# How to explain this simple phenomenon from a moving frame?

Let us suppose we have an electron moving with velocity $$\vec v$$ as shown in the fig. as seen from the ground frame. We know that when it enters the region with uniform magnetic field it will follow a circular trajectory.

But if the same charge is being observed from a frame moving with the charge then the charge will appear to be at rest while the region will move towards it. But since the charge is at rest it can't experience any magnetic force. So how do we explain its circular trajectory ?

Now one reason that a friend of mine told me is that there will be an electric field that will cause the charge to move but I didn't really get it.

I know that a time varying magnetic field produces a non conservative electric field but here in this case since the field is uniform everywhere so I don't think the magnetic field at any point is changing (when observed from moving frame). All that is happening is that each point is replaced by its neighbor point with same magnitude of magnetic field.

So how does it really work ?

• hmm..maybe a time varying magnetic field doesn't induce an electric field?
– JEB
Commented Aug 5, 2023 at 4:48
• @JEB then how to explain the path ;) ? Commented Aug 5, 2023 at 5:02
• tl;dr Is that the question that led Einstein and Lorentz develop SR? If so there are zillions of pedagogical papers in the internet, even from Feynman. Einstein's seminal SR paper from 1905 carries your question in the title and answers it. Commented Aug 5, 2023 at 6:59
• The electric field from the charges in the moving solenoid...the one that made the B field in the figure.
– JEB
Commented Aug 5, 2023 at 17:00
• In the charge's own reference frame, it is at rest; it does not have a circular trajectory. You seem to be trying to describe the particle in two different reference frames at once, as you are trying to describe the circular trajectory of a particle in one reference frame while treating it at rest in a different frame. In general, anytime you're in a non-inertial frame, you wind up with fictitious forces that have no physical explanation or cause - these forces can take arbitrary values and arise solely due to your particular choice of reference frame. Commented Aug 8, 2023 at 15:04

Questions like this are best handled using the standard tensor-based approach for the covariant formulation of classical electromagnetism. In this approach the electric and magnetic fields are combined into a single tensor as follows: $$F_{\mu \nu }=\left( \begin{array}{cccc} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \\ \end{array} \right)$$

In your specific case there is only $$B_z\ne 0$$ so $$F_{\mu \nu }=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -B_z & 0 \\ 0 & B_z & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

Now, you would like to see how this behaves in the charge's reference frame. The charge's reference frame is a rotating reference frame. So the transformation is $$t \rightarrow T$$$$x \rightarrow X \cos(\omega T) - Y \sin(\omega T)$$$$y \rightarrow Y \cos(\omega T) + X \sin(\omega T)$$$$z \rightarrow Z$$ where the cyclotron frequency is $$\omega$$.

Applying this transformation we get $$F_{M N}=\left( \begin{array}{cccc} 0 & X \omega B_z & Y \omega B_z & 0 \\ -X \omega B_z & 0 & -B_z & 0 \\ -Y \omega B_z & B_z & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

Compare this last expression to the first expression. We see that in this frame there is now an E-field. This E-field is proportional to the B-field and the cyclotron frequency. Furthermore, this E-field is directed radially. So this field is a real electromagnetic centripetal force which counteracts the fictitious centrifugal force and allows the charge to remain at rest in this non-inertial rotating frame.

Get["OGRe.m", Path -> NotebookDirectory[]]
TSetAllowOverwrite[True]

TNewCoordinates["Unprimed", {t, x, y, z}]
TNewCoordinates["Rotating", {T, X, Y, Z}]
"Unprimed" -> "Rotating", {t -> T,
x -> X Cos[\[Omega] T] - Y Sin[\[Omega] T],
y -> Y Cos[\[Omega] T] + X Sin[\[Omega] T], z -> Z}]
TNewMetric["Minkowski", "Unprimed", DiagonalMatrix[{-c^2, 1, 1, 1}], "\[Eta]"]

TShow@TNewTensor["EM tensor", "Minkowski",
"Unprimed", {1,  1},
{{0, -Ex, -Ey, -Ez},
{Ex,  0, -Bz,  By},
{Ey, Bz,   0, -Bx},
{Ez,-By,  Bx,   0}}, "F"]
TShow["EM tensor", {-1, -1}, "Unprimed"]
TShow["EM tensor", {-1, -1}, "Rotating"]

TShow@TNewTensor["EM tensor", "Minkowski",
"Unprimed", {1, 1},
{{0, -0,  -0, -0},
{0,  0, -Bz,  0},
{0, Bz,   0, -0},
{0, -0,   0,  0}}, "F"]
TShow["EM tensor", {-1, -1}, "Unprimed"]
TShow["EM tensor", {-1, -1}, "Rotating"]

• How did you apply the transformation? Commented Jul 21 at 9:08
• @Allure I used Mathematica
– Dale
Commented Jul 21 at 10:55
• Do you have a screenshot of the code used or similar? Commented Jul 21 at 11:03
• @Allure see edits
– Dale
Commented Jul 21 at 11:57

So the "magnetic field" is created by a solenoid with a current, moving electrons.

For the charge q moving to the right, some extra electrons seem to get packed to the lower part of the solenoid, and the electrons seem to be less densely packed on the upper part of the solenoid.

So the charge q thinks or feels that there is an electric field, and behaves accordingly.

Oh yes I almost forgot, there is induction too as charge q moves away from the left part of the solenoid, and towards the right part of the solenoid. Induction is an effect where a moving charge thinks that charges moving sideways across its path have squished and tilted electric fields.

See "A Charge Moving Perpendicular to a Wire" https://physics.weber.edu/schroeder/mrr/MRRtalk.html

Consider an inertial frame of reference a little distance away from $$q$$ moving at $$v$$ in the same direction that $$q$$ moves. Also, consider that this frame is electrically neutral so that no EM fields affect it. Therefore, from the viewpoint of the observer located in this frame, the electron is at rest WRT him. However, the observer asserts that the moving magnetic field (shown in the figure) would produce an electric field of magnitude:

$$E_y=\gamma v B'_z \space,$$

where $$B'_z$$ is the magnetic field measured in the rest frame of the source that generates the field such as a magnet or solenoid. The produced $$E_y$$ makes the charge recede from the observer while undergoing a constant force of $$F_y=E_yq$$ along $$y$$. Remember that this observer declines any circular trajectory but rather states that the charge accelerates away from him along $$y$$.

the field is uniform everywhere so I don't think the magnetic field at any point is changing

This is not the case. The magnetic field does change at infinity where the magnet edges are located. At infinity, $$\partial B/\partial t$$ can be defined/determined.

So how does it really work ?

As I understand it, it works through Lorentz transformation of the electromagnetic field, as Dale explained.

I know that a time varying magnetic field produces a non conservative electric field but here in this case since the field is uniform everywhere so I don't think the magnetic field at any point is changing

I mainly want to address the question: can it also be seen as an electric field generated by a changing magnetic field? I believe the answer is no. This can be shown by a simple example:

Consider an infinitely long straight wire with a steady current. (I mostly took this from Feynman lectures, volume II, 13-6). Far from the wire and for a particle with small velocity so that its movement is localized in a small region, the magnetic field involved becomes better and better approximation to a uniform field. Assume the current is caused by positive charges moving with velocity $$\vec{v}$$. For later use one has $$\rho_+ A v=I$$, where $$\rho_+$$ is the charge density of positive charges, $$A$$ is the cross-sectional area of the wire which is assumed to be very small, $$I$$ is the current of the wire.

One can calculate the magnetic force acted on a particle moving with the same velocity $$\vec{v}$$ at distance $$r$$ from the wire using $$B=\frac{\mu_0 I}{2 \pi r}$$ and the Lorentz force law. The result is $$F=q \rho_+ v^2 \frac{\mu_0}{2 \pi r} A , \,\,\,\,\,\,\,\,\,\,\,(1)$$ where $$q$$ is the charge of the particle. Take $$q$$ to be positive. The force is pointing from the particle to the wire. Use has been made of $$I = \rho_+ A v$$.

In the rest frame of the particle, the positive charges now are not moving, so is de-Lorentz-contracted by a factor $$\gamma=1/\sqrt{1-v^2/c^2}$$ and becoming $$\rho_+/\gamma$$, while the negative charges now are moving, so is contracted by a factor $$\gamma$$ and becoming $$\gamma\rho_-$$. Using that in the original frame $$\rho_- + \rho_+=0$$, here we have the total $$\rho=\rho_- \gamma v^2/c^2$$. Now using Gauss's law, one can get the electric field at distance $$r$$ from the wire. After multiplying the charge $$q$$, one get the magnitude of the force on the particle $$| q \gamma \rho_- v^2/c^2 A \frac{1}{2\pi r \epsilon_0}|$$. Using $$\epsilon_0 \mu_0 c^2=1$$, this becomes $$| \gamma q \rho_- v^2 \frac{\mu_0}{2\pi r}A |. \,\,\,\,\,\,\,\,\,\,\,(2)$$ The direction of the force is pointing from the particle to the wire, since the charge of the wire is negative and attracts the positively charge particle.

For normal speeds of charges in a wire, $$v \ll c$$, $$\gamma$$ is very close to 1. Thus apart from a tiny difference, (2) is essentially the same as (1), which is got in the rest frame of the wire where the force is magnetic in nature. Actually this tiny difference is due to the transformation of the force under Lorentz transformation so the result is the same (see Feynman lectures volume II 13-6).

So in this moving frame, the electric field is due solely to a conservative electric field produced solely by a static charge distribution. This is not explanable by a changing magnetic field.