# Question about light and blackholes

1. Nov 13, 2003

### magus

i was just wandering..... at the point where gravity is strong enough to be greater than the kinetic energy of which light possesses, do photons of light actually radiate some distance from the collapsed star then slow to rest and fall back to the surface, as a cannonball being shot straight up in Earth's atmosphere would, or at this point are the particles simply not capable of being emitted. or is their some other explanation of which i have not accounted for.
thx,
magus

2. Nov 13, 2003

### Ambitwistor

I wouldn't say that "gravity is strong enough to be greater than the kinetic energy [of light]". But I understand the gist of your question. In Newtonian gravity, light emitted from the surface of a "Newtonian black hole" will travel some distance away from the hole, then fall back. But in true relativistic gravity, light emitted at or within the event horizon of a black hole cannot travel outward at all: spacetime curves the path of the light so it always falls inward.

3. Nov 13, 2003

### magus

so it basically depends on which conception of gravity is actually correct?

4. Nov 13, 2003

### Ambitwistor

Yes. We know that Newton's theory of gravity is incorrect. We have good evidence that Einstein's theory works in many circumstances, but we don't have much in the way of direct tests of that theory near black holes. What indirect tests we do have still bear out Einstein's theory, so most astrophysicists would probably bet at the 95% level or higher that light cannot escape a black hole to any distance.

5. Nov 16, 2003

### Unkaspam

X-rays are able to escape (emit from) a black hole, and X-rays are a form of light. Do they travel faster than light to be able to escape the black hole?

6. Nov 16, 2003

### Ambitwistor

X-rays are not able to escape from a black hole. The X-rays we see indicating the presence of a black hole actually come from matter outside of the black hole.

7. Nov 16, 2003

### Unkaspam

Are you talking about the accretion disk?

Gas and dust surrounding a newborn star, a black hole, or any massive object help it to grow in size by attracting more and more material.

Does a black hole grow in size relative to the amount of material it attracts?

I understood that neutron stars bounce X-rays off their mass. Is it possible this can happen with a black hole? X-rays are said to have a very active wave. Could this help them to escape the intense gravity of a black hole? The same high frequency wave of the X-ray probably causes it to bounce off of the white dwarf or neutron star.

8. Nov 16, 2003

### Ambitwistor

Yes, or particles that are sucked from it and shot up along the magnetic poles.

A black hole grows in size relative to the amount of mass that falls into it.

Are you saying that the surface of a neutron star reflects X-rays??

What? "Active wave"? What is that?

X-rays can escape the gravity of a black hole if they're emitted outside of it and pointed in the right direction; so can light of any other frequency, equally well. Nothing can escape from inside a black hole.

9. Nov 17, 2003

### Herringbone

Hi,

I think you've got the right idea in "balancing" the potential energy of the cannonball with it's kinetic energy but remember, a photon's energy is not kinetic but is linearly related to it's color (frequency). E = hf

Consider the point where an object's gravitational field is strong enough such that it's escape velocity equals c (i.e. just inside the event horizon of a black hole).

Since photons always travel at the speed of light, they can't slow down and stop as a cannonball. They do lose energy however as they climb out of a gravitational field by experiencing a decrease in their frequency (they redshift).

A photon emitted within the event horizon wouldn't get out because its frequency would redshift to 0hz as all the photon's energy was consumed in climbing out of the gravitational field.

10. Nov 17, 2003

### Ambitwistor

This energy-conservation explanation can sometimes be misleading. It implies that a photon emitted inside a black hole can travel outwards, but will be redshifted out of existence before it gets to the horizon. In reality, it can't travel outwards at all.

11. Nov 17, 2003

### Herringbone

You're right. I should have said "... the photon's energy would be consumed as the photon attempted to climb out of the gravitational field."

12. Nov 17, 2003

### magus

thx for the insight, the explanations made sense, and just yesterday my physics instructor mentioned that that was probably what happened.

i now have a few more questions however. It was mentioned that the photon actually does not move outward at all. How does it redshift without moving at all, and what actually happens to the light when it reaches 0hz.

13. Nov 17, 2003

### Ambitwistor

There's a difference between "not moving outward" and "not moving". A photon always moves at the speed of light according to any local inertial observer.

Light never actually reaches 0 hz according to anybody; you'd have to travel away from it at the speed of light for it to be redshifted down to zero.

14. Nov 17, 2003

### magus

i think i have the idea that the photon simply vanishes, thus their is nothing to see.im not sure if thats right but its a guess.

15. Nov 17, 2003

### Herringbone

Well, actually I think somebody's said this since I remember reading it several times in the past. But I've been wrong before so I did some calculations and came up with this:

Since GR tells us that the ratio of a photon's gravitationally shifted frequency to the original frequency is:

$$\nu_s / \nu_o = \sqrt{1 - \frac{2GM}{c^2r}}$$

Where M is the mass of the gravitating body, r is the distance from it, G is the Grav. constant and c is the speed of light.

The photon redshifts to 0 Hz when:

$$\nu_s / \nu_o = 0$$

Which happens when:

$$M/r = \frac {c^2}{2G}$$

or where:

$$r = \frac {2GM} {c^2}$$

Which is the Schwarzschild radius (event horizon) of a black hole. So it seems to be a valid assumption that if any photon could be emitted within the event horizon, it would also redshift to 0 Hz. But that's just my opinion.

I believe there's another way for a photon to be redshifted down to 0 Hz too but that involves cosmological expansion which is not really the topic of this thread.

Cheers, HB

16. Nov 17, 2003

### Ambitwistor

Your formula applies to the redshifting of a light ray that is emitted at some r>rS and reaches an observer at infinity. It does not really make sense to apply it to a light ray that is emitted at r=rS, because such a light ray never leaves the horizon at all, let alone reaches an observer at infinity who can measure its wavelength.

17. Nov 17, 2003

### Unkaspam

Sorry, I mean energetic wave. X-rays have smaller/shorter wavelengths and so they have higher energy than ultraviolet waves or other light waves. Thank you Ambitwistor.

18. Nov 20, 2003

### Unkaspam

Gravitons escape the warp of a black hole with no problem and with no loss to gravitational energy.

Does that mean gravity travels faster than light (escaping the event horizon of a black hole?)

Last edited: Nov 20, 2003
19. Nov 20, 2003

### Ambitwistor

Virtual gravitons escape, but since they are not real, they do not carry any real energy, information, or anything else observable out of the black hole.

When people speak of "the speed of gravity", they mean the speed at which observable gravitational influences, i.e., changes in the gravitational field, propagate. According to that definition, the speed of gravity is the same as the speed of light.

20. Nov 22, 2003

### Unkaspam

Can a gravitational field be viewed simply as the sphere of influence or effect generated by a mass? Are there actual waves of gravity or simply a sphere of gravitationally influenced events that demonstrate the effects of a gravitational source?

How do we observe and verify something going faster than light?

Last edited: Nov 22, 2003