Question about light and blackholes

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  • #1
magus
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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
 

Answers and Replies

  • #2
Ambitwistor
841
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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
magus
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so it basically depends on which conception of gravity is actually correct?
 
  • #4
Ambitwistor
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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
Unkaspam
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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?
 
  • #6
Ambitwistor
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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
Unkaspam
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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.
 
  • #8
Ambitwistor
841
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Originally posted by Unkaspam
Are you talking about the accretion disk?

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


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

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


I understood that neutron stars bounce X-rays off their mass.

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

Is it possible this can happen with a black hole? X-rays are said to have a very active wave.

What? "Active wave"? What is that?

Could this help them to escape the intense gravity of a black hole?

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
Herringbone
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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.
 
  • #10
Ambitwistor
841
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Originally posted by Herringbone
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.

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
Herringbone
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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."
 
  • #12
magus
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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
Ambitwistor
841
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Originally posted by magus
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,

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.

and what actually happens to the light when it reaches 0hz.

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
magus
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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.:smile:
 
  • #15
Herringbone
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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:

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

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:

[tex]
\nu_s / \nu_o = 0
[/tex]

Which happens when:

[tex]
M/r = \frac {c^2}{2G}
[/tex]

or where:

[tex]
r = \frac {2GM} {c^2}
[/tex]

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
Ambitwistor
841
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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
Unkaspam
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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.
 
  • #18
Unkaspam
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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?)
 
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  • #19
Ambitwistor
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Originally posted by Unkaspam
Gravitons escape from black holes all the time, at no loss to their gravitational energy.

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.

Does that mean gravity travels faster than light (in order to escape the event horizon of a 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
Unkaspam
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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?
 
Last edited:
  • #21
Ambitwistor
841
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Are there actual waves of gravity or simply a sphere of gravitationally influenced events that demonstrate the effects of a gravitational source?

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.


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

Well, here is one way: send a wave past two detectors. See how long it takes between reaching one and reaching the other. Knowing the distance between them, determine its speed.
 
  • #22
Unkaspam
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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?
 
  • #23
Ambitwistor
841
1
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?

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.

So what I'm asking is does gravity emanate from a source as a wave, in the way that light radiates from a source,

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.)

or are we simply seeing the tell tale effects of gravity within a defined sphere of an inert source?

I still don't know what you're talking about.
 
  • #24
Unkaspam
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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.
 
  • #25
sheldon
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This is so heavy spicoli! Nothing escapes the black hole period. NOTHING NOTHING. Not even your thought waves:)
 
  • #26
jedijesus
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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.
 
  • #27
jedijesus
15
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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
 
  • #28
Ambitwistor
841
1


When light clashes with the event horizon of a black hole, it give off gamma rays.

No, it doesn't. It just falls in. When people say that black holes radiate X-rays and gamma rays and such, that radiation comes from matter which is outside of the black hole.
 

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