Does an electrical charge - at rest WRT a gravitational field - produce a magnetic field as well as an electrical one around itself?

Assume that a point electrical charge is at rest WRT an observer inside a uniform gravitational field. Using Einstein's equivalence principle (EEP), this scenario is equivalent to an electrical charge located in a uniformly accelerating shuttle. However, it is evident for an inertial observer outside the shuttle that there is a time-varying magnetic field as well as a time-varying electric field around the accelerated charge. However, EEP asserts that the same happens for a Schwarzschild observer as he looks at the charged particle located on a massive planet.

How can it be possible for the observers inside the shuttle and the one on the planet to detect no magnetic field, whereas the observer outside the shuttle as well as a Schwarzschild observer detects a magnetic field?

Is there any experiment being carried out showing that an electrical charge at rest inside a gravitational field produces no magnetic field around it?

I am aware of this Wiki article, however, it was not convincing to me that the paradox is apparent.

• WRT = with respect to? Means weightless free falling or means stationary to the earths surface? Jun 27 '20 at 19:18
• @HolgerFiedler stationary to the earth surface I meant. Jun 27 '20 at 20:55

Radiation of a charged particle as well as equivalence principle and gravitational field itself are all red herrings in this context. The question is completely answerable within the framework of special relativity.

If an observer inside the shuttle at a specific spacetime point is registering only electrical field $$\mathbf{E}$$, then the second observer at the same spacetime point but moving with the velocity $$\mathbf{v}$$ relative to the first would register both electric and magnetic fields (which we denote as $$\mathbf{E}'$$ and $$\mathbf{B}'$$) if the vector product $$\mathbf{v}\times \mathbf{E}$$ is nonzero according to the equations: $$\mathbf{E}_\|'=\mathbf{E}_\| \qquad \mathbf{E}_\bot'=\gamma \mathbf{E}_\bot ,\qquad \mathbf{B}'=-\gamma\frac1{c^2} \mathbf{v}\times\mathbf{E},$$ where $$\mathbf{E}_\bot$$ and $$\mathbf{E}_\|$$ are field components perpendicular and parallel to the velocity, $$\gamma=1/\sqrt{1-v^2/c^2}$$ is relativistic factor. This transformation law does not depend on the acceleration of either observer (only on their relative velocities) or on the presence of spacetime curvature (as long as it does not alter the measurements).

For the general transformation law of electric and magnetic fields between different frames see the Wikipedia article. The existence of such transformations is the evidence that within the framework of relativity (both special and general) there are no separate entities “electric field” and “magnetic field” but a unified concept electromagnetic field which is described by a new type of object electromagnetic field tensor. Electric and magnetic fields then form components of this tensor defined in a specific reference frame.

Relationship between electromagnetic tensor and its constituent electric and magnetic parts in different reference frames is analogous to the situation with vectors and vector components in different coordinate systems in Euclidean geometry. Let us say we have a vector $$\mathbf{A}$$ in an Euclidean space. It is an object existing without any specific coordinate system in this space. However, when we select one such coordinate system (let us assume it is cartesian), this vector would have components ($$A_x$$, $$A_y$$, $$…$$) within this system, and if we switch between different coordinate systems, vector components would transform according to some simple rules. And among those systems there would be, for example, coordinate system where our vector would have a zero $$x$$-component.

Likewise, electromagnetic field tensor at a given spacetime point is an object independent of any observers or the choice of coordinates. But, when one places a specific observer at this point this allows splitting of spacetime (another unified relativistic concept) into “space” and “time” parts (in a general relativity this could only be done locally, in a vicinity around this observer) and at the same time allows splitting of electromagnetic field tensor into an electric and magnetic components. If we choose another observer moving near the same spacetime point, the splitting of both spacetime and electromagnetic field would be different.

Continuing our analogy, a vector in an Euclidean space has an invariant, a specific quantity that would remain the same in all coordinate systems, namely its length: $$|A|=\sqrt{\mathbf{A}\cdot \mathbf{A} }=\sqrt{A_x^2+A_y^2+A_z^2}$$. Similarly, electromagnetic field tensor also has invariants, but two. In terms of its electric and magnetic components these are: $$P=|B|^2-\frac{|E|^2}{c^2}, \qquad Q = \frac{\mathbf{E}\cdot \mathbf{B}}{c},$$ meaning that in all possible reference frames at this spacetime point observers would measure the same $$P$$ and $$Q$$ as in any other. Since in OP's original reference frame there were no magnetic field then $$Q=0$$ and $$P<0$$, and so in all reference frames magnetic field (when present) would always be orthogonal to electric, electric field would always be present and its absolute value would exceed the value of $$|B| c$$.

The observers inside the shuttle/on the planet are not inertial observers: they are accelerating/in a gravitational field (which, as you correctly point out are equivalent statements). So even though the charged particle is accelerating with respect to an inertial frame, these observers do not see any radiation.

In contrast, the observer outside the shuttle/the observer at infinity in Schwarzschild spacetime, are inertial, so do detect radiation as would be expected from an accelerated charge.

• This answer paraphrases the wiki page you link; if you are not convinced by this argument it would be helpful for you to edit your post to explain exactly why. Jun 27 '20 at 11:13
• Then, what happens for the magnetic field lines from the standpoint of the Schwarzschild observer? Do they disappear near the observer located near the charge on the planet, whereas they start to grow near the Schwarzschild observer? Isn't it slightly awkward?! Jun 27 '20 at 11:21