Symmetric form of Einstein Field Equations?

Question

So normally, taking $c = 1$ and ${8\pi G = 1}$, and assuming the cosmological constant is negligible, the Einstein field equations read:

$$R_{\mu \nu} - \frac{1}2Rg_{\mu\nu} = T_{\mu \nu}.$$

However, there also exists a trace reversed form:

$$T_{\mu \nu} - \frac{1}2Tg_{\mu\nu} = R_{\mu \nu}.$$

Today while noodling around, I found that these two forms of the EFEs may be added and divided by 2 to yield what I'll refer to as the symmetric form of the EFEs:

$$R_{\mu \nu} - \frac{1}4Rg_{\mu\nu} = T_{\mu \nu} - \frac{1}4Tg_{\mu\nu} .$$

My questions:

1. Is there any interpretation to be had here?

In particular, what does subtracting 1/4 of the trace times the metric do?

2. Is this form the equation used in the literature anywhere?

Answer 1

We will start with the easy part:

In particular, what does subtracting 1/4 of the trace times the metric do?

This removes the trace of the symmetric tensor. So you now have an equation linking two symmetric traceless tensors. Since each tensor has one degree of freedom less, you end up with a less constraining set of equations. To get something that is fully equivalent to the Einstein Field Equations, you need to add the addition equation for the traces

$$ R =- T.$$

Answer 2

2. Is this form the equation used in the literature anywhere?

There is an interesting concept of 'traceless general relativity' in which Einstein's cosmological constant $\Lambda$ is not a fundamental constant, but a constant of integration - see the inspiring Willie WY Wong's explanation on https://qnlw.info/post/traceless-gr/ .