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Lorentz Transforming the electric field and the change of its directions

Physics Asked by Raeed Mundow on July 31, 2021

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:
$$ F^{munu} = Lambda^mu_alphaLambda^nu_beta F^{alphabeta} $$
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!

One Answer

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: $$ left(begin{array}{c}E_{x'}E_{y'}E_{z'}end{array}right)=left(begin{array}{c}E_xgamma,E_y-gamma,v,B_zgamma,E_z+gamma,v,B_yend{array}right),,~~ left(begin{array}{c}B_{x'}B_{y'}B_{z'}end{array}right)=left(begin{array}{c}B_xgamma,B_y+gamma,frac{v}{c^2},E_zgamma,B_z-gamma,frac{v}{c^2},E_yend{array}right),,~~gamma=frac{1}{sqrt{1-v^2/c^2}},. $$ It is correct that the field in the $x$-direction is unchanged.

Answered by Kurt G. on July 31, 2021

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