Calculating Einstein Equation from Lagrangian Eqn

[latentcorpse]

I have a 3d system with Lagrangian e_3^{-1} L_3 = -\frac{1}{2} R_3 + \delta_{ab} \partial_\rho q^a \partial^\rho q^b + \frac{1}{2H} V(q)

From this I want to calculate the Einstein equation by performing the Euler-Lagrange procedure. First of all, I move the 3d dreibein to the RHS and then I apply the E-L eqns. Using that \frac{\partial e_3}{\partial g^{\mu \nu}} = \frac{1}{2} e_3 g_{\mu \nu}, I see that

\frac{\partial}{\partial g^{\mu \nu}} (e_3 R_3) = e_3 R_{\mu \nu} + \frac{1}{2} e_3 R

Now, I don't want a Ricci scalar in the answer and I am unsure how to get rid of it. I tried looking at the trace of the Einstein equation R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = T_{\mu \nu} \Rightarrow R-\frac{3}{2} R = T \Rightarrow -\frac{1}{2} R =T and so if I can work out T then I can avoid a Ricci scalar being in the answer but I don't know how to calculate T let alone its trace.

Thanks.

 

[latentcorpse]

Yeah that's what I did and I'm fairly sure the technique is correct and I just cannot see the mistake that's giving me the wrong number. The Einstein eqn should be -\frac{1}{2} R_{\mu \nu} + \delta_{ab} \partial_\mu q^a \partial_\nu q^b + \frac{1}{2H} g_{\mu \nu} V(q)=0

Now when I vary with respect to g^{\mu \nu} I get \frac{\partial L}{\partial g^{\mu \nu}} = - \frac{1}{2} e_3R_{\mu \nu} - \frac{1}{4} e_3 R_3 + e_3 \delta_{ab} \partial_\mu q^a \partial_\nu q^b + \frac{1}{2} e_3 g_{\mu \nu} \delta_{ab} \partial_\rho q^a \partial^\rho q^b + \frac{1}{4H} g_{\mu \nu} V(q)

and \frac{\partial L}{\partial \partial_\lambda g^{\mu \nu}}=0.

The E-L eqn is - \frac{1}{2} e_3R_{\mu \nu} - \frac{1}{4} e_3 R_3 + e_3 \delta_{ab} \partial_\mu q^a \partial_\nu q^b + \frac{1}{2} e_3 g_{\mu \nu} \delta_{ab} \partial_\rho q^a \partial^\rho q^b + \frac{1}{4H} g_{\mu \nu} V(q)=0

Then I take the trace of the resulting E-L eqn to see -\frac{5}{4} e_3 R_3 + \frac{5}{2} \delta_{ab} \partial_\rho q^a \partial^\rho q^b + \frac{3}{4H} e_3 V(q)=0 Rearranging we find -\frac{1}{4} e_3 g_{\mu \nu}R_3 = -\frac{1}{2} e_3 \delta_{ab} g_{\mu \nu} \partial_\rho q^a \partial^\rho q^b -\frac{3}{20H} e_{3} g_{\mu \nu} V(q) If I substitute this into the original E-L eqn to try and get rid of the -\frac{1}{4} e_3 g_{\mu \nu} R term, all the unwanted terms cancel but I have the wrong factor in front of the potential term. Any ideas? Thanks.

 

[nrqed]

I have a 3d system with Lagrangian e_3^{-1} L_3 = -\frac{1}{2} R_3 + \delta_{ab} \partial_\rho q^a \partial^\rho q^b + \frac{1}{2H} V(q)

From this I want to calculate the Einstein equation by performing the Euler-Lagrange procedure. First of all, I move the 3d dreibein to the RHS and then I apply the E-L eqns. Using that \frac{\partial e_3}{\partial g^{\mu \nu}} = \frac{1}{2} e_3 g_{\mu \nu}, I see that

\frac{\partial}{\partial g^{\mu \nu}} (e_3 R_3) = e_3 R_{\mu \nu} + \frac{1}{2} e_3 R

This can't be right, the indices do not match.

Have you answered your question?