Bei's Study Notes

General Relativity 01
Last updated: 2017-05-01 22:09:22 PDT.

Reference: UCI OpenCourseWare - GR 02 (Playlist)

Reminders of Special Relativity

Special Relativity is a thoery of translating coordinates in planar spacetime, which has a constant measure \eta_{\alpha\beta} .

Lorentz Transformation

Any two inertial reference frames can be related by a Lorentz Transformation (or Poincaré transformation). Consider a photon is shot from a point O , then we can move the origin of the two frame to that point. Then we can rotate the frames so that their relative speed lays on the x axis. The location of the photon in these two frames can be defined as x, y, z, t and x', y', z', t' , then the photon's motion forms a straight line in each frames of reference. Since the speed of light is constant, we have these relations:

x^2 + y^2 + z^2 = c^2t^2 \\ x'^2 + y'^2 + z'^2 = c^2t'^2 \\

In Newtonian transformion, frame of references are related by Galilean transformation:

x' = x - vt \\ y' = y \\ z' = z \\ t' = t \\

Thus we have the notion of "absolute" time flow. This does not satisfy \text{(LInv)} . Lorentz transformation multiplies the time coordinate by a factor \gamma , which depends on |\mathbf v| :

x' = \gamma (x - vt) \\ y' = y \\ z' = z \\ t' = \gamma (t - vx/c^2) \\

Pluggin it into \text{(LInv)} :

\begin{align*} \gamma^2 x ^2 - 2\gamma^2 xvt + \gamma^2v^2t^2+y^2+z^2&=c^2\gamma^2t^2-2\gamma^2tvx+\gamma^2v^2x^2/c^2 \\ \gamma^2 x^2 - \gamma^2 x^2 + \gamma^2v^2t^2+c^2t^2&=c^2\gamma^2t^2 + \gamma^2v^2x^2/c^2 \\ \gamma^2 (x^2 + v^2t^2 - c^2t^2 - v^2x^2/c^2) &= x^2 - c^2t^2 \\ \gamma^2 (x^2 - c^2t^2) (1 - v^2/c^2) &= x^2 - c^2t^2 \\ \gamma^2 (1 - v^2/c^2) &= 1 \\ \gamma &= \frac{1}{\sqrt{1 - v^2/c^2}} \\ \end{align*}


In general, the Lorentz transformation \Lambda^\alpha{}_\beta relates two coordinates in different reference frames by

x'^\alpha = \Lambda^\alpha{}_\beta x^\beta\,.

\Lambda has a important property:

\Lambda^\alpha{}_\gamma \Lambda^\beta{}_\delta \eta_{\alpha\beta}= \eta_{\gamma\delta}

where \eta_{00} = -1, \eta_{0i} = \eta_{i0} = 0, \eta_{ij} = \delta_{ij} . It is similar to rotation in Euclidean space. It is usually seen as rotation in Minkovsky space.


From now on, we set c=1 .

Define proper time \tau that

d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2 = - \eta_{\alpha\beta} dx^\alpha dx^\beta \\

Also,

dx^\beta = \Lambda^\beta{}_\alpha dx^\alpha

, so the proper time in another frame is

\begin{align*} d\tau'^2 &= -\eta_{\alpha\beta} dx'^\alpha dx'^\beta \\ &= -\eta_{\alpha\beta} \Lambda^\alpha{}_{\gamma} dx^\gamma \Lambda^\beta{}_{\delta} dx^\delta \\ &= -\eta_{\gamma\delta} dx^\gamma dx^\delta \\ &= d\tau^2 \\ \end{align*}


General form of boost-only Lorentz transformations:

\begin{align*} \Lambda^0{}_0 &= \gamma \\ \Lambda^i{}_0 &= -\gamma v^i \\ \Lambda^0{}_j &= -\gamma v^j \\ \Lambda^i{}_j &= \delta_{ij} + (\gamma - 1)\frac{v_iv_j}{|\mathbf v|^2} \\ \end{align*}

If we define v^0 = \gamma v / (1-v) , we then have a unified form:

\begin{align*} \Lambda^\alpha{}_\beta &= \delta^{\alpha}{}_{\beta} - v^{\alpha}v^{\varepsilon}\eta_{\varepsilon\beta}\frac{1}{1+\gamma} \end{align*}

We can check if this obeys the identity:

\begin{align*} \Lambda^{\alpha}{}_{\gamma}\Lambda^{\beta}{}_{\delta} &= & \\ \end{align*}


Some 4-vectors

4-force F^\alpha

Define 4-force F^\alpha in that

  1. If the particle is momentarily at rest, then dt = d\tau , then the 4-force is

    f = (0, \mathbf F)

  2. Then apply Lorentz boost

\begin{align*} f'^\alpha &= \Lambda^\alpha{}_\beta f^\beta \\ f'^0 &= \Lambda^0{}_j f^j \\ &= \gamma \mathbf v \cdot F \\ f'^i &= \Lambda^i{}_j f^j \\ &= (\delta^i_{j} + v^iv_j\frac{\gamma-1}{v^2})f^j \\ \mathbf F' &= \mathbf F + \mathbf v\frac{\gamma-1}{v^2} (\mathbf v \cdot \mathbf F) \\ \end{align*}


4-velocity u^\alpha

u^\alpha = \frac{dx^\alpha}{d\tau} = (\gamma, \gamma \mathbf v) \\ u_\alpha u^\alpha = -\gamma^2 + \gamma^2 v^2 = -1


4-momentum p^\alpha

It can be defined as

p^\alpha = m u^\alpha\,,

then we get:

p = (\gamma m, \gamma \mathbf p)


Another way to generalize momentum: the momentum at rest is p = (p^0, 0) , applying Lorentz boost:

\begin{align*} p'^0 &= \gamma p^0 \\ p'^i &= \gamma v^i p^0 \end{align*}

When |v| \to c , \gamma \to 1 :

\begin{align*} p'^i &= v^i p^0 \end{align*}

This should become the clasicall limit where \mathbf p' = m\mathbf v , therefore p^0 = m .


p_\alpha p^\alpha = -m^2\\ E^2 - \mathbf p^2 = m^2


Scalars, Vectors, and Tensors

Contravariant vectors transform by Lorentz transformation; covariant vectors transform by inverse Lorentz transformation:

V^\alpha \to V'^\alpha = \Lambda^\alpha{}_\beta V^\beta\\ V_\alpha \to V'_\alpha = \Lambda_\alpha{}^\beta V_\beta\\

where

\Lambda_\alpha{}^\beta \equiv \eta_{\alpha\gamma}\eta^{\beta\delta}\Lambda^\gamma{}_\delta\,.

To prove this is indeed inverse Lorentz transformation:

\begin{align*} \Lambda_\alpha{}^\beta \Lambda^\alpha{}_\rho &= \eta_{\alpha\gamma}\eta^{\beta\delta}\Lambda^\gamma{}_\delta \Lambda^\alpha{}_\rho \\ &=\eta_{\rho\delta}\eta^{\beta\delta}\\ &=\delta^{\beta}{}_{\rho}\\ \end{align*}


Derivative with respect to a contravariant vector is covariant.