Lorentz Transforms of Electromagnetic Fields

[Glenn Rowe]

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.

 

[Dale]

Summary:: A simple(?) question about the cross product of the transformed electromagnetic fields.

if the perpendicular components in the transformed frame are parallel to each other,

Why would they be parallel to each other? As you show next they are not except in unusual circumstances, but I don’t understand why you thought that they are in the first place.

 

[Ibix]

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*}
$$

It isn't $\mathbf{n}\times\mathbf{B}_\perp$ in the second equation, though. It's $\mathbf{n}\times\mathbf{B}$, and similarly the third equation.

 

[Glenn Rowe]

I guess it is the notation that misled me. They are written as the components of E and B perpendicular to the motion in the transformed frame, so it would seem that they should be parallel.

 

[Dale]

I guess it is the notation that misled me. They are written as the components of E and B perpendicular to the motion in the transformed frame, so it would seem that they should be parallel.

Two vectors, each perpendicular to a third vector, are not necessarily parallel to each other. The set of all vectors perpendicular to the third vector form a plane. So the two vectors can be any pair of vectors in that plane. So the two vectors can have any angle between them.

 

[vanhees71]

I guess it is the notation that misled me. They are written as the components of E and B perpendicular to the motion in the transformed frame, so it would seem that they should be parallel.

Since $\vec{n} \times \vec{E}=\vec{n} \times \vec{E}_{\perp}$, it's ok.

 

[Glenn Rowe]

Two vectors, each perpendicular to a third vector, are not necessarily parallel to each other. The set of all vectors perpendicular to the third vector form a plane. So the two vectors can be any pair of vectors in that plane. So the two vectors can have any angle between them.

Of course. Stupid of me not to notice that. I guess my mind got stuck in a rut.

 

[Dale]

Of course. Stupid of me not to notice that. I guess my mind got stuck in a rut.

No worries. This stuff is not obvious, and that is why I asked for clarification about why you reached that conclusion. I needed to know where the trouble was out of all the many possibilities