If the gravity of a star can change the trajectory of a photon, is the speed of the photon reduced at the moment of departure from the stellar's surface? Can modern science calculate the measure of change?
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1Does this answer your question? Bending of light - photon's inertia instead of mass β Thermo's Second Law Nov 20 '20 at 14:19
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FYI: Gravitational redshift β Alexander Nov 20 '20 at 17:18
3 Answers
What you mean by "speed" is subtle in general relativity when comparing two disparate points. But, there are definitely a few facts that are universally agreed:
- Any observer sufficiently "close" to the light ray will always measure its local speed as $c$
- The light ray, far from an isolated black hole, will always have velocity $c$ long after it has escaped the black hole
- Spacetime geometry naturally divides into categories of lightlike, spacelike, and timelike, and light rays will always follow the lightlike path, and all observers will agree on this point. And in the case of special relativity, "lightlike" and "travelling at $c$" mean the same thing
So, with this in mind, it is probably clearest to just say "the light ray always travels at the speed of light, even if it bends" (and after all, even in newtonian mechanics, an orbiting planet in a circular orbit has constant speed, even if the direction of its velocity is always changing)

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When rising in a gravitational field, a photon loses energy and frequency, but not speed. This relates to the idea that a distant observer would say that time appears to run slower deep on a strong field. If there were some way to observe the motion of a photon from a great distance, such an observer would say that a photon deep in a field was moving slower. We can say that light cannot escape from a black hole because βatβ the event horizon a photon is moving so slow (from our point of view) that it would take forever to get out.

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From the perspective of the photon, it is traveling in a straight line through space with c, to an outside observer, the space that the photon is traveling in is curved, so it doesn't travel in a straight line, but still travels with c. This can be roughly modeled by an object swimming through a river, as the river is flowing. The object is crossing the river with a certain constant velocity, but the flow of the river is pushing it to the side as it is swimming across, changing the direction in which the object swims. The difference here though, is that the flow of the river is constant, so instead of a curve the object gets across in a straight but sloped line.
In general relativity, space "flows" in a similar manner to the river, but it is much more like an accelerating flow, so taking the model one step further would be imagining dropping a ball to the ground while you are in a train, if the train is moving at a constant velocity, the ball will hit the ground in a straight line from your perspective, inside the train. If someone outside the train observes the event, the ball will drop in a sloped line, much like the river. But if the train is instead accelerating, the ball will drop in a curved line, even from your own perspective, because you and the train are accelerating past the ball while it is dropping, which means the path of the ball gets curved. But that doesn't change the fact that the velocity of the ball dropping down is still constant. So still the photon travels with c, after it leaves the very dense stellar medium (inside the star it is slowed down vastly because of its interactions with the dense matter, it can take thousands of years for a photon emitted at the core of the Sun to get out).

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