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%TCIDATA{Created=Sat Nov 02 15:48:50 1996}
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\markright{{\sf Transformation of E+M Fields}}
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\begin{document}


\section{11.10\quad Transformation of E+M Fields.}

\subsection{11.10.1}

\quad \thinspace \textbf{a. }In practice, it is often more instructive and
useful to work with the fields $\mathbf{E}$ and $\mathbf{B,}$ rather than
with the tensor $F^{\alpha \beta }$\thinspace . Thus we give explicit
transformation formulae from the frame $K$ to $K^{\prime }$ moving with
velocity $\mathbf{v:}$%
\begin{equation}
\mathbf{E}_{\perp }^{\prime }=\gamma \left( \mathbf{E}_{\perp }+\frac{%
\mathbf{v}}c\times \mathbf{B}\right) \quad \quad \quad \quad \mathbf{E}%
_{/\!/}^{\prime }=\mathbf{E}_{/\!/}  \label{1}
\end{equation}
\begin{equation}
\mathbf{B}_{\perp }^{\prime }=\gamma \left( \mathbf{B}_{\perp }-\frac{%
\mathbf{v}}c\times \mathbf{E}\right) \quad \quad \quad \quad \mathbf{B}%
_{/\!/}^{\prime }=\mathbf{B}_{/\!/}  \label{2}
\end{equation}

\smallskip\ 

\textbf{b}. To derive (\ref{1}) we start from 
\[
F^{\prime \,\alpha \beta }=\frac{\partial x^{\prime \,\alpha }}{\partial
x^{\gamma }}\frac{\partial x^{\prime \,\beta }}{\partial x^{\delta }}%
F^{\gamma \delta }
\]
and specialize $\alpha =0,$ $\beta =b$ (space-like). Further, we take $%
\mathbf{v}$ along the common direction of $x_{1}$ and $x_{1}^{\prime }.$

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\newpage 

\subsection{11.10.2}

\quad \quad \textbf{a}. With the above choice of axes, only $\partial
x^{\prime \,1}/\partial x^{0},$ $\partial x^{\prime \,\,0}/\partial x^{1}$
and the diagonal terms $\partial x^{\prime \,\alpha }/\partial x^{\alpha }$
are non zero (see p. 11.3.2). Hence 
\begin{eqnarray*}
F^{\prime \,\,01} &=&\frac{\partial x^{\prime \,0}}{\partial x^{0}}\frac{%
\partial x^{\prime \,1}}{\partial x^{1}}F^{01}+\frac{\partial x^{\prime \,0}%
}{\partial x^{1}}\frac{\partial x^{\prime \,1}}{\partial x^{0}}F^{10} \\
F^{\prime \,\,02} &=&\frac{\partial x^{\prime \,0}}{\partial x^{0}}\frac{%
\partial x^{\prime \,2}}{\partial x^{2}}F^{02}+\frac{\partial x^{\prime \,0}%
}{\partial x^{1}}\frac{\partial x^{\prime \,2}}{\partial x^{2}}F^{12}\,.
\end{eqnarray*}

\smallskip \ 

\textbf{b. }The first equation gives $F^{\prime \,01}=F^{01}$ because 
\[
\left( 
\begin{tabular}{ll}
$\partial x^{\prime \,0}/\partial x^{0}$ & $\partial x^{\prime \,0}/\partial
x^{1}$ \\ 
$\partial x^{\prime \,1}/\partial x^{0}$ & $\partial x^{\prime \,1}/\partial
x^{1}$%
\end{tabular}
\right) =1\,.
\]
This is simply $E_{1}^{\prime }=E_{1}.$

\smallskip\ 

\textbf{c.} The second equation gives $E_{1}^{\prime }=\gamma
(E_{2}-(v/c)B_{3})$ and a similar equation for $F^{\prime \,03}$ gives $%
E_{3}^{\prime }=\gamma (E_{3}+(v/c)B_{2})$. We have thus
proved the transformation (\ref{1}) for the electric field (the proof is
general, because we are free to choose the axes to suit us). The
transformation (\ref{2}) on $\mathbf{B}$ follows from the usual duality
argument.

\smallskip\ 

\textbf{d.} The transformation formulae for the fields show that the
parallel components are unchanged, while in the transformation of
coordinates it is the $\perp $ components that remain unaffected. This shows
once again that $\mathbf{E}$ and $\mathbf{B}$ are \textsl{not} space
components of 4-vectors.

\newpage 

\subsection{11.10.3}

\quad \quad \textbf{a. }In comparing the fields seen by two observers in
relative motion, one must keep in mind that the observers will assign
different space and time coordinates to the point of observation. Failure to
be careful results in confusion.

\smallskip\ 

\textbf{b. }As an example, consider the fields at point $P$ of the figure
(on page 1) due to the charge $q$ moving with the origin of $K^{\prime }.$
The distance $PK^{\prime }$ is $r=\sqrt{b^{2}+v^{2}t^{2}}$ according to $K,$
but it is $r^{\prime }=\sqrt{b^{2}+v^{2}t^{\prime \,2}}$ according to $%
K^{\prime }$.

\smallskip\ 

\textbf{c. }The fields are most easily computed in the frame $K^{\prime }$,
where $q$ is at rest and only an $\mathbf{E}$ field is present. We have 
\begin{equation}
E_{1}^{\prime }=-\,\frac{qvt^{\prime }}{r^{\prime \,3}}\quad \quad \quad
E_{2}^{\prime }=\frac{qb}{r^{\prime \,3}}\quad \quad \quad E_{3}^{\prime }=0.
\label{3}
\end{equation}

\smallskip\ 

\textbf{d. }The fields in the $K$ frame are, according to (\ref{1}) and (\ref
{2}), 
\begin{equation}
E_{1}=E_{1}^{\prime }\quad \quad \quad E_{2}=\gamma E_{2}^{\prime }\quad
\quad \quad B_{3}=\gamma \beta E_{2}^{\prime }=\beta E_{2}  \label{4}
\end{equation}
(we have actually used the inverse equations, which are obtained from (\ref
{1}) and (\ref{2}) by $\mathbf{E\rightleftharpoons E}^{\prime }$, $\mathbf{%
B\rightleftharpoons B}^{\prime }$ and $\mathbf{v\rightarrow -v).}$

\smallskip\ 

\textbf{e.} These formulae solve the problem, but we must still express $%
r^{\prime }$ in terms of the $K$ coordinates and time. Simply, $t^{\prime
}=\gamma t,$ so that finally
\begin{eqnarray*}
E_{1} &=&\frac{q\gamma vt}{\left( b^{2}+\gamma ^{2}v^{2}t^{2}\right) ^{3/2}}
\\
E_{2} &=&\frac{q\gamma b}{\left( b^{2}+\gamma ^{2}v^{2}t^{2}\right) ^{3/2}}
\\
B_{3} &=&\frac{v}{c}\,\,\frac{q\gamma b}{\left( b^{2}+\gamma
^{2}v^{2}t^{2}\right) ^{3/2}}
\end{eqnarray*}

\end{document}
