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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
 

Offline yor_on

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How do photons lose energy?
« 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 »
 

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.

 

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.
 

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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How do photons lose energy?
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