# Einstein Equation

1. Aug 22, 2014

### 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.

2. Aug 23, 2014

### 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.

3. Aug 30, 2014

### nrqed

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