1. Jan 17, 2008

Ok I got a question.

So we know that in space light travels at a constant speed of 3X10^8m/s.
We also know that light cant escape a black hole gravity field.

So lets assume I got a light source just near a black hole (lets assume the gravity field is radial) ,my light source is transmiting light outwards fron the black hole,in the exact oposite direction of the gravity vector.

Now since we know the light cant escape than we know it will change direction and move to the black hole.

The paradox is:Since the light changed its direction it had to slow down,stop and start accelerating in the diffrent direction.Thus not compling with what we all learnt,that light moves at a constant speed.

Yes I know I look at this mostly on Newtonic physics,I would love to hear your quantom physics awnser to this.

2. Jan 17, 2008

### Xeinstein

Are you sure that light can't escape a black hole gravity field?
I think it's true only inside the event horizon

It won't slow down,stop and start accelerating in the different direction.
Inside the event horizon, light just follows the geodesic path which curves toward the center of black hole

3. Jan 17, 2008

### jdavel

"Since the light changed its direction it had to slow down...."

Not true. Even a massive object can follow a curved path without changing its speed. So can light.

4. Jan 17, 2008

### nonequilibrium

Yeah, you'd have to start emitting the light inside of the event horizon, but that would mean there was never any light to begin with, so no paradox. I presume that theoretically, placing a fixed laser juist outside the event horizon, perpendicular to the black hole's centerpoint, the light would not even bend and continue in the same speed.

5. Jan 17, 2008

I see some of you didnt actualy read my post.I specificly said that we are assuming that the gravity vectors are exacly radials (for example like the E of a sphere electrical charge) and that the light is traveling excaly against the direction of a singal vector.

mr.vodka got what Im talking about.

6. Jan 17, 2008

### Staff: Mentor

I assume you are familiar with the Cartesian distance formula, $d = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2 + (z_1-z_0)^2}$. This is but one of several Newtonian physics concepts you have to throw out when you talk about black holes. In particular, what you are implicitly doing is assuming that the "distance" between some point near a black hole and a point far from it is computed the same way you would compute the distance between two points far removed from the black hole.

Suppose your laser just is outside the event event horizon. It will take a long time for light emitted from that laser to reach an observer who is some distance from the black hole. The black hole stretches space-time.

7. Jan 17, 2008

### JesseM

Once inside the event horizon, the radial dimension becomes the time dimension to observers on the inside; the singularity lies not in any spatial direction, but in the future of these observers. So, the idea of an observer inside the event horizon trying to aim their laser in the outward radial direction is comparable to the idea of an observer outside a black hole trying to aim a laser backwards in time, it doesn't really make sense. The diagram here shows the future light cones for event on the worldline of an infalling observer, you can see that the light cones tip over in such a way that any light beam emitted by the oberver once he crosses the event horizon will always point inwards. This may be a little clearer in the diagram here (from this page) showing how light cones tip at various distances from the horizon, with this diagram illustrating both the wordline of an infalling particle (heavy line) and the worldlines of various photons (thin lines--in this coordinate system, note that the worldlines of photons emitted from the left edge of a light cone are always perfect diagonals).

Last edited: Jan 17, 2008
8. Jan 17, 2008

### pervect

Staff Emeritus
Note that the speed of light is always equal to c in special relativity. It is also always equal to c in General relativity, as long as one uses local clocks and rulers to measure that speed. This is important, because (to use the slightly over-simple popular explanation) clocks are known to tick at different rates depending on their location when one considers the effects of General relativity. It should be reasonably obvious that if clocks tick at different rates, one has to specify which clock to use to measure the time to compute the speed. There is, however, a choice of clock (and a corresponding choice of ruler) that makes the speed of light always constant and equal to c even in GR - this is a clock and ruler of some physical (jargon: timelike) observer at the same location as the light is.

Here's a specific example relevant to your question.

Suppose you are in a spaceship falling into a black hole. We will assum in this simple example that the spaceship is free-falling into the black hole and that its engines are off.

The front of the spaceship emits a particle of light (a photon, if you will, though you should think of it clasically and not quauntum mechanically) just as it enters the event horizon. The light particle is emitted away from the black hole.

The light particle will follow a very simple path. It will maintain constant Schwarzschild coordinates. To speak slightly losely, it will neither approach nor recede from the black hole, it will just "hang in space" - (using Schwarzschild coordinates as a reference).

The spaceship, however, can and must fall into the black hole.

The person on the spaceship knows the length of the spaceship, and has clocks on the front and rear of the spaceship which he has synchronized. He computes the travel time from when the light particle was emitted at the front, and when the light particle is received at the back. Taking the length of the spaceship divided by the time to traverse it gives the speed of the light particle relative to the spaceship. This measured speed will be exactly equal to 'c'.

Here's a more formal reference on the topic from the sci.physics.faq

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

Thus, talking about "speed" is ambiguous, there are multiple definitions of what it might mean. With the modern definition of speed, the speed of light is always constant in SR and in GR. With some other defintions of speed, definitions which are commonly used (though not "modern"), the speed of light is still always constant in SR, but with these alternate defintions, the speed is not necessarily constant in GR.

Last edited: Jan 17, 2008
9. Jan 17, 2008

### jdg812

Light CANT escape a black hole gravity only if the source of light is ON or BELOW an events horizon. If the source of light is a little bit ABOVE event horizon, then light escapes, but may get a large red shift.
"Just near a black hole" in plain English means "very close to the events horizon, but still outside of it". In this case light escapes the black hole gravity and there is no paradox.

Last edited: Jan 17, 2008
10. Jan 17, 2008

I see I got alot of reading to do :)

11. Jan 17, 2008

### DaveC426913

Event.
There's only one.

12. Jan 18, 2008

### jdg812

Yes, you are right!

13. Jan 20, 2008

You miss the point,forget about a black hole.Just imagine you got a gravity vector as strong as the one emited from a black hole and a light beam that is traveling in the oposite direction that feels its effect,its not a word paradox its a physics one :)

But you guys really helped me,and once I get some free time I will do some reading on the links you guys posted,thanks :)

14. Jan 20, 2008

### lightarrow

They have already answered you; I try to say things in another way, hope not to make a mistake, in case I hope to be corrected:

when you say that "light bends" you are actually saying: "if space were euclidean, then light would bend". Unfortunately we tend to always see things from an euclidean point of view. If your sheet of paper were not flat but had the real geometry, that is the one near the massive object, you would see that light actually doesn't bend at all.

Last edited: Jan 20, 2008
15. Jan 20, 2008

### pervect

Staff Emeritus
There isn't really any such thing as a "gravity vector" in GR. If you can rephrase your question in terms that make sense (such as the acceleration vector of some accelerating spaceship), we might be able to answer your question.

We can try this:

"Just imagine that you have a spaceship accelerating and emitting a light beam. If the light beam is emitted exactly in the direction that the spaceship is accelerating, will it ever fall backwards".

The answer to this question is that this won't/can't happen.

You may not regard that as a fair reinterpretation of your question, unfortunately, your question as it is written doesn't make sense, and probably contains some misunderstandings (on your part) of how gravity is handled in GR. As various posters have mentioned, in full GR, gravity is not "really" a force at all, you seem to be thinking of it in Newtonian terms, and expecting the Newtonian defintions to apply to GR. This is not correct and is confusing you - and rather than realizing that you are confused, you seem to be blaming the theory :-(.