# Bei's Study Notes

### General Relativity 10Last updated: 2016-08-25 18:41:25 PDT.

Reference: UCI OpenCourseWare - GR 11 (Playlist)

## Einstein's field equation

$$g_{00} = -(1+2\phi(x)) \quad(\text{Weak fields})\\$$

$$\Delta\phi = 4\pi G\,\rho\\ \Delta g_{00} = -8\pi G\,\rho\\$$

$$T_{00}=\rho$$

therefore

$$\Delta g_{00} = -8\pi G T_{00}$$

An extension might be:

$$G_{\alpha\beta}=-8\pi G T_{\alpha\beta}$$

where $G_{\alpha\beta}$ is an unknown object that relates to gravitational field.

Promotes to G.C.:

$$G_{\mu\nu}=-8\pi G T_{\mu\nu}$$

To determine $G_{\mu\nu}$, we write down the properties of $G_{\mu\nu}$:

1. $G_{\mu\nu}$ is a tensor.
2. $G_{\mu\nu}$ is made out of at most second derivatives of the metric tensor, e.g. $\left(\frac{\partial g}{\partial x}\right)^2$, or $\frac{\partial^2 g}{\partial x^2}$.
3. $G_{\mu\nu}=G_{\nu\mu}$.
4. $G^{\mu\nu}{}_{;\nu} = 0$
5. On non-relativistic limit, $G_{00}\to\Delta g_{00}$.

These requirements uniquely determines the form of $G_{\mu\nu}$.

$$G_{\mu\nu} = C_1 R_{\mu\nu}+ C_2 g_{\mu\nu} R$$

$$G^{\mu}{}_{\nu;\mu}=\left(\frac{1}{2}C_1+C_2\right)R_{;\nu}$$

therefore, $G_{\mu\nu}$ has the form $C_1(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)$. Take the non-relativistic limit, that is, $u^\mu \approx 0$ and the field is weak ($g_{\mu\nu} = \eta_{\mu\nu}+\delta h_{\mu\nu}$), and so is the affine connection and the Riemann tensor. Also, $R_{ij} \approx \frac{1}{2}g_{ij}R$.

$$R\equiv g^{\mu\nu} R_{\mu\nu}\approx R_{kk}-R_{00}=\frac{3}{2}R - R_{00}\\ R \approx 2R_{00}$$

$$G_{00} = C_1\left(R_{00}-\frac{1}{2}\eta_{00}R\right)=2C_1R_{00}\\ R_{\mu\nu}\approx \eta^{\lambda\sigma}R_{\lambda\mu\sigma\nu}\\ R_{00}\approx \eta^{\lambda\sigma}R_{\lambda 0\sigma 0}\\ R_{\lambda\mu\nu\kappa} = \frac{1}{2}\left[ \left(\frac{\partial^2g_{\lambda\nu}}{\partial x^\kappa\partial x^\nu}\right) _{[\lambda,\mu]} \right]_{[\nu,\kappa]} + o(\Gamma^2)\\ R_{0000} \approx 0\\ R_{i0j0} \approx \frac{1}{2}\frac{\partial^2g_{00}}{\partial x^i\partial x^j}\\ G_{00} \approx 2C_1(\frac{1}{2}g_{00}) = C_1 \Delta g_{00} \approx -8\pi G \rho\\ C_1 = 1$$

Conclusion (Einstein's field equation):

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=-8\pi G T_{\mu\nu}$$

Equivalent form:

\begin{align*} g^{\mu\nu}R_{\mu\nu}-\frac{1}{2}g^{\mu\nu}g_{\mu\nu}R&=-8\pi G T_{\mu\nu} g^{\mu\nu}\\ R-2R&=-8\pi G T^{\nu}{}_{\nu}\\ R&=8\pi G T^{\nu}{}_{\nu}\\ R_{\mu\nu}&=-8\pi G \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T^\lambda{}_\lambda\right)\\ \end{align*}

In vacuum, $T=0$, therefore $R_{\mu\nu}=0$, this is how gravitational waves are calculated, which does not imply Riemann tensor is $0$. In fact, even when $R_{\mu\nu}$ is $0$, $g_{\mu\nu}$ can still oscillate through spacetime.

(then a discussion about the cosmological constant).