# Lagrangian in reduced Horndeski Theory for i=2

I am trying to understand the calculations of the latest Charles Dalang's paper "Scalar and Tensor Gravitational Waves", arXiv:2009.11827.

Since I just learned basic general relativity, I found it hard to prove equation (13) in that paper.

Here is my calculation for the variation of scalar field $$\phi$$ for $${\Large\varepsilon}_{\phi}^{(2)}$$:

$$\delta S_2 = \int d^4x \delta(\sqrt{-g}G_2(\phi, X)) = \int d^4x \sqrt{-g} \delta(G_2(\phi, X))$$ where $$X = -\Large\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi$$. I think this can be solved by using Euler-Lagrange:

$$\frac{\partial G_2}{\partial \phi}=G_{2,\phi}$$

$$\nabla_{\mu} ({\frac{\partial G_2}{\partial(\nabla_{\mu} \phi)}})= \nabla_{\mu}(\frac{\partial G_2}{\partial X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)})= \nabla_{\mu}(G_{2,X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)})= ???$$ I can't solve this part.

Here the solution for $${\Large\varepsilon}_{\phi}^{(2)}$$ given in that paper (Eq. 13):

$${\Large\varepsilon}_{\phi}^{(2)} = G_{2,\phi} + G_{2,X}\square\phi - 2XG_{2,X\phi} + G_{2,XX}\phi^{,\mu}X_{,\mu}.$$

Actually I'm interested in finding $$\Large{\varepsilon}_{\phi}^{(2)}$$ and $$\Large{\varepsilon}_{\mu\nu}^{(4)}$$. I'd appreciate it so much if you give an explicit answer and details in the calculation to prove the equation.

• What is $i=2$ in title? Oct 18 '20 at 14:03
• the indeces of lagrange. actually, the form of action is $S_i = \int d^4x \sum_{i=2}^4 \mathcal{L}_i$ Oct 19 '20 at 6:57

$$\nabla_{\mu} \left({\frac{\partial G_2}{\partial(\nabla_{\mu} \phi)}}\right)= \nabla_{\mu}\left(\frac{\partial G_2}{\partial X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)}\right)= \nabla_{\mu}\left(G_{2,X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)}\right)= ?$$
Since in the bracket is not a tensor, we may change $$\nabla_\mu$$ with $$\partial_\mu$$, so then we have
$$\nabla_{\mu}\left(G_{2,X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)}\right)\longrightarrow \partial_{\mu}\left( G_{2,X}\frac{\partial X}{\partial(\partial_{\mu}\phi)} \right)$$ Now we can easily calculate $$\partial_\mu$$ term $$\partial_{\mu}\left( G_{2,X}\frac{\partial X}{\partial(\partial_{\mu}\phi)} \right) = (\partial_{\mu}G_{2,X})\frac{\partial X}{\partial(\partial_{\mu}\phi)} + G_{2,X}\partial_{\mu}\left(\frac{\partial X}{\partial(\partial_{\mu}\phi)} \right)$$
$$= \left(\frac{\partial G_{2,X}}{\partial \phi} \frac{\partial \phi}{\partial x^{\mu}} +\frac{\partial G_{2,X}}{\partial X} \frac{\partial X}{\partial x^{\mu}} \right)(-g^{\mu\nu}\partial_\nu \phi) + G_{2,X}\partial_\mu(-g^{\mu\nu}\partial_\nu \phi)$$
where $$\Large\frac{\partial X}{\partial(\partial_{\mu}\phi)}$$ $$= -g^{\mu\nu}\partial_\nu \phi$$
$$\left(\frac{\partial G_{2,X}}{\partial \phi} \frac{\partial \phi}{\partial x^{\mu}} +\frac{\partial G_{2,X}}{\partial X} \frac{\partial X}{\partial x^{\mu}} \right)(-g^{\mu\nu}\partial_\nu \phi) + G_{2,X}\partial_\mu(-g^{\mu\nu}\partial_\nu \phi) = G_{2,X\phi} \partial_\mu \phi (-g^{\mu\nu}\partial_\nu \phi) + G_{2,XX}X_{,\mu}(-g^{\mu\nu}\partial_\nu \phi) - G_{2,X}\square \phi$$ $$= 2XG_{2,X\phi} - G_{2,XX} \phi^{,\mu}X_{,\mu}-G_{2,X}\square \phi$$