Expert: Courtney Seligman - 6/12/2010

Question
If light has no rest mass, why is it attracted and captured by black holes?

Answer
Light isn't "attracted" by black holes (or gravity). But according to Einstein's theory of general relativity, neither is anything else. What we call the "force" of gravity is a curvature of space-time, such that in the future, things which are now a certain distance apart will be a little closer together, because the amount of space-time between them is smaller.

WARNING: MIND-BOGGLING (BUT HOPEFULLY EASY) PHYSICS AHEAD!!

BACKGROUND PHYSICS AND SPACE-TIME GEOMETRY:

To visualize this, think of space as a two-dimensional sheet (representing all positions in 3-D space), moving along a perpendicular to the sheet (representing time). Each second, the sheet moves forward one second in time, and anything in the sheet moves both that one second forward, and whatever amount sideways corresponds to its motion in space (during that second).

Now suppose that you have two points in space which are NOT moving relative to each other, save for their forward motion in time. You would expect that in space, they would stay in exactly the same place, so that their forward motion in time would be represented by two parallel lines -- one representing the motion of one object in time, and the other representing the motion of the other object in time. And since parallel lines are always the same distance from each other, the two objects should maintain a constant distance from each other in space, just as their parallel lines do in space-time. So we would observe them as maintaining a constant relationship, even though they are moving (forward in time).

But the above assumes that space-time is "flat" and "uniform", like the grid on a piece of rectilinear graph paper. Suppose that space-time is curved, like the surface of the Earth. To understand this example, suppose that the various positions in 3-D space are "mapped" onto a parallel of latitude, so that everywhere in the part of space we're looking at is somewhere on that parallel, and that time is "mapped" onto the meridians of longitude, so that moving forward in time is like moving along one of the meridians, toward the Pole.

In this example, at any given moment, objects which are not moving toward each other would be moving along parallel paths (meridians of longitude being parallel to each other), but because of the curvature of the Earth's surface, those parallels (the meridians) are gradually getting closer together, because the amount of space between them (the distance along the parallels of latitude) is gradually getting smaller, as you move toward the Pole (forward in time).

That is the way that Einstein's physics visualizes gravity. The presence of any mass -- even that of the Earth or Sun -- "curves" the fabric of space-time, so that objects which are moving along "straight-line" paths in that curved space-time end up closer together, in the future (in this case, we replace the phrase "straight-line" with "geodesic"). In the case of the Earth, anything near its surface is moving through space-time in such a way that each second, the amount of space between the geodesic the object is following is a little closer to the surface of the Earth.

To understand what I mean by "a little closer", we have to visualize the space-time "fabric" as having the same dimension for one second of time, as for 186,000 miles of space. In other words, moving forward one second carries you as far forward in time, as moving 186,000 miles in space carries you sideways in space (the 186,000 miles corresponding to one second being due to the speed of light being that distance per second).

In "flat", "empty" space-time, there would be no curvature of space-time, and as an object moves one second forward in time (equivalent to 186,000 miles in space), the amount of space between it and another object would remain constant, because the lines in space-time are absolutely parallel to each other.

But in the curved space-time near the Earth, moving one second forward curves the "parallels" of space-time closer together, by about 16 feet (the distance that an object dropped from rest would fall toward the surface of the Earth in one second). That 16 feet closer together isn't much curve, in a line which is equivalent to 186,000 miles of space. And if you take a shorter period of time, say a billionth of a second, then the amount of curvature would be that many times smaller. So the space-time curvature is very, very small (for us).

APPLICATION OF THE ABOVE TO THINGS FALLING TOWARD THE EARTH:

Now let's imagine things "falling" toward the Earth. In Newton's physics, they fall because they are pulled downward, by a force of gravity. That force is larger for more massive objects, because every object "falls" at the same rate, and to make more massive objects fall at the same rate, you need a larger force. Therefore, less massive objects should require smaller forces (they weigh less), and objects which have no mass at all (photons of light) should have no force acting on them, so they shouldn't fall at all. Or at least, that's what you would think, if you think "I see the thing fall, and it takes a force to make it fall, so there must be a force acting on it, and if there is no force acting on it, then it shouldn't fall."

But that's wrong. What we observe is that ALL things fall at the same rate, regardless of their weight or mass. So even if something had no weight or mass (that is, photons), it should still fall at the same rate. You wouldn't think it should, but if something has no mass, it doesn't need a force to make it fall, so it's "OK" to have it fall. The only problem is, how can you tell how fast it will fall, when the arithmetic involved divides zero by zero?

This is where Einstein's theory of gravity becomes useful. When objects fall, they don't fall because they are pulled downward by gravity. They fall because in the future, the parallels of space-time that they are moving along are 16 feet closer together, for each second they move forward in time. They don't fall because there is a force acting on them, but because the space-time near a massive object is "smaller" in the future than in the past. So in one second, anything -- asteroids, bowling balls, atoms, and even photons of light -- would fall toward the Earth exactly the same way.

Now there is a problem here. Namely, asteroids, bowling balls, atoms and such presumably have a small motion relative to the Earth, so we can wait a second or so, to see how they "fall", relative to the paths they would have taken, if they didn't fall. But light is always moving at 186,000 miles per second, so it wouldn't stay close to the Earth very long, and seeing it "fall" for even one second is out of the question. That's why I mentioned the idea of using only a billionth of a second, a few paragraphs ago.

Suppose we have a bunch of objects -- bowling balls, baseballs, atoms, and photons of light -- moving horizontally through the room that you are in. Now let's see what happens to them in one billionth of a second. Each would move forward one billionth of a second in time. Each would move in space, however far they could move in a billionth of a second (for light, that would be about ten feet; for everything else, a much smaller distance). AND each would "fall" toward the surface of the Earth by the amount that space-time "shrinks" in one-billionth of a second. EVERYTHING would fall toward the Earth exactly the same amount, because the shrinkage of space-time (some very tiny fraction of that 16 feet per second, reduced by at least a billion times) would be the same for all of them.

In other words, light would fall toward the Earth exactly the same amount, in a billionth of a second, as anything else would, in that same period of time. It doesn't make any difference that bowling balls have mass, and photons don't, because the "falling" has nothing to do with their mass. It is only an apparent change in position caused by the curvature of space-time, which is exactly the same for every object, regardless of its mass or speed in space. It is only its speed in time (which is the same for every object) which causes the apparent "falling", so everything falls the same.

AND (FINALLY), THE ANSWER TO YOUR QUESTION:

So, as stated in the very first paragraph, light doesn't "fall" toward black holes because the black hole's gravity "pulls" on its (zero) mass. It gets closer to the black hole because the black hole's very strong gravity curves space-time toward it by a very large amount (less than the speed of light outside the event horizon, equal to the speed of light at the event horizon, and more than the speed of light inside the event horizon).

If the light photon is outside the event horizon, moving straight outward (at the speed of light), the fact that the curvature of space-time is contracting the space between things at less than the speed of light means that the photon will slowly move away from the event horizon.

If the light photon is at the event horizon, moving straight outward at the speed of light, the fact that the curvature of space time is reducing the amount of space at the speed of light means that the photon will be "stuck" at that distance from the center of the black hole, unable to move outward at all, but not falling inward, either.

If the photon is INSIDE the event horizon, the fact that the curvature of space-time is faster than the speed of light means that even if the photon is heading straight outward, it will "fall" inward, at a rate equal to the inward curvature of space-time, minus the speed of light. And as a result, in a very short period of time, it will end up at the center of the black hole (the "singularity").