# Thread: Photon absorbtion - black holes vs color black

1. Newbie
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## Photon absorbtion - black holes vs color black

Hi all,

New to the forum, but I'm trying to resolve an idea (and annoyance, but that may be caused by my own ignorance) I've had for quite some time. Whenever I see or read something on black holes, the common attribute assigned to them is that their gravity is so strong, "not even light can escape". My problem is when you consider the definition of color. An object's color is determined by the wavelength of light which is reflected. So white reflects all wavelength photons and absorbs none, while black reflects none and absorbs all.

If in a black hole no light can escape, then it must be absorbed into the singularity. If a Styrofoam ball is painted back, it absorbs - or doesn't allow any photons to escape. So how is it that this statement is an attribute to describe strong gravity (or density)? A Styrofoam ball has very little gravitational attraction, yet no light can escape it if its black in color!

2. Newbie
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## Photons escaping black holes vs the color black

My second attempt to post this, so I apologize if it shows up twice, but I can't find my first posting.

I'm new here and have come to confirm a confusion I've had for some time now. Whenever I read or hear something on black holes, one common attribute is that they are so dense and have such a strong gravitational field that "not even light can escape". But then I consider the definition of color where it is all the photons of a range of wavelength that are not absorbed by the object. White then reflects all wavelength photons while black absorbs photons of all wavelengths.

So any object other than white could be said that it allows no light of a given set of wavelength to escape and black allows NO photons to escape (or no light). Furthermore, science defines black as an absent of color which also means an absence of photons. So what I fail to understand is how absorption of photons has any bearing on the strength of the gravitational force and why this attribute is so sensationalized. I'm hoping someone here can enlighten me if my logic is in error.

As an aside, I find I learn best when engaged in debate, so if I appear stubborn, understand it's my learning process!

3. Originally Posted by tomb523
My second attempt to post this, so I apologize if it shows up twice, but I can't find my first posting.
Hi tomb523, welcome to BAUT.

I merged your two threads/posts into one thread. The reason you didn't see your first post is that, as part of our anti-spam measures, the posts of newbies are held in a "moderation queue" until they can be approved by a moderator (even you can't see them). It is nothing personal, just something we have to do (we get a lot of spam). After several more posts, this will stop happening. And, depending on how many moderators are on-line, it can take from minutes to hours before a post is approved.

Good luck with your learning process. Stubborn is fine, just keep it polite.

4. Assuming your black painted ball is perfectly black, it will indeed absorb all the light that hits it. But you'll still detect radiation coming from it: thermal radiation based on its temperature, known as a blackbody spectrum. The incoming radiation will have an indirect effect, by changing the ball's temperature. By contrast, a black hole will emit essentially no radiation (other than Hawking radiation, which is actually still a thermal spectrum, but at a very low temperature for a typical black hole, probably slight enough to be unmeasurable).

5. Originally Posted by tomb523
Hi all,

New to the forum, but I'm trying to resolve an idea (and annoyance, but that may be caused by my own ignorance) I've had for quite some time. Whenever I see or read something on black holes, the common attribute assigned to them is that their gravity is so strong, "not even light can escape". My problem is when you consider the definition of color. An object's color is determined by the wavelength of light which is reflected. So white reflects all wavelength photons and absorbs none, while black reflects none and absorbs all.

If in a black hole no light can escape, then it must be absorbed into the singularity. If a Styrofoam ball is painted back, it absorbs - or doesn't allow any photons to escape. So how is it that this statement is an attribute to describe strong gravity (or density)? A Styrofoam ball has very little gravitational attraction, yet no light can escape it if its black in color!
Black objects can and do re-radiate energy they have absorbed, otherwise every black object would continue to build up energy until its heat exceeded the strength of its molecular bonds and it melted or exploded.

Here's a start-- http://en.wikipedia.org/wiki/Black_body

At Earth-ambient temperatures this emission is in the infrared region of the electromagnetic spectrum and is not visible. The object appears black, since it does not reflect or emit any visible light.

6. Order of Kilopi
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Photons do not escape from black hole due to the warping of spacetime, not due to what you are describing as absorption. Grey has given you a good explanation of thermal radiation. In the case of black holes, the path of light (or photons) once the photon is inside the event horizon, never passes outside the event horizon. All paths, whether of photons or of massive particles or objects will stay within the event horizon and will eventually (usually within a very short time) end up within the central singularity. As a result, all radiation entering the black hole appears to be absorbed, without being reradiated at a different frequency. As it would be with a normal object.

7. Though if a "black" hole absorbs all photons and doesn't allow any to escape, shouldn't it technically be an "invisible hole"? The only way we should be able to "see" one (accretion disk notwithstanding) is by the gravitational warping of light passing around it, right?

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Originally Posted by EDG
Though if a "black" hole absorbs all photons and doesn't allow any to escape, shouldn't it technically be an "invisible hole"? The only way we should be able to "see" one (accretion disk notwithstanding) is by the gravitational warping of light passing around it, right?
Right, and there should be a what appears to be an unlit disk, surrounded by the light passing around it. Which would look "black" as there would be no radiation either reflected or emitted from the disk.

9. Keep in mind that it's only called a "black hole" in a few languages. Even in English, an equivalent term is "collapsar".

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Originally Posted by tomb523
So what I fail to understand is how absorption of photons
has any bearing on the strength of the gravitational force
and why this attribute is so sensationalized.
Absorption is not the property of interest. Speed is.
It is sensational because light is the fastest thing there
is, and even it isn't fast enough to escape the gravity
of a black hole.

-- Jeff, in Minneapolis

11. Originally Posted by Tensor
Right, and there should be a what appears to be an unlit disk, surrounded by the light passing around it. Which would look "black" as there would be no radiation either reflected or emitted from the disk.
Why would it be a disk? Aren't the event horizons of black holes supposed to be spherical?
And why would it be black? If it's not reflecting or emitting anything, we can't see it. Something that doesn't emit or reflect radiation can't be black, it has to be invisible.

12. Originally Posted by tomb523
Hi all,

New to the forum, but I'm trying to resolve an idea (and annoyance, but that may be caused by my own ignorance) I've had for quite some time. Whenever I see or read something on black holes, the common attribute assigned to them is that their gravity is so strong, "not even light can escape". My problem is when you consider the definition of color. An object's color is determined by the wavelength of light which is reflected. So white reflects all wavelength photons and absorbs none, while black reflects none and absorbs all.

If in a black hole no light can escape, then it must be absorbed into the singularity. If a Styrofoam ball is painted back, it absorbs - or doesn't allow any photons to escape. So how is it that this statement is an attribute to describe strong gravity (or density)? A Styrofoam ball has very little gravitational attraction, yet no light can escape it if its black in color!

Welcome to BAUT forums Tomb523,

The problem is you are only thinking about visible light. A black StyrofoamTM ball still emits and reflects light. It will still emit its own infra red photons and will still reflect various other wavelengths like ultraviolet dependant on the material used to paint it with. In fact painting it black will cause it to emit more infra red photons do to the ball becoming hotter.

So while it might superficially seem like the 2 object have the same result the truth is far different when you look at the details. Note too I can make "white" light while only reflecting 3 frequencies of photons and not all the frequencies in the visible spectrum.

Hope this helps.

Wayne

13. Originally Posted by EDG
Why would it be a disk? Aren't the event horizons of black holes supposed to be spherical?
And why would it be black? If it's not reflecting or emitting anything, we can't see it. Something that doesn't emit or reflect radiation can't be black, it has to be invisible.
It would appear as a disk to us. It is a sphere but we can't see its topological features....so it will appear flat. The sun has a similar effect for us...if you look at the sun through certain filters it can appear to look like a disk facing you as you may not see features that make it look 3 dimensional.

The black hole looks black instead of invisible because if it appeared invisible then you'd be able to see light from behind it unaltered. What happens though is any light coming from behind it that we see is getting bent around the black hole and this has a very particular effect, called an Einstein ring, which is very different from an object that is invisible. Part of the problem is that to see the "black disk" you'd have to be pretty close, relatively speaking, to the black hole to be able to resolve the disk. Take a 1 solar mass black hole as an example. Think how close you'd have to be to resolve a 6km diameter black hole.

Go here http://apod.nasa.gov/htmltest/gifcity/rsgrow.html to see what you'd see in various interactions with a black hole. You'll see that the effect is very different from being invisible.

14. Oh right - now that makes more sense (well, insofar as these things make anything resembling "sense" ).

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So what is the dimension of black hole resistance? And what is its polarizability?

What is the absorption cross-section of a black hole of a certain Schwarzschild radius?

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Thanks everyone for your reply's. So if I understand everything that was said here, a black painted object absorbs only the visible wavelength photons (and possibly other non visible wavelength); however, it does not absorb all and what it does absorbed is given off again, at least in part as heat. So what this means is that for the photons that are absorbed by a black object are re-radiated, in part, as heat photons and possibly other non-visible wavelength photons which are not absorbed?

Still, I'm not sure I understand how gravity acts on Photons when photon have no mass since, as I understand it, gravity is an attraction between objects based on the mass of each object? Also, if this is a gravitational process, why are there cycles of 'feeding' and 'not feeding' on objects with mass; however, light can never escape it. This would also seem to violate the ideal that space is warped to only allow those paths to enter the black hole as the the mass objects such as planets and stars travel along those same paths as photons do.

Regarding speed of light being the factor - since the visible wavelength photon can't escape the black object either, that argument doesn't seem to add any credence to the argument.

17. Order of Kilopi
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The Newtonian understanding of gravity was that it acted on mass.
Einstein found that that was incorrect. Gravity acts on everything,
in proportion to its energy. Mass is the most concentrated form of
energy, so it was relatively easy to notice the gravitational attraction
between large masses, like the Earth and Moon. Things which are
massless, such as light, have comparatively little energy and so do
not have much gravitational effect on things around them. But they
do have some. And gravity certainly acts on light. The Sun's gravity
can measureably bend the paths of light rays going past it. I think
that radio signals from deep space probes such as those on Mars
have measurable deflections when they are behind the Sun. The
signals are slightly delayed by the curvature of the path. It is called
"Shapiro delay".

Anything that happens to fall past the event horizon of a black hole
can never escape. However, anything getting close to the horizon is
in danger. Any massive object closer than 3 times the Schwarzschild
radius will be drawn in by the gravity. Unless it uses a rocket to push
itself away before it reaches the event horizon, it will fall in.

The distance of 3 R is called "the last stable orbit".

Light that is moving parallel to the event horizon "surface" at a
distance of 1.5 times the Schwarzschild radius could theoretically
orbit the black hole forever in a circular orbit at that distance.
But if it deviated downwards in the slightest, it would spiral down
to its doom. If it deviated upwards in the slightest, it would spiral
away and escape.

The distance of 1.5 R is called "the photon sphere".

Light below 1.5 R can escape from the black hole only if it is
emitted above the event horizon (1 R) in an upward direction.
The closer it is to the event horizon when it is emitted, the
more nearly vertical its path must be in order to escape.

A massive object attempting to fly past a black hole cannot get
much inside the last stable orbit (3 R) without being drawn in.
No massive object can reach the photon sphere (1.5 R) and
escape without using rockets. At the event horizon and below,
rockets can slow your fall, but they won't enable you to escape.

-- Jeff, in Minneapolis

18. Originally Posted by tomb523
Still, I'm not sure I understand how gravity acts on Photons when photon have no mass since, as I understand it, gravity is an attraction between objects based on the mass of each object?
Also, if this is a gravitational process, why are there cycles of 'feeding' and 'not feeding' on objects with mass; however, light can never escape it. This would also seem to violate the ideal that space is warped to only allow those paths to enter the black hole as the the mass objects such as planets and stars travel along those same paths as photons do.
Regarding speed of light being the factor - since the visible wavelength photon can't escape the black object either, that argument doesn't seem to add any credence to the argument.
General Relativity defines the effect of gravity as occurring due to the curvature of spacetime in the presence of massive objects. Objects within a gravitational field follow these 'curves'. Photons follow 'null geodesics', or the shortest pathway that can be taken in a gravitational field. For black holes (BHs), this null geodesic wraps itself around the BH inside the event horizon. The photons aren't held due to gravity .. they are constrained by the extreme spacetime curvature.

The incredible heat given off in the vicinity of a black hole is due to the compressive effects on the matter as it falls towards the event horizon, but before it enters the event horizon, (the point of no return). It can be of the order of hundreds of millions of degrees and may occur sporadically, in response to the supply/decline of available material.

The BH itself however, can only generate heat through Hawking radiation. For a solar mass BH, the temperature is raised by only about one ten millionth of a degree above absolute zero. For larger BHs, the increase in temperature is even less, as a smaller percentage of mass is converted to Hawking radiation.

Hawking radiation itself is created by particle/antiparticle pairs that are predicted by Quantum Field Theory, and experimentally verified by the Casmir effect. If these pairs are formed near the event horizon of a BH, conservation of energy is still preserved as the particle/antiparticle pair are created and destroyed very rapidly. However if one of the particles, (the one with negative energy or energy less than the ground state), is pulled into the event horizon, energy is no longer conserved as the other particle escapes.
The BH emits Hawking Radiation to preserve the conservation of energy. The BH emits radiation with energy that is equal to the mass of the particle that passes through the horizon.

Hope this helps.

Cheers

19. Can I add a little explanation that might help without bringing the community crashing down upon me... we shall see...

Think of the Black Hole as a area of such mass density as to distort the space near to it.

By such a degree as to prohibit that light ever reaching you.. BUT, I do not 'see' a BH as black.

So much in falling mater orbiting so quickly.. So much pressure of density. It's going to be looking like a star..

But I do not know this.. Its just more of my often miss quoted misinformation.. or Marks logic..

The Name of a Black Hole is a mismatch.. for if you see it.. Black, you are now too close...to get away.

and then for a complexional and confused aspect.. All light reflected would be bright white..

No light reflected would be black.. being the absence of all colour..

but on the artists pallet the opposed is the fact.. Hmm...