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Author Topic: what would happen to a photon in this situation?  (Read 10253 times)

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?
 

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.
=

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 »
 

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?
 

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
 

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 »
 

Offline yor_on

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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 »
 

Offline 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.
=

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 »
 

Offline JohnDuffield

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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.
 

Offline yor_on

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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 »
 

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.
 

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Re: what would happen to a photon in this situation?
« Reply #34 on: 01/02/2015 17:04:13 »

 

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