# Question about light getting sucked into a black hole

1. Nov 23, 2009

### ranrod

Another noob question. The theory is that a black hole has such a strong gravity pull that light cannot escape. So a photon traveling too close to the black hole will get sucked in. Lets say there's a photon within the point of no return heading directly away from the black hole. Its velocity would get a vector towards the center of the black hole added, until it was headed right for it. Now lets say the same photon starts a little farther away from the black hole - just away from the point of no return. Gravity is still tremendous and if gravity still affects it, would the photon just slow down? You have the photon vector, headed directly away from the black hole, and the black hole vector balancing it out. If the photon does slow down, would it stay at that slowed down speed when it leaves the area of the black hole? Would it speed up again after it leaves that area? Say the photon is at a distance where the black hole's gravity pulls on it hard but doesn't suck it in, but it also can't escape. Could you have a photon in stasis?

The question is, if gravity affects photons (as it theoretically does), wouldn't you expect to see photons traveling at all sorts of different speeds?

Last edited: Nov 23, 2009
2. Nov 23, 2009

### mathman

I can't fully understand your comment. However, the black hole doesn't change the speed of the photon, but its direction, so that once inside it can't get out.

3. Nov 23, 2009

### ranrod

Thanks for the response. My example is a little confusing, but I was asking about a case where the photon's direction is headed directly away from the black hole, so that it's impossible to change its direction without changing its velocity (the photon's direction being parallel to the force of gravity).

4. Nov 23, 2009

### ranrod

...acknowledging some of the responses in related threads. light doesn't slow down or speed up, but its direction changes by the bending of space caused by large masses, such as the black hole.

In my example, since the photon is moving parallel and opposite to the vector from the photon to the center of the black hole, I don't see how the photon would change direction. Would light just escape in that case? In that example, it doesn't matter how you bent space, as long as you did so symmetrically, the photon would still point away from the black hole. Is that why some light does escape black holes? (the narrow symmetrical jets that supposedly shoot out of black holes perpendicular to the disc around it)

5. Nov 24, 2009

### EinsteinII

Hi Members,

My question is if blackhole is pulling light particles into it, it should happen somewhere close to the blackhole. And if there is a limit to that, then the space outside the limit should be luminescent. Is this really happens?

6. Nov 24, 2009

### Chalnoth

Well, basically, even though a light beam travels in the direction away from the black hole, the curvature of space-time is such that it nevertheless moves towards the singularity*.

One sort of simple way of looking at this is by looking at the escape velocity. Basically, if you are on an object like the Earth, and launch something with enough speed, it has enough energy to escape the Earth's gravity and keep on going. This is relatively easy to understand in terms of the potential energy. If the potential energy of a mass at infinity is zero, and the potential energy of the same mass on the surface of the Earth is some finite (but negative) number, then all I need to do is give it a kinetic energy that is large in magnitude than its negative potential energy, and it will be able to escape off to infinity before stopping. Note that when comparing the potential energy to the kinetic energy, the mass cancels out, and you find that the velocity has to exceed some pre-defined limit, based upon the mass of the object you want to escape from as well as how far away from its center you begin.

Now, come back to a black hole: when you generalize this simple picture to General Relativity, you find that if you have enough mass inside a given radius, then the escape velocity can well exceed that of light. And if the escape velocity is greater than the speed of light, then it is simply impossible for anything to escape. Even a light beam is inexorably drawn back towards the center.

* Note: this is a classical black hole. What is really inside the event horizon likely requires quantum gravity to describe, and we really don't know.

7. Nov 24, 2009

### jambaugh

It helps to think four dimensionally...(at least look include time as dimension).

Light traveling outward from a central point is also traveling forward in time. It is traveling along the edge of future light-cones. Mass is twisting the light-cones inward. You can think of the turning of the light not just as turning its spatial direction but its space-time direction. At the event horizon of a black hole the direction which is "forward in time" and "outward" at 1 light-second per second has been turned so that the event horizon itself is a light-cone which has been stretched into a "light-tube". Forward light-like pats are tangential tangential to the event horizon. Once past the event horizon there is no "outward" in terms of increasing r except going backward in time.

It is important to understand that when we speak of radial "distance" r near a black hole the "distance" is not really a distance in space but rather the equivalent of a distance for a given circumference (if the relevant circle were drawn in flat space). Think in terms of a flared horn (flared out to become a flat plane away from the center). Where it is most flared and almost flat you can draw a circle of a given circumference C and define r = C/pi given the flat space definition C=pi r. But moving inward you have to travel farther and farther along the horn to decrease C. The distance traveled is not how much r changed. Eventually you are traveling along a tube with C constant no matter how far you travel. But this horn analogy doesn't incorporate time.

As you move into the black hole the shape of space-time around you changes so that your motion is not one of traveling toward a central point (and forward in time) but rather traveling along a (hyper)-tube which is shrinking in circumference over time. To increase the circumferential radius r = C/pi you must "run the movie backward". Think of the horn above as one of an continuum of nested horns shrinking over time. You fall into the black hole past the horizon and you turn around and try to leave. The horns are shrinking faster than you can go back even at light-speed. There is no direction forward in time where r = C/pi is not decreasing.

Eventually the circumference of the tube is so small that your front and back are pressed up against each other and your left and right shoulders are touching and you are squished to a line. (tidal forces pull your head and feet to the long axis of the "tube"). But don't worry the tidal forces along the "tube" will have pulled you apart by then so you won't feel it. ;) That instant in your time when the r=C/pi = 0 is the singularity.

8. Nov 24, 2009

### ranrod

Thanks for your response. The examples you used were for objects with masses, do we think light has a mass?

9. Nov 24, 2009

### mathman

Light responds to gravity as if it was a mass, determined by its energy. One of the earliest tests of general relativity was the observation that starlight is bent when it passes close to the sun (1919 Eddington observation during a solar eclipse).

10. Nov 24, 2009

### ranrod

That would go back to my original question. If light is affected by gravity, like an object with mass would, do you see light going at different speeds as the photons get pushed and pulled by different gravitational forces (such as massive stars or black holes)?

11. Nov 24, 2009

### jimgraber

ranrod, the vertically-headed light from just outside the point of no return does escape, but is highly redshifted.
Best to all.
Jim Graber

12. Nov 24, 2009

### jimgraber

Not different speeds, but redshifted, or blue shifted.
Best Jim Graber

13. Nov 24, 2009

### Chalnoth

No, not at all. This changes the behavior slightly, but the argument listed above still holds in the full General-Relativistic case, it turns out.

14. Nov 24, 2009

### Chronos

A photon always travels at c - a fundamental tenant of relativity. jimbraber has correctly explained it. A photon has no mass, but has a mass equivilance. This is not apparent until the photon suffers a collision.

15. Nov 24, 2009

### DaveC426913

Gravity pulling on mass and on light is a Newtonian model and can lead to some odd results near black holes. It may be illuminating to understand the Einsteinian view of light and gravity.

In the Einsteinian view, gravity is nothing more than the curvature of space. Around a black hole, space curves quite sharply. The trajectories of mass and light are affected by this curvature of space. (This avoids the misleading idea that gravity is "pulling" on light as if it has mass.)