Some note on GR

1. In Cartesian coordinates, the Lagrangian density for a scalar field is given by:

$$\mathcal{L}(\phi, \partial_{\mu} \phi)=-\frac{1}{2} \delta^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi-4 \pi G \chi{(1+\phi)}$$

$\quad$where $\phi$ is a function of coordinates, and $\chi$ is the Newtonian constant.
$\quad$(a) Using the principle of stationary action , $\delta S=0$,derive the equation of motion for the scalar field from the action .
$\quad$(b) Based on the tilted gravity theory, if the scalar field represents the Newtonian potential, explain the physical meaning of $\chi$.

Answer:
（a）$S(\phi)=\int_{\Omega} \mathcal{L}(\phi, \partial_{\mu} \phi) d V$
Use the principle : $\delta S = 0$,thus,

$$\delta S =\int \delta \mathcal{L} \, dV = 0$$

Our goal is to make the form as :$\int[\quad]\delta \phi dV$,so it can easily think about this little math trick:
(notice that $\delta(\partial_{\mu}\phi) = \partial_{\mu}(\delta \phi)$)

$$\begin{aligned} \delta \mathcal{L} &= \frac{\partial \mathcal{L}}{\partial \phi}\delta \mathcal{\phi}+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\delta \mathcal{(\partial_{\mu}\phi)}\\\ &=\frac{\partial \mathcal{L}}{\partial \phi}\delta \mathcal{\phi}+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\partial_{\mu}\mathcal{(\delta\phi)}\\\ &=\frac{\partial \mathcal{L}}{\partial \phi}\delta \mathcal{\phi}+\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\delta\phi\right)-\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\right)\delta\phi \end{aligned}$$

We always require the boundary values to be zero,so :

$$\int_{\Omega}\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\delta\phi\right)dV \equiv 0$$

Here , the form of the principle of least action becomes:

$$\delta S = \int_{\Omega} \left(\frac{\partial \mathcal{L}}{\partial \phi}-\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\right)\right)\delta\phi dV$$

So we can derive the Euler-Lagrange equation as:

$$\frac{\partial \mathcal{L}}{\partial \phi}-\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\right)=0$$

Substituting the specific form of the Lagrangian density:

$$\begin{aligned} & \begin{array}{l} \therefore\left\{\begin{array}{l} \frac{\partial \mathcal{L}}{\partial \phi}=-4 \pi G \chi \\ \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}=-\partial^\mu \phi \end{array}\right. \end{array} \\ \end{aligned}$$

Thus:

$$-4\pi G \chi - \partial^\mu (-\partial_\mu \phi)= 0 \\\ \rightarrow \nabla^2 \phi = 4\pi G \rho$$

(b) According to Newton’s law of gravity, $\nabla^2 \phi = 4\pi G \rho$, which implies $\chi = \rho$.

$\rho$ is the source of the gravitational potential. In this case, the influence of $\chi$ on any scalar field $\phi$ can be summarized as follows:

1. $\chi$ serves as the source of the gravitational potential.
2. The distribution of $\chi$ directly impacts the distribution of $\phi$, with $\phi$ being interpretable as an excitation of $\chi$.

To be continued