Difference between revisions of "General Relativity Primer"

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\begin{document}
 
\begin{document}
  
\section*{ 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.
  
Lorentz invariance dictates that two inertial
+
\section*{Notation}
 +
 
 +
When spacetime can curve, it becomes important to define how you measure distance.  For this, we use the metric:
 +
\begin{equation}
 +
(\Delta s)^2=-(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2
 +
\end{equation}
 +
 
 +
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 $ds^2=0$ path. 
 +
 
 +
Let's pick two inertial observers.  Lorentz invariance dictates that two inertial
 
frame $(x, y, z, t)$ and $(x\p, y\p, z\p, t\p)$, with one  
 
frame $(x, y, z, t)$ and $(x\p, y\p, z\p, t\p)$, with one  
 
moving with respect to the other at velocity $\hat v=v\hat x$, are related by:
 
moving with respect to the other at velocity $\hat v=v\hat x$, are related by:
 
$$\begin{matrix} x\p=\gamma(x-vt),&y\p=y,&z\p=z,&t\p=\gamma\left(t-{v\over c^2}x
 
$$\begin{matrix} x\p=\gamma(x-vt),&y\p=y,&z\p=z,&t\p=\gamma\left(t-{v\over c^2}x
 
\right)\end{matrix}$$
 
\right)\end{matrix}$$
where $\gamma\equiv{1\over\sqrt{1-{v^2\over c^2}}}$.   
+
where $\gamma\equiv{1\over\sqrt{1-{v^2\over c^2}}}$.  The Lorentz transformation relating two frames of reference is straightforward, but a bit cumbersome to
Note, to give a taste of tensor forms, this all
+
write for an arbitrary velocity direction.  We can use matrix notation to clean things up a bit.
may be written as ${x\p}^\alpha = \Lambda^\alpha_\beta x^\beta+I_0^\alpha$.\par
+
 
 +
We can write the Lorentz transformation as a matrix acting on a vector. The Lorentz transformation will be denoted by the 2-index object $\Lambda^\mu_{\,\,\,\, \nu}$. The transformed four-vector is given by \begin{equation} (x^\prime)^\mu = \Lambda^\mu_{\,\,\,\, \nu} x^\nu . \end{equation} This is just matrix multiplication where
 +
\begin{equation}
 +
x^\mu = \left ( \begin{array}{c} t \\ x \\ y \\ z \end{array} \right )
 +
\end{equation}
 +
and, for example,
 +
\begin{equation}
 +
\Lambda^\mu_{\,\,\,\, \nu} = \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) .
 +
\end{equation}
  
Remember the Lorentz invariant interval, which is conserved between frames:
+
In the above example, $\mu$ and $\nu$ are indices denoting reference frames ($x^\mu$ and $x^\nu$, respectively.  We don't need the primed notation, because it is redundant: $(x^\prime)^\mu=x^\mu$.
$$ds^2=-c^2dt^2+(dx^2+dy^2+dz^2)$$
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Matrices map one coordinate system to another, with the top index ($\mu$ in $\Lambda_\nu^\mu$) indicating the output coordinate system and the lower index ($\nu$ in $\Lambda_\nu^\mu$) indicating
Light travels a $ds^2=0$ path.  In tensor form, this equation looks like:
+
the input coordinate system, and according to Einstein's notation, if you see the same index in the top and bottom (e.g. $\nu$ in $\Lambda_\nu^\mu x^\nu$), you sum over it and it goes away.  In this way,
$$ds^2=\eta_{\alpha\beta}dx^\alpha dx^\beta$$
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you can know, without knowing anything else about $\Lambda_\nu^\mu$, that it will operate on $x^\nu$ to put it into the $\mu$ frame.
 +
 
 +
We can also express our Lorentz invariant interval in this notation, where $ds^2=-c^2dt^2+(dx^2+dy^2+dz^2)$ becomes:
 +
\begin{equation}
 +
ds^2=dx^\alpha \eta_{\alpha\beta}dx^\beta
 +
\end{equation}
 
where $\eta_{\alpha\beta}$, the metric tensor, is given by:
 
where $\eta_{\alpha\beta}$, the metric tensor, is given by:
$$\eta_{\alpha\beta}\equiv\begin{pmatrix} -1&0&0&0\\0&1&0&0\\0&0&1&0\\ 0&0&0&1\\
+
\begin{equation}
\end{pmatrix}$$
+
\eta_{\alpha\beta}\equiv\begin{pmatrix} -1&0&0&0\\0&1&0&0\\0&0&1&0\\ 0&0&0&1\\
 +
\end{pmatrix}
 +
\end{equation}
 +
 
 +
Note how you can see, just from the Einstein summing conventions, that the result $ds$ is independent of any reference frame because it has no indices.
  
  
 
\end{document}
 
\end{document}
 
</latex>
 
</latex>

Revision as of 14:10, 17 January 2017

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:

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 path.

Let’s pick two inertial observers. Lorentz invariance dictates that two inertial frame and , with one moving with respect to the other at velocity , are related by:

where . 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 . The transformed four-vector is given by

This is just matrix multiplication where

and, for example,

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

We can also express our Lorentz invariant interval in this notation, where becomes:

where , the metric tensor, is given by:

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