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I'm trying to understand something from my Special Relativity notes.

Assume here that $\vec{E}$ and $\vec{B}$ are perpendicular and $|\vec{E}|>|\vec{B}|$.

Show that there is a frame $\bar{\Sigma}$ in which $\bar{\vec{B}}=0$.

This I understand, if we imagine having a boost in the normal direction to both $\vec{E}$ and $\vec{B}$ then $\vec{v} \times \vec{E}$ has magnitude $vE$ and $\vec{v} . \vec{B}=0$. Using the Lorentz Boosted Magnetic Field formula we can then find:

$$v=\frac{Bc^2}{E}$$

Show that the boost to reach this frame has velocity $\vec{v}=c^2 \frac{\vec{E}\times\vec{B}}{|\vec{E}|^2}$

In the details given by the lecturer: $\vec{v} = v\frac{\vec{E} \times \vec{B}}{EB}$

I don't understand where this formula comes from, any ideas ?

Many thanks!

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  • $\begingroup$ $\vec v\times\vec E=vE$ can't be right: the left is a vector and the right is a scalar. $\endgroup$
    – HTNW
    Commented Mar 22, 2023 at 16:59
  • $\begingroup$ @HTNW yes you are right I have made the edit $\endgroup$
    – bsaoptima
    Commented Mar 22, 2023 at 17:06
  • $\begingroup$ Considering the magnitude of E and B are in different units, $|E|> |B|$ doesn't really make sense. Should you not have $|E|>c|B|$? $\endgroup$
    – Triatticus
    Commented Mar 22, 2023 at 17:56

1 Answer 1

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The formulas are equivalent to your paragraph. $E\times B$ is in the direction of the normal of the plane spanned by $E,B$. Since $E,B$ are perpendicular, $$ |E\times B|=|E||B| $$ So $\frac{E\times B}{|E||B|}$ is merely the normalized normal vector.

Injecting your formula for $|v|$, you recover the formula for $v$.

Btw, before finding a boost, a good sanity check is to use the known Lorentz invariants of the EM field and see whether they are compatible with the sought transformation. Even if the components are not Lorentz invariant, the following expressions are ($c=1$): $$ E^2-B^2\quad E\cdot B $$

Hope this helps.

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  • $\begingroup$ @Ipz thank you for the response, why would be want to use the normalized normal vector here? $\endgroup$
    – bsaoptima
    Commented Mar 23, 2023 at 9:34
  • $\begingroup$ you said it yourself, you want "a boost in the normal direction to both 𝐸⃗ and 𝐵⃗", so while you calculated the necessary speed, if you want the full velocity, you'll need the normal unit vector. $\endgroup$
    – LPZ
    Commented Mar 23, 2023 at 10:00
  • $\begingroup$ oh yes that makes sense now, thank you! $\endgroup$
    – bsaoptima
    Commented Mar 23, 2023 at 14:15

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