Symmetric form of Einstein Field Equations?

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?

1. Is this form the equation used in the literature anywhere?
• If you take the trace of both sides you get $0=0$. Otherwise I don't see why this form of the equations is particularly special or interesting. Mar 23, 2022 at 22:13
• Yeah, that I knew. I suppose I'm looking for interpretive value. Like, it seems like subtracting 1/4 the trace may tell you have much the curvature deviates from if the curvature was evenly distributed in all directions?
– SSD
Mar 23, 2022 at 22:16
• None jumps out at me. There are an infinite number of ways to write any equation. Here you've added two equivalent forms of Einstein's equations, so presumably these equations are equivalent to Einstein's equations (unless something has gone horribly wrong). Randomly shuffling symbols without a concrete idea that you want to express is very unlikely to produce something meaningful. Mar 23, 2022 at 22:18
• Well the thing the makes this interesting is that the same operation is being applied to both the Ricci and the stress tensors. That's why I figured it may have something to say about interpretation (like my edited comment above). I wouldn't say I've randomly shuffled.
– SSD
Mar 23, 2022 at 22:20
• I see. This is unimodular gravity. You can derive it from an action principle where you assume the metric has unit determinant. It is equivalent to GR except that the cosmological constant appears as an integration constant. Some people think this is a possible resolution of the cosmological constant problem. Other people (including me) just think this is rewriting the same theory in different variables. But, I suppose I was too dismissive, since there is a name for this form of the equations. Mar 23, 2022 at 23:01

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

$$R =- T.$$

• Thanks! Would constraining the divergence of the stress tensor to equal 0 also produce the extra constraint necessary?
– SSD
Mar 23, 2022 at 23:14
1. 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/ .

• Very cool. Thanks for pointing me to Willie WY Wong's blog! Any idea as to what the traceless stress-energy tensor means, physically? Frankly, I don't even know what the trace of the stress energy tensor means.
– SSD
Mar 24, 2022 at 15:15
• I am not a great expert for this topic but I think it means that you have a massless matter as for example electromagnetic field, see en.wikipedia.org/wiki/… .
– JanG
Mar 24, 2022 at 15:42