Another way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good). The equivalence principle says that if you are in an enclosed box undergoing free fall in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field. So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerating upward, and the ball is simply being left behind (which checks better with the stresses you can easily detect on the objects around you and are not present on a falling ball, including the feeling you are receiving from your bottom right now).
Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally. In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that rising effect would accelerate. So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser. Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher than the point across from it where the light was emitted), and its path appears to curve downward. Ergo, light "falls."
Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because nothing is happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating.
Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit. Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass. All objects fall the same under Newtonian gravity. So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit. Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.
Answer from Ken G on Stack ExchangeAnother way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good). The equivalence principle says that if you are in an enclosed box undergoing free fall in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field. So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerating upward, and the ball is simply being left behind (which checks better with the stresses you can easily detect on the objects around you and are not present on a falling ball, including the feeling you are receiving from your bottom right now).
Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally. In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that rising effect would accelerate. So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser. Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher than the point across from it where the light was emitted), and its path appears to curve downward. Ergo, light "falls."
Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because nothing is happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating.
Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit. Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass. All objects fall the same under Newtonian gravity. So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit. Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.
There are a couple of ways one could approach your question:
Black holes are regions of space that have been deformed by a sufficiently concentrated mass. Light waves/particles always travel in a straight line at a constant velocity (
). Although a photon approaching a black hole will continue traveling in a straight line through space, space itself has curved so the photon's path will curve.
While photons don't speed up in the presence of a gravity well, they are affected by it in other ways. In specific, photons entering a gravity well are blue-shifted while photons leaving one are red-shifted. This red/blue-shifting happens because time passes slower within a gravity well than without. In all frames of reference, though, the speed of light remains constant. There's more info on this on the wiki.
Note: The question originally referred specifically to black holes. The above hold for any concentration of matter (of which black holes are an extreme example).
At least for light, you can write down an effective index of refraction in Schwarzschild coordinates. See for example section 5.1.3 of Perlick (2004) (direct link).
However, the idea of the "density of space" does not have any standard meaning. You would have to define it yourself -- and then this question just becomes a question of definitions.
Also, I would caution against thinking of space as anything physical. Within the context of general relativity, spacetime is the physical domain. "Space" is just a fairly arbitrary 3-dimensional surface within the 4-dimensional spacetime. Different observers or different conventions do not necessarily agree on a common choice of what "space" is.
Denser space, but less-dense time. The region fits in more ruler-markings than its surroundings, but fewer clock-ticks.
For a more formal description, we can say that the region around a gravity-source has positive curvature, and one of the ways of defining or identifying net positive curvature is by an excess of space in a bounded region compared to the expectations given by Euclidean geometry.
- With positive curvature, a defined perimeter distance and shape (e.g. a circle with known circumference) encloses more content than one would otherwise expect.
- With zero curvature, the perimeter contains the correct amount of space as predicted by Euclidean geometry, and is "flat".
- With negative curvature the perimeter surrounds less space than expected, and there's a space-deficit. Or stated another way, for a unit volume, there's an excess of perimeter-length.
With zero curvature, lines that are parallel in one region remain parallel everywhere, and never cross (Euclid's parallel postulate).
With positive curvature, the excess of space can be expressed in terms of refractive index, which converges lightrays so that the lines cross. In the case of a sphere (which shows uniform positive curvature), we can draw short parallel line-segments at the equator, all pointing North-South, and if we extend them, they'll cross each other at the poles. So positive curvature converges lightrays.
With negative curvature, initially-parallel rays diverge. If we start with a normal region of space, surgically excise two distant spherical volumes, and short-circuit the "bleeding edges" together outside normal space to create a wormhole connection (Visser, "spacetime surgery"), then when we track a pair of parallel lines leading into one wormhole mouth and out of the other, , when they emerge at the other side they'll be divergent. So a wormhole mouth has a spatial deficit, and a negative curvature.
The increase in spatial density around a gravitational source is expressed in the standard gravity-well diagrams, which show the extra quantity of normalised distance in the region, which has to extend out of the plane to fit it all in. It's also expressed by an increase in the amount of time that a signal takes to cross the region (Shapiro timelag).
However, it's not quite true to say that the effects of a gravitational field around a black hole are identical to those of "space being denser", because that's not quite the whole story. If it was, we could describe gravity using a scalar theory, in which we just apply a simple density value to each point in space. This doesn't work. It turns out that we also require a gravitational field to produce an asymmetry in the inward and outward velocities of light in the region (scalar-vector theory), or else things go bad. Also, if we want a rotating star to drag light around with its (rotational gravitomagnetism), we need the star's rotation to again produce an asymmetry in one-way light-velocities. With simple density values, we don't get the rotational drag effect.
So while your "increased-density" characterisation is basically correct, it's not quite the complete answer.