# Does CMBR include free photons?

1. Dec 8, 2011

### DrChinese

So here's a simple question for which the answer is not so simple using current theory: Do the photons from the CMBR exist free and independent of observation? Because some people say that photons are mediators of EM force but are otherwise abstractions. For example, this quoted from Mentz114 in an older thread:

"Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter. There is no evidence ( nor any way of getting any ) that photons exist in the EM field when it is not interacting with matter.

1. If the answer is "NO, there are no free photons" then that means that CMBR "knew" (at the time of emission) we would be here 13 billion years later to witness those particular photons. Pretty amazing feat. And yet fully consistent with QED.

2. If the answer is "YES, many CMBR photons are free and will never be detected" then this implies: a) that those photons could be considered "dark" (in the sense they will never land anywhere); and b) that in an accelerating/expanding cosmology, more and more energy from stars will end up as similarly "dark" over time.

It seems to me that an experiment would be possible to discern these 2 scenarios as well. Imagine a heat source in space surrounded by detectors. Now remove one of the detectors such that in one direction, the source is exposed to darkest space. In scenario 2 (Yes there are free photons), you expect no difference to the rest of the apparatus. But in scenario 1 (No free photons are possible), you might expect that photon emission would be suppressed in a particular direction (where there is no matter to absorb them). You ought to be able to account for that difference using conventional means.

Any thoughts?

2. Dec 8, 2011

### fleem

What we call a "photon" is simply a local interaction that occurs in a reference frame moving at c (within which the emitter and absorber are local). Specifically, the interaction will only take place when phases are correct in both a reference frame moving (at c) from the emitter to the absorber, and the reference frame moving from the absorber to the emitter (thus we see a wave-like nature in the interactions). So you can't have an interaction with only an emitter and no absorber, just as you can't have an interaction with only an absorber and not an emitter. Another way to say this is: We know photons can be transferred with no change in entropy, that absorbers aren't in the habit of absorbing photons that were never emitted, and that time is reversible if there's no change in entropy--so you can't reverse the time for an emission-only event without having an absorber that absorbed a photon that was never emitted. Its a little like one hand clapping.

3. Dec 8, 2011

### DrChinese

Nice description, and I think this view seems to be accepted by most. So would you say that some of the CMBR could be considered free photons? I think you are clearly saying NO. Therefore there is a sense in which those photons knew in advance we would be here to observe them billions of years later.

So do you think it would be possible to probe where matter will be in the future using some variation of a light source aimed at distant space? The idea is that photon emission would be suppressed in directions where there will be nothing to absorb the photon at any time in the future. Which would probably be most directions in our shrinking observable region, assuming a cosmology with an accelerating expansion.

4. Dec 8, 2011

### Bill_K

I'm a proponent of (1), i.e. that photons most certainly exist, but free photons exist as an idealization, a limiting case.

I believe the correct explanation is a Many Worlds scenario. It's not that the CMBR knew in advance whether and when the photon it emitted would be eventually absorbed. In fact it's the other way around, in that the emission and absorption are entangled, and the state of the emitter is unresolved until the absorption has taken place. The intervening 13 billion years makes this seem paradoxical. Yet the absorber does not know the state of the emitter until the moment the photon is received.

5. Dec 8, 2011

### DrChinese

So would you agree that in an accelerating expanding universe, there must be many directions in space (looking out from where we are) in which there will never exist any (future) matter to (potentially) absorb a photon emitted today?

6. Dec 8, 2011

### skippy1729

May I assume that "detected" includes both photons which register on a counter in some physicists lab and photons which are (eventually) absorbed by a star, galaxy or interstellar plasma? If so, then the question "are there CMBR photons which will never be detected" has an answer which depends on the cosmological model used:

spherical (so-called closed) FLRW: absorbtion is complete.

hyperbolic (so -called open) FLRW: absorbtion is never complete.

REF: P.C.W. Davies J. Phys. A: Gen. Phys., Vol. 5, December 1972 p. 1722-1737
P. E. Roe Mon. Not. R. Astr. Soc. (1969) 144, 219-230

The question was also extensively covered in the 1945 Wheeler-Feynman paper "Interaction with the Absorber as the Mechanism of Radiation" Rev. Mod. Phys. V. 17 (1945) p. 157-181. They considered the case of an infinite flat homogeneous universe. They conclude complete absorbtion. But see: https://www.physicsforums.com/showthread.php?t=230090&highlight=wheeler [Broken]

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7. Dec 8, 2011

### DrChinese

Thanks for the Davies reference, it can be read here:
http://cosmos.asu.edu/publications/papers/IsTheUniverseTransparentOrOpaque%209.pdf [Broken]

He mentions on top of 1726: "Put this way the problem is rather like Olber’s paradox in reverse." He also mention radiative damping, not sure if this relates.

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8. Dec 8, 2011

### qsa

Last edited by a moderator: May 5, 2017
9. Dec 8, 2011

### PatrickPowers

My first thought is that the photons have mass, so they can be detected without having been absorbed. My second thought is that the year 300,000 last scattering is equally in every direction, so the mass would be evenly distributed and thus hard to detect. My third thought is that in some directions there is dust and gas and so forth, so the CMB from those directions has been absorbed and the mass of the CMB is not evenly distributed and so can in theory be detected. This is fair game, as we are talking theory. The unabsorbed photons are measurable. In addition the photons are subject to gravitational lensing, so there is a second reason they are not evenly distributed.

I would say that photons exist as a potential. Consider a shell expanding at the speed of light in every direction. If 1% of that shell encounters absorbers, then 1% of the potential will become a different sort of potential.

This potential has mass, so it affects our world. Evidence may be gathered that they exist.

10. Dec 9, 2011

### fleem

Yes, I believe the answer is a resounding 'no'.

Of course, all emitters must know in advance what absorber will be used, and vice versa, for a given interaction so that conservation of energy is obeyed (its really a local interaction in IRFs traveling at c). If one electron didn't "pick" which of two other nearby electrons would be the recipient of a virtual photon mitigating repulsion, then both of those other electrons could decide to absorb that virtual photon even though they are each outside the other's light cones--which would violate energy conservation. Of course, the emitter doesn't really pick the absorber, rather they are always entangled and energy is transferred when the oscillator phase is right within those IRFs moving at c where everything is local. For this reason I wave my hands and say "everything is entangled and that explains Mach's principle".

I think an observer in an inertial reference frame always sees an isotropic density of the universe. And the anisotropy of the CMBR only indicates the universe is not yet at maximum entropy. When an observer is standing on a planet, however, that observer will see a few more photons from his otherwise isotropic emitter hit the planet or objects behind the planet because of the gravitational lensing. We are in the habit of saying this is because the photons are following curved paths in Newtonian space. But if there's no such thing as a photon in transit, then we can just as easily say more photons are hitting the planet (and the stuff behind the planet) because it is denser or appears denser in that reference frame, and the emitter is no longer isotropic.

So I think the question, 'can we detect future mass by noting the anisotropy of an otherwise isotropic emitter?', is the same as asking 'does gravity travel faster than light?'. The answer is, of course, 'no'.

I've wondered if some virtual photons also transfer a slight amount of negative momentum, then this whole thing would be the fundamentals of quantum gravity.

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11. Dec 9, 2011

### PatrickPowers

It ought to be possible to calculate the prob that a CMB photon would be absorbed before the end of the Universe. The chance is quite small in any given year, but if the Universe is infinite then the probability could be one. It depends on the acceleration of the expansion. Since so little is known about the future of that no figure can be calculated now. If the expansion is exponential I'd guess the probability of absorption is less than one.

12. Dec 9, 2011

### Q-reeus

Haven't bothered to read all the links supplied by various posters, but notice no-one here has pointed to what is surely a fundamental consideration. Cosmic expansion from BB till now has generated enormous redshift. An emitted photon back then has now an energy reduced by many orders of magnitude. So why should the emitter care a hoot about any 'future missed' photons - expansion alone has ensured that the universe as as a whole 'absorbs' any 'stray' photons - at least from the pov of energy/momentum. Not that teleological arguments strike me as being particularly sane in the first place. Awaiting savage repost.

13. Dec 9, 2011

### DrChinese

According to current estimates, our observable universe is shrinking fairly rapidly. That is looking to the past. Looking to the future, the portion of the observable universe that can ever see Earth (as it is today!) is much smaller. If I was guessing, I would say less than 1% of the original matter in the universe will ever have the opportunity to see us. Which means that a light beam headed to deep space is not too likely to run into anything in many/most directions. On the other hand, light headed to Earth or a lab wall would.

So here is a question. If there are paths for which an emission and subsequent absorption are possible [X+Y], versus ones which there are far fewer [X] (because the Y paths are aimed to empty space where there is no absorber): I would have to believe that either:

a) the likelihood of an emission event is less in [X], ; or

b) the likelihood of emission is the same but the X paths are more likely to lead to an absorption (i.e. they will be somewhat brighter).

How are these cases not physically distinguishable? (I realize that there is probably a reason, I just can't figure it out.)

14. Dec 9, 2011

### skippy1729

Dear Dr. Chinese, Could you rephrase or clarify this post? I just can't seem to get it through my head.

Skippy

15. Dec 9, 2011

### DrChinese

This may be a little better.

I have a laser out in space, and aim it towards the sun. There is an absorber available for every photon emitted because the sun is not transparent.

I now aim the laser to darkest space. It is an area in which the light could never, due to the ongoing expansion of the universe, hope to encounter other matter which would absorb any of the photons. According the view that QED requires both an emitter and absorber to exist for a photon quantum to be exchanged, no light at all can be emitted from the laser (ideal case).

Therefore, according to conservation of energy, the laser itself will get hotter or will use less energy than when aimed at the sun. I should be able to detect either of these cases. Either that, or I conclude a) the universe is not expanding as described by current theory (i.e. there is an absorber because the universe will eventually contract); or b) QED does not accurately describe photons (i.e. "free" photons exist from the CMBR).

(OK, I know intuitively this must be wrong but I am asking why.)

16. Dec 9, 2011

### apeiron

More recent papers by Lineweaver (working with Davies and Davis) suggests a role for event horizons. They seem to suggest that at the heat death, all that remains of the CMB is the blackbody radiation of the cosmological event horizon of a de Sitter universe - photons with a wavelength of the visible universe (ie: ~36 billion light years, which is pretty cold!).

So there is some kind of concrete terminus in this it seems. CMB photons cannot keep redshifting away forever without some interruption. In a dark energy dominated universe, ultimate holographic limits arise.

The horizons are emitters of blackbody radiation of course. So this may not deal with the question of whether the CMB consists of free photons, or the horizons are also an absorber. But it would seem "free" photons would have to be free to cross over the horizon. So holographic considerations would seem to make the horizons also the ultimate absorber.

See what you think. Fig 6.2 and 6.3 make the basic case.
http://msowww.anu.edu.au/~charley/papers/LineweaverChap_6.pdf

A more far out approach perhaps is this from Jack Sarfatti who combines the holographic principle with a Wheeler-Feynman retrocausal approach. Here it would seem the future cosmological event horizon is actively involved in giving the CMB its reality. It is the future screen that allows the radiation of the big bang to be emitted.

See http://journalofcosmology.com/SarfattiConsciousness.pdf from p21. Yes, ignore the consciousness nonsense. But it seems you are sympathetic to retrocausal approaches to non-locality and future cosmological event horizons would seem to have to be the ultimate absorbers of the CMB, or at least some kind of definite limit on "free photons".

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17. Dec 10, 2011

### Q-reeus

Make it even easier and cleaner. Forget the sun. Just an absorbing patch; say a circular disk coated with carbon black. On a clear night, hold the disk at arms length, and aim the laser pointer alternately at the disk, or 'empty space'. A garden-variety multimeter could easily pick up any power fluctuations. Dramatically so if the retrocausal argument is true, and the accelerated expansion of universe is not a temporary thing, and my argument in #12 - of which apeiron's first two paragraphs in #16 is imo the logical extension, is incorrect. To make the experiment easier again, forget even the multimeter. Just place a sheet of clear plastic or glass ahead of the laser pointer. There will be small but easily visually observable back reflection of the laser beam - which should vary dramatically as per retrocausality argument. [Simpler again - no need for absorbing disk. Just alternately point laser at ground or sky. How easy is that!]

So what are you waiting for - go to it! Willing to bet big \$'s no power fluctuations will be forthcoming.

Last edited: Dec 10, 2011
18. Dec 10, 2011

### PatrickPowers

The chance that a deep-space photon is absorbed during a unit of time is greater than zero. The expansion of the universe doesn't matter: as the universe expands the photon grows larger, canceling out the reduction in density of absorbers. The chance of absorption remains constant. Over infinite time, the probability of absorption is one.

19. Dec 10, 2011

### Q-reeus

Of course - but how much greater than zero is the question.
Not saying that is necessarily wrong, but your particular assertions needs some justification. How can you say redshift exactly compensates for expanded volume of universe when absorption will be a sensitive function of frequency re interstellar dust clouds for instance. I would off-hand think the trend to be exacerbation of inbalance, not cancellation.
On general grounds agree that a photon could not survive for infinite time. Base that on the need for some kind of either infinitely recycling 'mono-verse', or multiverse that on the largest time scales is probably best thought of as a self-contained 'gas' of universes in quasi-thermal equilibrium - individual universes having limted lifespans. That heads towards cosmological speculation - but then the OP's scenario calls for such imo.

Last edited: Dec 10, 2011
20. Dec 10, 2011

### PatrickPowers

The point is that when you have infinity on your side then ANY definite number greater than zero will do. If the chance of absorption in a billion years is greater than one in a jillion, that's enough. One in a jillion squared or cubed is enough. One in a jillion to the jillionth power a jillion times over is enough. You can use a number hugely less than the actual minimum probability. It doesn't matter.

The only way the probability can be less than one is the probability over time decreases quickly enough that the series converges to less than one. But I don't see any reason the probability should decrease with time since the photon expands along with the universe.

Last edited: Dec 10, 2011