Bremsstrahlung Radiation A thought experiment.
Consider an electron falling into a black hole. From an external observer to the electron and the black hole, the electron accelerates, and should give off Bremsstrahlung radiation
From the electron's frame of reference, it is travelling along a geodesic in free fall, and so is not accelerating at all so doesn't generate Bremsstrahlung radiation.
Which is the correct situation and why?
 A: You don't need a black hole for this thought experiment: just drop an electron from a height on the surface of the Earth, and you have exactly the same problem.
The Equivalence Principle of General Relativity claims that such a system should be indistinguishable from an accelerated electron.
However, Maxwell's Equations tell us that accelerated charges emit radiation, but no such radiation appears to be observed.
This "thought experiment" has been studied since 1909, and Wikipedia even has an article about it, including a resolution. The bottom line is that such a charge does indeed radiate. Essentially, while in the charge's rest frame it would appear to not radiate, when one moves into the lab frame, this transformation is not a Lorentz Transformation, and it leads to a radiating solution in the lab frame.
A: Philip's answer is great, I'd just add a couple of things to it.
Generally, it's important to point out that radiation is not an observer-independent phenomenon. For example, ignoring gravity, an accelerating particle will radiate according to an inertial observer, but not according to an observer who is co-accelerating. Maxwell's equations are not invariant under a transformation from an inertial to an accelerating frame.
To expand on the reason to this phenomenon a little bit, the accelerating observer experiences a Rindler horizon - anything behind that horizon cannot reach them as long as they continue to accelerate. A co-accelerating charge emits radiation into the spacetime region behind the observer's horizon, as explained in this paper (https://arxiv.org/abs/physics/0506049).
Analogously, a particle in free fall will not radiate according to a freely falling observer but will do so according to one who is stationary in the gravitational field. Meanwhile a stationary charge in that field will radiate according to the freely falling observer but not according to a similarly stationary observer.
