Suppose I have a uniform magnetic field through all of space $$\textbf{B}(x,y,z)=\hat{\textbf{z}}$$ and a charge $q$ moving at a velocity of $v\hat{\textbf{x}}$. In this frame of reference, a magnetic force will act on $q$, pushing it up (or down, but let's assume up). Suppose instead I change my frame of reference so that $q$ is stationary at the origin. In this frame of reference, there is no force to act on $q$ and it shouldn't move. Naturally, this seems inconsistent.

The normal explanation given for this from what I've seen is that the magnetic field changes into an electric field upon a change of reference so that there is still a force in the second frame of reference and the results agree. I have some problems with this though:

This doesn't explain the source of the $\textbf{E}$ field in the second frame. Where does it come from? According to Maxwell's equations, $\textbf{E}$ fields can only be produced by time-varying magnetic fields or charges. There are no charges (apart from $q$) and no time-varying magnetic fields in either frame of reference - so Maxwell's equations seem to predict a zero electric field in either case. Am I wrong to say that Maxwell's equations must hold in all inertial frames of reference? How would both Maxwell's equations be satisfied and the same results be obtained upon transformation to the second frame of reference?

Let's just ignore Maxwell's equations even. Here's another reason it seems inconsistent: What if we started our analysis in the second frame of reference instead, where $q$ is stationary? In this frame, we don't expect $q$ to move upwards. Then presumably, after transforming the $\textbf{E}$ and $\textbf{B}$ fields into the first reference frame, you should still see the same result (no upwards movement). So explaining the problem as "Lorentz transforms make $\textbf{B}$ into $\textbf{E}$" doesn't seem to cut it, you could just as well do the analysis backwards and get contradictory results.

Maybe you could argue that due to the ultimate symmetry of the field, there is no way to distinguish any movement of the charge so that it's meaningless to talk about this. This just feels like a cop-out to me. It would also seem to be easily evaded, since what if I just defined $\textbf{B}$ to be zero outside of $(-\infty,\infty)\times[-1,1]$? In this case, any upward movement of the charge would be noticeable, and we still have the same problem that we started with.

EDIT: Regarding the original source of $\textbf{B}$, let's just say that it's created by either a large solenoid or two permanent magnets so that $\textbf{B}$ is approximately uniform in the region of interest. A finite $\textbf{B}$ field like this will create induced $\textbf{E}$ fields at the fringes when it moves, but this shouldn't affect the local uniform region.

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    $\begingroup$ "This doesn't explain the source of the $E$ field in the second frame." - You haven't given a source for the $B$ field in the first frame, either! Why is the $B$ field allowed to magically exist but the $E$ field is not? $\endgroup$ – ACuriousMind Dec 1 '18 at 16:57
  • $\begingroup$ Ok let's say it's created by two infinitely long permanent magnets. Or just consider the limit as the length of the two permanent magnets approaches infinity. All my argument requires is symmetry of the magnetic field along the x-axis so that there are no induced E fields in either frame of reference. $\endgroup$ – hddh Dec 1 '18 at 17:02
  • $\begingroup$ In fact, I suspect the argument would even just work with one solenoid that is large enough so that $\textbf{B}$ is approximately uniform in the area of interest. Any induced $\textbf{E}$ fields due to $\textbf{B}$ changing at the fringes of a moving region shouldn't affect the local region of the approx. uniform $\textbf{B}$ field. $\endgroup$ – hddh Dec 1 '18 at 17:06
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    $\begingroup$ Note that Maxwells equations are differential equations, and you can add a constant $E$ or $B$ field to any solution and it will still be a solution. Thus, Maxwells equations do not predict $E=0$ as you claim, they simply predict $E$ being constant across space and time. $\endgroup$ – Eddy Dec 1 '18 at 17:18
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    $\begingroup$ if you have a magnet the magnet is moving in the reference frame in which the charge is at rest, thus generating an electric field $\endgroup$ – Wolphram jonny Dec 1 '18 at 17:23

You claim that the absence of charges or time-varying $\mathbf B$ fields implies that the $\mathbf E$ field is zero, but this is only true if we add the boundary condition that $\mathbf E=0$ at spatial infinity. The existence of a constant, uniform $\mathbf E$ field is perfectly consistent with Maxwell's equations, in precisely the same way as your initial $\mathbf B$ field is.

In any real situation, of course, your $\mathbf B$ field will be generated by sources of some description, and will not be constant or spatially uniform. The observer in the particle's rest frame will therefore see a time-varying $\mathbf B$ field and an induced $\mathbf E$ field. Maxwell's equations are preserved, and there is no contradiction between the observers.


You need to look into the math of this particular problem. We restrict ourselves to the classical interpretation. Let us have an inertial coordinate system $O\hat{x}\hat{y}\hat{z}$ and in it - a constant magnetic field $\vec{B}$ directed along the vertical axis $O\vec{z}$. You suggest that the magnetic field is described by the vertical basis unit vector $\vec{B} = \hat{z}$.

Furthermore, you have a particle $Q$ of charge $q$ and mass $m=1$ (for simplicity) moving in the magnetic field with initial velocity $\vec{v}(0) = v_0\, \hat{x}$. By $\, \vec{r} = \vec{r}(t) = x(t)\, \hat{x} + y(t) \, \hat{y} + z(t) \, \hat{z}$ denote the position vector of the particle $Q$ at time $t$.

The Lorentz law of motion states that the particle's position $\vec{r} = \vec{r}(t)$ changes with time $t$ according to the second order system of differential equations \begin{align} &\frac{d^2 \vec{r}}{dt^2}\, = \, q\frac{d \vec{r}}{dt} \times \vec{B}\, = \, q\frac{d \vec{r}}{dt} \times \hat{z}\\ & \frac{d \vec{r}}{dt}(0)\, =\, v_0 \,\hat{x}\\ & \vec{r}(0) = \vec{r}_0 \end{align} Now, observe that according to Lorentz's law, when a particle's velocity is zero, i.e. $\frac{d\vec{r}}{dt} = \vec{0}$, then the Lorentz force $q\frac{d \vec{r}}{dt} \times \vec{B} = \vec{0}$ so the said particle is stationary, that is, no force acts on it.

Next, you switch from the original inertial coordinate system $O\hat{x}\hat{y}\hat{z}$ to a non-inertial coordinate system $Q\hat{x}\hat{y}\hat{z}$ with origin at the moving charged particle $Q$ and coordinate axes, let's say, always staying parallel to the axes of $O\hat{x}\hat{y}\hat{z}$. Denote the position of any object in the non-inertial system $Q\hat{x}\hat{y}\hat{z}$ by $\vec{R} = \vec{R}(t) = X(t)\, \hat{x} + Y(t) \, \hat{y} + Z(t) \, \hat{z}$. Then the change of coordinates is given by the vector formula $$\vec{r} = \vec{R} + \vec{a}$$ where $\vec{a} = \vec{a}(t)$ is the position of the particle $Q$ at time $t$ with respect to $O\hat{x}\hat{y}\hat{z}$. Then $\vec{a} = \vec{a}(t)$ satisfies the equations $$\frac{d^2\vec{a}}{dt^2} = q \frac{da}{dt} \times \vec{B}$$ Consequently, we would like to represent Lorentz' s law in the new coordinate system $Q\hat{x}\hat{y}\hat{z}$.,

First, by differentiating the change of coordinates $\vec{r} = \vec{R} + \vec{a}$ with respect to time $t$, find that $$\frac{d\vec{r}}{dt} = \frac{d\vec{R}}{dt} + \frac{d\vec{a}}{dt}$$ and by differentiating one more time obtain $$\frac{d^2\vec{r}}{dt^2} = \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2}$$ Second, the Lorentz's law yields \begin{align} \vec{0} &= \frac{d^2 \vec{r}}{dt^2} \, - \, q\frac{d \vec{r}}{dt} \times \vec{B} = \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2} \, -\, q\left(\frac{d\vec{R}}{dt} + \frac{d\vec{a}}{dt}\right)\times \vec{B} = \\ &= \frac{d^2\vec{R}}{dt^2} + \frac{d^2\vec{a}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, - \, q\frac{d\vec{a}}{dt}\times \vec{B} = \\ &= \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, + \,\left(\frac{d^2\vec{a}}{dt^2} \, - \, q\frac{d\vec{a}}{dt}\times \vec{B} \right) =\\ & = \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \, + \, \vec{0} = \\ & = \frac{d^2\vec{R}}{dt^2} \, -\, q\frac{d\vec{R}}{dt} \times \vec{B} \end{align} Thus, the Lorentz's law in the non-inertial reference frame $Q\hat{x}\hat{y}\hat{z}$ is $$\frac{d^2\vec{R}}{dt^2} \, = \, q\frac{d\vec{R}}{dt} \times \vec{B}$$ which is exactly the same as the Lorentz's law in the inertial reference frame $O\hat{x}\hat{y}\hat{z}$. In the non-inertial frame though, the particle $Q$ is stationary, i.e. its velocity is zero. And as observed, whenever the velocity $\frac{d\vec{R}}{dt} = \vec{0}$, the Lorentz force of the form $q\frac{d\vec{R}}{dt} \times \vec{B} = \vec{0}$, and so the trajectory of $Q$ should be a fixed point (more precisely, the origin of the coordinate system) in the non-inertial system. As you can see, there is no evidence of a an electric field in the non-inertial coordinate system.

Now, if you introduce a non-inertial coordinate system attached at $Q$, which twists and rotates around $Q$, then things change substantially. Then, the fictitious forces of the non-inertial system could manifest as an additional term to the the magnetic field force and could play the role of an electric field in the Lorentz's law $$\frac{d^2\vec{r}}{dt^2} = q\vec{E}(\vec{r}, t) + q\frac{d\vec{r}}{dt}\times \vec{B}(\vec{r}, t)$$


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