# how to get a linear ricci flow equation??

one could use linear approximation $$g_{ab}=\eta_{ab}+h_{ab}$$ to get linear ricci flow equation (2). How to do that? is any process ? I am studying general relativity , i just use the linear approximation : $$g_{ab}=\eta_{ab}+\gamma_{ab}$$, and calculate the christoffl symbol : $$\Gamma_{a b}^{(1) c}=\frac{1}{2} \eta^{c d}\left(\partial_{a} \gamma_{b d}+\partial_{b} \gamma_{a d}-\partial_{d} \gamma_{a b}\right)$$,finally, get the Ricci tensor:$$R_{a b}^{(1)}=\partial^{c} \partial_{(a} \gamma_{b) c}-\frac{1}{2} \partial^{c} \partial_{c} \gamma_{a b}-\frac{1}{2} \partial_{a} \partial_{b} \gamma$$ but there are two extra term :$$\partial^{c} \partial_{(a} \gamma_{b) c}$$ and $$-\frac{1}{2} \partial_{a} \partial_{b} \gamma$$, which not match the linearization ricci flow equation(2). what is wrong with me?

• You need to calculate the Gateaux derivative of the operator as shown in this Q&A: it is usually not difficult, even if perhaps someone may consider it a bit tedious. If I were more versed in differential geometry, I'd post an answer following the same lines of the one referred. Sep 30, 2019 at 6:48
• i have modified my question, why am I wrong? Oct 4, 2019 at 14:28
• @Daniele Tampieri Oct 10, 2019 at 1:07