I know that the magnetism is just electricity and due to length contraction and the change in charge density moving charges experience what we call magnetic force. so my question is can we calculate that force using electricity stuff (like electric field and the permittivity of free space) instead of using magnetic field?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Yes, we can and you know this. $\endgroup$– Vladimir KalitvianskiCommented Dec 21, 2018 at 10:51
-
1$\begingroup$ @VladimirKalitvianski Not always. I challenge you to find a reference frame where an ellipse-shaped current loop can be described only in terms of an electric field. $\endgroup$– probably_someoneCommented Dec 21, 2018 at 11:12
-
1$\begingroup$ Apply the Lorentz transformation to the electromagnetic field tensor, or the 4-potential (phi, A_i). $\endgroup$– user196418Commented Dec 21, 2018 at 12:23
-
1$\begingroup$ Next to a electrically neutral wire with a current $I$ the magnetic field $B$ is big. It may act on a moving charge according to the Lorentz force $F\propto v\times B$. In a co-moving reference frame the charge is still ($v'=0$) so the force from the wire is purely electric $F'\propto E'$ which is possible due to non zero charge density $\rho'$ in this reference frame. $\endgroup$– Vladimir KalitvianskiCommented Dec 21, 2018 at 16:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
I dont think you can, because $E^2-B^2$ is an invariant, and if $|B|>|E|$ in one reference frame it will be the same in all reference frames, so you cannnot always eliminate $ B $.
Another, intuitive way to see that, even if they are related they are also independent is that if magnetic monopones would exist, then you would have a source of magnetic fields that cannot be eliminated, as the magnetic charge would also be a constant in any reference frame.
-
1$\begingroup$ The quantity $E^2-B^2$ is a local function and vary from one place to another. The statement to prove is simple: can one find a reference frame where the magnetic force is absent and there is an electric force solely? The answer is yes. Instantly the force on a charge can be reduced to an electric one even if there is some magnetic field too. One makes the magnetic force zero in any co-moving RF due to $v'=0$. $\endgroup$ Commented Dec 21, 2018 at 17:20
-
$\begingroup$ oh, then I missinterpreted the question, I thought it meant to eliminate the magnetic field. But even if you are right about the locality, as soon as you have three or more charges moving in the correct different directions, you can no longer eliminate the magnetic force from all of them, or am I missing something? $\endgroup$– user65081Commented Dec 21, 2018 at 17:27
-
$\begingroup$ No, one cannot eliminate the magnetic field in all the space, even instantly, so you are right. In case of a straight neutral wire with $I$, the wire looks charged in a moving RF and the charge density $\rho'$ depends on velocity and velocity sign ;-) $\endgroup$ Commented Dec 21, 2018 at 17:29