Imagine a static electric field shaped as a sinusoid created by an array of electrodes on an infinite line. Does an observer flying parallel inside this e-field measure an h-field ?

  • $\begingroup$ Some sort of diagram might help. In general yes, if you have a static E-field in one frame of reference then in a different moving frame of reference there will be both an E-field and B- (or H-) field. $\endgroup$
    – ProfRob
    Nov 17, 2015 at 16:52

1 Answer 1


A bit of special relativity is required to understand this. Electric and magnetic fields form a 4-vector with their potentials:

$$ A^{\mu} = (\phi, \vec{A}) $$

Where $ \phi $ is the scalar electric potential and $ \vec{A} $ is the magnetic potential. In the case you described, this would be:

$$ A^{\mu} = (\phi, 0) $$

Where $ \phi $ is constant in time and $ \vec{A} = 0 $. The observer, assuming they are moving in an inertial frame at velocity $ v $ in the x-direction, has undergone a Lorentz boost. In the observers frame, they measure $ A'^{\mu} $:

$$ A'^{\mu} = \Lambda^{\mu}_{\nu} A^{\nu} $$

Using Einstein notation. Just think of this as a linear transformation acting on the 4-vector $ A^{\nu} $ (the original 4-potential). Therefore the elements of the new 4-vector are, using $ \Lambda $ from the link provided, produced upon using the matrix:

$$ \lambda^{\mu}_{\nu} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Which results in a new potential:

$$ A'^{\mu} = (\gamma \phi, -\beta \gamma \phi, 0, 0) $$

As you can see the 4-potential to the observer moving at constant velocity $ v $ has picked up a magnetic potential component in the direction of motion. The magnetic potential looks to him to be:

$$ \vec{A} = (-\beta\gamma\phi, 0, 0) $$

Which has a non-zero curl, and therefore produces a magnetic field. Notice that both terms have a factor of $ \gamma $, but only the second term has the factor of $ \beta $. For non-relativistic speeds $ \gamma \approx 1 $ and $ \beta = v/c \approx 0 $, which reduces the potentials back to their places.


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