For a boost in the x direction,

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

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

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

where

\[\gamma = \frac{1}{\sqrt{1-v^2}}\]

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

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

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.

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,

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

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

\[\frac{\omega_1}{\omega_2} = \sqrt{\frac{1-\frac{2M}{r_2}}{1-\frac{2M}{r_1}}}\]

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

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