# Bei's Study Notes

### General Relativity 05Last updated: 2016-08-14 19:21:06 PDT.

Reference: UCI OpenCourseWare - GR 06 (Playlist)

### Equation of free fall from Lagrangian

Last time we wrote down the equation of free fall:

$$\frac{d^2x^\lambda}{d\tau^2} + \Gamma^\lambda{}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0\\ \\ \text{or}\\ \\ g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} + \Gamma_{\lambda\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0$$

where $\Gamma$ is the Christoffel symbol of the first kind, which can be written as a function of metric tensor:

$$\Gamma_{\lambda\mu\nu} = \frac 12 \left( \frac{\partial g_{\nu\lambda}}{\partial x^\mu} + \frac{\partial g_{\mu\lambda}}{\partial x^\nu} - \frac{\partial g_{\mu\nu}}{\partial x^\lambda} \right)$$

Today we are going to derive the equation of free fall from the variational principal.

Action:

$$A = \int_A^Bd\tau\\$$

the principal is that the equation of motion can be derived from:

$$\delta A = 0$$

By this, it means a particle goes from $A$ to $B$ from a trajectory $x^\mu(p)$ where $p \in [0, 1]$ is an parameter. If we perturb the trajectory with an arbitrary finite variation $x^\mu(p)$ multiply by an infinitesimal factor of $\Delta$, then the action remains the same. We can write it down:

\begin{align*} \delta \int_A^Bd\tau &= \delta \int_A^Bdp \frac{d\tau}{dp}\\ \\ d\tau^2&=-g_{\mu\nu}dx^\mu dx^\nu \\ \delta d\tau&=\delta\sqrt{-g_{\mu\nu}dx^\mu dx^\nu} \\ &=\frac{1}{2d\tau}\delta(-g_{\mu\nu}dx^\mu dx^\nu) \\ &=\frac{1}{2d\tau}( -\delta g_{\mu\nu}dx^\mu dx^\nu -g_{\mu\nu}\left( d\delta x^\mu dx^\nu + dx^\mu d\delta x^\nu \right) ) \\ &=\frac{1}{2d\tau}( -\frac{\partial g_{\mu\nu}}{\partial x^\lambda}\delta x^\lambda dx^\mu dx^\nu -g_{\mu\nu}d\delta x^\mu dx^\nu -g_{\mu\nu}dx^\mu d\delta x^\nu ) \\ \end{align*}

The boundary term is $0$ and integrate by partial:

\begin{align*} -g_{\mu\nu}d\delta x^\mu dx^\nu &\to d(g_{\mu\nu}dx^\nu) \delta x^\mu \\ &= (dg_{\mu\nu}dx^\nu + g_{\mu\nu}d^2x^\nu) \delta x^\mu \\ &= (dg_{\lambda\nu}dx^\nu + g_{\lambda\nu}d^2x^\nu) \delta x^\lambda \\ &= \left( \frac{\partial g_{\lambda\nu}}{\partial x^\mu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right) \delta x^\lambda \\ \\ -g_{\mu\nu}dx^\mu d\delta x^\nu &= \left( \frac{\partial g_{\lambda\mu}}{\partial x^\nu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right) \delta x^\lambda \\ \end{align*}

Plug it in,

\begin{align*} \delta d\tau &\to \frac{1}{2d\tau}\left( -\frac{\partial g_{\mu\nu}}{\partial x^\lambda}dx^\mu dx^\nu + \frac{\partial g_{\lambda\nu}}{\partial x^\mu}dx^\mu dx^\nu + \frac{\partial g_{\lambda\mu}}{\partial x^\nu}dx^\mu dx^\nu + 2g_{\lambda\sigma}d^2x^\sigma \right)\delta x^\lambda \\ &= \frac{1}{d\tau}\left( \Gamma_{\lambda\mu\nu}dx^\mu dx^\nu + g_{\lambda\sigma}d^2x^\sigma \right)\delta x^\lambda \\ &= \left( \Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} \right)\delta x^\lambda \\ \end{align*}

\begin{align*} 0 = \delta \int_A^Bd\tau &= \int_A^Bd\tau \left( \Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} \right)\delta x^\lambda\\ \end{align*}

Since $\delta x^\lambda$ is arbitrary, we get again the equation of free fall:

$$\Gamma_{\lambda\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} + g_{\lambda\sigma}\frac{d^2x^\sigma}{d\tau^2} = 0$$

By definition, this is trajectory of the stationary path (in terms of proper time) in 4 dimentional spacetime. This is the Geodesics. Therefore we replaced gravitational force with the geometry of spacetime.

There has a clear consequence with QM because of path integral:

$$Amp(A\to B)\propto \int d \exp\left({\frac{i}{\hbar} \int d\tau mc^2A}\right)$$

### Newtonian approximation

For slow moving particles:

\begin{align*} \frac{d^2x^\lambda}{d\tau^2} + \Gamma^\lambda{}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}&\approx \frac{d^2x^\lambda}{dt^2}\left(\frac{dt}{d\tau}\right)^2 + \Gamma^\lambda{}_{00}\left(\frac{dt}{d\tau}\right)^2\\ \frac{d^2x^\lambda}{dt^2} + \Gamma^\lambda{}_{00} &\approx 0 \end{align*}

For weak, stationary fields:

\begin{align*} g_{\alpha\beta} &=\eta_{\alpha\beta}+h_{\alpha\beta},\quad\quad|h_{\alpha\beta}| \ll 1\,,\dot h = 0\\ g^{\alpha\gamma} &=\eta^{\alpha\gamma}+h'^{\alpha\gamma} \\ g^{\alpha\gamma}g_{\alpha\beta}&= \delta^\gamma{}_\beta+\eta^{\alpha\gamma}h_{\alpha\beta}+\eta_{\alpha\beta}h'^{\alpha\gamma}+o(h^2)\\ 0 &= \eta^{\alpha\gamma}h_{\alpha\beta}+\eta_{\alpha\beta}h'^{\alpha\gamma}\\ &=h^\gamma{}_{\beta}+h'^\gamma{}_{\beta}\\ g^{\alpha\gamma} &= \eta^{\alpha\beta}-h^{\alpha\beta}\\ \\ \Gamma^{\lambda}{}_{00}&\approx\frac{1}{2}\eta^{\sigma\lambda}( \partial_0 g_{\sigma 0} + \partial_0 g_{\sigma 0} - \partial_\sigma g_{0 0} )\\ &=\frac{1}{2}\eta^{\sigma\lambda}( \partial_0 h_{\sigma 0} + \partial_0 h_{\sigma 0} - \partial_\sigma h_{0 0} )\\ &=-\frac{1}{2}\eta^{\sigma\lambda}\partial_\sigma h_{0 0}\\ &=-\frac{1}{2}(0, \vec\nabla h_{0 0})^\lambda\\ \end{align*}

If both conditions are met:

$$\vec{a}= \frac{1}{2}\vec\nabla h_{0 0}$$

This looks a lot like the the Newtonian gravitation:

$$\vec{a} = -\vec\nabla \phi$$

where $\phi$ is the gravitational potential (which is always negative),

$$h_{00} = -2\phi + const$$

At the boundary, the gravitation field is $0$, the $h_{00}$ should also be $0$, therefore the constant should be $0$ as well.

$$h_{00} = -2\phi$$

\begin{align*} \phi &= -\frac{mG}{|r|}\\ g_{00} &= -1-2\phi\\ &= \frac{2mG}{|r|}-1 \end{align*}

Remark: 1. $g_{00}$ goes to $0$ at the event horizon. 2. The boundary gravitational field may not be $0$.

In the case of stronger gravitational field $\phi$ (which is dimensionless):

1. $10^{-4}$ at the surface of a white dwarf;
2. $10^{-6}$ at the surface of the sun;
3. $10^{-9}$ at the surface of the earth;
4. $10^{-39}$ at the surface of an proton;

--

### Time Dilation

$$d\tau = \sqrt{-\eta_{\alpha\beta} d\xi^\alpha d\xi^\beta} = \sqrt{-g_{\alpha\beta} dx^\alpha dx^\beta}$$

We can caluculate $dt/d\tau$ which is the factor of time dilation:

\begin{align*} \frac{dt}{d\tau}&=\left(-g_{\alpha\beta} \frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}\right)^{-\frac 12}\\ \frac{dx^\mu}{dt} &= (1, \vec v)\\ \end{align*}

If the $\vec v=0$:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{-g_{00}}}$$

comparing two clocks:

$$\frac{dt_1}{dt_2} = \frac{\sqrt{-g_{00}(x_2)}}{\sqrt{-g_{00}(x_1)}}$$

The frequency of the two stationary clock:

$$\frac{dv_2}{dv_1} = \frac{\sqrt{-g_{00}(x_2)}}{\sqrt{-g_{00}(x_1)}}$$

In the previous conditions:

\begin{align*} \frac{dv_2}{dv_1} &\approx 1 + \phi(x_2) - \phi(x_1) \\ &= 1 + MG \frac{|r_1| - |r_2|}{|r_2r_1|} \\ \end{align*}