This is a two part question about the Lorentz transformation of the electromagnetic field, the electric field to specific. The Lorentz transformation will be a simple boost in the x direction.

first question: Can I transform the Electric field without the need of Electromagnetic tensor. for example, instead of using: $$ F^{\mu\nu} = \Lambda^\mu_\alpha\Lambda^\nu_\beta F^{\alpha\beta} $$ can I use: $$ E^\mu = \Lambda^\mu_\alpha E^\alpha $$ assuming I add a zero to the electric field to turn in into 4 vector. $$ (0,E^1,E^2,E^3)$$ because when I use this approach I don't get the relation: $$ E^{'}_\Vert = E_\Vert $$

This brings me to my second question, the last relation I mentioned doesn't make sense to me. does it assume the x' is in the same direction as x? isn't Lorentz transformation basically a rotation? I can derive it easily by transforming the Tensor and getting $$ F^{1'0'} = F^{10} $$ but this just shows that the Electric field in new coordinates in the x' direction has the same value as the electric field in the original coordinates in the x axis right?

Sorry for the long post, I would appreciate any input!

  • 1
    $\begingroup$ It is straightforward to check your proposal is incorrect. Just compute both and note they do not match. $\endgroup$ Jun 24 at 20:21

The formula $E^{\mu'}={\Lambda_\alpha}^{\mu'} E_\alpha$ is not correct. See Transformation of Electro Magnetic Field how the electromagnetic fields transforms. In particular: the transformed field in the $y$- and $z$-direction is a linear combination of the $E$ and $B$ field: $$ \left(\begin{array}{c}E_{x'}\\E_{y'}\\E_{z'}\end{array}\right)=\left(\begin{array}{c}E_x\\\gamma\,E_y-\gamma\,v\,B_z\\\gamma\,E_z+\gamma\,v\,B_y\end{array}\right)\,,~~ \left(\begin{array}{c}B_{x'}\\B_{y'}\\B_{z'}\end{array}\right)=\left(\begin{array}{c}B_x\\\gamma\,B_y+\gamma\,\frac{v}{c^2}\,E_z\\\gamma\,B_z-\gamma\,\frac{v}{c^2}\,E_y\end{array}\right)\,,~~\gamma=\frac{1}{\sqrt{1-v^2/c^2}}\,. $$ It is correct that the field in the $x$-direction is unchanged.

  • $\begingroup$ Thanks for the reply, I know this much. but does this mean that the transformation I mentioned isn't mathematically correct? it's still a vector field which should be transformed. (I forgot to mention this but in the example I can simply request B=0 in the original reference frame) $\endgroup$ Jun 25 at 12:05
  • $\begingroup$ In SR and GR not everything that looks like a "vector" is one. I recommend for example Sean Carroll's book on GR. As you know the electromagnetic field is described by the Faraday tensor $F$ and this transforms as $F_{\mu'\nu'}={\Lambda_{\nu'}}^\mu{\Lambda_{\mu'}}^\nu F_{\mu\nu}\,.$ It is a good exercise to work through this by having $E$ and $B$ explicitly in $F$. $\endgroup$
    – Kurt G.
    Jun 25 at 14:02
  • $\begingroup$ In other words: what you get with the "mathematically correct" operation ${\Lambda_\alpha}^{\mu'}E^\alpha$ is something a physicist would not be very interested in. $\endgroup$
    – Kurt G.
    Jun 25 at 14:15
  • $\begingroup$ Thank you so much I'll look into it $\endgroup$ Jun 25 at 15:58

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