- #1
blueyellow
Homework Statement
Using the transform of the electromagnetic tensor F between frames,
F'=RFR^{T}
verify that:
i) the perpendicular component of the magnetic field in the frame, S', moving with velocity v with respect to the frame S, can be found from the transform of B_{\bot} in S by
B'_{\bot}=\gamma B_{\bot} - \frac{1}{c^2}\gammav \timesE
and that
ii) the parallel component of the E-field, E(parallel), remains unchanged.
Remember that the parallel and perpendicular components are with respect to the direction of the velocity v of the S' frame in S.
The attempt at the solution:
\begin{equation*}<br /> \mbox{R}=\left[<br /> \begin{array}{rrrr}<br /> \gamma&0&0&i\beta\gamma\\<br /> 0&1&0&0\\<br /> 0&0&1&0\\<br /> -i\beta\gamma&0&0&\gamma\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
\begin{equation*}<br /> \mbox{F}=\left[<br /> \begin{array}{rrrr}<br /> 0&B_{3}&-B_{2}&\frac{-i}{c}E_{1}\\<br /> -B_{3}&0&B_{1}&\frac{-i}{c}E_{2}\\<br /> -B_{2}&-B_{1}&0&\frac{-i}{c}E_{3}\\<br /> \frac{i}{c}E_{1}&\frac{i}{c}E_{2}&\frac{i}{c}E_{3}&0\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
F'=
\begin{equation*}<br /> \left[<br /> \begin{array}{rrrr}<br /> \gamma&0&0&i\beta\gamma\\<br /> 0&1&0&0\\<br /> 0&0&1&0\\<br /> -i\beta\gamma&0&0&\gamma\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
\begin{equation*}<br /> \left[<br /> \begin{array}{rrrr}<br /> 0&B_{3}&-B_{2}&\frac{-i}{c}E_{1}\\<br /> -B_{3}&0&B_{1}&\frac{-i}{c}E_{2}\\<br /> -B_{2}&-B_{1}&0&\frac{-i}{c}E_{3}\\<br /> \frac{i}{c}E_{1}&\frac{i}{c}E_{2}&\frac{i}{c}E_{3}&0\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
\begin{equation*}<br /> \left[<br /> \begin{array}{rrrr}<br /> \gamma&0&0&-i\beta\gamma\\<br /> 0&1&0&0\\<br /> 0&0&1&0\\<br /> i\beta\gamma&0&0&\gamma\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
=
\begin{equation*}<br /> \left[<br /> \begin{array}{rrrr}<br /> 0&\beta_{3}\gamma - \frac{\beta\gamma E_{2}}{c}&-i\beta_{2}\gamma - \frac{\beta\gamma E_{3}}{c}&\frac{\beta^{2}\gamma^{2}E_{1}}{c} - \frac{\beta\gamma^{2}E_{a}}{c}\\<br /> -\beta_{3}\gamma + \frac{\beta\gamma E_{2}}{c}&0&\beta_{1}&-i\beta\gamma\beta_{3} - \frac{i\gamma E_{2}}{c}\\<br /> \beta_{2}\gamma + \frac{\beta\gamma E_{3}}{c}&\beta_{1}&0&-i\beta\gamma\beta_{2} - \frac{i\gamma E_{3}}{c}&\\<br /> \frac{i E_{1}\gamma^{2}}{c} - \frac{i\beta^{2}\gamma E_{1}}{c}&-i\beta\beta_{3}\gamma + \frac{i\gamma E_{2}}{c}&-\beta\beta_{2}\gamma - \frac{i\gamma E_{3}}{c}&0\\<br /> \end{array}<br /> \right]<br /> \end{equation*}
But I do not see any perpendicular signs in the matrix above. And the minus and cross signs are not there either. So how is one supposed to go from that matrix to the equation:
B'_{\bot}=\gamma B_{\bot} - \frac{1}{c^2}\gammav \timesE
I have looked through textbooks and the internet, but nowhere, it seems, will tell me how to do this question or a similar question.
Please help.
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