth question of yours in a few minutes where you have made a rather minor edit to the title. Please do not make minor edits to your posts simply to bump them on the active page. $\endgroup$ – ACuriousMind Jun 29 '18 at 20:49
  • $\begingroup$ I could be wrong but I'm pretty sure that the above actions are only invariant under diffeomorphisms if you assume your diffeomorphism is orthonormal (ie. orthogonal unitary), anything else will change your action I believe. This includes the Einstein Hilbert action as well. $\endgroup$ – R. Rankin Nov 23 '18 at 6:25
  • $\begingroup$ Vladimir Fock wrote in his book that A. Einstein discovered in fact laws of gravity and that a "general relativity" is empty of sense since, cast in a tensor form, any equation is "covariant" under variable changes. $\endgroup$ – Vladimir Kalitvianski Jan 27 at 17:18
0
$\begingroup$

General Relativity can be formulated as a locally Lorentz invariant theory, indeed. See Weinberg's book, for example.

The diffeomorphism invariance is a way of stating that the local laws are keeping their form everywhere on the manifold (kind of translational invariance). About all theories could be formulated in a way to be diffeomorphic invariant since their physical laws retain their form everywhere on their manifold : a law coulnd't be called a "law" if it isn't the same relation between local observables for all locations in spacetime! This is why diffeomorphism invariance is almost a triviality.

Take any manifold, especially a non-homogeneous one. An observer shouldn't be able to learn his local position on the manifold by making local measurements of local observables. A local law constraining these observables have the same form/shape/relation as for any other location on the manifold (or it's not a "law"!).

What is not clear about this interpretation of the diffeomorphism invariance, is its implicit and very subtle relation with Einstein's equivalence principle (EEP). I think there is some mixing with EEP and the idea that laws should stay invariant under local active translations (i.e. a "law" is the same relation everywhere). This implicit mixing isn't very surprising since EEP is also a "law" (or a "meta-law") that should be valid everywhere on the manifold.

$\endgroup$
  • $\begingroup$ I think the content of EEP is in identifying the deviation in the form of the laws under a general diffeomorphism from its form within Poincare transformations with gravity. Would you agree? $\endgroup$ – Dvij Mankad Jun 24 '18 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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