# Lorentz Transforming the electric field and the change of its directions

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!

• It is straightforward to check your proposal is incorrect. Just compute both and note they do not match. 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.
• 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 $F$. Jun 25 at 14:02
• 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. Jun 25 at 14:15