I hear this a lot and obviously know about the Veritasium video. When the current is not flowing through the wire, we know that a positive charge would not be repelled. When current is flowing through the wire, the stationary positive charge experiences no repulsion as given by the Lorentz force; but surely, the electrons would be contracted with respect to the positive ions at rest in the stationary positive charge's frame of reference as they are moving relative to it (we know the default case where no current was flowing, the charges were of equal distribution to cause 0 repulsion, so now they can't be); so according to special relativity, it would attract? This is where the dichotomy lies for me.

Also I don't really understand people saying that magnetism is completely a figment of special relativity, because what about intrinsic magnetism (ie spin and orbital angular momentum) in bar magnets where the lorentz force still applies?

  • $\begingroup$ I've also seen the channel, the dude tried to convince you by applying length contraction in electron then proton reference frame there will be magnetic force at work. $\endgroup$
    – user6760
    Feb 29, 2020 at 11:21

3 Answers 3


Magnetism is not solely a relativistic effect.

The key point people are trying to make when they talk about this is that the electric field $E$ and magnetic field $B$ are not relativistically invariant quantities: different observers will measure different values of them. The simplest example is a single electric charge; in its rest frame there is no magnetic field, but it generates a magnetic field in a frame where it is moving.

In the one-charge case, there is a frame where there is no magnetic field. However, in a more complicated system such as two charges moving towards each other, or a permanent magnet, there is no frame transformation that can get rid of the magnetic field.


I'll answer to your question giving to you the concrete calculation of the magnetic field seen by an observer when a point charge moves with a velocity $v$, and for simplicity i'll make it a straight-line path.

First of all, the electromagnetic field is given by a rank $2$ tensor, the EM-tensor $F^{\mu\nu}$. Being a tensor, under a Lorentz transformation, transforms as

$$ F^{\prime\mu\nu} = \frac{\partial x^{\prime\mu}}{\partial x^\alpha}\frac{\partial x^{\prime\nu}}{\partial x^\beta} F^{\alpha\beta} $$

Taking the simplest Lorentz transformation, a boost along the $x$-axis, you can easily see that the components of the electric and magnetic field, in the frame $K^\prime$ mix in this way

$$ E_x^\prime = E_x \qquad B_x^\prime = B_x\\ E_y^\prime = \gamma(E_y-\beta B_z) \qquad B_y^\prime = \gamma(B_y+\beta E_z)\\ E_z^\prime = \gamma(E_z+\beta B_y) \qquad B_z^\prime = \gamma(B_z-\beta E_y) $$

From this we find the illuminating fact that in a frame where the charge is moving, you'll measure a magnetic field as well as an electric field. In fact suppose that the charge $q$, centred in the frame $K^\prime$, is moving along the $x$-axis with a velocity $v$, and the distance of closest approach to the observer is $b$. Morover suppose that the two frames coincide at time $t=t^\prime=0$. Name the point at which the observes measures the fields in his frame $P$ which has coordinates in the moving frame $$P^\prime = (-vt^\prime, b, 0) $$ and it's at a distance $$r^\prime = \sqrt{b^2+(vt^\prime)^2} $$ The electric field in the frame of the charge is clearly

$$ E_x^\prime = -\frac{qvt^\prime}{r^{\prime 3}}\qquad E_y^\prime = \frac{qb}{r^{\prime 3}} \qquad E_z^\prime = 0$$

the magnetic field is zero everywhere. Using Lorentz transformation we can write this fields in the coordinate of the frame $K$

$$ E_x^\prime = -\frac{q\gamma vt}{(b^2+\gamma^2 v^2 t^2)^{3/2}} \\ E_y^\prime = \frac{qb}{(b^2+\gamma^2 v^2 t^2)^{3/2}} $$

Then, by using the field transformation found before we get our searched result

$$ E_x = E_x^\prime = -\frac{q\gamma vt}{(b^2+\gamma^2 v^2 t^2)^{3/2}}\\ E_y = \gamma E_y^\prime = \frac{\gamma qb}{(b^2+\gamma^2 v^2 t^2)^{3/2}}\\ \color{red}{B_z = \gamma\beta E_y^\prime = \beta E_y} = \frac{\beta\gamma qb}{(b^2+\gamma^2 v^2 t^2)^{3/2}} $$

And here you go

A person in a frame $K$ at rest with respect to a uniformly moving charge, whose frame is $K^\prime$, will measure a magnetic field as well as an electric field.

A more in depth treatment of the subject can be found on Jackson's book, obviously!


If you look at the classical Lorentz force relation you will notice that Lorentz force is velocity dependent. One simple example is a proton beam or $\alpha$ particle beam. In a stationary frame (or lab frame) the beam is moving very fast, hence there is a huge magnetic force that keeps the beam together. But in the frame of the $\alpha$ particles the electric forces should be dominant and they should fly apart. This was the paradox that led Einstein to relativity.

Let there be a lorentz transformation $\textbf{x}\rightarrow\textbf{x'}$. Where $\textbf{x}$ is the lab frame spacetime coordinates and $\textbf{x'}$ is the spacetime coordinates of a frame moving with the particle beam. The transformation can be expressed as $$x'^{\mu} = \Lambda^{\mu}_{\nu} x^{\nu}$$ The electromagnetic four potential will also transform according to the Lorentz transformation. $$A'^{\mu}=\Lambda^{\mu}_{\nu} A^{\nu}$$ From here the electric and magnetic fields can be determined. When this transformation is considered the net forces will be same because the fields are now Lorentz invariant.

Hope this helps.


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