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

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!