# Difference between revisions of "General Relativity Primer"

General Relativity really takes an entire class to do it justice. Here, we just cover some important concepts and results without really going into detail. GR is notorious for getting bogged down in notation. We’ll give you a taste of the notation, but we’ll try to avoid the bog.

## Notation

When spacetime can curve, it becomes important to define how you measure distance. For this, we use the metric:

${\displaystyle (\Delta s)^{2}=-(c\Delta t)^{2}+(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}\,\!}$

As we know from special relativity that our metric, when defined this way, is invariant under Lorentz transformations, which means that all inertial observers can agree on it, and light travels a ${\displaystyle ds^{2}=0}$ path.

Let’s pick two inertial observers. Lorentz invariance dictates that two inertial frame ${\displaystyle (x,y,z,t)}$ and ${\displaystyle (x^{\prime },y^{\prime },z^{\prime },t^{\prime })}$, with one moving with respect to the other at velocity ${\displaystyle {\hat {v}}=v{\hat {x}}}$, are related by:

${\displaystyle {\begin{matrix}x^{\prime }=\gamma (x-vt),&y^{\prime }=y,&z^{\prime }=z,&t^{\prime }=\gamma \left(t-{v \over c^{2}}x\right)\end{matrix}}\,\!}$

where ${\displaystyle \gamma \equiv {1 \over {\sqrt {1-{v^{2} \over c^{2}}}}}}$. The Lorentz transformation relating two frames of reference is straightforward, but a bit cumbersome to write for an arbitrary velocity direction. We can use matrix notation to clean things up a bit.

We can write the Lorentz transformation as a matrix acting on a vector. The Lorentz transformation will be denoted by the 2-index object ${\displaystyle \Lambda _{\,\,\,\,\nu }^{\mu }}$. The transformed four-vector is given by

${\displaystyle (x^{\prime })^{\mu }=\Lambda _{\,\,\,\,\nu }^{\mu }x^{\nu }.\,\!}$

This is just matrix multiplication where

${\displaystyle x^{\mu }=\left({\begin{array}{c}t\\x\\y\\z\end{array}}\right)\,\!}$

and, for example,

${\displaystyle \Lambda _{\,\,\,\,\nu }^{\mu }=\left({\begin{array}{cccc}\gamma &-v\gamma &0&0\\-v\gamma &\gamma &0&0\\0&0&1&0\\0&0&0&1\end{array}}\right).\,\!}$

In the above example, ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are indices denoting reference frames (${\displaystyle x^{\mu }}$ and ${\displaystyle x^{\nu }}$, respectively. We don’t need the primed notation, because it is redundant: ${\displaystyle (x^{\prime })^{\mu }=x^{\mu }}$. Matrices map one coordinate system to another, with the top index (${\displaystyle \mu }$ in ${\displaystyle \Lambda _{\nu }^{\mu }}$) indicating the output coordinate system and the lower index (${\displaystyle \nu }$ in ${\displaystyle \Lambda _{\nu }^{\mu }}$) indicating the input coordinate system, and according to Einstein’s notation, if you see the same index in the top and bottom (e.g. ${\displaystyle \nu }$ in ${\displaystyle \Lambda _{\nu }^{\mu }x^{\nu }}$), you sum over it and it goes away. In this way, you can know, without knowing anything else about ${\displaystyle \Lambda _{\nu }^{\mu }}$, that it will operate on ${\displaystyle x^{\nu }}$ to put it into the ${\displaystyle \mu }$ frame.

We can also express our Lorentz invariant interval in this notation, where ${\displaystyle ds^{2}=-c^{2}dt^{2}+(dx^{2}+dy^{2}+dz^{2})}$ becomes:

${\displaystyle ds^{2}=dx^{\alpha }\eta _{\alpha \beta }dx^{\beta }\,\!}$

where ${\displaystyle \eta _{\alpha \beta }}$, the metric tensor, is given by:

${\displaystyle \eta _{\alpha \beta }\equiv {\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}\,\!}$

Note how you can see, just from the Einstein summing conventions, that the result ${\displaystyle ds}$ is independent of any reference frame because it has no indices.