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%TCIDATA{Created=Mon Nov 18 16:23:01 1996}
%TCIDATA{LastRevised=Fri Nov 29 09:12:30 1996}

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\markright{{\sf Motion in uniform, static fields     }}
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\begin{document}


\section{12.3\quad Motion in Uniform, Static Fields.}

\subsection{12.3.1}

\quad \thinspace \textbf{a. }The problem of a charged particle moving in
static, uniform $\mathbf{E}$ and $\mathbf{B}$ fields can be solved
explicitly. We show first how the general solution is obtained, then discuss
several particular cases.

\smallskip\ 

\textbf{b}. For the general treatment, it is best to start from the
Einstein-Lorentz equations in the covariant form 
\begin{equation}
m\frac{du_\alpha }{d\tau }=eF_{\alpha \beta }\frac{u^\beta (\tau )}c
\label{1}
\end{equation}
and solve for $u^\alpha \left( \tau \right) .$ Then one can also solve $%
dx^\alpha /d\tau =u^\alpha (\tau ).$

\smallskip\ 

\textbf{c. }In this fashion the solution appears in the parametric form $%
u^\alpha =u^\alpha (\tau ),\,x^\alpha =x^\alpha (\tau ).$ One can then, in
some cases, eliminate explicitly $\tau $ in favor of $t,$ from $%
u^0=cdt/d\tau =c\gamma ,$ or by inverting $t=t\left( \tau \right) .$

\smallskip\ 

\textbf{d. }The system of linear equations (\ref{1}) has solutions of the
form 
\begin{equation}
u^\alpha (\tau )=u^\alpha e^{-i\omega \tau }  \label{2}
\end{equation}
with 
\begin{equation}
-im\omega g_{\alpha \beta }u^\beta =\frac ecF_{\alpha \beta }u^\beta
\label{3}
\end{equation}

\smallskip\ 

\textbf{e. }The system of linear algebraic equations (\ref{3}) is soluble
for four eigenfrequencies $\omega _{\left( \lambda \right) }$ (not
necessarily all distinct). The four adjustable coefficients in the general
solution 
\[
u^{\alpha }(\tau )=\sum_{\lambda }u_{\left( \lambda \right) }^{\alpha }\exp
(-i\omega _{\left( \lambda \right) }\tau ) 
\]
are determined by the initial conditions. One can assign only three of the
four initial $u^{\alpha }(0)$ arbitrarily, since it must be $\mathsf{u\cdot u%
}\equiv u^{\alpha }u_{\alpha }=c^{2}$ at all times. The further integration
of $dx^{\alpha }/d\tau =u^{\alpha }(\tau )$ is elementary.\newpage

\subsection{12.3.2}

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\quad \quad \textbf{a. }For explicit solutions, it is convenient to choose $%
\mathbf{B}$ along the $x_{2}$ axis (say) and $\mathbf{E}$ in the $x_{1}x_{2}$
plane. Then 
\[
F_{\alpha \beta }=\left( 
\begin{tabular}{llll}
$\quad 0$ & $E_{\perp }$ & $E_{/\!/}$ & \thinspace \thinspace $0$ \\ 
$-E_{\perp }$ & $\,\,\,\,0$ & $\,\,\,\,0$ & $B$ \\ 
$-E_{/\!/}$ & $\,\,\,\,0$ & $\,\,\,\,0$ & $\,\,0$ \\ 
$\quad 0$ & $-B$ & $\,\,\,\,0$ & $\,\,0$%
\end{tabular}
\right) \,, 
\]
where $E_{/\!/}\equiv E_{2},$ $E_{\perp }\equiv E_{1}.\quad (/\!/$ and $%
\perp $ to $\mathbf{B).}$

\smallskip\ 

\textbf{b.} The explicit form of (\ref{3}) is 
\begin{eqnarray}
-im\omega u^{0} &=&(e/c)\left( E_{\perp }u^{1}+E_{/\!/}u^{2}\right)
\label{3'} \\
im\omega u^{1} &=&(e/c)\left( -E_{\perp }u^{0}+Bu^{3}\right)  \nonumber \\
im\omega u^{2} &=&(e/c)\left( -E_{/\!/}u^{0}\right)  \nonumber \\
im\omega u^{3} &=&(e/c)\left( -Bu^{1}\right)  \nonumber
\end{eqnarray}

\smallskip\ 

\textbf{c. }In writing down the solutions, it should be kept in mind that $%
u^{0}=c\gamma $ and $\mathbf{u}=\gamma \mathbf{v,}$ with $\gamma =\left(
1-v^{2}/c^{2}\right) ^{-1/2}.$ This is true at all times (see 1g), so that $%
\mathsf{u\cdot u}=c^{2}.$

\newpage

\subsection{12.3.3}

\smallskip\ 

\textbf{a. }As a first particular case, take $\mathbf{E}=0.$ Then we find
solutions with $\omega =0$ and $u^{1}=u^{3}=0$ and solutions with $\omega
=\pm \omega _{L}=\pm eB/mc$ and $u^{0}=u^{2}=0.$ The general solution, is: 
\[
u^{0}\left( \tau \right) =\gamma c,\;u^{2}\left( \tau \right) =\gamma
v_{/\!/}, 
\]
\[
\,u^{1}\left( \tau \right) =\gamma v_{\perp }\sin \left( \omega _{L}\tau
-\alpha \right) ,\,\,\,u^{3}\left( \tau \right) =\gamma v_{\perp }\cos
\left( \omega _{L}\tau -\alpha \right) .\, 
\]
with constants $v_{/\!/},$ $v_{\perp }$, $\alpha $, and $\gamma =\dfrac{1}{%
\sqrt{1-\left( v_{/\!/}^{2}+v_{\perp }^{2}\right) /c^{2}}}\,.$

\smallskip\ 

\textbf{b. }Integrating $u^{0}=cdt/d\tau =c\gamma $ gives simply $t=\gamma
\tau .$ Thus the frequency as seen in the laboratory is not $\omega _{L}$
but $\omega _{L}/\gamma .$ This is the familiar synchrotron motion when $%
u_{/\!/}=0.$

\smallskip\ 

\textbf{c. }The path is, generally, a helix with axis along $\mathbf{B}$ and
gyration radius $a$ such that $\gamma v_{\perp }=\omega _{L}a,$ or $%
cp_{\perp }=eBa.$ In numbers, $p_{\perp }($MeV/c)$=3\times 10^{-4}Ba$
(gauss-cm).

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\newpage

\subsection{12.3.4}

\quad \quad \textbf{a. }As a second example, take $\mathbf{B}=0$ and $%
\mathbf{E}$ (for instance) along the $x^{1}$ axis $\left( E_{\perp
}=E,\,E_{/\!/}=0\text{ in (\ref{3'})}\right) .$ This is motion with a
uniform acceleration $\mathbf{a}^{\prime }=e\mathbf{E/}m$ in the frame of
the moving particle (plus a transverse motion, if any).

\smallskip\ 

\textbf{b. }Proceeding from (\ref{3'}), we find solutions with $\omega =0$
and $u^0=u^1=0$ and solutions with $\omega =\pm ia^{\prime }/c=\pm ieE/mc$
and $u^2=u^3=0$. The general solution, with $/\!/$ and $\perp $ referring
now to the direction of $\mathbf{E,}$ can be written in terms of the phase $%
\phi =a^{\prime }\tau /c-\alpha =\left( eE/mc\right) \tau -\alpha ,$ with
constant $\alpha $, as follows: 
\[
\mathbf{u}_{\perp }\left( \tau \right) =\mathbf{u}_{\perp },\quad u^0\left(
\tau \right) =\sqrt{c^2+\mathbf{u}_{\perp }^2}\cosh \phi ,\quad u^1\left(
\tau \right) =\sqrt{c^2+\mathbf{u}_{\perp }^2}\sinh \phi . 
\]

\smallskip\ 

\textbf{c. }It should be noted that $\mathbf{u}_{\perp }$ is constant, but $%
\mathbf{v}_{\perp }=d\mathbf{x}_{\perp }/dt$ is not. Rather, $\mathbf{v}%
_{\perp }\left( \tau \right) =c\mathbf{u}_{\perp }/u^0\left( \tau \right) .$

\smallskip\ 

\textbf{d. }Similarly, although $u^1\left( \tau \right) $ appears to
increase without bounds as $\phi \rightarrow \infty ,$ we see that in fact $%
v_{/\!/}\left( \tau \right) =cu^1\left( \tau \right) /u^0\left( \tau \right)
=c\tanh \phi $ approaches $c$ as $\phi \rightarrow \infty .$

\smallskip\ 

\textbf{e. }The relation between $t$ and $\tau $ can be obtained by
integrating $cdt=u^0\left( \tau \right) d\tau .$ We find $a^{\prime }t/c=%
\sqrt{1+u_{\perp }^2/c^2}\left( \sinh \phi +\sinh \alpha \right) $ (fixing $%
t=0$ at $\tau =0).$ This can be plugged back in the results obtained above
to complete the discussion. Here we remark only that for $\alpha =0,$
corresponding to $v_{/\!/}=0$ at $t=0,$ one has simply $u^1(t)=a^{\prime }t.$
This resembles $v_{/\!/}\left( t\right) =a^{\prime }t$ of Newtonian free
fall, but in fact $v_{/\!/}\left( t\right) =a^{\prime }t/\sqrt{1+\left(
a^{\prime }t/c\right) ^2}.$

\smallskip\ 

\textbf{f. }Many of these results were obtained by an (apparently) unrelated
procedure in discussing the problem of an accelerating space ship.

\newpage

\subsection{12.3.5}

\quad \quad \textbf{a. }We now tackle the general case. It is easy to show
directly that the eigenfrequencies of (\ref{3'}) are given by 
\begin{equation}
\omega ^2=\frac{e^2}{2m^2c^2}\left[ B^2-E^2\pm \sqrt{\left( B^2-E^2\right)
^2+4\left( \mathbf{B\cdot E}\right) ^2}\right]  \label{4}
\end{equation}
where we have written $\mathbf{B\cdot E}$ for $BE_{/\!/}$ and $E^2$ for $%
E_{/\!/}^2+E_{\perp }^2.$

\smallskip\ 

\textbf{b. }We notice that the eigenvalues are expressed in terms of the
Lorentz invariants $2\left( B^2-E^2\right) =F^{\alpha \beta }F_{\alpha \beta
}$ and $-4\mathbf{E\cdot B=}F^{\alpha \beta }\frak{F}_{\alpha \beta
}=\epsilon _{\alpha \beta \gamma \delta }F^{\alpha \beta }F^{\gamma \delta
}. $ This could have been anticipated, since the Lorentz transformations are
akin to orthogonal transformations and leave invariant the secular equation
of the matrix $F_\alpha ^{\,\,\,\,\beta }.$ In fact, (\ref{4}) could be
derived without recourse to the explicit equations (\ref{3'}), directly from
(\ref{3}).

\smallskip\ 

\textbf{c. }It is clear now that only patience stands in the way of
obtaining a complete solution in every possible case.

\smallskip \ 

\textbf{d. }We shall consider only the practically important case of
``crossed $\mathbf{E}$ and $\mathbf{B}$ fields'', where $\mathbf{B\cdot E=}0%
\mathbf{,}$ or $E_{/\!/}=0$ in (\ref{3'}). In this case the frequencies are
simply $\omega =0$ (double root) and $\omega =\pm (e^2/mc)\left(
B^2-E^2\right) ^{1/2}.$ This is quite similar to the cases already discussed
and one can in fact recover the simple equations by using $Bu^3-Eu^0$ and $%
Eu^3-Bu^0$ as new variables.

\smallskip\ 

\textbf{d. }The form of the new variables suggests that we perform a Lorentz
transformation on the equations before solving.

\quad - For $B>E,$ we go to a frame K$^{\prime }$ drifting with velocity $%
cE/B.$ in the $x^{3}$ direction. In this frame $E^{\prime }=0,$ $B^{\prime
}=\left( B^{2}-E^{2}\right) ^{1/2}.$ In general, the drift velocity is $%
c\left( \,\mathbf{E}\times \mathbf{B}\right) /B^{2}.$

\quad - For $E>B,$ we go to a frame moving with velocity $cB/E$ in the $x^{3}
$ direction. In this frame $B^{\prime }=0,$ $E^{\prime }=\left(
E^{2}-B^{2}\right) ^{1/2}.$ In general, the drift velocity is $c\left( \,%
\mathbf{E}\times \mathbf{B}\right) /E^{2}.$

The motion is conveniently parametrized by the the particle's proper time $%
\tau $ or by the time $t^{\prime }$ in the drifting frame K$^{\prime }$. In
other words, it is easy to get $\mathbf{x}(\tau )$ or $\mathbf{x}(t^{\prime
}),$ but it is not a good idea to compute $d\mathbf{x}/dt$ and try to
integrate it to get $\mathbf{x}(t).$ The $t$-dependence can be obtained
implicitly from $\mathbf{x}(\tau )$ and $t(\tau ),$ or from $\mathbf{x}%
(t^{\prime })$ and $t(t^{\prime })$

\smallskip

\newpage

\subsection{12.3.6}

\quad \quad \textbf{a. }Here are explicit formulas for the case $E<B.$. We
start from the $\tau \,$-\thinspace dependent equations corresponding to (%
\ref{3'}): 
\begin{eqnarray}
du^{0}/d\tau  &=&(eE/mc)\,u^{1} \\
du^{1}/d\tau  &=&-(e/mc)\left( Bu^{3}-Eu^{0}\right)   \nonumber \\
du^{2}/d\tau  &=&0  \nonumber \\
du^{3}/d\tau  &=&(eB/mc)\,u^{1}  \nonumber
\end{eqnarray}
and we go to the drifting frame K$^{\prime }$ by defining 
\[
u^{\prime \,0}=\,\frac{Bu^{0}-Eu^{3}}{\sqrt{B^{2}-E^{2}}}\quad \quad \quad
u^{\prime \,3}=\frac{Bu^{3}-Eu^{0}}{\sqrt{B^{2}-E^{2}}}
\]
We take $u^{2}=0$ for simplicity and we find 
\[
du^{\prime \,0}/d\tau =0\;\;\;\;\;du^{1}/d\tau =-\omega u^{\prime \,3}\quad
\quad du^{\prime \,3}/d\tau =\omega u^{1}
\]
with $\omega =\dfrac{e}{mc}\,\sqrt{B^{2}-E^{2}}.$ With a suitable choice of
integration constants: 
\[
u^{\prime \,0}=\gamma ^{\,\prime }c\quad \quad u^{1}=\gamma ^{\,\prime
}v^{\prime }\cos \omega \tau \quad \quad u^{\prime \,3}=\gamma ^{\,\prime
}v^{\prime }\sin \omega \tau ,
\]
where $v^{\prime }$ is the speed in the frame K$^{\prime }$ and $\gamma
^{\,\prime }=\left[ 1-(v^{\prime }/c)^{2}\right] ^{-1/2}$ as usual. A
further integration gives 
\[
x^{\prime \,0}=\gamma ^{\,\prime }c\tau \quad \quad x^{1}=-\,a\cos \omega
\tau \quad \quad x^{\prime \,3}=a\sin \omega \tau ,
\]
where $a=\gamma ^{\,\prime }v^{\prime }/\omega .$ The transformation from $%
\left( u^{0},\,u^{3}\right) $ to $\left( u^{\prime \,0},\,u^{\prime
\,3}\right) $ is a Lorentz boost with velocity $c(E/B).$ We can apply the
inverse transformation to obtain $\left( x^{0},\,x^{3}\right) $ from $\left(
x^{\prime \,0},\,x^{\prime \,3}\right) ,$ with the result 
\begin{eqnarray*}
x^{0} &=&\,\frac{Bx^{\prime \,0}+Ex^{\prime \,3}}{\sqrt{B^{2}-E^{2}}}=\frac{%
B\gamma ^{\,\prime }c\tau +Ea\cos \omega \tau }{\sqrt{B^{2}-E^{2}}} \\
x^{3} &=&\,\frac{Bx^{\prime \,3}+Ex^{\prime \,0}}{\sqrt{B^{2}-E^{2}}}=\frac{%
Ba\cos \omega \tau +E\gamma ^{\,\prime }c\tau }{\sqrt{B^{2}-E^{2}}}
\end{eqnarray*}
We may prefer to introduce the time $t^{\prime }=$ $x^{\prime \,0}/c=$ $%
\gamma ^{\prime }\tau $ and the corresponding frequency $\omega ^{\prime
}=\omega /\gamma ^{\prime }$ in the drifting frame. Or we may use as
parameter the invariant phase $\phi =\omega \tau =\omega ^{\prime }t^{\prime
}$ and write the trajectory in the form 
\[
\frac{\omega ^{\prime }}{c}\,x^{3}=\frac{v^{\prime }/c}{\sqrt{1-b^{2}}}%
\,\cos \phi +\frac{b\phi }{\sqrt{1-b^{2}}}\quad \quad \quad \frac{\omega
^{\prime }}{c}\,x^{1}=-\,\frac{v^{\prime }}{c}\,\sin \phi 
\]
where $b=E/B.$ The graph shows the trajectories corresponding to $\dfrac{%
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