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This is a question related to the an answer for this question regarding boosts which make $\vec{E}$ parallel to $\vec{B}$ in the boosted frame.

I do understand the approach and the cases 1a and 1b. However, I don't see how we arrive at the condition for $\alpha$ for case 2.

I tried to make use of the cross product of the parallel vectors $$ \vec{E'} \times \vec{B'} = 0 $$

but could not find the solution. I also tried to argue that the boosted $\vec{E'}$ and $\vec{B'}$ must have the same components for $\vec{E}$ and $\vec{B}$. However, this leads to the following two equations: $$ 1 - \alpha B^2 = (\vec{E} \cdot \vec{B}) \alpha $$ $$ 1 - \alpha E^2 = (\vec{E} \cdot \vec{B}) \alpha $$

and results in $\alpha = 0$ which is not a valid solution. And yes, this approach is wrong because only the direction must be the same and not the length. But I fail to express this in math. What did I miss?

Note, I have to open a new question as I cannot add comments (yet).

Qmechanic
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tasty
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1 Answers1

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We have $$ \vec{E'} = \gamma(\vec{E}+\vec{v'}\times\vec{B}) = (1 - \alpha B^2) \vec{E}+\alpha (\vec{E} \cdot \vec{B}) \vec{B} \\ \vec{B'} = \gamma(\vec{B}-\vec{v'}\times\vec{E}) = (1 - \alpha E^2) \vec{B}+\alpha (\vec{E} \cdot \vec{B}) \vec{E} $$ If $\vec{E}'$ and $\vec{B}'$ are parallel, then there exists a $\lambda$ such that $\vec{E}' = \lambda \vec{B}'$. In particular, this means that (taking $\vec{E}$ and $\vec{B}$ as basis vectors for our space) the coefficients of $\vec{E}$ and $\vec{B}$ in each expression have the same ratio; they need not : $$ \frac{1 - \alpha B^2}{\alpha \vec{E} \cdot \vec{B}} = \lambda = \frac{\alpha \vec{E} \cdot \vec{B}}{1 - \alpha E^2} $$ This equation can be rearranged to read $$ 1 - \alpha(B^2 + E^2) = \alpha^2 \left[ \left(\vec{E} \cdot \vec{B}\right)^2 - E^2 B^2 \right] = -\alpha^2 (\vec{E} \times \vec{B})^2 $$ using the identity $(\vec{A} \times \vec{B})^2 = A^2 B^2 - (\vec{A} \cdot \vec{B})^2$.

The final equation in the previous answer follows from these last identities.

Michael Seifert
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  • Dear Michael, two notes : First, the velocity $\vec{v'}$ in your equations must be unprimed since it is that of the system S' with respect to S. It's a vector in S. Second, making the hypothesis that $:\vec v=\alpha(\vec E \times\vec B):$ is the reason for missing the fact that there exists a continuous 1-parametric set of solutions (boosts) with velocities $\vec{\mathrm w}$ not parallel to $:(\vec E \times\vec B)$, see here Wigner Rotation of static E & M fields is dizzying – Frobenius Feb 03 '24 at 09:15
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    @Frobenius: The $\vec{v}'$ notation is what was used in the answer the OP was asking about, so that's what I used here. – Michael Seifert Feb 03 '24 at 13:26