# Can GR be reformulated in terms of invariant observables?

## Question

So recently I was thinking about this: How many scalars are available in $$4$$ dimensions in General Relativity (without being redundant)? For example, with metric we can construct the following scalar:

$$g^{\mu \nu} g_{\mu \nu} = 4$$ is the same as: $$(g^{\mu \nu} \otimes g^{\rho \kappa}) \cdot (g_{\mu \nu} \otimes g_{\rho \kappa} ) = 16$$

We also have scalars like curvature, torsion, inner product of the riemann tensor with itself, etc.

## Motivation

My motivation for doing so is as follows: GR is currently through (rank $$2$$ symmetric) tensors formulated as: $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$ Hence any solution of the above automatically satisfies:
$$(R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}) (R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu}) = \bigg(\frac{8 \pi G}{c^4}\bigg)^2 T_{\mu \nu} T^{\mu \nu}$$

But note the later equation is written purely in invariant observables. I was wondering if General Relativity could also be written purely in terms of observables? If not how many short are we? Can the remaining variables be expressed as something invariant and not a scalar (not sure if it would be a tensor either)?

• relevant: Lovelock's theorem. – AccidentalFourierTransform Oct 29 '18 at 19:46
• @AccidentalFourierTransform If the reformulation was successful wouldn't the equations be the same just the fundamental variables be different? – More Anonymous Oct 29 '18 at 20:03
• There are many cases in which it is useful to determine how quantities change with reference frame, because those quantities are experimentally relevant. Which specific parts of GR are you proposing to recast? – probably_someone Oct 29 '18 at 20:57