Question about light and blackholes

[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

 

[magus]

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

 

[Unkaspam]

Originally posted by 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.

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?

 

[Unkaspam]

Originally posted by 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.

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.

 

[Herringbone]

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.

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.

 

[Herringbone]

In reality, it can't travel outwards at all.

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."

 

[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.

 

[magus]

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

 

[Herringbone]

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.

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:

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

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:

<br /> \nu_s / \nu_o = 0<br />

Which happens when:

<br /> M/r = \frac {c^2}{2G}<br />

or where:

<br /> r = \frac {2GM} {c^2} <br />

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

 

[Unkaspam]

Originally posted by Ambitwistor
What? "Active wave"? What is that?

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.

 

[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?)

 

[Unkaspam]

Originally posted by Ambitwistor
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.

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?

 

[Unkaspam]

Originally posted by Ambitwistor
I'm not sure what the difference is. Take, say, electromagnetic radiation (light): "Are there actual waves of electromagnetism, or simply a sphere of electromagnetically influenced events that demonstrate the effects of an electromagnetic source?" The answer to this question is the same as the answer to the corresponding gravitational question, but I don't know whether that answer is "yes" or "no" because I don't understand the question.

When a wave of water moves a leaf of seaweed we can say the mass of water and its motion are the cause of the event. When an object falls to the ground is this caused by a wave of gravity or a condition created by a source of gravity?

To put it another way, when there is an eclipse and the Earth is in shadow, there is no "shadow wave" emanating from the moon, the shadow is simply an effect of the moon blocking the sun, there are no "shadowtons" or shadow waves.

So what I'm asking is does gravity emanate from a source as a wave, in the way that light radiates from a source, or are we simply seeing the tell tale effects of gravity within a defined sphere of an inert source?

 

[Unkaspam]

Originally posted by Ambitwistor
A static gravitational field (such as, more or less, the Earth's gravitational field) doesn't have any gravitational waves, but objects still fall. (Likewise, the electric field around a point charge doesn't have any electromagnetic waves, but charges are still attracted or repelled from it.) Gravitational waves are changes in the gravitational field, just like electromagnetic waves (light) are changes in the electromagnetic field.



I think you are confusing some issues.

Gravitational waves can emanate from a source as a wave, analogous to how light (electromagnetic waves) radiates from a source. But electromagnetic waves are not responsible for, say, the electrostatic attraction (or repulsion) between two charges, nor are gravitational waves intrinsically responsible for the attraction of two masses. Nothing has to "emanate" from a mass or charge, in the sense of some effect propagating at some speed through space, in order for one body to influence another. (But if you change the source, then the effects of that change will propagate out in terms of changes in the field at successively more distant points.)



I still don't know what you're talking about.

You've managed to answer my question anyways, thanks.

 

[sheldon]

This is so heavy spicoli! Nothing escapes the black hole period. NOTHING NOTHING. Not even your thought waves:)

 

[jedijesus]

Originally posted by 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,
the point of which you are speaking, is called the event horizon. light particles are not able to withstand the graivty of the sigularity. Thus they are sucked in. They don't go out and then get sucked into a black hole, they just go into the black hole. Now when they are sucked in, they give of a nasty gamma burst that is detectible. They give this off because of the energy that is involved with clashing into a singulary which is like smashing into Earths atmosphere. And that is what can detect it, with other obsevations of course.

 

[jedijesus]

the event horizon is the edge of a black hole. the gavity there is greater than you can imagine. if light is going into the black hole, you are not going to see it. When light clashes with the event horizon of a black hole, it give off gamma rays. Thats it, that is the only way to see it. If you could optically see a black hole, it wouldn't be a black hole