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\begin{document}

\title{Phys 131 Study Guide}


\author{Alec Aivazis}
\affil{Affiliation not available}



\date{\today}

\maketitle




\section{Special Relativity}

\subsection{Lorenz Boosts}

For a boost in the x direction, 

\begin{equation}
\begin{split}
t' = \gamma ( t - v x) \\
x' = \gamma (x - vt)  \\
y' = y \\
z' = z \\
\end{split}
\end{equation}

\subsection{Fun with 4 Vectors}

\begin{equation}
\begin{split}
\mathbf{u} = (\gamma, \gamma v) \\
\mathbf{u} \cdot \mathbf{u} = -1 \\
\mathbf{p_{\gamma}} \cdot \mathbf{p_{\gamma}} = 0  \\
E_{obs} = - p \cdot \mathbf{u} \\
\mathbf{p}^2 = m^2 \\
E^2 = p^2 + m ^2 \\
\end{split}
\end{equation}


\subsection{More special relativity}

\begin{equation}
\begin{split}
l' =  \frac{l}{\gamma} \\
\Delta x' = \gamma (\Delta x - v \Delta t) \\
\Delta t' = \gamma (\Delta t - x \Delta x) \\
u' = \frac{u - v}{1- uv}
\end{split}
\end{equation}

where 

\begin{equation}
\gamma = \frac{1}{\sqrt{1-v^2}}
\end{equation}

\subsection{Principle of Equivalence}

elevators in a feely falling elevator on earth experience the same physics as someone who is not in the elevator. From this you can get the bending of light

weight in elevator $ = m(g+a) $ so if $a = -g$ then $w = 0$ so gravity is not a real force and can be seen as a curvature in spacetime



\section{Spacetime Geometry}

\begin{equation}
\begin{split}
ds^2 = g_{\alpha \beta} \frac{d}{d x}^{\alpha} \frac{d}{d x}^{\beta}\\
\end{split}
\end{equation}

If the time component is bigger than the space component than its time like. If the space component is bigger than the time than its spacelike. If they're equal than its null.

\subsection{Gravitational Redshift}

In a static universe, t is ignorable so there is a killing vector $\xi = (1,0,0,0)$. $\xi \cdot \mathbf{u}$ is conserved along the trajectory of the particle. Therefore, 

\begin{equation}
\begin{split}
\frac{E_{\rm{obs}}}{E_e} & = \frac{g_{tt}(e)}{g_{tt}(obs)} \\
\frac{\omega_{\inf}}{\omega_{e}} & = \sqrt{ \left( 1-\frac{2 M}{r_e} \right)}
\end{split}
\end{equation}

to compare to positions, find the shift with respect to infinity to 1 and then 2 and then take the ratio to find $\frac{\omega_1}{\omega_2}$:


\begin{equation}
\frac{\omega_1}{\omega_2} = \sqrt{\frac{1-\frac{2M}{r_2}}{1-\frac{2M}{r_1}}}
\end{equation}

\subsection{Schwarzchild Metric}

\begin{equation}
ds^2 = -(1-\frac{2GM}{r}) dt^2 + (1-\frac{2GM}{r})^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta d \phi^2)
\end{equation}

\section{Orbits of a Photon}
From the schwarzchild metric,

\begin{equation}
\begin{split}
e = -\xi \cdot u &= W_{\rm{eff}}(r) \frac{d t}{d \lambda} \\
l = \eta \cdot u &= r^2 \sin^2 \theta \frac{d \phi}{d \lambda}
\end{split}
\end{equation}

are conserved quantities along the path of the photon. 

setting $u \cdot u = 0$ since its a light ray with $\theta = \frac{\pi}{2}$ gives

\begin{equation}
\begin{split}
u \cdot u = g_{\alpha \beta} \frac{d x^\alpha}{d \lambda}\frac{d x^\alpha}{d \lambda} \\
\frac{1}{b^2} = \frac{1}{l^2}\left( \frac{dr}{\lambda}\right)^2 + W_{\rm{eff}}(r)
\end{split}
\end{equation}

where

\begin{equation}
\begin{split}
b^2 &= \frac{l^2}{e^2} \\
W_{\rm{eff}}(r) &= \frac{1}{r^2}(1-\frac{2M}{r})
\end{split}
\end{equation}

Another useful equation:

\begin{equation}
\delta \phi_{\rm{def}} = \frac{4 G M }{c^2 b}
\end{equation}

where b is the impact parameter

\section{Grav Lenses}

Diagram is in figure \ref{fig:gravlense}.

\begin{equation}
\theta D_{S} = \beta D_S + \alpha D_{LS}
\end{equation}

Because $b \approx \xi$ and $\xi \approx \theta D_L$ we can use the small angle approx. Therefore, $\theta = \beta + \frac{\theta_{E}^2}{\theta}$. Where $\theta_E$ is the Einstein Angle:

\begin{equation}
\theta_E = \left[ 2 R_S \frac{D_{LS}}{D_{S}D_{L}} \right] ^\frac{1}{2}
\end{equation}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/grav-lense/grav-lense}
\caption{{\label{fig:gravlense}%
}}
\end{center}
\end{figure}

\section{Past Light Cones}

eddington finkelstein

Make the following transformation to schwarzchild to remove coordinate singularity
$t = v - r-2 M \log \mid \frac{r}{2M} -1 \mid$

result:
\begin{equation}
\mathrm{d}s^2 = -\left(1-\frac{2 G M}{r}\right)\mathrm{d}v^2 + 2\mathrm{d}v\mathrm{d}r + r^2 \mathrm{d}\Omega^2
\end{equation}

where $\mathrm{d}\Omega = \mathrm{d}\theta^2 + sin^2 \theta \mathrm{d} \phi^2$

\subsection{Important Notes}

incoming light rays are at 45 degrees regardless of potion

if you're inside the singularity (r = 2M) then outgoing rays

  curve to the positive side on the outside
  
  curve to the negative side if inside
  
  light cones are at the intersection

\subsection{Gravitational Collapse}

see fig \ref{fig:collapse}. horizon at r = 2 M\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/collapse/collapse}
\caption{{\label{fig:collapse}
The story of a collapsing star sending signals%
}}
\end{center}
\end{figure}

\section{Cosmology}

the flat Robinson-Walker metric:

\begin{equation}
ds^2 = -dt^2 + a^2 (t) (dx^2 + dy^2 + dz^2)
\end{equation}

geodesics are of the form $x(t) = \sqrt{t_0  t}$ 

$\rightarrow$ look like parabolas. Light cones are area in between. Everyone has a common history so we can be causally connected (much bigger than in a flat spacetime).

\subsubsection{More Goodies}

\begin{equation}
\begin{split}
\frac{d x}{d t} &= \frac{1}{a} \implies x(t) = \int \frac{\mathrm{d} t}{a(t)}\\
1 + z &= \frac{\lambda_0}{\lambda} = \frac{\omega_0}{\omega} = \frac{a(t_0 )}{a(t_e )} \\
v &= H_0 d \\
H_0 &= H(t_0) = \frac{\dot{a}(t_0)}{a(t_0)}
\end{split}
\end{equation}

where $H_0$ is Hubble's constant

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