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toothdust
2008-Dec-06, 08:00 PM
If a black hole is a gravitationally acting object, how then is it able to trap even massless particles, such as photons?

TheQuixoticMan
2008-Dec-06, 08:33 PM
I think that what you're encountering is the difference between how gravity is defined by Newton and by Einstein. With Newtonian ideas, gravity is simply mass acting on mass in a static universe, but Einstein has a different view of the force. Now, I don't have too much experience here, so take what I say with a grain of salt, but as far as I understand it, Relativity states that gravity acts upon space-time itself, and not as a direct object to object basis. Rather, gravity warps space-time, and matter and energy travel along that warped space time on geodesics, being curved surfaces, rather than straight lines. Think of it in terms of the planet. We don't actually walk in straight lines across the planet's surface, we walk on the geodesics that run across it. In relativity, the shortest distance between two points is not a straight line: it is a curve. In the same way, as photons approach a large gravity well, their path is bent by it. They aren't necessarily drawn into it, but they're redirected by it, like your car being redirected over a speed-bump. Black holes have such strong gravity that they warp space-time in on itself. The curve of the geodesic is so severe that there is no path the photons can make to get away from the black hole.

Did that make any sense whatsoever?

mugaliens
2008-Dec-06, 09:32 PM
Gravity bends spacetime. It doesn't matter whether it's massless, a melon or a moon - Earth's gravity will warp them inward. The only difference between a photon and the Moon is that the Moon imparts an additional gravitational attraction on the Earth. Thus, the Moon and the Earth accelerate a bit faster than would a melon and the Earth, or the massless particle and the Earth.

toothdust
2008-Dec-06, 09:32 PM
An answer from someone with experience would be nice.

korjik
2008-Dec-06, 09:36 PM
E=h(nu)=mc^2
h(nu)/c^2=m

F=GMm/r^2

to put it into english, energy divided by the speed of light squared is equivalent to a mass, which is then put into the gravitational force equation.

toothdust
2008-Dec-06, 09:38 PM
So basically, photons (as well as the rest of the light spectrum) fall into orbit around the BH somewhere below the event horizon?

korjik
2008-Dec-06, 09:49 PM
orbit is the wrong word. technically, if the black hole is not rotating, a photon could orbit on the event horizon, but that would require a infinitely unlikely set of circumstances to occur.

the main problem is that once something is inside the event horizon, all lines end up in the singularity. any orbit that dipped inside the horizon would mean the photon ends up in the singularity.

On the other hand, flybys are possible. If the photon has the right trajectory, it could go past on a hyperbolic orbit.

Ken G
2008-Dec-06, 09:53 PM
If a black hole is a gravitationally acting object, how then is it able to trap even massless particles, such as photons?A black hole is defined by its ability to trap even massless photons, that's the "black" part. You are asking for a theoretical understanding of how such an object might work, so you first have to specify what theory you are interested in using. So far you've been given answers from both the Newtonian view, and the Einsteinian view, and they were all pretty good answers.

Presuming that you want our best current theory, which is general relativity, that theory says that gravity alters something fundamental about space and time such that all objects in free space, whether they be massless or otherwise, in some sense "think" they are moving in straight lines, but we perceive their paths as curved when we compare their path to a Euclidean notion of how space and time "ought to" behave. Basically, we build our notions of how space works in an accelerated environment (there is a net upward force on just about everything around you right now), so we fail to recognize that it works differently in the unaccelerated environment of space. And the "disconnect" between our expectation, and the reality, extends to photons-- and is especially severe when gravity is strong, like in a black hole.

grant hutchison
2008-Dec-06, 11:04 PM
orbit is the wrong word. technically, if the black hole is not rotating, a photon could orbit on the event horizon ...It turns out that the closest possible circular orbit for photons is farther out than the event horizon, at 1.5 times the Schwarzschild radius, in the so-called "photon sphere".

Grant Hutchison

alainprice
2008-Dec-07, 12:46 AM
I'm with Grant. Any particle which passes beyond the event horizon will not attain a stable orbit. They all head for the singularity.

creativedreams
2008-Dec-07, 01:35 AM
maybe a black hole is a "big bang" creating a smaller universe?

Apparently you have asked this question in a number of other threads. If you are suggesting an ATM and are prepared to defend it I would suggest you take that there, if this is just a question then as Toothdust has suggested you should start a separate Q&A thread.

Sticks

toothdust
2008-Dec-07, 01:37 AM
maybe a black hole is a "big bang" creating a smaller universe?

That is not what this thread is about (although that thought has crossed my mind before). Please start a separate thread if you have questions.

toothdust
2008-Dec-07, 01:45 AM
It turns out that the closest possible circular orbit for photons is farther out than the event horizon, at 1.5 times the Schwarzschild radius, in the so-called "photon sphere".

Grant Hutchison

Wow, I was just tossing that out there. I didn't think it was possible for a photon to actually attain any kind of orbit.

I am still confounded on how bent space affects photons. I know it happens: The Hubble uses gravity lenses to see into deep space.

Would it be correct to assert photons travel on a medium, which is space-time itself?

Two questions:

If gravity can affect photons, do photons potentially have an incredibly small amount of mass (a fraction of that of neutrinos)?

Is there any way massive stars could be affecting their own speed of light by applying their own gravity to escaping photons?

Ken G
2008-Dec-07, 01:53 AM
Would it be correct to assert photons travel on a medium, which is space-time itself?
I think that is very much the way Einstein himself thought of it.

If gravity can affect photons, do photons potentially have an incredibly small amount of mass (a fraction of that of neutrinos)? There is no direct connection between the first and second part of that question-- the former part does not bear on the answer to the latter (which is unknown, but in the theory of relativity, the mass is exactly zero).


Is there any way massive stars could be affecting their own speed of light by applying their own gravity to escaping photons?They do create what is known as "gravitational redshift". Observers near the star would still measure the speed as c, it appears from the successes of relativity, but it would be seen as redder and redder the farther from the star you go (the Sun does the same thing-- remember high mass stars generally have weak surface gravity). The effect is quite small.

mugaliens
2008-Dec-07, 01:53 AM
E=h(nu)=mc^2
h(nu)/c^2=m

F=GMm/r^2

to put it into english, energy divided by the speed of light squared is equivalent to a mass, which is then put into the gravitational force equation.

Yes, but...

...if the particle is massless, that particular equation (Newton's Law) falls apart!

Instead, you must use:

θ = GM/rc2

However, even this only holds for distant point sources. For closer work, you must use a different formula, but be forewarned, as it's a bit tricky (http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Approximate_f ormula_for_the_bending_of_light)...

toothdust
2008-Dec-07, 02:18 AM
There is no direct connection between the first and second part of that question-- the former part does not bear on the answer to the latter (which is unknown, but in the theory of relativity, the mass is exactly zero).

How, then, does gravity affect a massless object? Have photons ever been tested for mass?


They do create what is known as "gravitational redshift". Observers near the star would still measure the speed as c, it appears from the successes of relativity, but it would be seen as redder and redder the farther from the star you go (the Sun does the same thing-- remember high mass stars generally have weak surface gravity). The effect is quite small.

Hmm. Again, how is said star affecting photons it has released if they have no mass? Has this been considered as an explanation for the observed Hubble redshift (please don't boot me to ATM!)? The effect may be small, as you say, but over intergalactic distances?

I did not know that the more massive a star, the less surface gravity. Please explain!

Ken G
2008-Dec-07, 02:32 AM
How, then, does gravity affect a massless object? There is nothing in general relativity that requires the object have mass, the mass is needed for the source of the gravity, not the response to it. The response is simply governed by the geometry of spacetime, which is why all objects fall together, regardless of their mass. The usual explanation is like an ant crawling on an apple. The ant needs no mass to follow the contours of the apple, it is just a geometric effect.


Have photons ever been tested for mass? Certainly. The problem is, you can never get anything but an upper limit-- the mass has to be below a tiny upper limit.


Again, how is said star affecting photons it has released if they have no mass?Why would they need mass to be redshifted? There are several ways you can describe gravitational redshift. One is that space itself is falling into the star, and being radially stretched in the process (that's a tidal effect). The radial stretching of the space also stretches the photon wavelength, just like it does in the Big Bang. That casts it as something that happens during propagation. Or, you can imagine that time itself is slowed by the gravity of the star (and sure enough, people at the surface, were that possible, would indeed age more slowly than we), so that the light isn't red-shifted, it was always red, due to what was going on with time where it was emitted. The explanations vary with the coordinates chosen, all you can say is the light shows up redder than you'd otherwise expect.

Has this been considered as an explanation for the observed Hubble redshift (please don't boot me to ATM!)?Yes, you can view cosmological redshifts as a gravitational effect. However, when most people use the term "gravitational redshift", they mean something over and above just the cosmological redshift, even though the latter can be cast as a form of the former.


The effect may be small, as you say, but over intergalactic distances?Yes, the gravity of all the stars could in principle do that, but it turns out you need even more mass than in the stars-- you need "dark matter".


I did not know that the more massive a star, the less surface gravity. Please explain!
Surface gravity depends on mass and radius, and massive stars are also bigger.

Hornblower
2008-Dec-07, 02:34 AM
How, then, does gravity affect a massless object? Have photons ever been tested for mass?



Hmm. Again, how is said star affecting photons it has released if they have no mass? Has this been considered as an explanation for the observed Hubble redshift (please don't boot me to ATM!)? The effect may be small, as you say, but over intergalactic distances?

I did not know that the more massive a star, the less surface gravity. Please explain!
To understand the gravitational behavior of photons according to Einstein et. al. you need to forget some of what you learned, or thought you learned, about pre-20th century classical physics. I will not elaborate further because many others have answered your question at least as well as I could, if not better, in this forum and in others.

A star with say 10 solar masses typically is bloated to upwards of 10 times the diameter. Eddington and others have done the calculations to explain this. The inverse square relation thus yields well under 1/10 the solar gravitation at the photosphere.

grant hutchison
2008-Dec-07, 02:36 AM
Wow, I was just tossing that out there. I didn't think it was possible for a photon to actually attain any kind of orbit.Orbits in the photon sphere are unstable: any tiny deviation from perfect circularity increases over time, and the photon escapes outwards or falls inwards. But the better the match to circularity, the more times the photon goes around in roughly the same region.

Grant Hutchison

George
2008-Dec-07, 02:58 AM
Orbits in the photon sphere are unstable: any tiny deviation from perfect circularity increases over time, and the photon escapes outwards or falls inwards. But the better the match to circularity, the more times the photon goes around in roughly the same region.
Any idea of their flux distribution? Would a majority of the photons somehow be orbiting along the bh's equitorial plane and traveling in the same spin directin? If so, then I suppose one side would appear red from redshifting and the opposite side would present no light.

grant hutchison
2008-Dec-07, 03:04 AM
Any idea of their flux distribution? Would a majority of the photons somehow be orbiting along the bh's equitorial plane and traveling in the same spin directin? If so, then I suppose one side would appear red from redshifting and the opposite side would present no light.For a rotating black hole, there are actually two photon spheres, one (inner) prograde and one (outer) retrograde, the split being induced by frame dragging.
Some interesting orbits (http://www.physics.nus.edu.sg/~phyteoe/kerr/) can occur between these two delimiting spheres.

Grant Hutchison

toothdust
2008-Dec-07, 03:08 AM
There is nothing in general relativity that requires the object have mass, the mass is needed for the source of the gravity, not the response to it. The response is simply governed by the geometry of spacetime, which is why all objects fall together, regardless of their mass. The usual explanation is like an ant crawling on an apple. The ant needs no mass to follow the contours of the apple, it is just a geometric effect.

Hmm. Ok.


Certainly. The problem is, you can never get anything but an upper limit-- the mass has to be below a tiny upper limit.

Was a tiny upper limit found? Is the upper limit zero?


Why would they need mass to be redshifted? There are several ways you can describe gravitational redshift. One is that space itself is falling into the star, and being radially stretched in the process (that's a tidal effect). The radial stretching of the space also stretches the photon wavelength, just like it does in the Big Bang. That casts it as something that happens during propagation. Or, you can imagine that time itself is slowed by the gravity of the star (and sure enough, people at the surface, were that possible, would indeed age more slowly than we), so that the light isn't red-shifted, it was always red, due to what was going on with time where it was emitted. The explanations vary with the coordinates chosen, all you can say is the light shows up redder than you'd otherwise expect.

So which is it? Space falling, space stretching, or time slowed? Or all of the above? If gravity is just a curvature of space-time, would it not be a uniform "surface" anymore? Or does the analogy of the ball on a rubber sheet fit here, where the rubber sheet is more stretched the closer you get to the ball? Even if this is the case, why don't the photons just ride the curvature out of the gravity well as they do in say passing through a gravity lens?

From my understanding, curved space is just that; curved. Not curving, unless an object was rapidly gaining mass.


Yes, you can view cosmological redshifts as a gravitational effect. However, when most people use the term "gravitational redshift", they mean something over and above just the cosmological redshift, even though the latter can be cast as a form of the former.

Yes, the gravity of all the stars could in principle do that, but it turns out you need even more mass than in the stars-- you need "dark matter".

Almost sounds like you are saying that the presence of dark matter eliminates the need to invoke expanding space to explain the cosmological redshift. I doubt this is what you are saying, but on the surface it seems that way.


Surface gravity depends on mass and radius, and massive stars are also bigger.

Ah yes. I was assuming a same volume/more massive. A physics friend of gave an analogy that if the sun turned into a black hole, the planets would just keep orbiting as they always have. Only if you got really close to it would things get ugly.

Thank you for your replies by the way.

toothdust
2008-Dec-07, 03:09 AM
For a rotating black hole, there are actually two photon spheres, one (inner) prograde and one (outer) retrograde, the split being induced by frame dragging.
Some interesting orbits (http://www.physics.nus.edu.sg/~phyteoe/kerr/) can occur between these two delimiting spheres.

Grant Hutchison

wow. Interesting indeed.

toothdust
2008-Dec-07, 03:27 AM
Just had another thought.

If you were to somehow isolate and steady a photon, and place it into a vertical vacuum tube on a gravitational body, would the photon fall downwards? Has this ever been tested/ is this possible to test?

George
2008-Dec-07, 03:33 AM
For a rotating black hole, there are actually two photon spheres, one (inner) prograde and one (outer) retrograde, the split being induced by frame dragging.
Some interesting orbits (http://www.physics.nus.edu.sg/~phyteoe/kerr/) can occur between these two delimiting spheres. Wild!

George
2008-Dec-07, 03:36 AM
Just had another thought.

If you were to somehow isolate and steady a photon, and place it into a vertical vacuum tube on a gravitational body, would the photon fall downwards? Has this ever been tested/ is this possible to test? It is well known Einstein contemplated ideas of catching up to a photon, but he realized it can't be done.

Nevertheless, one of his early predictions was that light from a star would appear to be deflected due to the gravitational field of that star. Solar eclispe photographs showed that light is affected by gravity.

toothdust
2008-Dec-07, 03:44 AM
It is well known Einstein contemplated ideas of catching up to a photon, but he realized it can't be done.

Nevertheless, one of his early predictions was that light from a star would appear to be deflected due to the gravitational field of that star. Solar eclispe photographs showed that light is affected by gravity.

I was thinking more along the lines of isolating a photon from a nearby source, such as a flashlight. No need to chase it through space if you have the source in your hand.

Ken G
2008-Dec-07, 03:48 AM
Was a tiny upper limit found? Is the upper limit zero?An upper limit is never zero, but I couldn't tell you what it is. My first Google hit, at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html , says "The new limit is 7 10-17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3 10-27 eV, but there is some doubt about the validity of this method."


So which is it? Space falling, space stretching, or time slowed? Or all of the above? Instead of "all of the above", I'd say "any of the above". For each coordinatization you choose to make your picture, you get a different answer, but you don't get multiple answers for the same coordinatization. This is a common issue-- if you were watching a sports event, and I asked which direction a player had to go to score, you could say "toward the left" or "toward the right" depending on which team you meant and which side of the field you were watching from, but never "either way"-- the game wouldn't work.


If gravity is just a curvature of space-time, would it not be a uniform "surface" anymore? Or does the analogy of the ball on a rubber sheet fit here, where the rubber sheet is more stretched the closer you get to the ball?Thinking in terms of analogy is always risky-- if there were a simpler way to express the theory, then the theory itself, then that would be the theory instead. Analogies are good for getting a sense of a theory, but not for answering questions about the theory, and so questions about the analogy are especially unimportant.


Even if this is the case, why don't the photons just ride the curvature out of the gravity well as they do in say passing through a gravity lens?There is too much curvature to ride out-- that's the difference with a gravity lens.

From my understanding, curved space is just that; curved. Not curving, unless an object was rapidly gaining mass. The problem with statements about "curved space" is that they really aren't faithful representations of the theory. The theory can talk about curvature of spacetime, but then the time is in their too. Curvature, when time is included, is a lot like dynamical changes-- curving of space, in other words.

Almost sounds like you are saying that the presence of dark matter eliminates the need to invoke expanding space to explain the cosmological redshift. I doubt this is what you are saying, but on the surface it seems that way.The way I look at it is, the expansion is an initial condition, but its effect is to weaken the pervasive gravity of the dark matter. As such, photons are absorbed in a weaker gravity environment then they were emitted, and that's what is normally a description of gravitational redshifting, though the term is not generally used that way in cosmology.

Jeff Root
2008-Dec-07, 03:51 AM
toothdust,

One thing everyone learns that you need to unlearn is that mass is the
source of gravity. The source of gravity is actually energy. By the familiar
formula E=mc^2, a little bit of mass contains a lot of energy. That means
a drop of water "generates" more gravity than does all the light in a huge,
overlit office building. But get enough light together in one place, and it
would have a significant gravity field.

As Ken says, that does not mean that light has mass. E=mc^2 assumes
that light has no mass. But any quantity of light will have an equivalent
mass-energy which can be calculated and used in Newton's gravity law
just as if it were mass.

An article in the May 1976 issue of Scientific American titled "The Mass
of the Photon" by Goldhaber and Nieto explains the history of attempts
to measure photon mass, and says:


Limits on the mass of the photon have been improved by a
factor of a billion since Robison's first test of Coulomb's law
more than 200 years ago. The best limits based on direct
measurements are deduced from the failure of experiments
to detect any exponential decay in large magnetic fields.
Measurements of Jupiter's magnetic field, obtained by the
Pioneer 10 spacecraft ... imply that the mass of the photon
cannot exceed 8x10^-49 gram.
It also suggests observations of the Galaxy's gravitational field
which could reduce the maximum possible mass even further.

-- Jeff, in Minneapolis

toothdust
2008-Dec-07, 04:37 AM
An upper limit is never zero, but I couldn't tell you what it is. My first Google hit, at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html , says "The new limit is 7 10-17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3 10-27 eV, but there is some doubt about the validity of this method."

So the "mass" is actually measured in energy, electron Volts (eV)? Using E=mc^2, what does this translate to in mass?


Instead of "all of the above", I'd say "any of the above". For each coordinatization you choose to make your picture, you get a different answer, but you don't get multiple answers for the same coordinatization. This is a common issue-- if you were watching a sports event, and I asked which direction a player had to go to score, you could say "toward the left" or "toward the right" depending on which team you meant and which side of the field you were watching from, but never "either way"-- the game wouldn't work.

So its a matter of perspective. Makes sense.


Thinking in terms of analogy is always risky-- if there were a simpler way to express the theory, then the theory itself, then that would be the theory instead. Analogies are good for getting a sense of a theory, but not for answering questions about the theory, and so questions about the analogy are especially unimportant.

Let me try again. When speaking about the gravitational redshift from the radial stretching of space/space "falling in" to the star, is this falling of space into the star causing the radial stretching.

As for the analogy, notice I said rubber sheet, not rubber ball. I will use the term elastic sheet.

I think the sheet analogy is a good one, makes visualizing (2D anyways) gravity wells very easy. Now if I am understanding you correct, you are saying that the space in the gravity well of a star is stretching, as opposed to just stretched. Is there a difference between the two?

Say I am observing light coming from a star with said stretched space around it (elastic sheet). Would a good way to think of this to be that the reason said light is redshifted is because it is having to cross more space (stretched elastic sheet) relative to the unstretched (unstretched elastic sheet) space I am occupying? Wouldn't this just equate to a longer distance to cross, in effect only creating a longer delay effect in observing the light as opposed to when it was originally emitted?

If your confused by what I am asking, imagine how I feel!


There is too much curvature to ride out-- that's the difference with a gravity lens.

I was speaking of a star, not a black hole.


The problem with statements about "curved space" is that they really aren't faithful representations of the theory. The theory can talk about curvature of spacetime, but then the time is in their too. Curvature, when time is included, is a lot like dynamical changes-- curving of space, in other words.

So "curved space" is just a lay representation of the theory?


The way I look at it is, the expansion is an initial condition, but its effect is to weaken the pervasive gravity of the dark matter. As such, photons are absorbed in a weaker gravity environment then they were emitted, and that's what is normally a description of gravitational redshifting, though the term is not generally used that way in cosmology.

Can you please expand on what you mean in bold. Again, not trying to go ATM, but if the observed Hubble redshift can be explained by the presence of dark matter through its effect on gravitational redshifting, doesn't this eliminate the need for expansion? I thought the Hubble redshift was the reason for invoking an expanding universe? Or does theory (GR?) require expansion?

Edit: Lots of stuff.

toothdust
2008-Dec-07, 04:52 AM
toothdust,

One thing everyone learns that you need to unlearn is that mass is the
source of gravity. The source of gravity is actually energy. By the familiar
formula E=mc^2, a little bit of mass contains a lot of energy. That means
a drop of water "generates" more gravity than does all the light in a huge,
overlit office building. But get enough light together in one place, and it
would have a significant gravity field.

As Ken says, that does not mean that light has mass. E=mc^2 assumes
that light has no mass. But any quantity of light will have an equivalent
mass-energy which can be calculated and used in Newton's gravity law
just as if it were mass.

I am so utterly confused.:doh: I am familiar with the concept that all matter is energy. So if all that light in the office building were transferred into mass, say through plant photosynthesis, you would have a gain in plant material equal to the drop of water (or whatever amount of water said arbitrary office building light would generate)? Basically saying that any amount of photons has a potential mass?


An article in the May 1976 issue of Scientific American titled "The Mass
of the Photon" by Goldhaber and Nieto explains the history of attempts
to measure photon mass, and says:

It also suggests observations of the Galaxy's gravitational field
which could reduce the maximum possible mass even further.

-- Jeff, in Minneapolis

How can it even have a maximum at all if GR states that light can have NO mass? Even if it were 1 x 10^-100,000, that would still be an amount of mass.

toothdust
2008-Dec-07, 05:00 AM
Yes, but...

...if the particle is massless, that particular equation (Newton's Law) falls apart!

Instead, you must use:

θ = GM/rc2

However, even this only holds for distant point sources. For closer work, you must use a different formula, but be forewarned, as it's a bit tricky (http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Approximate_f ormula_for_the_bending_of_light)...

Yikes. Not even going to try to wrap my head around that....yet.

Jeff Root
2008-Dec-07, 05:46 AM
So if all that light in the office building were transferred into mass,
say through plant photosynthesis, you would have a gain in plant
material equal to the drop of water (or whatever amount of water
said arbitrary office building light would generate)? Basically saying
that any amount of photons has a potential mass?
Yes, I think that's exactly right. The energy of light is converted
into other forms of energy when it is absorbed, and those other
forms of energy (chemical bonds, electrostatic charge, and heat,
for example) have tiny masses associated with them.



How can it [light] even have a maximum [mass] at all if GR states
that light can have NO mass? Even if it were 1 x 10^-100,000,
that would still be an amount of mass.
The article describes attempts to measure the mass of the photon.
So far those measurements are consistent with a mass of zero.
However, they are also consistent with a very, very small mass.
Only measurement can say for sure what the correct value is, but
this kind of measurement can never be infinitely precise, so it can
never measure a value of exactly zero. In the unlikely event that
photons do have mass, a variation of Maxwell's equations called
Proca's equations would be needed to describe light's behavior, in
place of Maxwell's equations.

Einstein's relativity theories are based on the assumption that
Maxwell's equations correctly describe the behavior of light.
So far, that appears to be the case. If they are wrong, they
certainly are not wrong by much, and Proca's additions would
be a very tiny correction.

-- Jeff, in Minneapolis

Tensor
2008-Dec-07, 06:04 AM
Toothdust, one thing that seems to be missing from the explanations is that the photon has a zero rest mass. That is, if you are in the same rest frame as a photon, the photon will have zero mass. The problem with this is that a photon always travels at c. Anything with mass cannot travel at c. In relativity, the rest frame of a photon is not a valid frame of reference.

Now, even though it has a rest mass of zero, a photon will have an energy as it travels. That energy can be found by using the formula E = hf where E is the energy, h is Planck's constant and f is the frequency of the photon. This energy is usually measured as momentum using the following formula p = h/w where p is the momentum, h is Planck's constant and w is the wavelength of the photon.

Very seldom is the energy of a single photon used in relativity. Most of the time, what is used is called the energy-momentum four-vector. This basically takes the flow of momentum through the three spatial coordinates. The other part of the vector is the total energy through that coordinate. If there is mass involved, it is converted into energy using E = mc2. So what you have is the total energy at a point and the total flow of momentum through that point from the combination of the three spatial coordinates.

Ken G
2008-Dec-07, 07:32 AM
So the "mass" is actually measured in energy, electron Volts (eV)? Using E=mc^2, what does this translate to in mass?
The mass of an electron is half an MeV, so the larger upper limit translates to a mass of about 10-22 the mass of the electron.
Real small.


Let me try again. When speaking about the gravitational redshift from the radial stretching of space/space "falling in" to the star, is this falling of space into the star causing the radial stretching.I will try again also-- explanations of "causes" are quite often dependent on the coordinate system, or reference frame, one chooses in which to give that explanation. If a rocket moving away from us sends us a signal, and we receive it redshifted, did the redshift happen when the light was emitted, when we received it, or continuously as it propagated? Answer: depends on how you choose to coordinatize the problem (which is like saying, it depends on the reference frame of the observer to whom you pose the question).



I think the sheet analogy is a good one, makes visualizing (2D anyways) gravity wells very easy. Now if I am understanding you correct, you are saying that the space in the gravity well of a star is stretching, as opposed to just stretched. Is there a difference between the two? Yes, the stretched sheet is a bit of a swindle, because it really only works for light, and should be ants crawling along, instead of balls rolling. The problem is, the usual way is really just using a component of the actual gravitational acceleration as if it was some kind of geometric effect, but it just isn't. It's basically a lie. But it does work as a geometric effect for finding the geodesics-- the paths that light takes.



Say I am observing light coming from a star with said stretched space around it (elastic sheet). Would a good way to think of this to be that the reason said light is redshifted is because it is having to cross more space (stretched elastic sheet) relative to the unstretched (unstretched elastic sheet) space I am occupying? Wouldn't this just equate to a longer distance to cross, in effect only creating a longer delay effect in observing the light as opposed to when it was originally emitted?The longer delay has to be longer for subsequent wavefronts in the wavetrain that defines the frequency of the light. That's why a curving (stretching in space) sheet works better than a curved (stretched) sheet.

So "curved space" is just a lay representation of the theory?
And not a very good one at all. The curvature is of spacetime-- what is happening in space alone depends on how you coordinatize it (your reference frame).



Can you please expand on what you mean in bold. Yes. The strength of gravity (here I mean the gravitational potential) depends on the distances between the masses, so as the universe expands, the pervasive gravity decreases. That gravity comes from dark matter, and dark energy (in the new view). But still, it is emitted in a stronger gravity environment than it is absorbed in, so that can be viewed as gravitational redshift, and it seems to me should be so viewed, in the standard "comoving" coordinates used in cosmology.


Again, not trying to go ATM, but if the observed Hubble redshift can be explained by the presence of dark matter through its effect on gravitational redshifting, doesn't this eliminate the need for expansion? No, because if there is no expansion, there is no change in the strength of gravity-- as though the light we emitted and absorbed at the same gravitational potential, yielding no redshift.

galacsi
2008-Dec-07, 11:58 AM
I think , from what I learn , some time ago, that you must distinguish to cases :

First light from outside the black hole and passing by will be deflected.

Then light from the inside of the black hole will be red-shifted by gravity. From may be X rays to visible light then to infra red then to radio waves . The speed of light will still be "c".But I cannot see how these rays can be prevented to escape.
Is that if they come from the very heart of the singularity , they can be red-shifted to zero energy ?

mugaliens
2008-Dec-07, 01:49 PM
Any idea of their flux distribution? Would a majority of the photons somehow be orbiting along the bh's equitorial plane and traveling in the same spin directin? If so, then I suppose one side would appear red from redshifting and the opposite side would present no light.

Negative. As was mentioned, the orbits in the photon sphere are unstable. It's a location of static stability, but dynamic instability. Thus, the slightest bit off perfect... And it's either all in, or all out. Thus, it's more void of photons than other locations.

Think of a photon sphere as a ridgeline of a map in elevation relief. Put marbles on the map and shake. The marbles tend to congregate in the valleys, though occasionally they'll cross a ridgeline.

mugaliens
2008-Dec-07, 01:52 PM
Yikes. Not even going to try to wrap my head around that....yet.

No worries - it's easy!

θ = GM/rc2

θ = angle of deflection in radians
G = gravitational constant
M = mass in kg of object bending the light
r = distance in meters from center of object to the photon's closest point of approach
c = speed of light in m/s