When gravity bends light, does the light still propagate orthogonally to its $\vec E$ and $\vec B$ fields?

Question

An ordinary photon travels perpendicularly to the direction of its oscillating E & B vector fields (i.e. $\vec{v} \propto \vec{E} \times \vec{B}$). Let's say $\vec{E}$ is oscillating "in-out" of the page, $\vec{B}$ is oscillating "up down", and so $\vec{v}$ propagates to the right. Now turn on a strong uniform downward gravitational field. The photon bends downward, and now $\vec{v}$ is at some angle with respect to the horizontal and has a downward component. I can think of two possibilities: (1) $\vec{E}$ and $\vec{B}$ are still "in-out" and "up-down", and so $\vec{v}$ is no longer $\propto \vec{E} \times \vec{B}$. Or, (2) we still have $\vec{v} \propto \vec{E} \times \vec{B}$ and so $\vec{E}$ and $\vec{B}$ have rotated as the photon bent and no longer oscillate "in-out" and "up-down."

Scenario (2) seems consistent with Maxwell's equations. But scenario (1) seems consistent with the equivalence principle (i.e. can exactly replace uniform gravitational field with an accelerating reference frame). If I imagine there is no gravity and I watch a right-moving light beam from an upward-accelerating reference frame, I would still see the light beam bend down, but (I think?) I would still see the fields oscillating "in-out" and "up-down," no longer orthogonal to the perceived direction of propagation of the light beam.

Which scenario is correct, and how is it internally consistent? Thanks!

Answer 1

(1) is wrong; (2) is right. Observations in an accelerating frame would also give (2).

I say this based on general knowledge; I have not done the calculation, the details of which would be tricky. But you can imagine a case where the light got bent through a large angle around a massive object and then went on its way in a straight line. The fields would certainly end up transverse, and they have to get to those final directions somehow.

Answer 2

The equivalence principle means that in each part of the motion light behaves exactly as normal in an inertial frame. You do not "replace uniform gravitational field with an accelerating reference frame", you get rid of the gravitational field locally by using an inertial reference frame.