what would happen to a photon in this situation?

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Offline jeffreyH

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what would happen to a photon in this situation?
« on: 27/01/2015 02:01:12 »
If we have a photon that is traveling on a path that takes it exactly towards the centre of gravity of a black hole how would the gravity affect its speed? When a photon is moving away from a source its wavelength is red shifted. Would we get a blue shift and an increase in energy? How much of this would be kinetic?

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Offline Atomic-S

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Re: what would happen to a photon in this situation?
« Reply #1 on: 27/01/2015 04:42:52 »
Quote
When a photon is moving away from a source its wavelength is red shifted.
It depends. What is the reference frame of the photon's observer?

The speed of the photon as viewed by any observer near it will be simply c.  The speed of the photon as seen by a different observer could be calculated by noting the relationship between space and time measured by that observer vs. space and time as it applies in the vicinity of the photon.  I believe the answer to that, in the situation you propose, is that a clock near the black hole will run slower than one far away, so that in a given interval of far-away-observer time, the photon will have undergone, as seen by that observer, less time, meaning that because it is traveling at c in its own reference frame, it will have traveled a shorter distance in its frame than the clock of the observer says it should. However, we also have to take into account the difference in spatial measures. I understand that a given length near the black hole, measured radially by a local observer, will measure longer than it will appear to be to a distant observer.  That, if correct (and I am not absolutely certain), would indicate that not only does the photon travel a lesser distance in the lesser time, but the distance that it does travel will appear less to the remote observer than it will locally. both of these phenomena would result in the photon appearing, to the remote observer, to slow down as it approached the black hole, eventually approaching an apparent speed of zero.

 

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Offline Atomic-S

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Re: what would happen to a photon in this situation?
« Reply #2 on: 27/01/2015 04:52:00 »
As pertains to the energy relationships:  An observer near the hole observes the clock far from the hole running faster than his, so that a wave emitted from far will appear to be in sync with that faster clock, and will therefore have a higher frequency than a wave emitted by a like apparatus located  near the hole. this is a blue-shift. Because the number of photons will remain unchanged by traveling, and because the quantum energy increases with frequency, the light upon arrival near the hole will have more energy as seen by the near observer than it did when it began its journey from afar as seen by the far observer.
 

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Offline yor_on

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Re: what would happen to a photon in this situation?
« Reply #3 on: 27/01/2015 11:05:46 »
yep, it's relationships, set in an idea of conservation laws keeping count. So if we assumed that we actually could observe a photon from several positions in a SpaceTime, we would get different definitions. It's a good example of what I call using a 'container model' to describe it. In reality that 'photon' will interact just once, in its annihilation. To that you can add a 'recoil' observed as whatever it leaves react to it.
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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #4 on: 28/01/2015 01:08:50 »
Another thing that came to mind with regard to photons and black hole is the mass in the accretion disk and perhaps the ergosphere. Gravitational fields that are in opposition will partially cancel. This is most evident in a cavity at the centre of gravity of a massive body. Depending upon the amount of mass in the ergosphere or accretion disk there will be a gravitational field that can in effect cancel some of the gravitation of the black hole at the event horizon. This could mean that the effective horizon may shrink and that light could in effect exit the horizon due to this cancelling effect. It may be a miniscule effect but any effect should show some decrease in the radius of the horizon.

This does not mean that the black hole will necessarily lose mass but that photons generated just inside the horizon may be able to escape. I am really not sure if this is right but I can't off the top of my head think of why it shouldn't be that way. If this were true then that puts a dent in the current view of information loss. If we had two black holes orbiting each other at fairly close range then this cancellation could reduce the horizon of each by a significant amount.
« Last Edit: 28/01/2015 01:11:01 by jeffreyH »

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #5 on: 28/01/2015 12:29:06 »
If we have a photon that is traveling on a path that takes it exactly towards the centre of gravity of a black hole how would the gravity affect its speed?
It slows down. Yes. Check out the coordinate speed of light.

When a photon is moving away from a source its wavelength is red shifted. Would we get a blue shift and an increase in energy? How much of this would be kinetic?
People say the ascending photon is redshifted, but actually, it doesn't change. Conservation of energy applies. In similar vein people say the descending photon is blueshifted, but again, it doesn't change, and conservation of energy applies. You can work this out by sending a 511keV photon into that black hole. Ask yourself this: by how much does the black hole mass increase? The answer is 511keV/c≤. Not a gazillion tons. Also ask yourself this: if you accelerate towards a 511keV photon, does it get blueshifted? The answer is that actually, the photon doesn't change a jot. Your measurement of its frequency and energy changes because you moved. But again, conservation of energy applies.

Quote from: jeffreyh
If this were true then that puts a dent in the current view of information loss.
You should read about Friedwardt Winterberg's firewall.  I think it destroys the current view of information loss. Meh, I'm an IT guy, and something of an amateur relativist. I am not impressed with cosmologists talking about information as if it was some physical thing.   

what is the ctually, it doesn't change. uis

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Offline PmbPhy

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Re: what would happen to a photon in this situation?
« Reply #6 on: 28/01/2015 16:04:50 »
Quote from: jeffreyH
If we have a photon that is traveling on a path that takes it exactly towards the centre of gravity of a black hole how would the gravity affect its speed?
The coordinate speed of the photon will change as it moves towards the event horozon.

Quote from: jeffreyH
When a photon is moving away from a source its wavelength is red shifted.
Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.

For details on potential and kinetic energy of a photon see
http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #7 on: 29/01/2015 02:20:18 »
Quote from: jeffreyH
If we have a photon that is traveling on a path that takes it exactly towards the centre of gravity of a black hole how would the gravity affect its speed?
The coordinate speed of the photon will change as it moves towards the event horozon.

Quote from: jeffreyH
When a photon is moving away from a source its wavelength is red shifted.
Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.

For details on potential and kinetic energy of a photon see
http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

Thanks Pete. I asked this because I really don't know. I haven't had time to look at your page but I will later and respond.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #8 on: 29/01/2015 03:33:17 »
I have just read the first part of your page and I noted the two time axes. This is actually a method I hadn't thought of. Do you believe that? I will read more later. The hour is late.

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #9 on: 29/01/2015 12:08:46 »
The coordinate speed of the photon will change as it moves towards the event horozon.
See what Einstein said:

"Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable".

Rather counter-intuitively, the descending photon slows down, and the ascending photon speed up. The "modern" interpretation of general relativity says the speed doesn't change, but this is not in line with what you might call the Einstein interpretation of general relativity.

Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.
Good man.

For details on potential and kinetic energy of a photon see http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm
I noticed the total energy of a photon moving through a gravitational field is constant. That's nice to see. I wish more people knew it. And that it's the same for a falling brick. 

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #10 on: 29/01/2015 23:33:16 »
Quote from: jeffreyH
If we have a photon that is traveling on a path that takes it exactly towards the centre of gravity of a black hole how would the gravity affect its speed?
The coordinate speed of the photon will change as it moves towards the event horozon.

Quote from: jeffreyH
When a photon is moving away from a source its wavelength is red shifted.
Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.

For details on potential and kinetic energy of a photon see
http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

Well your page has cleared up a lot of questions for me. Especially on the conservation of energy in a gravitational field. Would you mind if I used the equations?

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Offline dlorde

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Re: what would happen to a photon in this situation?
« Reply #11 on: 30/01/2015 09:42:25 »
.. This could mean that the effective horizon may shrink and that light could in effect exit the horizon due to this cancelling effect. It may be a miniscule effect but any effect should show some decrease in the radius of the horizon.
...
This does not mean that the black hole will necessarily lose mass but that photons generated just inside the horizon may be able to escape.
I can see why the horizon might shrink, but light would still not escape from it - but it would be able to escape from where the horizon would have been be if it hadn't shrunk. As I understand it, the horizon is defined as the point where light can't escape.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #12 on: 30/01/2015 13:27:13 »
.. This could mean that the effective horizon may shrink and that light could in effect exit the horizon due to this cancelling effect. It may be a miniscule effect but any effect should show some decrease in the radius of the horizon.
...
This does not mean that the black hole will necessarily lose mass but that photons generated just inside the horizon may be able to escape.
I can see why the horizon might shrink, but light would still not escape from it - but it would be able to escape from where the horizon would have been be if it hadn't shrunk. As I understand it, the horizon is defined as the point where light can't escape.

Yes that is the point. If we have two equivalent black holes that are in the process of merging the horizons of both may shrink radically, releasing tell tale gamma ray bursts. As galaxies merge it may be that gamma ray bursts are releases of trapped energy from temporarily cancelled fields inside the horizon. The cancellation allowing high energy photons to escape. How fast would this phenomena happen?

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Offline PmbPhy

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Re: what would happen to a photon in this situation?
« Reply #13 on: 30/01/2015 16:52:06 »
Jeff - Take a look at the chapters of the new version of Exploring Black Holes. They're online at
http://www.eftaylor.com/exploringblackholes/

Especially the chapter GlobalLightBeams141029v2.pdf

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Offline PmbPhy

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Re: what would happen to a photon in this situation?
« Reply #14 on: 30/01/2015 16:53:35 »
.. This could mean that the effective horizon may shrink and that light could in effect exit the horizon due to this cancelling effect. It may be a miniscule effect but any effect should show some decrease in the radius of the horizon.
...
This does not mean that the black hole will necessarily lose mass but that photons generated just inside the horizon may be able to escape.
I can see why the horizon might shrink, but light would still not escape from it - but it would be able to escape from where the horizon would have been be if it hadn't shrunk. As I understand it, the horizon is defined as the point where light can't escape.

Yes that is the point. If we have two equivalent black holes that are in the process of merging the horizons of both may shrink radically, releasing tell tale gamma ray bursts. As galaxies merge it may be that gamma ray bursts are releases of trapped energy from temporarily cancelled fields inside the horizon. The cancellation allowing high energy photons to escape. How fast would this phenomena happen?
There is a formula to determine what the value of the remaining black hole is. Obviously its a function of the two radii and is larger than either.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #15 on: 30/01/2015 17:15:24 »
.. This could mean that the effective horizon may shrink and that light could in effect exit the horizon due to this cancelling effect. It may be a miniscule effect but any effect should show some decrease in the radius of the horizon.
...
This does not mean that the black hole will necessarily lose mass but that photons generated just inside the horizon may be able to escape.
I can see why the horizon might shrink, but light would still not escape from it - but it would be able to escape from where the horizon would have been be if it hadn't shrunk. As I understand it, the horizon is defined as the point where light can't escape.

Yes that is the point. If we have two equivalent black holes that are in the process of merging the horizons of both may shrink radically, releasing tell tale gamma ray bursts. As galaxies merge it may be that gamma ray bursts are releases of trapped energy from temporarily cancelled fields inside the horizon. The cancellation allowing high energy photons to escape. How fast would this phenomena happen?
There is a formula to determine what the value of the remaining black hole is. Obviously its a function of the two radii and is larger than either.

Thanks for the information Pete I will be reading it while I wait for a new book on Lagrangians and Hamiltonians. I may be back asking question!

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Offline PmbPhy

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Re: what would happen to a photon in this situation?
« Reply #16 on: 30/01/2015 17:34:56 »
Quote from: jeffreyH
Thanks for the information Pete I will be reading it while I wait for a new book on Lagrangians and Hamiltonians.
What book is that?

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #17 on: 30/01/2015 18:26:50 »
Thanks for the information Pete I will be reading it while I wait for a new book on Lagrangians and Hamiltonians. I may be back asking question!
I recommend some caution. It's a renowned text, but there are... issues. I think the best way to show this is with why doesnít the light get out?
 
You're standing on a gedanken planet holding a laser pointer straight up. The light doesn't curve round, or slow down as it ascends, or fall down. It goes straight up. Now I wave my magic wand and make the planet denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive, and take it to the limit such that it's a black hole. At no point did the light ever curve round, or slow down as it ascends, or fall down. So why doesn't the light get out?
 
Note that some people might answer with the waterfall analogy. Itís wrong, because a gravitational field is a region of inhomogeneous space which we model as curved spacetime. See this Einstein quote and this paper. A gravitational field alters the motion of light and matter through space, but it doesnít suck space in. We do not live in a Chicken-Little world where the sky is falling in.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #18 on: 30/01/2015 19:28:57 »
Thanks for the information Pete I will be reading it while I wait for a new book on Lagrangians and Hamiltonians. I may be back asking question!
I recommend some caution. It's a renowned text, but there are... issues. I think the best way to show this is with why doesnít the light get out?
 
You're standing on a gedanken planet holding a laser pointer straight up. The light doesn't curve round, or slow down as it ascends, or fall down. It goes straight up. Now I wave my magic wand and make the planet denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. I make the planet even denser and more massive, and take it to the limit such that it's a black hole. At no point did the light ever curve round, or slow down as it ascends, or fall down. So why doesn't the light get out?
 
Note that some people might answer with the waterfall analogy. Itís wrong, because a gravitational field is a region of inhomogeneous space which we model as curved spacetime. See this Einstein quote and this paper. A gravitational field alters the motion of light and matter through space, but it doesnít suck space in. We do not live in a Chicken-Little world where the sky is falling in.

I never take anything at face value. I will read it without reference to anything else simply because I don't want to start with a slanted view. There are real difficulties with gravitation that no one has an answer to. The main difficulty is in experimental validation for some of the more extreme macroscopic or microscopic environments.

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #19 on: 31/01/2015 15:41:30 »
I don't think there's much of a problem with gravitation myself, or that nobody has answers. But I think there is a problem wherein what's taught just doesn't square with what Einstein said. Pete wrote about it in this paper. I share his sentiment about "Einstein's general relativity" and "modern general relativity", but I think the issues are greater than he appreciates. I recommend that you understand why doesn't the light get out? IMHO it really gets to the heart of things.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #20 on: 31/01/2015 16:09:07 »
Quote from: jeffreyH
Thanks for the information Pete I will be reading it while I wait for a new book on Lagrangians and Hamiltonians.
What book is that?

The book is "A Student's Guide to Lagrangians and Hamiltonians" by Patrick Hamill.

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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #21 on: 31/01/2015 16:13:27 »
I don't think there's much of a problem with gravitation myself, or that nobody has answers. But I think there is a problem wherein what's taught just doesn't square with what Einstein said. Pete wrote about it in this paper. I share his sentiment about "Einstein's general relativity" and "modern general relativity", but I think the issues are greater than he appreciates. I recommend that you understand why doesn't the light get out? IMHO it really gets to the heart of things.

Why do you make such evasive statements? Go on then why doesn't the light get out. No need to be all mysterious. If you believe I am assuming something erroneous or I am not aware of all the facts the honest thing to do is be direct. What are you getting at John?

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #22 on: 31/01/2015 16:26:13 »
See this? Look at the second paragraph:



That's Einstein talking about the speed of light varying. He says that this is why light curves. He says it's the cause of gravity. Now take a look at Professor Tom Moore's answer to the question:

"As the planet's mass approaches the black hole limit, the signal emitted from the surface will seem to move more and more slowly away from the surface (and will also be seen to be increasingly red-shifted as observed from infinity). When the surface of the planet coincides with the black hole's event horizon, the signal will stop moving outward from the surface (and the redshift observed at infinity will go to infinity). So light no longer escapes." 

The light doesn't get out because it's stopped. That fits with what Einstein said. However you will not find anything like this in E F Taylor's book.
« Last Edit: 31/01/2015 16:28:38 by JohnDuffield »

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Offline yor_on

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Re: what would happen to a photon in this situation?
« Reply #23 on: 31/01/2015 16:42:24 »
As I said before John, is doesn't matter for SR. SR is part of GR too, although you will see gravity and the equivalence principle joining. It also depends on whether you expect there to be one perfect description of a SpaceTime, which then should be GR, as far as I read you? My thoughts of it is that GR is where we are, but also that SR may be closer to some initial parameter(s). Take a look at 'the clock postulate' for some arguments around it. Using my way to define a clock, as equivalent to splitting 'c' in equal chunks, its a simple argument, and that is what I like. As simple arguments as possible :)
=

As for you describing the light being stopped as the way Einstein saw it. Nope, he saw it in terms of observer dependencies. In those terms the light being 'stopped' according to the 'far observer' is equivalent to the light 'propagating' for the local observer. He actually got questioned on that one once as I remember, don't have the reference here though, and then explained his thinking of it through just observer dependencies. You can exchange observer dependencies for different 'frames of reference' if you like.

The point is that the 'frame' we use for physics is local, always local. We don't use a clock at some event horizon, to define local time ticking. Neither do we use such for repeatable experiments. So defining it your way becomes a observer dependency, in where you, as you use your local clock and ruler to define your observation, exchange the one creating your definition (local clock and ruler) for a clock and ruler you defined elsewhere.

Locally light doesn't stop anywhere.

Hmm, that one can be questioned :) Can't it? Let's put it this way. Locally defined light is 'c'. And that has nothing to do with whether I conserve energy equivalent to some light quanta into a 'super cold' cloud of atom(s), to then be released as 'light' at some later moment. Light is at 'c' from the 'instant' it exist, until it annihilates at a detector, no acceleration involved.
=

What Einstein did, and a lot of other physicists and mathematicians still do, is treating 'time' as illusionary. And as your local clock here is what defines the clock at the event horizon, it doesn't make sense arguing that it 'stops', as the time you used to measure that clock in now would be described as 'illusionary' too. By setting your local clock to 'c' though, it becomes a reality. A local reality, equivalent to how we define physics, created through locally done 'repeatable experiments', tested to hold true over varying SpaceTime positions. 
« Last Edit: 31/01/2015 17:49:51 by yor_on »
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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #24 on: 31/01/2015 18:09:05 »
As for you describing the light being stopped as the way Einstein saw it. Nope, he saw it in terms of observer dependencies. In those terms the light being 'stopped' according to the 'far observer' is equivalent to the light 'propagating' for the local observer..
That's what people say, but it just doesn't square with Einstein saying a gravitational field is a place where the speed of light varies. If the light is stopped it's stopped. The local observer won't see it propagating. The light is stopped. He can't see. He's stopped too. If you put a stopped observer in front of a stopped clock, he doesn't see it ticking normally "in his frame". He doesn't see anything. 

The point is that the 'frame' we use for physics is local, always local. We don't use a clock at some event horizon, to define local time ticking. Neither do we use such for repeatable experiments. So defining it your way becomes a observer dependency, in where you, as you use your local clock and ruler to define your observation, exchange the one creating your definition (local clock and ruler) for a clock and ruler you defined elsewhere. Locally light doesn't stop anywhere.
If that was true, the vertical light beam would get out of the black hole. Then it wouldn't be a black hole. This is why the question above cuts to the heart of the matter. 

Hmm, that one can be questioned :) Can't it? Let's put it this way. Locally defined light is 'c'. And that has nothing to do with whether I conserve energy equivalent to some light quanta into a 'super cold' cloud of atom(s), to then be released as 'light' at some later moment. Light is at 'c' from the 'instant' it exist, until it annihilates at a detector, no acceleration involved.
And at the event horizon, the coordinate speed of light is zero. So c is zero.   

What Einstein did, and a lot of other physicists and mathematicians still do, is treating 'time' as illusionary. And as your local clock here is what defines the clock at the event horizon, it doesn't make sense arguing that it 'stops', as the time you used to measure that clock in now would be described as 'illusionary' too. By setting your local clock to 'c' though, it becomes a reality. A local reality, equivalent to how we define physics, created through locally done 'repeatable experiments', tested to hold true over varying SpaceTime positions.
Gravitational time dilation at the event horizon is infinite. Has that light clock at the event horizon ticked yet? No. And it never ever will.

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Offline yor_on

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Re: what would happen to a photon in this situation?
« Reply #25 on: 31/01/2015 18:31:08 »
I'll try to look up the reference later John. Been some time since I saw that one, although I'm sure I've mentioned it sometime, at TNS. As for " If the light is stopped it's stopped. The local observer won't see it propagating. The light is stopped." :) No. Light has a speed, and it is 'c'. Make a experiment, measure it. You could argue that it is observer dependent though, but that's not the same as proposing that it 'stops'. Locally you will get 'c', and if we want to be very specific about it we'll define through SR and Maxwell's equations, as it seems Einstein first did. As for coordinate speeds :)

You use proper time and proper distance to measure 'c', and you do it locally. That meaning that you use your local clock and ruler to measure it in. Should I read you as stating that this won't hold true outside a event horizon?
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Offline yor_on

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Re: what would happen to a photon in this situation?
« Reply #26 on: 31/01/2015 18:41:27 »
I can see all sort of complications if that was true, a universe of varying 'force fields' of time. Some of them quite deadly, dissolving reality as where no light propagates, we too should stop. Also you have to remember that with different speeds, we will define different clocks to that 'clock' hanging on the event horizon. And as there is no resolution to what would consist of a correct 'frame of reference', defining all other frames (in this commonly agreed on 'container universe') Every 'uniformly moving frame' can then alternatively be used as the 'correct' one, no matter its 'speed/velocity' relative that 'inertial' clock, set above the event horizon.
=

Had to make it a little more understandable here, 'uniform frame' may make sense to me as I write:) But maybe not later. And you can actually find a 'golden standard' for all frames, but only as defined locally. That's the turning stone for me, defining it. I'm using a local constant called 'c', defining it as equivalent to a clock, so also making your local 'time'. Then I just presume that this must hold true everywhere, as it is what repeatable experiments actually builds on. That when you locally do a experiment, letting others do it too at other places and time, all finding the outcome to be the same. We then call it repeatable, and use it to define physics. And they all did it locally, never worrying about what their 'frame of reference' relative their possible uniform speed, or their mass would do to it. It's not that easy though, but as a simplification of arguments it has to do :)

It's a corner stone of physics actually that physics should be the same throughout a universe. If it's wrong then physics are wrong too, including repeatable experiments.
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Remember a rather cool SF though, that treated the universe that way :) as the local 'clock' speeded up, as we (Earth) got out of a slower 'force field' inhibiting us. It made Orwells 'animal farm' look as idiots :) And us as instant Einsteins. ( The funniest thing with this book is that it somehow presumed 'c' to be equivalent to that speeding clock though. If it didn't we should have become morons, as 'c' actually would have slowed down, relative the 'speeding clock'. So, he was before me with the idea of a equivalence, or maybe just missed its implications:) Or, maybe not? That one was trickier than I thought, and can be seen different ways. Splitting the clock from 'c', it should leave the speed of information the same, although the 'time' 'c' took to propagate slower. Ah well..
« Last Edit: 31/01/2015 20:20:29 by yor_on »
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Offline yor_on

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Re: what would happen to a photon in this situation?
« Reply #27 on: 31/01/2015 19:49:40 »
I'll give you another argument John. Assume that different uniform speeds gives you 'real' time dilation(s) relative everything else. Then consider a 'black box scenario' in where you have detectors surrounding a light bulb, measuring red respective blue shifts. It's 'drifting' at a uniform motion in a geodesic. You then accelerate it to some other 'faster' uniform motion, to once again measure the lightbulbs blue respective red shift. You only measure when there's no 'forces', as 'gravity', acting upon you, which then excludes all moments of acceleration. Would you expect to measure a blue or red shift? Should it be different for different uniform motions?

Put that together with my proposal of uniform motion as giving you 'time dilation(s)'  relative all other 'clocks' in a universe.

If you don't expect different blue/red shifts inside that box, due to different uniform motions, but still expect time dilations due to uniform motions. Where does that leave your 'local clock'? Can it vary?
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Offline PmbPhy

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Re: what would happen to a photon in this situation?
« Reply #28 on: 31/01/2015 23:05:08 »
Quote from: jeffreyH
The book is "A Student's Guide to Lagrangians and Hamiltonians" by Patrick Hamill.
Great. Let me know what you think of it when you get it. I have the same book myself. I downloaded it off the internet, printed it out and bound it myself (it's a little hobby of mine). I save a lot of money on my physics textbooks like that. That particular one can be downloaded at

http://bookzz.org/book/2339781/d3b971

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Offline JohnDuffield

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Re: what would happen to a photon in this situation?
« Reply #29 on: 01/02/2015 12:50:33 »
Yor on: you're missing the obvious: there is no time flowing inside an optical clock. When that clock goes slower, it's because light goes slower. See http://arxiv.org/abs/0705.4507 re the tautology wherein we define our second and our metre using the local motion of light, then use them to measure the local motion of light.

Quote from: yor_on
I'll give you another argument John. Assume that different uniform speeds gives you 'real' time dilation(s) relative everything else. Then consider a 'black box scenario' in where you have detectors surrounding a light bulb, measuring red respective blue shifts. It's 'drifting' at a uniform motion in a geodesic. You then accelerate it to some other 'faster' uniform motion, to once again measure the lightbulbs blue respective red shift. You only measure when there's no 'forces', as 'gravity', acting upon you, which then excludes all moments of acceleration. Would you expect to measure a blue or red shift? Should it be different for different uniform motions...
I'm sorry, this is confusing SR time dilation and GR time dilation. The GR time dilation is very simple: it's there because the speed of light varies with gravitational potential. As does the speed of all electromagnetic phenomena. Light goes slower when its lower, and so do you, so you don't notice that it goes slower.

« Last Edit: 02/02/2015 22:34:29 by JohnDuffield »

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Re: what would happen to a photon in this situation?
« Reply #30 on: 01/02/2015 13:03:38 »
What I'm referring to is an absence of red and blue shifts, those you measure in time, it's one of the parameters. You then combine that with theoretical proofs of time dilations in a uniform motion, and you get a situation in where there is no proof of your local clock changing 'pace' inside that black box, combined with 'real' time dilation(s), as defined relative other clocks.

(Where do you see GR in this?

To me it's about SR, different uniform motions, and the question of whether there will be blue/redshifts inside that box, depending on those relative motions. I clearly stated that the moments of acceleration should be excluded from the measurements.)
« Last Edit: 01/02/2015 14:46:43 by yor_on »
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Re: what would happen to a photon in this situation?
« Reply #31 on: 01/02/2015 13:10:35 »
If you want to insist on believing in 'slow time(s)' existing, then that's your privilege. Myself I've used this example and another in where we find different motion presenting different observers with a different 'clock rate' of the clock above a event horizon.
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The real thing being questioned if you would accept my reasoning is the way we define a universe. What you're fighting for here is the idea of a 'container universe' as I see it. But that one doesn't hold to me.
« Last Edit: 01/02/2015 13:19:01 by yor_on »
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Re: what would happen to a photon in this situation?
« Reply #32 on: 01/02/2015 13:17:19 »
If you want to insist on believing in 'slow time(s)' existing, then that's your privilege.
Huh? Gravitational time dilation is not in doubt. See this interview where David Wineland of NIST says they can see one optical clock run slower than another when it's a mere 30cm lower.

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Re: what would happen to a photon in this situation?
« Reply #33 on: 01/02/2015 13:23:14 »
I'm not discussing NIST John. And I'm not doubting time dilations, neither from uniform motion, accelerations, or Gravity. I gave you some examples, as 'though experiments' to why I don't agree with your interpretation.
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You can look at my point of view this way. I set 'c' to a clock. Locally defined that clock doesn't change although a acceleration becomes a twisty argument to follow. I use 'locality', in the manner of how I read relativity, which is what repeatable experiments do too. From that follows that whatever idea(s) we have of a 'container universe' should be questionable. What joins us are 'c', gravity, and the 'dimensions' we find. To that you can add parameters as relative motion, accelerations and 'mass'. They all slide the 'container' we see into other configurations. As another argument, Would one really define the energy one spend locally, accelerating, to being in any way comparable to the energy needed to 'shrink' a universe in front of oneself?  In the end it seems to be about communication and information, to me that is.

And when it comes to communication I see no better description of it than 'c', it contains the clock and it contains the information. And yes, it presumes a ruler as a constant too, all locally defined.

If you look at it my way then scaling is what it is about, namely QM. That scaling holds true at every SpaceTime position you can think of, and it should present us the same information. Scaling is one way to define what a 'locality' may be about, the other is the ones in where we idealize. As with ones 'local clock' situated at some idealized 'position', defining ones 'time'.
« Last Edit: 01/02/2015 15:40:31 by yor_on »
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Offline jeffreyH

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Re: what would happen to a photon in this situation?
« Reply #34 on: 01/02/2015 17:04:13 »
Quote from: jeffreyH
The book is "A Student's Guide to Lagrangians and Hamiltonians" by Patrick Hamill.
Great. Let me know what you think of it when you get it. I have the same book myself. I downloaded it off the internet, printed it out and bound it myself (it's a little hobby of mine). I save a lot of money on my physics textbooks like that. That particular one can be downloaded at

http://bookzz.org/book/2339781/d3b971

The first few pages were very illuminating. I want to get on to the Calculus of Variations but have a bit of reading to do first.