Black Holes and Gravitational Red Shift

Perpetual Student · 4th November 2009, 05:18 PM

Since gravitational red shift is determined by the formula:

$z= \dfrac{1}{\sqrt{1-\frac{2GM}{rc^2}}} -1$ ,

it seems that the red shift would continue to increase as the mass of the object increases and also increase as the radius decreases. As we continue to change these two variables in this way, we continue to increase the wavelength, the limit of which would appear to be a flat line. Why then do we ever get a black hole instead of lower and lower frequency radio waves?

edd · 4th November 2009, 05:48 PM

Originally Posted by Perpetual Student

Since gravitational red shift is determined by the formula:

$z= \dfrac{1}{\sqrt{1-\frac{2GM}{rc^2}}} -1$ ,

it seems that the red shift would continue to increase as the mass of the object increases and also increase as the radius decreases. As we continue to change these two variables in this way, we continue to increase the wavelength, the limit of which would appear to be a flat line. Why then do we ever get a black hole instead of lower and lower frequency radio waves?

The above expression diverges when r=2GM/c^2. This is for a black hole, when redshift is infinite. A light wave with infinite wavelength is no light wave at all.

I'm not sure where your confusion is, but increasing mass does not increase redshift in tandem, as r becomes ever more entwined in the two. Essentially when you hit the point you have a black hole, M and r increase together to keep the expression for z the 'same' - divergent.

ben m · 4th November 2009, 05:55 PM

Hi PS, I think you're mistaken in thinking that the limit is a flat line. What limit are you talking about exactly?

When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. When R < 2GMc^2, the quantity in the square root is negative; the equation has no solution any more and photons do not escape at all.

Perpetual Student · 4th November 2009, 06:21 PM

Originally Posted by ben m

Hi PS, I think you're mistaken in thinking that the limit is a flat line. What limit are you talking about exactly?

Obviously, I'm confused. Sorry for the ambiguity. The limit I was referring to is the wavelength. See below.

Quote:

When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. When R < 2GMc^2, the quantity in the square root is negative; the equation has no solution any more and photons do not escape at all.

OK, let's take a given wavelength of light, say (red) 700nm. As M increases above, z increases and 700nm becomes a greater wavelength. The limit is infinity, which would translate to a straight line, whatever that means for electromagnetic waves. So, the light would continue to emanate from the source (as we increase M), in longer and longer wavelengths. How and why does it ever stop emanating and become a black hole?

MattusMaximus · 4th November 2009, 06:24 PM

You're confusing the wavelength of the escaping light with whether or not it escapes in the first place. They aren't the same thing.

Perpetual Student · 4th November 2009, 07:14 PM

Originally Posted by MattusMaximus

You're confusing the wavelength of the escaping light with whether or not it escapes in the first place. They aren't the same thing.

OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating?

Roboramma · 4th November 2009, 07:25 PM

Originally Posted by Perpetual Student

OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating?

Wouldn't light whose wavelength, well, stopped waving and became a straight line have zero energy?

Reality Check · 4th November 2009, 07:34 PM

Originally Posted by Perpetual Student

OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating?

The light cannot pass the escape velocity of the black hole - see the Wikipedia article on black holes.
Basically a Newtonian black star has a radius where the escape velocity of light has to be greater then the speed of light.
This corresponds to the Schwarzschild radius in general relativity with the change that the radius r is no longer easy to define due to the curvature of space-time.

ben m · 4th November 2009, 07:45 PM

Originally Posted by Perpetual Student

OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating?

A light wave with infinite frequency isn't a wave at all, it carries zero energy, zero momentum, zero Poynting flux. It doesn't escape from the black hole because on the escaping trajectory it doesn't exist.

Remember that the redshift equation is not the entire equation of motion---it's just a way of keeping track of the behavior of particles that DO escape to infinity. If you look at the actual equation of motion, it'll tell you "how" the particles fail to escape. (More or less, if you get below the event horizon and fire a photon "outward", its actual trajectory goes inwards. There is no direction in which you can aim your photon-gun that makes them go anywhere but inward.)

Dave Rogers · 5th November 2009, 04:21 AM

Originally Posted by Perpetual Student

OK, let's take a given wavelength of light, say (red) 700nm. As M increases above, z increases and 700nm becomes a greater wavelength. The limit is infinity, which would translate to a straight line, whatever that means for electromagnetic waves. So, the light would continue to emanate from the source (as we increase M), in longer and longer wavelengths. How and why does it ever stop emanating and become a black hole?

Mathematically speaking, it stops because the red shift doesn't go to infinity as the quantity (mass divided by radius) goes to infinity; it goes to infinity when (mass divided by radius) is still finite. The energy of the photon is inversely proportional to its wavelength, so at that point the energy becomes zero. In effect, there's no longer a photon to escape.

Dave

sol invictus · 5th November 2009, 05:11 AM

Originally Posted by Perpetual Student

OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above.

Actually it does slow the speed of light, depending on exactly how you define "speed" in a curved space. For example: put a mirror near the horizon and hold it there. A distant experimenter sends a pulse of laser light at the mirror and measures the time it takes to come back. That time increases without bound as you move the mirror closer and closer to the horizon.

Quote:

How and why does it ever stop emanating and become a black hole?

Look at your formula again. There's a black hole for all values of M>0, because for all M>0 there's a radius where the redshift goes to infinity. Remember, that radius is not the radius of the hole - is that what you were thinking? It's the radial coordinate at which the light is emitted. r=0 is the center of the hole, inside the horizon.

edd · 5th November 2009, 06:46 AM

Originally Posted by ben m

A light wave with infinite frequency isn't a wave at all, it carries zero energy, zero momentum, zero Poynting flux. It doesn't escape from the black hole because on the escaping trajectory it doesn't exist.

Nitpick: infinite wavelength or zero frequency.

Perpetual Student · 5th November 2009, 09:38 AM

Thanks for all the responses (this is such a great forum!).
I'm still not able to put all the above information together, but I'm getting there -- some of it appears to be contradictory, obviously because of my lack of knowledge in this area. The mathematics is simple enough -- it's the physical reality that is confusing. A little more study may help.

Perpetual Student · 5th November 2009, 09:46 AM

Originally Posted by Reality Check

The light cannot pass the escape velocity of the black hole - see the Wikipedia article on black holes.
Basically a Newtonian black star has a radius where the escape velocity of light has to be greater then the speed of light.
This corresponds to the Schwarzschild radius in general relativity with the change that the radius r is no longer easy to define due to the curvature of space-time.

Thanks for the link. The escape velocity concept seems to be clear enough. The problem I see here is that light will leave a gravitational object at c if it's an asteroid or the sun -- it does not leave slower from a more massive body -- so why would escape velocity be a factor since the only thing that happens is that wavelength is lengthened from a more massive body?

Perpetual Student · 5th November 2009, 09:57 AM

Originally Posted by ben m

...
When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. ...

So is that it? As 2 GM -> rc^2, the object is becoming a black hole?
So, if we have a collapsing massive body, as r shrinks, the light emanating from the object red shifts to the point where we have very low frequency radio waves ultimately flat lining to infinity and that defines becoming a black hole?

s.i.: can I not assume r is the radius of the object and the light is emanating from the surface?

ben m · 5th November 2009, 10:30 AM

Originally Posted by Perpetual Student

So is that it? As 2 GM -> rc^2, the object is becoming a black hole?
So, if we have a collapsing massive body, as r shrinks, the light emanating from the object red shifts to the point where we have very low frequency radio waves ultimately flat lining to infinity and that defines becoming a black hole?

Yep, that's basically correct.

Quote:

s.i.: can I not assume r is the radius of the object and the light is emanating from the surface?

For any distance R from the center, calculate the amount of mass M that's *inside* that radius. (For example, if you want to calculate the gravitational properties of "sea level on Earth", R would be 6378 km and M would be the mass of everything below sea level---mantle, core, ocean, but not Mount Everest and not the atmosphere.). If you can find an R anywhere which exceeds the redshift condition you found, then that R is an event horizon and everything inside it is "a black hole". Under most circumstances, you're right---the place the event horizon is most likely to appear is at the surface, since that's normally where GM/R is largest.

But in (for example) a massive type-II supernova, there's a very dense bit in the middle of the star where 2GM/Rc^2 > 1, but there's lots of gas beyond that (including what you would think of as the star's "surface", the visible part) at larger radii for which 2GM/Rc^2 < 1.

Perpetual Student · 5th November 2009, 11:24 AM

Originally Posted by ben m

Yep, that's basically correct.

For any distance R from the center, calculate the amount of mass M that's *inside* that radius. (For example, if you want to calculate the gravitational properties of "sea level on Earth", R would be 6378 km and M would be the mass of everything below sea level---mantle, core, ocean, but not Mount Everest and not the atmosphere.). If you can find an R anywhere which exceeds the redshift condition you found, then that R is an event horizon and everything inside it is "a black hole". Under most circumstances, you're right---the place the event horizon is most likely to appear is at the surface, since that's normally where GM/R is largest.

But in (for example) a massive type-II supernova, there's a very dense bit in the middle of the star where 2GM/Rc^2 > 1, but there's lots of gas beyond that (including what you would think of as the star's "surface", the visible part) at larger radii for which 2GM/Rc^2 < 1.

Got it! Thanks.

4th November 2009, 05:18 PM	#1
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Black Holes and Gravitational Red Shift Since gravitational red shift is determined by the formula: $z= \dfrac{1}{\sqrt{1-\frac{2GM}{rc^2}}} -1$ , it seems that the red shift would continue to increase as the mass of the object increases and also increase as the radius decreases. As we continue to change these two variables in this way, we continue to increase the wavelength, the limit of which would appear to be a flat line. Why then do we ever get a black hole instead of lower and lower frequency radio waves?
	__________________ $\xi$

4th November 2009, 05:48 PM	#2
edd Muse Join Date: Nov 2007 Posts: 533	Originally Posted by Perpetual Student Since gravitational red shift is determined by the formula: $z= \dfrac{1}{\sqrt{1-\frac{2GM}{rc^2}}} -1$ , it seems that the red shift would continue to increase as the mass of the object increases and also increase as the radius decreases. As we continue to change these two variables in this way, we continue to increase the wavelength, the limit of which would appear to be a flat line. Why then do we ever get a black hole instead of lower and lower frequency radio waves? The above expression diverges when r=2GM/c^2. This is for a black hole, when redshift is infinite. A light wave with infinite wavelength is no light wave at all. I'm not sure where your confusion is, but increasing mass does not increase redshift in tandem, as r becomes ever more entwined in the two. Essentially when you hit the point you have a black hole, M and r increase together to keep the expression for z the 'same' - divergent.
edd Muse Join Date: Nov 2007 Posts: 533	__________________ When I look up at the night sky and think about the billions of stars out there, I think to myself: I'm amazing. - Peter Serafinowicz

4th November 2009, 05:55 PM	#3
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460	Hi PS, I think you're mistaken in thinking that the limit is a flat line. What limit are you talking about exactly? When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. When R < 2GMc^2, the quantity in the square root is negative; the equation has no solution any more and photons do not escape at all.
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460

4th November 2009, 06:21 PM	#4
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Originally Posted by ben m Hi PS, I think you're mistaken in thinking that the limit is a flat line. What limit are you talking about exactly? Obviously, I'm confused. Sorry for the ambiguity. The limit I was referring to is the wavelength. See below. Quote: When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. When R < 2GMc^2, the quantity in the square root is negative; the equation has no solution any more and photons do not escape at all. OK, let's take a given wavelength of light, say (red) 700nm. As M increases above, z increases and 700nm becomes a greater wavelength. The limit is infinity, which would translate to a straight line, whatever that means for electromagnetic waves. So, the light would continue to emanate from the source (as we increase M), in longer and longer wavelengths. How and why does it ever stop emanating and become a black hole?
	__________________ $\xi$

4th November 2009, 06:24 PM	#5
MattusMaximus Intellectual Gladiator Join Date: Jan 2006 Location: In the midst of a vast, beautiful & uncaring universe Posts: 7,123	You're confusing the wavelength of the escaping light with whether or not it escapes in the first place. They aren't the same thing.
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4th November 2009, 07:14 PM	#6
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Originally Posted by MattusMaximus You're confusing the wavelength of the escaping light with whether or not it escapes in the first place. They aren't the same thing. OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating?
	__________________ $\xi$

4th November 2009, 07:25 PM	#7
Roboramma Illuminator Join Date: Feb 2005 Location: Shanghai Posts: 3,374	Originally Posted by Perpetual Student OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating? Wouldn't light whose wavelength, well, stopped waving and became a straight line have zero energy?
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4th November 2009, 07:34 PM	#8
Reality Check Illuminator Join Date: Mar 2008 Location: New Zealand Posts: 4,365	Originally Posted by Perpetual Student OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating? The light cannot pass the escape velocity of the black hole - see the Wikipedia article on black holes. Basically a Newtonian black star has a radius where the escape velocity of light has to be greater then the speed of light. This corresponds to the Schwarzschild radius in general relativity with the change that the radius r is no longer easy to define due to the curvature of space-time.
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4th November 2009, 07:45 PM	#9
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460	Originally Posted by Perpetual Student OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. So, how does a black hole stop light from emanating? A light wave with infinite frequency isn't a wave at all, it carries zero energy, zero momentum, zero Poynting flux. It doesn't escape from the black hole because on the escaping trajectory it doesn't exist. Remember that the redshift equation is not the entire equation of motion---it's just a way of keeping track of the behavior of particles that DO escape to infinity. If you look at the actual equation of motion, it'll tell you "how" the particles fail to escape. (More or less, if you get below the event horizon and fire a photon "outward", its actual trajectory goes inwards. There is no direction in which you can aim your photon-gun that makes them go anywhere but inward.)
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460

5th November 2009, 04:21 AM	#10
Dave Rogers The Unbanned Join Date: Jan 2007 Location: Notlob Posts: 8,013	Originally Posted by Perpetual Student OK, let's take a given wavelength of light, say (red) 700nm. As M increases above, z increases and 700nm becomes a greater wavelength. The limit is infinity, which would translate to a straight line, whatever that means for electromagnetic waves. So, the light would continue to emanate from the source (as we increase M), in longer and longer wavelengths. How and why does it ever stop emanating and become a black hole? Mathematically speaking, it stops because the red shift doesn't go to infinity as the quantity (mass divided by radius) goes to infinity; it goes to infinity when (mass divided by radius) is still finite. The energy of the photon is inversely proportional to its wavelength, so at that point the energy becomes zero. In effect, there's no longer a photon to escape. Dave
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5th November 2009, 05:11 AM	#11
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 5,153	Originally Posted by Perpetual Student OK, but gravity does not slow the speed of light (obviously), instead it increases the wavelength by an amount determined by the equation above. Actually it does slow the speed of light, depending on exactly how you define "speed" in a curved space. For example: put a mirror near the horizon and hold it there. A distant experimenter sends a pulse of laser light at the mirror and measures the time it takes to come back. That time increases without bound as you move the mirror closer and closer to the horizon. Quote: How and why does it ever stop emanating and become a black hole? Look at your formula again. There's a black hole for all values of M>0, because for all M>0 there's a radius where the redshift goes to infinity. Remember, that radius is not the radius of the hole - is that what you were thinking? It's the radial coordinate at which the light is emitted. r=0 is the center of the hole, inside the horizon.

5th November 2009, 06:46 AM	#12
edd Muse Join Date: Nov 2007 Posts: 533	Originally Posted by ben m A light wave with infinite frequency isn't a wave at all, it carries zero energy, zero momentum, zero Poynting flux. It doesn't escape from the black hole because on the escaping trajectory it doesn't exist. Nitpick: infinite wavelength or zero frequency.
edd Muse Join Date: Nov 2007 Posts: 533	__________________ When I look up at the night sky and think about the billions of stars out there, I think to myself: I'm amazing. - Peter Serafinowicz

5th November 2009, 09:38 AM	#13
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Thanks for all the responses (this is such a great forum!). I'm still not able to put all the above information together, but I'm getting there -- some of it appears to be contradictory, obviously because of my lack of knowledge in this area. The mathematics is simple enough -- it's the physical reality that is confusing. A little more study may help.
	__________________ $\xi$

5th November 2009, 09:46 AM	#14
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Originally Posted by Reality Check The light cannot pass the escape velocity of the black hole - see the Wikipedia article on black holes. Basically a Newtonian black star has a radius where the escape velocity of light has to be greater then the speed of light. This corresponds to the Schwarzschild radius in general relativity with the change that the radius r is no longer easy to define due to the curvature of space-time. Thanks for the link. The escape velocity concept seems to be clear enough. The problem I see here is that light will leave a gravitational object at c if it's an asteroid or the sun -- it does not leave slower from a more massive body -- so why would escape velocity be a factor since the only thing that happens is that wavelength is lengthened from a more massive body?
	__________________ $\xi$ Last edited by Perpetual Student; 5th November 2009 at 11:23 AM.

5th November 2009, 09:57 AM	#15
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Originally Posted by ben m ... When 2GM = Rc^2, the denominator is zero and the redshift is infinite; any photon attempting to escape from a distance R winds up at literally zero energy. ... So is that it? As 2 GM -> rc^2, the object is becoming a black hole? So, if we have a collapsing massive body, as r shrinks, the light emanating from the object red shifts to the point where we have very low frequency radio waves ultimately flat lining to infinity and that defines becoming a black hole? s.i.: can I not assume r is the radius of the object and the light is emanating from the surface?
	__________________ $\xi$

5th November 2009, 10:30 AM	#16
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460	Originally Posted by Perpetual Student So is that it? As 2 GM -> rc^2, the object is becoming a black hole? So, if we have a collapsing massive body, as r shrinks, the light emanating from the object red shifts to the point where we have very low frequency radio waves ultimately flat lining to infinity and that defines becoming a black hole? Yep, that's basically correct. Quote: s.i.: can I not assume r is the radius of the object and the light is emanating from the surface? For any distance R from the center, calculate the amount of mass M that's inside that radius. (For example, if you want to calculate the gravitational properties of "sea level on Earth", R would be 6378 km and M would be the mass of everything below sea level---mantle, core, ocean, but not Mount Everest and not the atmosphere.). If you can find an R anywhere which exceeds the redshift condition you found, then that R is an event horizon and everything inside it is "a black hole". Under most circumstances, you're right---the place the event horizon is most likely to appear is at the surface, since that's normally where GM/R is largest. But in (for example) a massive type-II supernova, there's a very dense bit in the middle of the star where 2GM/Rc^2 > 1, but there's lots of gas beyond that (including what you would think of as the star's "surface", the visible part) at larger radii for which 2GM/Rc^2 < 1.
ben m Graduate Poster Join Date: Jul 2006 Posts: 1,460

5th November 2009, 11:24 AM	#17
Perpetual Student Graduate Poster Join Date: Jul 2008 Location: USA Posts: 1,263	Originally Posted by ben m Yep, that's basically correct. For any distance R from the center, calculate the amount of mass M that's inside that radius. (For example, if you want to calculate the gravitational properties of "sea level on Earth", R would be 6378 km and M would be the mass of everything below sea level---mantle, core, ocean, but not Mount Everest and not the atmosphere.). If you can find an R anywhere which exceeds the redshift condition you found, then that R is an event horizon and everything inside it is "a black hole". Under most circumstances, you're right---the place the event horizon is most likely to appear is at the surface, since that's normally where GM/R is largest. But in (for example) a massive type-II supernova, there's a very dense bit in the middle of the star where 2GM/Rc^2 > 1, but there's lots of gas beyond that (including what you would think of as the star's "surface", the visible part) at larger radii for which 2GM/Rc^2 < 1. Got it! Thanks.
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