- #1
Glenn Rowe
Gold Member
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- TL;DR Summary
- A simple(?) question about the cross product of the transformed electromagnetic fields.
The Lorentz transformations of electric and magnetic fields (as given, for example in Wikipedia) are
$$
\begin{align*}
\bar{\boldsymbol{E}}_{\parallel} & =\boldsymbol{E}_{\parallel}\\
\bar{\boldsymbol{E}}_{\perp} & =\gamma\left(\boldsymbol{E}_{\perp}+\beta\boldsymbol{n}\times\boldsymbol{B}_{\perp}\right)\\
\bar{\boldsymbol{B}}_{\parallel} & =\bar{\boldsymbol{B}}_{\parallel}\\
\bar{\boldsymbol{B}}_{\perp} & =\gamma\left(\boldsymbol{B}_{\perp}-\beta\boldsymbol{n}\times\boldsymbol{E}_{\perp}\right)
\end{align*}
$$
where ##\boldsymbol{E}_{\parallel}## is a field parallel to the relative motion of the two Lorentz frames and ##\boldsymbol{E}_{\perp}## is perpendicular to the motion. The unit vector ##\boldsymbol{n}## is parallel to the direction of motion, and ##\beta## is the relative speed, with ##\gamma\equiv\frac{1}{\sqrt{1-\beta}^{2}}## as usual.
Now it would seem that if the perpendicular components in the transformed frame are parallel to each other, that is, if ##\bar{\boldsymbol{E}}_{\perp}## is parallel to ##\bar{\boldsymbol{B}}_{\perp}##, then we should have ##\bar{\boldsymbol{E}}_{\perp}\times\bar{\boldsymbol{B}}_{\perp}=0##. If we write this out, and use the fact that ##\boldsymbol{E}_{\perp}## is parallel to ##\boldsymbol{B}_{\perp}##, and that ##\boldsymbol{n}\times\boldsymbol{E}_{\perp}## is parallel to ##\boldsymbol{n}\times\boldsymbol{B}_{\perp}## then we get the condition
$$
\begin{align*}
\bar{\boldsymbol{E}}_{\perp}\times\bar{\boldsymbol{B}}_{\perp} & =-\gamma^{2}\boldsymbol{E}_{\perp}\times\left(\beta\boldsymbol{n}\times\boldsymbol{E}_{\perp}\right)+\gamma^{2}\beta\left(\boldsymbol{n}\times\boldsymbol{B}_{\perp}\right)\times\boldsymbol{B}_{\perp}\\
& =-\gamma^{2}\beta\left(E_{\perp}^{2}+B_{\perp}^{2}\right)\boldsymbol{n}=0
\end{align*}$$
where I've used the triple vector product formula
$$\begin{equation}
\boldsymbol{a}\times\left(\boldsymbol{b}\times\boldsymbol{c}\right)=\left(\boldsymbol{a}\cdot\boldsymbol{c}\right)\boldsymbol{b}-\left(\boldsymbol{a}\cdot\boldsymbol{b}\right)\boldsymbol{c}
\end{equation}$$
and the fact that ##\boldsymbol{n}## is perpendicular to ##\boldsymbol{E}_{\perp}## and to ##\boldsymbol{B}_{\perp}##.
This would appear to require that either ##\beta=0## (no motion) or that the perpendicular components of both fields are zero, which we clearly can't require in general. Can anyone see what I'm missing? Are the transformed perpendicular components not actually parallel to each other? Thanks.
$$
\begin{align*}
\bar{\boldsymbol{E}}_{\parallel} & =\boldsymbol{E}_{\parallel}\\
\bar{\boldsymbol{E}}_{\perp} & =\gamma\left(\boldsymbol{E}_{\perp}+\beta\boldsymbol{n}\times\boldsymbol{B}_{\perp}\right)\\
\bar{\boldsymbol{B}}_{\parallel} & =\bar{\boldsymbol{B}}_{\parallel}\\
\bar{\boldsymbol{B}}_{\perp} & =\gamma\left(\boldsymbol{B}_{\perp}-\beta\boldsymbol{n}\times\boldsymbol{E}_{\perp}\right)
\end{align*}
$$
where ##\boldsymbol{E}_{\parallel}## is a field parallel to the relative motion of the two Lorentz frames and ##\boldsymbol{E}_{\perp}## is perpendicular to the motion. The unit vector ##\boldsymbol{n}## is parallel to the direction of motion, and ##\beta## is the relative speed, with ##\gamma\equiv\frac{1}{\sqrt{1-\beta}^{2}}## as usual.
Now it would seem that if the perpendicular components in the transformed frame are parallel to each other, that is, if ##\bar{\boldsymbol{E}}_{\perp}## is parallel to ##\bar{\boldsymbol{B}}_{\perp}##, then we should have ##\bar{\boldsymbol{E}}_{\perp}\times\bar{\boldsymbol{B}}_{\perp}=0##. If we write this out, and use the fact that ##\boldsymbol{E}_{\perp}## is parallel to ##\boldsymbol{B}_{\perp}##, and that ##\boldsymbol{n}\times\boldsymbol{E}_{\perp}## is parallel to ##\boldsymbol{n}\times\boldsymbol{B}_{\perp}## then we get the condition
$$
\begin{align*}
\bar{\boldsymbol{E}}_{\perp}\times\bar{\boldsymbol{B}}_{\perp} & =-\gamma^{2}\boldsymbol{E}_{\perp}\times\left(\beta\boldsymbol{n}\times\boldsymbol{E}_{\perp}\right)+\gamma^{2}\beta\left(\boldsymbol{n}\times\boldsymbol{B}_{\perp}\right)\times\boldsymbol{B}_{\perp}\\
& =-\gamma^{2}\beta\left(E_{\perp}^{2}+B_{\perp}^{2}\right)\boldsymbol{n}=0
\end{align*}$$
where I've used the triple vector product formula
$$\begin{equation}
\boldsymbol{a}\times\left(\boldsymbol{b}\times\boldsymbol{c}\right)=\left(\boldsymbol{a}\cdot\boldsymbol{c}\right)\boldsymbol{b}-\left(\boldsymbol{a}\cdot\boldsymbol{b}\right)\boldsymbol{c}
\end{equation}$$
and the fact that ##\boldsymbol{n}## is perpendicular to ##\boldsymbol{E}_{\perp}## and to ##\boldsymbol{B}_{\perp}##.
This would appear to require that either ##\beta=0## (no motion) or that the perpendicular components of both fields are zero, which we clearly can't require in general. Can anyone see what I'm missing? Are the transformed perpendicular components not actually parallel to each other? Thanks.