How do photons lose energy?

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Douglas Sykora

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How do photons lose energy?
« on: 06/02/2009 02:30:02 »
Douglas Sykora asked the Naked Scientists:
   
Consider a single photon that was emitted 13.3 billion years ago from the last scattering surface.  

It is a visible light photon.  As it travels across the universe, it arrives at earth today as a much lower energy photon because of the expansion of the universe the photon's wavelength has increased by about a factor of 1000.  

The single photon is still a single photon.  It has not divided into multiple photons, but it is much lower in energy (by a factor of about 1000) due to the fact that it has a much longer wavelength.  

Where did the energy go?

Does conservation of energy apply to this photon?  If not, why not?

Does it have something to do with the reference frame we are in here on Earth?

What is the relationship of our reference frame to the reference frame 13.3 billion years ago when this photon was emitted?  I have been struggling with this for well over a year and I have not found a good answer.  Please help.

What do you think?

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Offline Vern

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How do photons lose energy?
« Reply #1 on: 06/02/2009 03:05:23 »
The conservation of energy does apply. The photon could have lost energy any number of ways; one; the speed of recession of the galaxy from which it came may have Doppler shifted it; or two; it may have encountered charged particles in its journey and gave up energy to them. It is a speculative notion, but any time a photon's path is bent and it encounters a charged particle, it must give up energy to that charged particle. This must cause it to shift further toward the red.

Don't take the speculative notion to school; it is purely speculative; it is my own pet hypothesis.
« Last Edit: 06/02/2009 03:07:22 by Vern »

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Offline yor_on

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How do photons lose energy?
« Reply #2 on: 06/02/2009 07:24:23 »
First of all Douglas. If your photon had a specific lightquanta I would expect it to have it still.
The expansion is about space, not particles.
Your photon is called 'timelsess' for a reason.
It doesn't care about 'our' count of the years, the only time 'passed' for that photon will be at its impact.

That's one of the strangest things with this Universe, that time doesn't 'count the same' for different objects.

Even thought you can notice a beam of light, you are not really noticing the photons traveling.
You are noticing their interaction with 'matter/atoms' as they gets absorbed into electron shells and 're emitted' as 'new' photons.
If I'm correct that should mean that in the absence of matter you will never be able to observe any light at all.
There are some ideas trying to treat a photon when meeting the electron shell as being 'reflected' in some mysterious way, I don't think so though.
But that doesn't really matter for your question.

Your light never tires, although we might:)
And 'conservation of energy' is for the universe as a whole.
It's not needed for explaining 'timeless' photons.

And this I would call a 'main stream' image:)
« Last Edit: 06/02/2009 07:29:06 by yor_on »
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Offline LeeE

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How do photons lose energy?
« Reply #3 on: 06/02/2009 12:58:01 »
That's a very good and well thought out question Douglas.

As you mentioned 13.3 billion years ago and the last scattering surface you're talking about the CMBR here.  You also ask if it may be something to do with the reference frame  we are in, here on Earth, and this is the answer.  In short, the effect of the expansion of the universe has been to introduce a relative velocity between the frames, as they are now further apart than they used to be, and if you take this into consideration everything adds up again.  If you were to re-align one of the frames (in practice, the observers frame is the only one that can be changed) by removing the difference between them i.e. reducing their relative velocities to zero,  you'd then see the energy that seemed to have been lost.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Vern

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How do photons lose energy?
« Reply #4 on: 06/02/2009 13:29:37 »
I agree LeeE; a Doppler shifted photon has not lost energy; I was thinking of things that might actually reduce the energy of a photon.

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lyner

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How do photons lose energy?
« Reply #5 on: 06/02/2009 15:31:25 »
I agree LeeE; a Doppler shifted photon has not lost energy; I was thinking of things that might actually reduce the energy of a photon.
It has lost energy in as far as it can't transfer as much energy to what it hits as when it left home.

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Offline Vern

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« Reply #6 on: 06/02/2009 17:10:39 »
Ok; I see now that you're right. I was typing without thinking [:)]

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Offline LeeE

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« Reply #7 on: 06/02/2009 20:16:35 »
I agree LeeE; a Doppler shifted photon has not lost energy; I was thinking of things that might actually reduce the energy of a photon.
It has lost energy in as far as it can't transfer as much energy to what it hits as when it left home.

... which is because what it hits will be in a different referential frame to the one that it started it's journey from.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Vern

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« Reply #8 on: 07/02/2009 00:54:26 »
Note to self: Self; count slowly to ten before starting to respond to a post [:)]

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Offline chris

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How do photons lose energy?
« Reply #9 on: 07/02/2009 23:09:13 »
I don't understand this answer relating to referential frames. The photon has less energy now than it one did, if I measure it. So where is that missing energy?

Chris
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Offline yor_on

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How do photons lose energy?
« Reply #10 on: 08/02/2009 00:02:20 »
Frames is definitely confusing :)

But 'red shift', if photons is of a specific, not changing, light quanta, must be a 'relative' effect.
If we have two objects moving away from each other (frame_A frame_B), and light is 'time less' internally.
 
Then red shift, which consists of a longer wavelength, or if one like, fewer photons per distance counted, or time slice, will be a relation.
That 'relation' is then expressed as red shift.
The faster that frame_A moves away from the light chasing it, the more red shifted that light will appear from frame_A.

When it 'catch up' to frame_A there will be fewer photons per time slice hitting that object.
But the photons will still be of the same energy content.
As photons are 'time less' internally they have no reason to 'loose' any energy as time isn't there internally.

I'm not sure how to describe it in a wave way though.
Phase velocity?
And what should I see as the group velocity here?

Another strange thing about it is that if we turn the table around and have two frames (A and B) moving towards each other.
And then measure the light coming towards you at frame_A from frame_B, then that light still will be moving at 'only C' in space.
The only thing we will notice is a so called 'blue shift'.

So it doesn't matter if we move towards or moving from that light.
It will not change 'speed' when observed, the only effect we will notice is a red / blue shift relative us observing.

(So yes Lightarrow, I definitely see your point made before:)

But I would really like to see someone describe it as waves, with as little math as possible.
As I don't know how to describe this that way.
« Last Edit: 08/02/2009 00:06:41 by yor_on »
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Offline Vern

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How do photons lose energy?
« Reply #11 on: 08/02/2009 00:21:00 »
There is no problem describing the red shift as waves. The wave length is longer. The distance from trough to trough is greater. If the causing factor is a Doppler shift, the greater distance from trough to trough is because the generator of the wave is moving away.

There is speculation that a photon of light moving through space for billions of years might lose energy by some other process. One such might be Halton Arp's Tired Light scheme.
« Last Edit: 08/02/2009 01:20:14 by Vern »

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Offline yor_on

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« Reply #12 on: 08/02/2009 00:48:24 »
Vern, if you look at it as a wave instead of particles then where are the (light quanta's) 'energy' placed?

You have two descriptions, phase and group velocity, describing a wave.
As far as I understand that include the whole wave approach?
Well that and the frequency:)

Phase velocity is the one giving a FTL 'look a like' if I get it right.
And group velocity is the one carrying the 'information' here.
That one (group velocity) won't ever travel faster than 'C' in space.

It's easier to understand the concept of 'energy' when looking at photons as particles to me.
But when I look at it as waves I find the 'energy's focus' harder to define.
So I would really like to see the concept of red and blue shift explained.

Is that unclear?
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Offline LeeE

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How do photons lose energy?
« Reply #13 on: 08/02/2009 01:21:50 »
I don't understand this answer relating to referential frames. The photon has less energy now than it one did, if I measure it. So where is that missing energy?

Chris

It isn't missing, you just haven't accounted for the fact that you are moving away from the source of the photon; the energy that appears to be missing is accounted for in your velocity away from the photon's source.  If you were approaching the photon's source instead of receding from it, you'd think it had gained energy rather than losing it.

Yor_on; you write some interesting stuff but I find it very difficult to read because you seem to be inserting a newline/return at the end of each sentence.  When I try to read this my eyes keep on having to jump back and forth across the screen, which tires them, and I keep losing my place in your text, making it difficult to keep track of what you're saying.

No offence intended, and you can format your posts however you like, but bare in mind that if you want people to read what you write you should try to make it easy for them.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Vern

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How do photons lose energy?
« Reply #14 on: 08/02/2009 01:30:52 »
Quote from: yor_on
Vern, if you look at it as a wave instead of particles then where are the (light quanta's) 'energy' placed?


A photon is both a wave and a particle; we all know that; but we all do not agree upon just how to visualize that. I like to visualize things in physics. Most physicists like to make principles and laws and theories and not visualize them. So you hear things like; nothing can go faster than light because of the theory of relativity. That is completely wrong. Things can't go faster than light because of some fundamental property of the things. The theory of relativity just describes the process.

So when I visualize a photon; I can visualize it as both a particle and a wave at the same time. All I need do is consider it as being two electromagnetically saturated points surrounded by electric and magnetic fields that extend outward from the points forever into space.

The fields give me the wave; the points give me the particle. There are many other benifits
that derive from the vision. For example, I can know the fundamental cause of relativity phenomena. I can answer the question: What is mass? And many others.

 
« Last Edit: 08/02/2009 01:35:21 by Vern »

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Offline yor_on

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« Reply #15 on: 08/02/2009 02:09:04 »
Ok Vern, point taken.
I know that you define it both ways.

But I would like to see a 'main stream' view too.
Is there anyone who would like to describe red and blue shift.

using words :)
As waves using frame_A frame_B?

----

LeeE, I'm just used to write that way, it may be a bad habit but:)
I find it easier to see how I thought when I 'chop' it up.
Maybe you're right though?
« Last Edit: 08/02/2009 02:15:07 by yor_on »
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Offline Vern

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« Reply #16 on: 08/02/2009 02:22:20 »
Quote from: yor_on
But I would like to see a 'main stream' view too.
Is there anyone who would like to describe red and blue shift.
I don't think I was deviating from the mainstream view when I described the red and blue shift. It if is Doppler, it is because the light originated in something that is moving relative to us. If it is moving away, it will be red shifted because a light wave takes time to form. As the wave develops, the originator moves away, so the peaks and troughs must spread further apart.

That seems like a simple concept; are you having trouble with that?

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Offline yor_on

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« Reply #17 on: 08/02/2009 02:40:00 »
Vern I can describe it that way too, but I still can't visualize how the 'energy' is thought to be delivered.

And speaking about trouble:)
When I find myself confused I'm afraid I get like a bull terrier, I get stuck until I'm through::))

This is what really made me start to wonder though.

"If we imagine the wave profile as a solid rigid entity sliding to the right, then obviously the phase velocity is the ordinary speed with which the actual physical parts are moving.

However, we could also imagine the quantity "A" as the position along a transverse space axis, and a sequence of tiny massive particles along the x axis, each oscillating vertically in accord with A0 cos(kx - wt). In this case the wave pattern propagates to the right with phase velocity vp, just as before, and yet no material particle has any lateral motion at all.

This illustrates that the phase of a traveling wave form may or may not correspond to a particular physical entity. It's entirely possible for a wave to "precess" through a sequence of material entities, none of which is moving in the direction of the wave. In a sense this is similar to the phenomenon of aliasing in signal processing.

What we perceive as a coherent wave may in fact be simply a sequence of causally disjoint processes (like the individual spring-mass systems) that happen to be aligned spatially and temporally, either by chance or design, so that their combined behavior exhibits a wavelike pattern, even though there is no actual propagation of energy or information along the sequence."

http://www.mathpages.com/HOME/kmath210/kmath210.htm
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Offline Vern

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« Reply #18 on: 08/02/2009 03:44:39 »
Okay; this is much more clear and I agree with you. There need not be the physical movement of any material thing for the wave to progress. And as at least one of our local peers suggests there might need not be anything happening between the originating event and the observing event when a photon goes from place to place. But I need to think more on this to even get a visual image of it in my mind.

Any thing we can contrive to explain a present observation must also be compatible with all the past observations that have ever been made. So I think of the Fourier transforms and how barriers placed in the path of light cause distortions and wonder whether that means that there must be some electromagnetic disturbance that is physically present in the space between the source of light and the detector.

So; I am learning; I hope. But I do find some very interesting ideas abounding here.

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Offline LeeE

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« Reply #19 on: 08/02/2009 12:30:43 »
Yor_on, I don't see it as being right or wrong;  You have a perfect right to say what you want in the way that you want to say it, but I do think it's true that if you want people to read what you say you should try to make it easy for them.  I do want to read what you say, but sometimes find it harder work than it need be.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Vern

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« Reply #20 on: 08/02/2009 15:16:40 »
Okay; it is a new day and maybe thought will happen a little better now. On the matter of whether a red shifted photon has lost energy; if it is the result of Doppler shifting, it has not. The photon is the same as it always was.

If the red shift is the result of Halton Arp's Tired Light scheme, then the photon has lost energy. In this case the thing from which the photon came need not be moving away. This is the scenario I like. I like it mainly because we need not dispose of the physical laws of nature if that is the case. If we like alternative views, we immediately introduce an avalanche of problems with the physical laws which seem to contradict the views.
« Last Edit: 08/02/2009 15:23:37 by Vern »

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Offline yor_on

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« Reply #21 on: 09/02/2009 17:02:13 »
LeeE, I will test:)

But I'm not sure it will be easier to read, the type of English I really like to write is the one with all those 'strange' words, eloquent sort of. I'm quite in love with the richness of the English language:) which doesn't necessarily mean that it gets more understandable.

And then there is English English, (New Hebrides English, Falklands English, Hong Kong English, Irish English, Scottish English), Australian English, New Zealand English, Indian English, American English.

I guess that list can go on for quite a bit, I've probably missed quite a lot of countries. So English have a lot of different 'traditions' when it comes to how to express and use it:)

That's what makes it into one of the most important languages today. Once you had Latin (Europa), but now I would say it's English.
« Last Edit: 09/02/2009 17:08:06 by yor_on »
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Offline yor_on

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« Reply #22 on: 09/02/2009 17:22:03 »
Yes Vern, if that idea would be right the photon would become easier to understand. But you know that I'm a 'follower' of Einsteins model. And there he describes photons as being 'time less' internally.

There are arguments against 'tired light' that makes sense to me, one need to see that some build on other main stream assumptions though, but they all seem to fit together. I have this link http://www.astro.ucla.edu/~wright/tiredlit.htm that discuss some of the arguments against.
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Offline LeeE

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« Reply #23 on: 10/02/2009 17:21:18 »
Yor_on:  I found that really easy to read and understand:)  The sentences that related to each other were grouped together and so the relation, and therefore meaning, between them was clear to me.

Certainly use exotic or strange words - they're fun - and carry on trying to be eloquent too, as eloquent writing is a pleasure in itself to read.
...And its claws are as big as cups, and for some reason it's got a tremendous fear of stamps! And Mrs Doyle was telling me it's got magnets on its tail, so if you're made out of metal it can attach itself to you! And instead of a mouth it's got four arses!

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Offline Vern

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« Reply #24 on: 10/02/2009 18:01:34 »
Quote from: yor_on
There are arguments against 'tired light' that makes sense to me, one need to see that some build on other main stream assumptions though, but they all seem to fit together. I have this link http://www.astro.ucla.edu/~wright/tiredlit.htm that discuss some of the arguments against.
I notice that Halton Arp seems to have abandoned the Tired Light model. His latest articles offer a new explanation. I'm not sure I can sign on to his latest scheme which proposes that the matter in newly created galaxies is less energetic.

I agree there are some very convincing arguments for an expanding universe.

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Offline djsykora

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How do photons lose energy?
« Reply #25 on: 14/02/2009 02:50:33 »
I received an email from the Naked Scientists today informing me that they put my question on this forum on my behalf and I am glad that they did.  So I joined the forum and I read through all of the responses in this thread about five times because it took me several times to either figure out what each comment meant or to formulate what it is that I donít understand and of course to come up with additional questions. 

I would like to thank all those that responded to this question and to all those that may respond to my additional questions.  I really do appreciate it. 

Here is a summary of some clarifications, things I understand and what still puzzles me.

1)    Conservation of energy applies to this photon.  Energy is neither created nor destroyed.
2)   The perceived loss of energy is due to the fact we are moving in a reference frame relative to the reference frame the photon was emitted in.  If I was in a rocket ship traveling toward the photon as it reached earth and my speed was very high I would perceive this photon blue shifted and at a higher energy than if I was standing on earth.  So energy is not the same in two reference frames.  This must be similar to length of an object or the passage of time not being the same in two reference frames. 
3)   Does the original reference frame still exist?  If so, where is it?  How far is it from us? How fast is it moving away from us?  Is it accelerating away from us?  What has the relative velocity between these two reference frames been as a function of time?  t = 0 to t = 13.3 billion years.  Can this be calculated using the Hubble constant and the distance of 13.3 billion light years?  If so how is this calculation made?  When I tried this I came up with an answer that we are moving at nearly the speed of light away from that original reference frame.  Thus I concluded my calculation is faulty.
4)   I am wondering if there could be additional factors contributing to the loss of the photonís energy.  Factors that are not related to the reference frames.  Is it possible that the photon is losing energy because it is moving out of a high density cosmos and into a less dense cosmos?  Could there be a gravity well that this photon is climbing out of.   Was their larger gravity forces acting on this photon when the universe was young compared to the present lower density universe?  Letís assume this photon is not absorbed and reemitted via interactions with matter.  Letís assume the only interaction this photon has is with the expanding universe and any possible gravity well it might be climbing out of. 
5)   This photon does not experience the passage of time.  As far as it is concerned it has just been emitted from the last scattering surface, the CMBR (cosmic microwave background radiation), even though it took 13.4 billion years to get here. 
6)   The photon must have zero energy in its own reference frame because its wavelength must be infinitely long.  But if this is true does the photon even exist in its own reference frame.  I would think zero energy means nonexistence. 
7)   I didnít really follow the phase velocity and group velocity descriptions. 

Cheers

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Offline yor_on

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« Reply #26 on: 14/02/2009 16:35:25 »
I received an email from the Naked Scientists today informing me that they put my question on this forum on my behalf and I am glad that they did.  So I joined the forum and I read through all of the responses in this thread about five times because it took me several times to either figure out what each comment meant or to formulate what it is that I donít understand and of course to come up with additional questions. 

I would like to thank all those that responded to this question and to all those that may respond to my additional questions.  I really do appreciate it. 

Here is a summary of some clarifications, things I understand and what still puzzles me.

1)    Conservation of energy applies to this photon.  Energy is neither created nor destroyed.
2)   The perceived loss of energy is due to the fact we are moving in a reference frame relative to the reference frame the photon was emitted in.  If I was in a rocket ship traveling toward the photon as it reached earth and my speed was very high I would perceive this photon blue shifted and at a higher energy than if I was standing on earth.  So energy is not the same in two reference frames.  This must be similar to length of an object or the passage of time not being the same in two reference frames. 
3)   Does the original reference frame still exist?  If so, where is it?  How far is it from us? How fast is it moving away from us?  Is it accelerating away from us?  What has the relative velocity between these two reference frames been as a function of time?  t = 0 to t = 13.3 billion years.  Can this be calculated using the Hubble constant and the distance of 13.3 billion light years?  If so how is this calculation made?  When I tried this I came up with an answer that we are moving at nearly the speed of light away from that original reference frame.  Thus I concluded my calculation is faulty.
4)   I am wondering if there could be additional factors contributing to the loss of the photonís energy.  Factors that are not related to the reference frames.  Is it possible that the photon is losing energy because it is moving out of a high density cosmos and into a less dense cosmos?  Could there be a gravity well that this photon is climbing out of.   Was their larger gravity forces acting on this photon when the universe was young compared to the present lower density universe?  Letís assume this photon is not absorbed and reemitted via interactions with matter.  Letís assume the only interaction this photon has is with the expanding universe and any possible gravity well it might be climbing out of. 
5)   This photon does not experience the passage of time.  As far as it is concerned it has just been emitted from the last scattering surface, the CMBR (cosmic microwave background radiation), even though it took 13.4 billion years to get here. 
6)   The photon must have zero energy in its own reference frame because its wavelength must be infinitely long.  But if this is true does the photon even exist in its own reference frame.  I would think zero energy means nonexistence. 
7)   I didnít really follow the phase velocity and group velocity descriptions. 

Cheers


1. The photon is a timeless object internally, so Conservation of energy doesn't apply to it I think.
And it is timeless as it is massless, that's also what allows it an 'instant acceleration'.
---
Although (thinking again:) If you are referring to its interactions with matter that principle (Conservation of energy) seems to hold true.
-----


2. " And the energy of a single photon is related only to its frequency.  The frequency is unchanged during all the transitions.

Energy = (Plancks constant) * (frequency)

It is also possible to re-write the energy (or frequency) in terms of the velocity and wavelength of the light. The velocity changes during the transitions, going from fast to slow and back to fast as the photons leave the substance.  However, the wavelength of the light also changes by the same factor during the transitions going from long to short and
back to long.

E = (Planck's constant) * (Velocity) / (wavelength)

   = (Planck's Constant) * (Velocity/Index of Refraction) /
(Wavelength/Index of Refraction)

The Index of Refraction cancels itself out and you are left with the same before and after any transition.  All in all, the energy remains constant.

From Michael S. Pierce
Materials Science Division
Argonne National Laboratory "

When you discuss reference frames you are correct in that, when observing the results of your observation, it will be a description of relations between your frame of observation compared to whatever 'frame of reference' you are observing. Although I'm finding 'frames' a very difficult concept to define:)

You are wrong in saying that the "energy is not the same in two reference frames". the photons energy will always be the same, no matter if you observe a photon coming from frame_A towards your frame_B, or if you stand at frame_A instead, observing that same photon leaving your frame_A. The 'red/blueshift' perceived is a relation between your 'frames' and also a 'symmetry' in that it won't matter what frame (_A or _B) you define as, for example, receding or approaching in your observation. Any choice is as valid, and therefore the red/blue shift will be the same, no matter what frame you observe this relation from.

3. That 'original frame of reference' is long gone, that we can observe 'old' light at all is a direct consequence of that light not aging, if there was 'tired light' or light had a mass, then one could expect light to age and even 'die out' before reaching us.
If you want to calculate the size of our universe it will depend on what premises you choose.
Read http://www.thenakedscientists.com/forum/index.php?topic=20354.msg227700#msg227700

4. There is no 'photon energy loss'. The photon either exist or it doesn't. It will be its interactions with matter that defines it, nothing else as far as I have seen.

5. Yep, I agree, with the exception that if it had a 'consciousness' the first and last thing it ever should notice is its own annihilation as it interact with matter.

6.First, it depends on how you want to define 'photons'? As 'Waves' or 'particles'? Or both? Or neither? 
If you see it as a object traveling (somehow) in our macroscopic 'spacetime' interacting it sure has a energy quanta, and it's not zero. The nearest 'thing' having energy but somehow 'hiding/neutralizing' it seems to be 'vacuum' aka space, not photons as I see it.

7. I agree, it's a devious subject to grasp :) just when I think I understand it slips through my fingers again.
http://en.wikipedia.org/wiki/Group_velocity and look at the gifs, there you'll get a visual description.

Those are my views naturally, but I tried to stay inside 'main stream' definitions here.
Cheers.
« Last Edit: 14/02/2009 20:58:19 by yor_on »
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Offline big-bang-skeptic

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How do photons lose energy?
« Reply #27 on: 10/07/2010 04:44:04 »
Sorry for joining this thread after it has been lying dormant for so long, however, I simply typed "do photons lose energy in space" into Google and this is where I ended up. On this topic, I prefer Vern's viewpoint expressed early in the thread, being that light can lose energy through contact with charged particles and the longer the distance that light has to travel, the more charged particles that it will contact and thus lose energy along the way. In losing energy, the photon's wavelength will become shifted more towards the lower energy, red end of the spectrum and the further the light has to travel, the greater the wavelength shift.

Those that expressed the standard scientific viewpoint that it is the expanding universe and the doppler effect that causes a wavelength shift are following a self-fullfilling prophecy. After all, it is the increasing wavelength shift with distance that is used to support the expanding universe theory!!!

If we start again with a clean slate and an open mind, the alternative possibility is worth pursueing that the universe is not expanding as fast as estimated and that at least some of the wavelength shift of light with distance is attributable to charged particles in space. The density of charged particles may be homogenous or not, but I suspect that homogenous is a likely possibility.

I would be interested in any serious rebuttal to this theory since my mind is open to finding the truth.


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Offline sciconoclast

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How do photons lose energy?
« Reply #28 on: 15/07/2010 23:38:17 »
Interesting Topic.

     What happens during gravitational Doppler shift when light entering a gravitational filed is blue shifted and light leaving a gravitational field is red shifted.  This effect is of course confirmed by astronomical observations as well as the earthbound experiments such as the Pound and Rebka experiment.   Is there a transfer of energy between the gravitational field and the photon or is this also the result of changes in reference frames?
       
                         Thanks in advance for the answer.

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Offline yor_on

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How do photons lose energy?
« Reply #29 on: 23/07/2010 10:34:24 »
I have to admit it.

I love photons :)
Slippery they are, difficult to understand but so loveable. What they are is intimate connected to our concept of 'motion and distance'. We see them as moving objects, we also see unchanging distances. At least we thought we did, but if the theory of relativity is correct distance is a property of the frame you measure it of. And it's not as simple as to say that 'Yeah, when I at last get that Porsche the distance will shrink! :)

Or looked at another way, that statement is all true..

Distance is a relation to something we call 'speed'. Like with that Porsche, but it's more than that. As I understands it your frame of reference will define the distance. If you travel near lightspeed, you not only travel faster. The truth, as I understands it, is that from your moving frame of reference all distances do shrink, for real, no joke. And that is a very hard statement to digest. But if it is true then it seems to me that what we call motion also becomes questionable. And if both what we experience as distance and motion are questionable ideas, what then about that photon that 'moves' faster than anything else we know, transferring energy. What does it see, will the universe and its distances exist for such an object, or is it no movement at all from that frame of reference?

Assume that a Photon represents an absolute negation of distances, as seen from its own frame of reference, how many photons are then needed to 'fill up'  a universe? I think we have the idea of motion and distance subtly wrong. Also remember that without distance it becomes meaningless to discuss speed, as speed is what an object take moving from A to B as measured by a clock of some kind (time).

Without a distance there is no time, and so time seems to me only to come into place when distances appear as they always do from our perspective, and the perspective of everything made out from invariant mass (matter). And that is due to that matter never can reach the speed of light. So to me it makes sense wondering as that scientist/mathematician did above..

"If we imagine the wave profile as a solid rigid entity sliding to the right, then obviously the phase velocity is the ordinary speed with which the actual physical parts are moving.

However, we could also imagine the quantity "A" as the position along a transverse space axis, and a sequence of tiny massive particles along the x axis, each oscillating vertically in accord with A0 cos(kx - wt). In this case the wave pattern propagates to the right with phase velocity vp, just as before, and yet no material particle has any lateral motion at all.

This illustrates that the phase of a traveling wave form may or may not correspond to a particular physical entity. It's entirely possible for a wave to "precess" through a sequence of material entities, none of which is moving in the direction of the wave. In a sense this is similar to the phenomenon of aliasing in signal processing.

What we perceive as a coherent wave may in fact be simply a sequence of causally disjoint processes (like the individual spring-mass systems) that happen to be aligned spatially and temporally, either by chance or design, so that their combined behavior exhibits a wavelike pattern, even though there is no actual propagation of energy or information along the sequence."

But even that take 'time' as seen from our frames of reference observing it. I'm not even sure there are individual moving photons at all :)There definitely are so from our frames ,observing, and thinking otherwise becomes very strange but I still wonder. Time needs not only the clock it seems, it also needs distance to make sense of it, and what makes distances and time seems to me to be invariant mass.

:)

Yep.
« Last Edit: 23/07/2010 10:37:06 by yor_on »
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