A Physical Basis of Lovelock's Theorem: GR & Equivalence Principle

[bhobba]

This came up in another thread.

GR more or less follows directly from Lovelock's Theorem. You simply assume the metric has a Lagrangian. Where does that leave other things like the Equivalence principle?

Thanks
Bill

 

[martinbn]

I wouldn't say that GR follows from Lovelock's theorem. Only the field equations follow from the theorem. By the way what is the precise statement of the theorem?

 

[bhobba]

I wouldn't say that GR follows from Lovelock's theorem. Only the field equations follow from the theorem. By the way what is the precise statement of the theorem?

Mate you are overworking me. I had to dig up my copy of Lovelock and Rund, Tensors Differential Forms and Variational Principles. The exact statement is on page 321 but I will explain it in my own words rather than give a direct transcription.

Given any Lagrangian in the metric Tensor of the form L1 + L2 where L1 only involves the metric and up to its second derivatives (you can prove you must go at least to the second derivatives) and L2 is the interaction Lagrangian between the field and what its interacting with, then the only possible equations of motions are the EFE's ie Euv = Tuv where Euv is the Einstein Tensor. (yes I have left out the cosmological constant for simplicity and used units so there is no k in front of the stress energy tensor - it should be kTuv + λGuv).

Note that's the only assumption that went into it - no explicit equivalence principle etc. But - and this is crucial - it only works in 4 dimensions.

The question is why is the metric a dynamical variable - writing the equation of motion of a free particle in general coordinates leads of course to dt = GuvXuXv via a little calculus from dt = NuvXuXv. This means of course the metric determines the motion of particles so acts like a gravitational field - but implying it has it own Lagrangian - now that while almost smacking you in the face is an assumption. My suspicion is its the key one. But I could be wrong

Thanks
Bill

 

[jim mcnamara]

This might help:
https://en.wikipedia.org/wiki/Lovelock's_theorem

The links at the bottom of the page presumably provide a more rigorous derivation - I cannot get to some of them, so that is a guess.
https://arxiv.org/pdf/1106.2476.pdf Sec 2.4.1 is a discussion of the theorem as well.

 

[PAllen]

I would say the principle of equivalence is involved in setting the mathematical framework needed to even state Lovelock’s theorem. That is a manifold with Minkowskian metric, and Minkowski space tangent plane can be seen as an embodiment of the EP. Further, the idea that the EFE are essentially unique long predated Lovelock. His theorem simply formalizes arguments that go back to Hilbert.