Proving the Vacuum Field Equations are Trivial for n=3

[Airsteve0]

I am trying to prove a concept that was presented in my course but I am having a bit of an issue with understanding the final result. The concept was that in 3 dimension the vanishing of the Riemann tensor leaves only the trivial solution for the vacuum equations. I started by subbing Eq.(1) into Eq.(2) and solving for R (I took the trace in case you were wondering - my result was 6$\Lambda$) and then subbed then this answer back into solve for $R_{\alpha\beta}$ (answer was 2$\Lambda$$g_{\alpha\beta}$). I then subbed both of these into (3) and solved. However, as you can see my answer is not zero. If anyone could point out a mistake or if I just haven't properly conceptualized the answer I would appreciate the help, thanks.

(1) $G_{\alpha\beta}$=$R_{\alpha\beta}$-$\frac{1}{2}$$g_{\alpha\beta}$R

(2) $G_{\alpha\beta}$+$\Lambda$$g_{\alpha\beta}$=0

(3) $R_{\alpha\beta\gamma\delta}$ = $g_{\alpha\gamma}$$R_{\beta\delta}$+$g_{\beta\delta}$$R_{\alpha\gamma}$-$g_{\alpha\delta}$$R_{\beta\gamma}$-$g_{\beta\gamma}$$R_{\alpha\delta}$-$\frac{R}{2}$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

= $\Lambda$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

 

[Ben Niehoff]

in 3 dimension the vanishing of the Riemann tensor leaves only the trivial solution for the vacuum equations.

I assume you mean "vanishing of the Einstein tensor", not the Riemann tensor.

= $\Lambda$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

In the presence of a non-zero cosmological constant, the "trivial solution" is not flat space; it is de Sitter (positive lambda) or anti-de Sitter (negative lambda) space.

Also, I'm not sure the statement you are trying to prove is true in the case of nonzero lambda. The BTZ black hole is not trivial, for example.

 

[Airsteve0]

yes, sorry I did mean the Einstein tensor

 

[haushofer]

I think the statement is that in three dimensions the riemann and ricci tensor have an equal amount of components, see e.g. Carroll's notes. That means that the vacuum equations with Lamda=0, namely that the ricci tensor vanishes, imply that the riemann tensor vanishes. This is also clear from your expressions. Hence no gravitational waves. See e.g. Carlip's book or notes on 2+1 gravity, or Nakahara. Things change when you introduce a Lamda, which is also clear from your expressions. The vacuum equations then no longet imply that the ricci tensor vanishes, but that it is proportional to the metric.

 

[sardar]

First of all, it is important to note that the vacuum field equations are a set of equations that describe the curvature of spacetime in the absence of matter or energy. These equations are given by the Einstein field equations, which relate the curvature of spacetime (represented by the Riemann tensor) to the energy and momentum content of the universe (represented by the stress-energy tensor).

In the case of n=3 dimensions, the Riemann tensor can be expressed in terms of the Ricci tensor and the scalar curvature, as shown in equation (3). Substituting this expression into the vacuum field equations (equation 2) yields a simple relation between the Ricci tensor and the cosmological constant (represented by \Lambda). This is because, in 3 dimensions, the Riemann tensor is completely determined by the Ricci tensor and the scalar curvature.

Now, it is important to understand that the Ricci tensor is a symmetric 2nd-order tensor, meaning that its components are symmetric with respect to interchange of indices. This means that for every component R_{\alpha\beta}, there is a corresponding component R_{\beta\alpha} with the same value. Therefore, when you solve for R_{\alpha\beta} in terms of \Lambda, you are essentially solving for all possible components of the Ricci tensor. This means that your result of 2\Lambda g_{\alpha\beta} is actually a set of 9 equations, one for each combination of \alpha and \beta.

Finally, when you substitute this result into equation (3), you are essentially multiplying it by the symmetric factor (g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma}), which will result in a symmetric tensor. However, since \Lambda is a constant, it can be pulled out of this expression and will cancel with the \Lambda in the vacuum field equations (equation 2). This leaves you with a trivial solution, as the remaining terms in equation (3) will cancel out.

In conclusion, your result is not wrong, but it is not the full picture. The vacuum field equations in 3 dimensions are indeed trivial, with the only solution being a constant cosmological constant \Lambda. This is because, in 3 dimensions, the Riemann tensor is completely determined by the Ricci tensor and the scalar curvature, and therefore the only non-zero