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In inertial frame $\mathcal{O}$, a region of space-time is filled with constant electric field $\vec{E}$ and magnetic field $\vec{B}$. Another inertial frame $\mathcal{O}'$ has 3-velocity $\vec{V}$ relative to $\mathcal{O}$. What is the electromagnetic field $\left(\vec{E}', \vec{B}'\right)$ measured in $\mathcal{O}'$? Express the result in terms of $\vec{E}$, $\vec{B}$, $\vec{V}$, dot product ($\cdot$) and cross product ($\times$).

This is to get the general formula for the boost transformation of the electromagnetic fields. I know the general form of the Lorentz boost transformation. So, obviously the solution for this problem seems to be applying this boost transformation to the electromagnetic field tensor $F^{uv}$. That is, for the boost transformation $\Lambda^u_v$, calculate $F'^{ab}=\Lambda^a_u \Lambda^b_vF^{uv}$. But this seems like a tremendous amount of work... Is there any more efficient solution than this? Could anyone suggest me?

Qmechanic
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Keith
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2 Answers2

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Yes of course. There are three different methods at least

  • using transformation of the electromagnetic field tensor $F^{\mu\nu}$ (which is TOO LONG!!!)
  • using the transformation of 4-vector potential $\mathbf A^\mu$
  • using the Lorentz force $\mathbf{F}=q \mathbf{E}+q \mathbf{v}\times\mathbf{B}$
BioPhysicist
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El-Mo
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The general transformation I know is just $$\vec{E}' = \vec{E}_{\parallel} + \gamma(\vec{E}_{\perp} + \vec{v}\times \vec{B})\ , $$ where $\vec{E}_{\parallel}$ and $\vec{E}_\perp$ are the components parallel and perpendicular to the velocity difference between the frames. Hence $$\vec{E}' = \frac{\vec{v}}{v^2}\vec{E}\cdot \vec{v} +\gamma\left(\vec{E} - \frac{\vec{v}}{v^2}\vec{E}\cdot \vec{v} + \vec{v}\times \vec{B}\right) $$ $$\vec{E}' = \gamma\left(\vec{E} + \vec{v}\times \vec{B}\right) - \frac{(\gamma - 1)}{v^2}(\vec{E}\cdot \vec{v})\vec{v}\ , $$ with something similar for the B-field: $$\vec{B}' = \gamma(\vec{B} - \vec{v}\times \vec{E}) - \frac{(\gamma - 1)}{v^2}(\vec{B}\cdot \vec{v})\vec{v}\ . $$

See for example here.

ProfRob
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