# A rotating disc with a charged edge between Special and General relativity

Consider we have a rotating disc (rotating, with a constant angular velocity w, relative to an inertial observer A). The disc has its edge charged uniformly with a total charge Q. No doubt that A can detect loops of a magnetic field, by his pocket compass, due to the constant electric current loop generated by the rotating disc charged edge. A is observing a current = Q/T = Q w/ 2 Pi

My question is about the observer B, in the same disc plane, which is rotating around the disc axis with the same angular velocity w, relative to A.

Would B able to detect a magnetic field by her pocket compass?

If we look to the problem from Special Relativity point of view, the charged disc edge is moving relative to B (same omega, but different distance from the disc axis, so relative speed), so her compass may detect a magnetic field.

But what if B considered herself, according to GR, constant relative to all disc points, including disc edge, considering that she would expect centrifugal and Coriolis fields? She should think that her compass is not moving relative to the disc edge, so the compass shouldn't detect a magnetic field!

Is she correct or wrong?

• What is ""constant", relative to the disc plane,"? – Cryo Feb 2 at 20:42
• @Cryo Observer B and the charges distributed around the disc plan edge – Ahmed Kamal Kassem Feb 4 at 4:11
• I can't tell without doing calculations. I think the non-inertial observer will detect magnetic field, but its a hunch. Perhaps it would be better to simplify the problem. A ring of spinning charge at the origin, aligned with z-axis, observer B on XY-plane spinning with the same angular velocity as the ring. I would suggest to first find the electric and magnetic field for a stationary observer and then check whether $B^2-E^2/c^2>0$ (see en.wikipedia.org/wiki/…). If it is then all observers will observe some magnetic field, ... – Cryo Feb 4 at 17:03
• if not, more work is needed – Cryo Feb 4 at 17:03
• Would you accept an answer involving a rotating spherical shell instead? The inertial-frame fields are much easier to derive for that one, making an exact calculation possible in principle. – Michael Seifert Feb 4 at 18:36

I started Googling the basic results for a rotating spherical shell and found that someone had beat me to this very calculation:

Electromagnetic Fields of a Rotating Shell of Charge, Kirk T. McDonald

In the paper, the author calculates the fields in an inertial frame, and then provides a transformation to a rotating frame to find the magnetic and electric fields as observed in that frame. The author treats the more general case where a shell with charge $$Q$$ and radius $$a$$ rotates with angular velocity $$\pmb{\omega}$$, while the rotating frame rotates at $$\pmb{\omega}'$$; the case requested by the OP is simply the case $$\pmb{\omega} = \pmb{\omega}'$$.

The result inside the shell is actually remarkably simple: in the limit of slow rotations, the magnetic field measured by a corotating observer inside the shell is exactly the same as the magnetic field measured by an inertial observer inside the shell. The reason for this is pretty straightforward: the transformation between the fields for two inertial frames, in the limit of slow velocities, is $$\mathbf{B}' \approx \mathbf{B} - \frac{\mathbf{v}}{c} \mathbf{E}.$$ Inside the shell, $$\mathbf{E} = 0$$ and so $$\mathbf{B}' = \mathbf{B}$$. Thus, if an inertial observer measures a magnetic field inside the shell, so will the corotating observer.

The reason there is still a field inside the shell, even though the rotating observer sees no currents, is simply that Maxwell's equations do not hold in a rotating reference frame. The above-linked paper presents Maxwell's equations in a (slowly) rotating reference frame in Eqs. (31–34) of the above-linked notes. In the absence of bound sources, and in the stationary case (as we have here), they become $$\nabla \cdot \mathbf{E}' = 4 \pi \left[ \rho' - \frac{\mathbf{v}}{c^2} \cdot \mathbf{J}' + \frac{\pmb{\omega}' \cdot \mathbf{B}}{2 \pi c} \right], \\ \nabla \times \mathbf{B}' = \frac{4 \pi}{c} \left[\mathbf{J}' + \mathbf{v} \rho' + \frac{\pmb{\omega}' \times \mathbf{E}'}{4 \pi} - \frac{\omega'}{4 \pi} \frac{\partial \mathbf{E}'}{\partial \phi'}\right],$$ where $$\mathbf{v} = \pmb{\omega}' \times \mathbf{r}$$ and $$\phi'$$ (The other two of Maxwell's equations still hold; the notes have $$\pmb{\omega}$$ instead of $$\pmb{\omega}'$$, but I believe this is an error.) In the rotating frame, while $$\mathbf{J}' = 0$$, there are still contributions to the magnetic field from the other terms on the right-hand side of the last equation, and thus the magnetic field does not vanish. (I believe that the last two terms cancel in the present case, since they correspond to the rate of change of $$\mathbf{E}$$ with respect to $$t$$; but the $$\rho' \mathbf{v}$$ term certainly does not.)

The references in the linked article may also be of interest; in particular, you may wish to consult the same author's notes on electrodynamics in rotating frames, as well as the following papers:

• L.I. Schiff, "A Question in General Relativity", Proc. Nat. Acad. Sci. 25, 391 (1939)
• C.T. Ridgely, "Applying relativistic electrodynamics to a rotating material medium", Am. J. Phys. 66, 114 (1998)
• C.T. Ridgely, "Applying covariant versus contravariant electromagnetic tensors to rotating media", Am. J. Phys. 67, 414 (1999)
• My intuition says that this residual magnetic field is due to Maxwell equations do not hot in non-inertial frames of reference. I feel strongly that the co-moving observer shouldn't notice any magnetic field in the general case since there is no "relative-current" from GR point of view. Also, I didn't understand the author' statement "a nonzero magnetic field is observed in that frame, which can be attributed to the “fictitious” currents.a nonzero magnetic field is observed in that frame, which can be attributed to the “fictitious” currents." what is “fictitious” current?! – Ahmed Kamal Kassem Feb 5 at 0:46
• @AhmedKamalKassem: I believe that it refers to the extra terms that appear on the right-hand side of Gauss's Law and Ampere's Law, which I wrote out above (for a special case.) The analogy is to "fictitious forces" that appear when Newton's Laws are written out in a non-inertial frame. The author of the notes I linked appears to have gotten the terminology from the Schiff paper I refer to at the end of the answer. – Michael Seifert Feb 5 at 1:32

So, I will work out the answer for a rotating ring of charge of radius $$R$$, which can be described by current density:

$$\mathbf{J}=I\int_0^{2\pi} R.d\phi\,\left(\begin{array}\\-\sin\phi\\\cos\phi\\0\end{array}\right) \delta^{(3)}\left(\mathbf{r}-R\left(\begin{array}\\\cos\phi\\\sin\phi\\0\end{array}\right)\right)$$

Where $$I$$ is the current in the ring. This is for an inertial observer and is in Cartesian coordinates. This can be fed into Biot-Savart law to find the magnetic field ($$\mu_0$$ is vacuum permeability):

$$\mathbf{B}\left(\mathbf{r}\right)=\frac{\mu_0}{4\pi}\int d^3 r' \frac{\mathbf{J}\left(\mathbf{r'}\right)\times\left(\mathbf{r}-\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|^3}$$

For an observer (A) sitting on the XY-plane, at distance $$\rho$$ away from the origin, the magnetic field will be z-polarized:

$$B_z\left(\rho\right)=\frac{I\mu_0}{4\pi}\int_0^{2\pi}Rd\phi.\frac{R-\rho\cos\phi}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}=\frac{I\mu_0}{4\pi R}K\left(\rho,\,R\right)$$

Where

$$K\left(\rho,\,R\right)=\int_0^{2\pi}d\phi.\frac{R^2\left(R-\rho\cos\phi\right)}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}$$

is simply a short-hand for the unitless integral which I don't know how to evaluate (other than in the limit or numerically).

Let the full charge of the loop be $$Q$$, let the angular velocity of rotation be $$\omega$$. So $$I=Q\omega/2\pi$$ and:

$$B_z\left(\rho\right)=\frac{Q\omega\mu_0}{8\pi^2 R}K\left(\rho,\,R\right)$$

Next the electric field from the ring of charge density:

$$\rho=\frac{Q}{2\pi}\int^{2\pi}_0 d\phi \,\delta^{(3)}\left(\mathbf{r}-R\left(\begin{array}\\\cos\phi\\\sin\phi\\0\end{array}\right)\right)$$

For observer $$A$$ the electric field will be purely radial ($$\epsilon_0$$ is vacuum permittivity):

$$E_{\rho}\left(\rho\right)=\frac{Q/2\pi}{4\pi\epsilon_0}\int_0^{2\pi}d\phi. \frac{\rho-R\cos\phi}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}$$

again, introducing:

$$S\left(\rho,R\right)=\int_0^{2\pi}d\phi. \frac{R^2\left(\rho-R\cos\phi\right)}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}$$

the electric field becomes

$$E_{\rho}\left(\rho\right)=\frac{Q}{8\pi^2 R^2\epsilon_0}\,S\left(\rho,R\right)$$

Now consider:

$$\frac{B_z}{E_\rho/c}=\frac{\omega R}{c}\,\times\frac{K\left(\rho,R\right)}{S\left(\rho,R\right)}$$

Where $$c$$ is the speed of light. Playing with $$r\approx R$$ shows that for $$K/S\sim 1$$, so the key quantity here is $$\omega R/c$$ which is the ratio of the speed with which the charges move in the loop to the speed of light.

Clearly $$\omega R/c<1$$, but for large angular velocities it is, in principle, possible to have $$c B_z/E_\rho>1$$. Then all observers going through that point in space will observe non-zero magnetic field. The logic is based on $$B^2-E^2/c^2$$ invariant of the electromagnetic field

Under normal settings the electric and magnetic fields seen by the inertial observer will be, assuming the observer is displaced from the origin along the x-axis:

\begin{align} E_x &= E \\ B_z &\approx \frac{\omega R}{c} E \end{align}

From here you should be able to apply Lorentz tranforms to approximate the electromagnetic field in the instantaneous inertial frame of the rotating observer B

Noticed that ratio $$K/S$$ diverges for $$r \uparrow R$$, so if the observer is sitting on the loop you could probably have a situation $$c B_z/E_\rho>1$$ even for relatively mild rotation speeds