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%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Created=Wednesday, August 23, 2000 15:54:58}
%TCIDATA{LastRevised=Tuesday, October 16, 2001 12:06:41}
%TCIDATA{}
%TCIDATA{}
%TCIDATA{CSTFile=LaTeX article (bright).cst}
\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\begin{document}
\begin{center}
{\LARGE Hour Exam No.1\medskip }
\end{center}
Please attempt all of the following problems before the due date. All
problems count the same even though some are more complex than others.%
\vspace{0.25in}
{\LARGE Problem 1\medskip }
At Minkowski coordinate time, $t$, an object is located at the Minkowski
position coordinates%
\[
x\left( t\right) =\sqrt{t^{2}+1};\quad y\left( t\right) =z\left( t\right)
=0.
\]
Using $c=1$ units, find
\begin{enumerate}
\item[{\protect\LARGE a.}] the object's ordinary Newtonian velocity vector
components.\bigskip
\end{enumerate}
{\large Answer 1a}{\LARGE \medskip }
$v_{x}=\frac{dx}{dt}=\frac{d}{dt}\left( \sqrt{t^{2}+1}\right) =\allowbreak
\frac{1}{\sqrt{\left( t^{2}+1\right) }}t$
$v_{y}=v_{z}=0.$\bigskip
\begin{enumerate}
\item[{\protect\LARGE b.}] the components of the object's four-velocity
vector.\bigskip
\end{enumerate}
{\large Answer 1b}{\LARGE \medskip }
Use the formula $u=u^{0}\left( e_{0}+\vec{v}\right) $or%
\begin{eqnarray*}
u^{0} &=&\frac{1}{\sqrt{1-v^{2}}} \\
u^{i} &=&u^{0}v^{i}
\end{eqnarray*}
$v^{2}=v_{x}{}^{2}+v_{y}{}^{2}+v_{z}{}^{2}=\left( \frac{1}{\sqrt{\left(
t^{2}+1\right) }}t\right) ^{2}=\allowbreak \frac{1}{t^{2}+1}t^{2}$
$1-v^{2}=1-\frac{1}{t^{2}+1}t^{2}=\allowbreak \frac{1}{t^{2}+1}$
$u^{0}=\frac{1}{\sqrt{\frac{1}{t^{2}+1}}}=\allowbreak \sqrt{\left(
t^{2}+1\right) }$
$u^{1}=u_{x}=u^{0}v_{x}=\sqrt{\left( t^{2}+1\right) }\frac{1}{\sqrt{\left(
t^{2}+1\right) }}t=t$%
\begin{eqnarray*}
u^{0} &=&\sqrt{\left( t^{2}+1\right) } \\
u_{x} &=&t \\
u_{y} &=&u_{z}=0
\end{eqnarray*}%
\pagebreak
{\LARGE Problem 2\medskip }
Consider a two-dimensional spacetime manifold where we are using the
coordinates $t,x$ to locate events and the corresponding holonomic basis
vectors%
\[
\partial _{t}=\frac{\partial }{\partial t},\quad \partial _{x}=\frac{%
\partial }{\partial x}
\]%
to span each tangent space. A different coordinate system $t^{\prime
},x^{\prime }$ also locates events in this spacetime where
\[
t^{\prime }=t;\quad x^{\prime }=x-\sqrt{t^{2}+1},
\]%
and the corresponding holonomic basis vectors
\[
\partial _{t^{\prime }}=\frac{\partial }{\partial t^{\prime }},\quad
\partial _{x^{\prime }}=\frac{\partial }{\partial x^{\prime }}
\]%
also span each tangent space.
\begin{enumerate}
\item[{\protect\LARGE a.}] Notice that $\partial _{t^{\prime }}$ is not
equal to $\partial _{t}$ even though $t^{\prime }=t$. Express $\partial
_{t^{\prime }}$ in terms of $\partial _{t}$ and $\partial _{x}$.\bigskip
{\large Answer 2a}{\LARGE \medskip }
\end{enumerate}
From the chain rule for partial derivatives,
$\partial _{t^{\prime }}=\frac{\partial }{\partial t^{\prime }}=\frac{%
\partial t}{\partial t^{\prime }}\frac{\partial }{\partial t}+\frac{\partial
x}{\partial t^{\prime }}\frac{\partial }{\partial x}$
To evaluate these partials, we need to solve for $x,y$ in terms of $%
x^{\prime },t^{\prime }$.
$x^{\prime }=x-\sqrt{t^{2}+1}$
$x=x^{\prime }+\sqrt{t^{2}+1}=x^{\prime }+\sqrt{t^{\prime 2}+1}$
$t=t^{\prime }$
$\frac{\partial t}{\partial t^{\prime }}=1,\qquad \frac{\partial x}{\partial
t^{\prime }}=\frac{\partial }{\partial t^{\prime }}\left( x^{\prime }+\sqrt{%
t^{\prime 2}+1}\right) =\frac{t^{\prime }}{\sqrt{t^{\prime 2}+1}}$
(Notice that if you use SN--Maple to do this, it gets confused by the
primes.)
\[
\partial _{t^{\prime }}=\frac{\partial }{\partial t}+\frac{t^{\prime }}{%
\sqrt{t^{\prime 2}+1}}\frac{\partial }{\partial x}
\]
\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE b.}] Express $\partial _{x^{\prime }}$ in terms of $%
\partial _{t}$ and $\partial _{x}$.\bigskip
\end{enumerate}
{\large Answer 2b}{\LARGE \medskip }
$\partial _{x^{\prime }}=\frac{\partial }{\partial x^{\prime }}=\frac{%
\partial x}{\partial x^{\prime }}\frac{\partial }{\partial x}+\frac{\partial
t}{\partial x^{\prime }}\frac{\partial }{\partial t}=\frac{\partial }{%
\partial x}$\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE c.}] Quick answer: What differential forms correspond
to the basis dual to the basis vectors $\partial _{t}$ and $\partial _{x}?$%
\bigskip
\end{enumerate}
{\large Answer 2c}{\LARGE \medskip }
\[
dt,dx
\]%
Check this (optional):%
\begin{eqnarray*}
\partial _{t}\cdot dt &=&\frac{\partial t}{\partial t}=1 \\
\partial _{x}\cdot dx &=&\frac{\partial x}{\partial x}=1 \\
\partial _{t}\cdot dx &=&\frac{\partial x}{\partial t}=0 \\
\partial _{x}\cdot dt &=&\frac{\partial t}{\partial x}=0
\end{eqnarray*}%
\vspace{0.5in}\pagebreak
{\LARGE Problem 3}\vspace{0.25in}
Suppose that the metric tensor on a spacetime has the form%
\[
g=-dt\otimes dt+dx\otimes dx
\]%
and you decide to use the ``null'' coordinates%
\begin{eqnarray*}
u^{1} &=&t+x \\
u^{2} &=&t-x
\end{eqnarray*}%
and the corresponding ``null basis'' vectors
\[
e_{1}=\frac{\partial }{\partial u^{1}},\quad e_{2}=\frac{\partial }{\partial
u^{2}}.
\]
\begin{enumerate}
\item[{\protect\LARGE a.}] Express the null basis vectors in terms of the
basis vectors $\partial _{t}$ and $\partial _{x}$ that go with the
coordinates $x,t$.\bigskip
\end{enumerate}
{\large Answer 3a}{\LARGE \medskip }
From the chain rule for partial derivatives,$\bigskip $
$e_{1}=\frac{\partial }{\partial u^{1}}=\frac{\partial t}{\partial u^{1}}%
\frac{\partial }{\partial t}+\frac{\partial x}{\partial u^{1}}\frac{\partial
}{\partial x},\qquad e_{2}=\frac{\partial }{\partial u^{2}}=\frac{\partial t%
}{\partial u^{2}}\frac{\partial }{\partial t}+\frac{\partial x}{\partial
u^{2}}\frac{\partial }{\partial x}\smallskip $
Invert the relation to get $x,t$ in terms of $u^{1},u^{2}\smallskip $
$t=\frac{1}{2}\left( u^{1}+u^{2}\right) ,\qquad x=\frac{1}{2}\left(
u^{1}-u^{2}\right) \smallskip $
$\frac{\partial t}{\partial u^{1}}=\frac{1}{2},\qquad \frac{\partial x}{%
\partial u^{1}}=+\frac{1}{2},\qquad \frac{\partial t}{\partial u^{2}}=\frac{1%
}{2},\qquad \frac{\partial x}{\partial u^{2}}-\frac{1}{2}$
\begin{eqnarray*}
e_{1} &=&\frac{1}{2}\left( \partial _{t}+\partial _{x}\right) \\
e_{2} &=&\frac{1}{2}\left( \partial _{t}-\partial _{x}\right)
\end{eqnarray*}
\begin{enumerate}
\item[{\protect\LARGE b.}] Find the metric components $g_{11},g_{12},g_{22}$
in the null basis.
\end{enumerate}
{\large Answer 3b}{\LARGE \medskip }
$g_{11}=\frac{1}{2}\left( \partial _{t}+\partial _{x}\right) \cdot \frac{1}{2%
}\left( \partial _{t}+\partial _{x}\right) =\frac{1}{4}\left( \partial
_{t}\cdot \partial _{t}+\partial _{x}\cdot \partial _{x}\right) =\frac{1}{4}%
\left( -1+1\right) =0\smallskip $
$g_{12}=\frac{1}{2}\left( \partial _{t}+\partial _{x}\right) \cdot \frac{1}{2%
}\left( \partial _{t}-\partial _{x}\right) =\frac{1}{4}\left( \partial
_{t}\cdot \partial _{t}-\partial _{x}\cdot \partial _{x}\right) =\frac{1}{4}%
\left( -1-1\right) =-\frac{1}{2}\smallskip $
$g_{22}=\frac{1}{2}\left( \partial _{t}-\partial _{x}\right) \cdot \frac{1}{2%
}\left( \partial _{t}-\partial _{x}\right) =\frac{1}{4}\left( \partial
_{t}\cdot \partial _{t}+\partial _{x}\cdot \partial _{x}\right) =\frac{1}{4}%
\left( -1+1\right) =0$
\pagebreak
{\LARGE Problem 4}\vspace{0.5in}
An electromagnetic field two-form is given by%
\[
f=r^{-2}dt\wedge dr
\]%
where $t$ is the usual Minkowski time function and $r=\sqrt{x^{2}+y^{2}+z^{2}%
}$ is a radius coordinate. In the following, use the basis vectors%
\[
\partial _{0}=\frac{\partial }{\partial t},\quad \partial _{1}=\frac{%
\partial }{\partial r}
\]%
and their dual basis forms and assume $c=1$ units.\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE a.}] Find the components $F_{00},F_{01},F_{10},F_{11},$
of $f$.\bigskip
\end{enumerate}
{\large Answer 4a}{\LARGE \medskip }
$f=r^{-2}dt\wedge dr=r^{-2}\left( dt\otimes dr-dr\otimes dt\right) $
The components are\smallskip
$F_{00}=f\left( \partial _{0},\partial _{0}\right) =r^{-2}\left( dt\otimes
dr-dr\otimes dt\right) \left( \partial _{0},\partial _{0}\right) $\smallskip
$=r^{-2}\left( dt\left( \partial _{0}\right) dr\left( \partial _{0}\right)
-dr\left( \partial _{0}\right) dt\left( \partial _{0}\right) \right) $%
\smallskip
$=r^{-2}\left( 1\times 0-0\times 1\right) =0$\smallskip
Copy the above three lines several times and just change the subscripts to
get the rest.\smallskip
$F_{01}=f\left( \partial _{0},\partial _{1}\right) =r^{-2}\left( dt\otimes
dr-dr\otimes dt\right) \left( \partial _{0},\partial _{1}\right) $\smallskip
$=r^{-2}\left( dt\left( \partial _{0}\right) dr\left( \partial _{1}\right)
-dr\left( \partial _{0}\right) dt\left( \partial _{1}\right) \right) $%
\smallskip
$=r^{-2}\left( 1\times 1-0\times 0\right) =r^{-2}$\smallskip
$F_{10}=f\left( \partial _{1},\partial _{0}\right) =r^{-2}\left( dt\otimes
dr-dr\otimes dt\right) \left( \partial _{1},\partial _{0}\right) $\smallskip
$=r^{-2}\left( dt\left( \partial _{1}\right) dr\left( \partial _{0}\right)
-dr\left( \partial _{1}\right) dt\left( \partial _{0}\right) \right) $%
\smallskip
$=r^{-2}\left( 0\times 0-1\times 1\right) =-r^{-2}$\smallskip
$F_{11}=f\left( \partial _{1},\partial _{1}\right) =r^{-2}\left( dt\otimes
dr-dr\otimes dt\right) \left( \partial _{1},\partial _{1}\right) $\smallskip
$=r^{-2}\left( dt\left( \partial _{1}\right) dr\left( \partial _{1}\right)
-dr\left( \partial _{1}\right) dt\left( \partial _{1}\right) \right) $%
\smallskip
$=r^{-2}\left( 1\times 1-1\times 1\right) =0$\smallskip
\[
\left(
\begin{array}{cc}
F_{00} & F_{01} \\
F_{10} & F_{11}%
\end{array}%
\right) =\left(
\begin{array}{cc}
0 & r^{-2} \\
-r^{-2} & 0%
\end{array}%
\right)
\]%
\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE b.}] Find the components $F^{1}{}_{0}$ and $%
F^{0}{}_{1} $ of the related tensor.\bigskip
\end{enumerate}
{\large Answer 4b}{\LARGE \medskip }
$F^{1}{}_{0}=F_{10}=-r^{-2}$\smallskip
$F^{0}{}_{1}=-F_{01}=-r^{-2}$\vspace{0.5in}
\begin{enumerate}
\item[{\protect\LARGE c.}] Identify the sort of object that would make this
electromagnetic field.\bigskip
\end{enumerate}
{\large Answer 4c}{\LARGE \medskip }
You can probably guess that this thing is a point charge. However, to get
the sign and amount of the charge right, you should recall the form of the
force law that we are using: For a test particle of charge $e$ and mass $m$%
\[
\frac{dp}{d\tau }=\frac{e}{m}g^{-1}\left( f\left( p\right) \right)
\]%
or, in terms of components%
\[
\frac{dp^{\alpha }}{d\tau }=\frac{e}{m}g^{\alpha \sigma }f_{\sigma \rho
}p^{\rho }=\frac{e}{m}F^{\alpha }{}_{\rho }p^{\rho }
\]%
For a test charge at rest, $p^{r}=0$, $p^{0}=m,$ and $\tau =t$
\[
\frac{dp^{1}}{dt}=eF^{1}{}_{0}=-er^{-2}
\]%
Because $e_{1}$ points radially, we find a radial force of $-\frac{e}{r^{2}}$%
. Compare this result to the usual form of Coulomb's law:%
\[
k\frac{Qe}{r^{2}}=-\frac{e}{r^{2}}
\]%
The charge that generates this field must be%
\[
Q=-\frac{1}{k}
\]%
or about $-1.1\times 10^{-10}$ Coulombs.
If you want the units to come out right, you need to attach the correct
units $\left( \text{N/C}\right) $ to the field two-form.\bigskip
{\LARGE Problem 5 was the same as problem 3.}\pagebreak
{\LARGE Problem 6\medskip }
A set of observers are sitting on a flat turntable that is rotating with
angular velocity $\omega $. Each observer is located at a fixed pair of
Cartesian coordinates $x^{\prime },y^{\prime }$ that rotate with the disk.
In terms of the non-rotating Minkowski coordinates $t,x,y$ the position of
an given observer at $x^{\prime },y^{\prime }$ is
\begin{eqnarray*}
x &=&x^{\prime }\cos \omega t-y^{\prime }\sin \omega t \\
y &=&x^{\prime }\sin \omega t+y^{\prime }\cos \omega t
\end{eqnarray*}
Using $c=1$ units,
\begin{enumerate}
\item[{\protect\LARGE a.}] find the four-velocity vector $u$ of the observer
at%
\begin{eqnarray*}
x^{\prime } &=&r \\
y^{\prime } &=&0
\end{eqnarray*}%
in terms of the non-rotating Minkowski basis vectors $\partial _{t},\partial
_{x},\partial _{y}$\bigskip
\end{enumerate}
{\large Answer 6a}{\LARGE \medskip }
First find the Newtonian velocity components of an object at constant $%
x^{\prime },y^{\prime }$.
$v_{x}=\frac{dx}{dt}=\frac{d}{dt}\left( x^{\prime }\cos \omega t-y^{\prime
}\sin \omega t\right) =\allowbreak -x^{\prime }\left( \sin \omega t\right)
\omega -y^{\prime }\left( \cos \omega t\right) \omega $
$v_{y}=\frac{dy}{dt}=\frac{d}{dt}\left( x^{\prime }\sin \omega t+y^{\prime
}\cos \omega t\right) =\allowbreak x^{\prime }\left( \cos \omega t\right)
\omega -y^{\prime }\left( \sin \omega t\right) \omega $
The value of $v^{2}$ we know should be $\left( x^{\prime 2}+y^{\prime
2}\right) \omega ^{2}=r^{2}\omega ^{2}.$
Now use the formulas for the four-velocity components
$u^{0}=\frac{1}{\sqrt{1-v^{2}}}=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}$
$u_{x}=u^{0}v_{x}=\frac{-x^{\prime }\left( \sin \omega t\right) \omega
-y^{\prime }\left( \cos \omega t\right) \omega }{\sqrt{1-r^{2}\omega ^{2}}}=-%
\frac{\omega r}{\sqrt{1-r^{2}\omega ^{2}}}\sin \omega t$
$u_{y}=u^{0}v_{y}=\frac{x^{\prime }\left( \cos \omega t\right) \omega
-y^{\prime }\left( \sin \omega t\right) \omega }{\sqrt{1-r^{2}\omega ^{2}}}=%
\frac{\omega r}{\sqrt{1-r^{2}\omega ^{2}}}\cos \omega t$
and put these together with the basis vectors to form the four-velocity%
\[
u=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( \partial _{t}-\omega r\sin
\omega t\partial _{x}+\omega r\cos \omega t\partial _{y}\right)
\]%
\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE b.}] The co-rotating coordinate system consists of $%
t^{\prime }=t,x^{\prime },y^{\prime }$. Note that it is still the same $t$.
Express the co-rotating holonomic basis vectors $\partial _{x^{\prime }}$
and $\partial _{y^{\prime }}$at%
\begin{eqnarray*}
x^{\prime } &=&r \\
y^{\prime } &=&0
\end{eqnarray*}%
in terms of the non-rotating Minkowski basis vectors $\partial _{t},\partial
_{x},\partial _{y}$\bigskip
\end{enumerate}
{\large Answer 6b}{\LARGE \medskip }
First notice that the partials with respect to $x^{\prime },y^{\prime }$ are
holding $t^{\prime }$ and therefore $t$ constant and the same is true of the
partials with respect to $x,y$. Thus $t$ and $t^{\prime }$ are just
constants at this point.
$\partial _{x^{\prime }}=\frac{\partial }{\partial x^{\prime }}=\frac{%
\partial x}{\partial x^{\prime }}\frac{\partial }{\partial x}+\frac{\partial
y}{\partial x^{\prime }}\frac{\partial }{\partial y}=\cos \omega t\partial
_{x}+\sin \omega t\partial _{y}$
$\partial _{y^{\prime }}=\frac{\partial }{\partial y^{\prime }}=\frac{%
\partial x}{\partial y^{\prime }}\frac{\partial }{\partial x}+\frac{\partial
y}{\partial y^{\prime }}\frac{\partial }{\partial y}=-\sin \omega t\partial
_{x}+\cos \omega t\partial _{y}$
Notice that the values of $x^{\prime }$ and $y^{\prime }$ do not actually
matter.\vspace{0.25in}
\begin{enumerate}
\item[{\protect\LARGE c.}] Suppose that the observer at%
\begin{eqnarray*}
x^{\prime } &=&r \\
y^{\prime } &=&0
\end{eqnarray*}%
uses the basis vectors
\begin{eqnarray*}
e_{0} &=&u \\
e_{1} &=&\partial _{x^{\prime }} \\
e_{2} &=&\partial _{y^{\prime }}
\end{eqnarray*}%
Find the components of the metric tensor in this basis.\bigskip
\end{enumerate}
{\large Answer 6c}{\LARGE \medskip }
Collect the results of parts a,b to express the $e_{i}$ in terms of the
Minkowkski basis vectors.
$e_{0}=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( \partial _{t}-\omega r\sin
\omega t\partial _{x}+\omega r\cos \omega t\partial _{y}\right) $
$e_{1}=\cos \omega t\partial _{x}+\sin \omega t\partial _{y}$
$e_{2}=-\sin \omega t\partial _{x}+\cos \omega t\partial _{y}$
Take dot products using the usual Minkowski metric tensor.
$g_{00}=e_{0}\cdot e_{0}=u\cdot u=-1$
$g_{01}=e_{0}\cdot e_{1}$
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( \partial _{t}-\omega r\sin
\omega t\partial _{x}+\omega r\cos \omega t\partial _{y}\right) \cdot \left(
\cos \omega t\partial _{x}+\sin \omega t\partial _{y}\right) $
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( -\omega r\sin \omega t\partial
_{x}\cdot \cos \omega t\partial _{x}+\omega r\cos \omega t\partial _{y}\cdot
\sin \omega t\partial _{y}\right) $
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( -\omega r\sin \omega t\cos
\omega t+\omega r\cos \omega t\sin \omega t\right) =0$
$g_{02}=e_{0}\cdot e_{2}$
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( \partial _{t}-\omega r\sin
\omega t\partial _{x}+\omega r\cos \omega t\partial _{y}\right) \cdot \left(
-\sin \omega t\partial _{x}+\cos \omega t\partial _{y}\right) $
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( -\omega r\sin \omega t\partial
_{x}\cdot \left( -\sin \omega t\partial _{x}\right) +\omega r\cos \omega
t\partial _{y}\cdot \cos \omega t\partial _{y}\right) $
$=\frac{1}{\sqrt{1-r^{2}\omega ^{2}}}\left( \omega r\sin ^{2}\omega t+\omega
r\cos ^{2}\omega t\right) =\frac{\omega r}{\sqrt{1-r^{2}\omega ^{2}}}$
$g_{11}=e_{1}\cdot e_{1}=\left( \cos \omega t\partial _{x}+\sin \omega
t\partial _{y}\right) \cdot \left( \cos \omega t\partial _{x}+\sin \omega
t\partial _{y}\right) $
$=\left( \cos ^{2}\omega t+\sin ^{2}\omega t\right) =1$
$g_{12}=e_{1}\cdot e_{2}=\left( \cos \omega t\partial _{x}+\sin \omega
t\partial _{y}\right) \cdot \left( -\sin \omega t\partial _{x}+\cos \omega
t\partial _{y}\right) $
$=\cos \omega t\partial _{x}\cdot \left( -\sin \omega t\partial _{x}\right)
+\sin \omega t\partial _{y}\cdot \cos \omega t\partial _{y}$
$=-\sin \omega t\cos \omega t+\sin \omega t\cos \omega t=0$
$g_{22}=e_{2}\cdot e_{2}=\left( -\sin \omega t\partial _{x}+\cos \omega
t\partial _{y}\right) \cdot \left( -\sin \omega t\partial _{x}+\cos \omega
t\partial _{y}\right) $
$=\left( -\sin \omega t\partial _{x}\right) \cdot \left( -\sin \omega
t\partial _{x}\right) +\cos \omega t\partial _{y}\cdot \cos \omega t\partial
_{y}$
$=\sin ^{2}\omega t+\cos ^{2}\omega t=1$%
\[
\left(
\begin{array}{ccc}
g_{00} & g_{01} & g_{02} \\
g_{10} & g_{11} & g_{12} \\
g_{20} & g_{21} & g_{22}%
\end{array}%
\right) =\left(
\begin{array}{ccc}
-1 & 0 & \frac{\omega r}{\sqrt{1-r^{2}\omega ^{2}}} \\
0 & 1 & 0 \\
\frac{\omega r}{\sqrt{1-r^{2}\omega ^{2}}} & 0 & 1%
\end{array}%
\right)
\]%
\vspace{0.5in}
{\LARGE Added Note:}
At the point $x^{\prime }=r,y^{\prime }=0$ the basis vectors $e_{1}=\partial
_{x^{\prime }},e_{2}=\partial _{y^{\prime }}$ can be expressed in terms of
polar coordinates on the turntable as
$e_{1}=\partial _{r}$
$e_{2}=r^{-1}\partial _{\varphi }$
$\partial _{\varphi }=re_{2}$
so that the corresponding metric components are%
\[
\left(
\begin{array}{ccc}
g_{00} & g_{0r} & g_{0\varphi } \\
g_{r0} & g_{rr} & g_{r\varphi } \\
g_{\varphi 0} & g_{\varphi r} & g_{\varphi \varphi }%
\end{array}%
\right) =\left(
\begin{array}{ccc}
-1 & 0 & \frac{\omega r^{2}}{\sqrt{1-r^{2}\omega ^{2}}} \\
0 & 1 & 0 \\
\frac{\omega r^{2}}{\sqrt{1-r^{2}\omega ^{2}}} & 0 & r^{2}%
\end{array}%
\right)
\]%
and the standard form of the metric is%
\[
ds^{2}=-dt^{2}+2\frac{\omega r^{2}}{\sqrt{1-r^{2}\omega ^{2}}}dtd\varphi
+r^{2}d\varphi ^{2}
\]%
The cross-term in $dtd\varphi $ is the signal that this is the metric of a
rotating system.
\end{document}
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