- #1
Airsteve0
- 80
- 0
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\Lambdag_{\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}+\Lambdag_{\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})
(1) G_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R
(2) G_{\alpha\beta}+\Lambdag_{\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})
