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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.

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  • $\begingroup$ I don't understand what you're doing, but $T$ is the Noether current given by translational symmetry in the Lagrangian. $\endgroup$ – user21433 Aug 22 '14 at 20:50
  • $\begingroup$ Seems to me that to calculate $T$ you should take the variation of the matter part of the Lagrangian (two last terms) with respect to the metric. Follow the usual procedure from the textbooks. $\endgroup$ – firtree Aug 23 '14 at 0:30
  • $\begingroup$ $T_{\mu\nu}$, of course. $\endgroup$ – firtree Aug 23 '14 at 0:40
  • $\begingroup$ 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 [tex]-\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[/tex] $\endgroup$ – user11128 Aug 23 '14 at 10:18
  • $\begingroup$ 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 [tex]-\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[/tex] Now when I vary with respect to [tex]g^{\mu \nu}[/tex] I get [tex]\frac{\partial L}{\partial g^{\mu \nu}} = - \frac{1}{2} e_3R_{\mu \nu} - 1/4 e_3 R_3 + e_3 \delta_{ab} \partial_\mu q^a \partial_\nu q^b + 1/2 e_3 g_{\mu \nu} \delta_{ab} \partial_\rho q^a \partial^\ $\endgroup$ – user11128 Aug 23 '14 at 10:28

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