We will explain Einstein’s general theory of relativity, from which we begin to explore the origin of our universe — a touch of the modern cosmology.
copyright © jxu@ustb.edu.cn
The core of general relativity (GR) lies in the following Einstein’s field equation (EFE)
\begin{eqnarray}
G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu},
\end{eqnarray}
where $T_{\mu\nu}=\rho v_\mu v_\nu$ is the stress-energy tensor, with $G$ and $\Lambda$ the gravitational and cosmological constants, respectively. The metric tensor $g_{\mu\nu}$ characterizes the space-time geometry, through the following infinitesimal interval (or line element)
\begin{eqnarray}
ds^2=g_{\mu\nu}dx^\mu dx^\nu.
\end{eqnarray}
The Einstein tensor $G_{\mu\nu}$ is defined as
\begin{eqnarray}
G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}
\end{eqnarray}
with the Ricci tensor
\begin{eqnarray}
R_{\mu\nu}=\partial_\lambda\Gamma_{\mu\nu}^\lambda-\partial_\nu\Gamma_{\mu\lambda}^\lambda+\Gamma_{\lambda\sigma}^\lambda\Gamma_{\mu\nu}^\sigma-\Gamma_{\mu\sigma}^\lambda\Gamma_{\lambda\nu}^\sigma,
\end{eqnarray}
and the Ricci scalar $R=g^{\mu\nu}R_{\mu\nu}$. Here $\Gamma_{\mu\nu}^\lambda$ is the Christoffel symbol that is connected to the metric tensor as
\begin{eqnarray}
\Gamma_{\mu\nu}^\lambda=\frac{1}{2}g^{\lambda\sigma}\left(\partial_\mu g_{\nu\sigma}+\partial_\nu g_{\mu\sigma}-\partial_\sigma g_{\mu\nu}\right).
\end{eqnarray}
Basically, the EFE states that the mass object (or equivalently the energy) determines spacetime curvature, which in turn, guides the mass object according to the following geodesic equation
\begin{eqnarray}
\frac{\partial^2 x^{\beta}}{\partial t^2}+\Gamma_{\mu\nu}^\beta\frac{\partial x^{\mu}}{\partial t}\frac{\partial x^{\nu}}{\partial t}=0.
\end{eqnarray}
This creates exactly the gravitational force. Thus, Einstein’s genius idea provides a geometrical interpretation of Newton’s theory of gravity.
As you might eventually realize, this seemingly simple EFE is extremely difficult to solve. One needs to deal with not a single, but TEN coupled non-linear differential equations! So far, we have only several exact solutions to EFE. The simplest nontrivial example might be the following Schwarzschild solution with the line element
\begin{eqnarray}
ds^2=-\left(1-\frac{2GM}{c^2r}\right)c^2dt^2+\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2+r^2d\Omega^2,
\end{eqnarray}
and the metric tensor
\begin{eqnarray}
g_{\mu\nu} =
\begin{pmatrix}
-\left(1-\frac{2GM}{c^2r}\right) & 0 & 0 & 0 \\
0 & \left(1-\frac{2GM}{c^2r}\right)^{-1} & 0 & 0 \\
0 & 0 & r^2 & 0 \\
0 & 0 & 0 & r^2\sin^2\theta
\end{pmatrix}.
\end{eqnarray}
This solution has a singularity at the Schwarzschild radius $r_s=2GM/c^2$, where not even the light can escape from it. It is just the event horizon of a Schwarzschild black hole.
Now let’s explore a little bit about modern cosmology. I will show you the Friedmann–Lemaitre–Robertson–Walker solution with
\begin{eqnarray}
ds^2=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2d\Omega^2\right),
\end{eqnarray}
where $a>0$ ($a<0$) for an expanding (contracting) universe, and $k>0$ ($k<0$) for a closed (open) universe with positive (negative) space curvature. Then we have the metric tensor
\begin{eqnarray}
g_{\mu\nu} =
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & \frac{a^2}{1-kr^2} & 0 & 0 \\
0 & 0 & a^2r^2 & 0 \\
0 & 0 & 0 & a^2r^2\sin^2\theta
\end{pmatrix},
\end{eqnarray}
with $dx^\mu=(cdt,dr,d\theta,d\phi)$. Since the $g_{\mu\nu}$ here is diagonal, the Christoffel symbol becomes
\begin{eqnarray}
\Gamma_{\mu\nu}^\lambda=\frac{1}{2}g^{\lambda\lambda}\left(\partial_\mu g_{\nu\lambda}+\partial_\nu g_{\mu\lambda}-\partial_\lambda g_{\mu\nu}\right),
\end{eqnarray}
and for a nonzero element, at least two of the three indices $\lambda, \mu, \nu$ are equal. So we find the following Christoffel symbols
\begin{eqnarray}
\Gamma_{\lambda\lambda}^\lambda=\frac{1}{2}g^{\lambda\lambda}\partial_\lambda g_{\lambda\lambda}&\Rightarrow &\Gamma_{00}^0=\Gamma_{22}^2=\Gamma_{33}^3=0, \ \Gamma_{11}^1=\frac{kr}{a^2}g_{11},\\
\Gamma_{\mu\mu}^\lambda=-\frac{1}{2}g^{\lambda\lambda}\partial_\lambda g_{\mu\mu}&\Rightarrow & \Gamma_{ii}^0=\frac{\dot{a}}{ca}g_{ii},\ \Gamma_{00}^1=0, \ \Gamma_{22}^1=\Gamma_{33}^1=-\frac{1-kr^2}{a^2r}g_{22/33},\\
&&\Gamma_{00}^2=\Gamma_{11}^2=0,\ \Gamma_{33}^2=-\sin\theta\cos\theta,\ \Gamma_{\mu\mu}^3=0,\\
\Gamma_{\mu\lambda}^\lambda=\Gamma_{\lambda\mu}^\lambda=\frac{1}{2}g^{\lambda\lambda}\partial_\mu g_{\lambda\lambda}&\Rightarrow & \Gamma_{0i}^i=\frac{\dot{a}}{ca},\ \Gamma_{10}^0=0, \ \Gamma_{12}^2=\Gamma_{13}^3=\frac{1}{r},\\
&&\Gamma_{20}^0=\Gamma_{21}^1=0,\ \Gamma_{23}^3=\frac{\cos\theta}{\sin\theta},\ \Gamma_{3\lambda}^\lambda=0.
\end{eqnarray}
Considering an uniform isotropic universe, the stress-energy tensor $T_{\mu\nu}$ must be diagonal, then according to the EFE, the nonzero components of the Ricci tensor are
\begin{eqnarray}
R_{ii}=\partial_\lambda\Gamma_{ii}^\lambda-\partial_i\Gamma_{i\lambda}^\lambda+\Gamma_{\lambda\sigma}^\lambda\Gamma_{ii}^\sigma-\Gamma_{i\sigma}^\lambda\Gamma_{\lambda i}^\sigma,
\end{eqnarray}
which gives
\begin{eqnarray}
R_{00}&=&-\partial_0\frac{\dot{a}}{ca}\times 3-\left(\frac{\dot{a}}{ca}\right)^2\times 3=-3\frac{\ddot{a}}{c^2a},\\
R_{11}&=&\frac{1}{c^2}\partial_t\left(\frac{\dot{a}a}{1-kr^2}\right)+\frac{2}{r}+\frac{3}{c^2}\frac{\dot{a}^2}{1-kr^2}-\frac{2}{r^2}-\frac{2}{c^2}\frac{\dot{a}^2}{1-kr^2}+\frac{2k}{1-kr^2}\\
&=&\frac{1}{1-kr^2}\left(2\frac{\dot{a}^2}{c^2}+\frac{\ddot{a}a}{c^2}+2k\right), \\
R_{22}&=&\frac{1}{c^2}\partial_t\left(\dot{a}ar^2\right)+\partial_r\left(-r+kr^3\right)+1-kr^2+\frac{1}{c^2}\dot{a}^2r^2\\
&=&r^2\left(2\frac{\dot{a}^2}{c^2}+\frac{\ddot{a}a}{c^2}+2k \right)\nonumber\\
R_{33}&=&\frac{1}{c^2}\partial_t\left(\frac{\dot{a}}{a}a^2r^2\sin^2\theta\right)-\partial_r\left(\frac{1-kr^2}{a^2r}a^2r^2\sin^2\theta\right)+\frac{1}{c^2}\dot{a}^2r^2\sin^2\theta+\sin^2\theta-k^2r^2\sin^2\theta\nonumber\\
&=&r^2\sin^2\theta\left(2\frac{\dot{a}^2}{c^2}+\frac{\ddot{a}a}{c^2}+2k\right).
\end{eqnarray}
Thus we find the curvature scalar
\begin{eqnarray}
R&=&g^{\mu\nu}R_{\mu\nu}\nonumber\\
&=&3\frac{\ddot{a}}{c^2a}+\frac{1}{a^2}\left(2\frac{\dot{a}^2}{c^2}+\frac{\ddot{a}a}{c^2}+2k\right)\times3\nonumber\\
&=&\frac{1}{c^2a^2}\left(6\ddot{a}a+6\dot{a}^2+6kc^2\right)\nonumber\\
&=&\frac{6}{c^2}\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^2}{a^2}+\frac{kc^2}{a^2}\right).
\end{eqnarray}
By definition, the first component of the Einstein tensor
\begin{eqnarray}
G_{00}&=&R_{00}-\frac{1}{2}Rg_{00}\nonumber\\
&=&-\frac{3}{c^2}\frac{\ddot{a}}{a}+\frac{1}{2}\frac{6}{c^2}\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^2}{a^2}+\frac{kc^2}{a^2}\right)\nonumber\\
&=&\frac{3}{c^2}\left(\frac{\dot{a}^2}{a^2}+\frac{kc^2}{a^2}\right).
\end{eqnarray}
Then from the EFE, we have
\begin{eqnarray}
G_{00}+\Lambda g_{00}=\frac{3}{c^2}\left(\frac{\dot{a}^2}{a^2}+\frac{kc^2}{a^2}\right)-\Lambda=\frac{8\pi G}{c^4}T_{00}=\frac{8\pi G}{c^4}\rho c^2,
\end{eqnarray}
Thus, we arrive the Friedmann equation
\begin{eqnarray}
\left(\frac{\dot{a}}{a}\right)^2+\frac{kc^2}{a^2}=\frac{8\pi G}{3}\rho+\frac{\Lambda c^2}{3}.
\end{eqnarray}
For an almost flat (what we have observed so far) universe without cosmological constant $k=\Lambda=0$, and at the early stage with radiation dominated $\rho\sim\rho_{rad}\sim a^{-4}$ (c.f., $\rho_{matter}\sim a^{-3}, \rho_\Lambda=const$) , we have
\begin{eqnarray}
\dot{a}\propto\frac{1}{a}\Rightarrow a\propto\sqrt{t}.
\end{eqnarray}
This predicts an expanding universe with a Big Bang singularity at $t=0$. At the Big Bang, our universe holds infinite dense of matters with $\rho\to\infty$ at just a single point!