Let's suppose we have a charge $q$ at rest. Its velocity is $\vec{v}=(0, 0, 0)$. We place the charge inside a uniform magnetic field $\vec{B}=(0,0,B_{z})$. Because the charge has no velocity, it will not experience any force. But if we give the charge some initial velocity with respect to the $x$ axis, so that $\vec{v}=(v_{x},0,0)$, it will experience a magnetic force on the $y$ axis: $F=q(0,v_{x}B_{z},0)$.

In the proper frame of the charge, the magnetic field is moving with $\vec{v}=(-v_{x},0,0)$ and if we apply the Lorentz transformation of the magnetic field, we will get that: $B_{z}'=\gamma\left(B_{z}-\frac{v}{c^2}E_{y}\right)$. My question is this: Since the charge will be accelerated only if there is a relative motion with respect to the magnetic field source and since the magnetic field as seen from the charge's perspective under Lorentz transformation contains an electric field component on the direction of the Lorentz force, can we say that in proper frame of the charge, the magnetic Lorentz force is in fact the electric component of the Lorentz force?

Another way to put it is: If the charge is accelerated only if there is a relative motion wrt the magnetic field, it seems to me that the magnetic field acts upon the charge only if there is an electric field component in the proper frame of the charge. The argument is that the electric field has the same direction as the magnetic Lorentz force.

Another related question: Would there still be a magnetic Lorentz force acting on the charge if the magnetic field source moves along with the particle with the same velocity $\vec{v}$? I think the answer is no.

  • $\begingroup$ What do you mean by "really"? $\endgroup$
    – ACuriousMind
    Oct 26, 2017 at 12:24
  • $\begingroup$ When the charge has no relative velocity wrt $\vec{B}$, the M.F. has no effect on the charge. But when there is a relative velocity, the M.F. will deflect the charge with a force $\vec{F}=q\left(\vec{v}\times \vec{B}\right)$. But that $\vec{B}$ "seen" from the proper frame of the charge is in fact a combination of E.F. and M.F. Because in that particular case that I described above, $B_{z}$ has a $E_{y}$ component in the charge's frame and that coincides with the direction of lorentz force, isn't the E.F. responsible for accelerating the charge? (from the charge's perspective) $\endgroup$
    – Nemo
    Oct 27, 2017 at 8:11

2 Answers 2


What really 'acts' on the charge is the electromagnetic field (represented via the Faraday tensor). The motivation is following - as you noted, one observer can 'see' an electric field while the other sees the magnetic field (although the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ remains invariant, up to a corresponding Lorentz transformation). Therefore to covariantly describe the interaction you need something more complex that connects both electric and magnetic components more intimately - the electromagnetic tensor.

  • $\begingroup$ Yep. I was imprecise. Meant to say that the forces measured in both frames are connected via a Lorentz transformation of forces. $\endgroup$
    – DrLRX
    Oct 26, 2017 at 12:12

You are visualising 3D as 1D. Also when 2 frames with 3 components are involved you cannot based your analysis by looking at only one component. Note that even if say x' and x of the frames may be relatively at rest. the unit vectors do changes with respect to the other. When Bx is constant By and Bz can still change. Also note that Lorentz force is invariant under a coordinate transformation.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Nov 16, 2022 at 0:46

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