<Bows respectfully to superguru>

Some time ago there was some discussion that the universe might be toroidal rather than spherical, not that one can ever define such a shape from the outside! Does this impact on the thinking above, or is the toroidal universe concept now passé?

Hi PWS, no 'supergurus' around anywhere near here!

AFAIK, the WMAP data sets do not support a toroidal cosmos, which is a relief to me, because I have no idea how to model a cosmos based upon it.

-J

Hi Jorrie,

I couldn't get the link from [1] to work, can you check it.

I think it's worth mentioning that the energy of a photon is E=hc/λ meaning that the photon actually has less energy when it arrives then when it started.

Also, I'd like to hear your take on De Broglie waves and gravitational redshift. I've looked around and can't find anything discussing it but it must occur.

Roger

Hi Roger, tx for pointing out the broken link - fixed it.

Yes, the photons lose both linear momentum and energy as they travel. The 'lost' energy goes into expansion, I guess.

I think De Broglie waves may fit better under the next Blog topic that I'm planning - on particle energies in an expanding cosmos - but yes, there may be a link between the fact that particles lose momentum during cosmic expansion and the redshift of photons. A sort of 'particle redshift' in terms of De Broglie wavelengths. Interesting!

-J

A very 'enlightening' thread! I also bow.

"Yes, the photons lose both linear momentum and energy as they travel. The 'lost' energy goes into expansion, I guess."

Maybe you've discovered the Dark Energy. As the light (photons) get red-shifted they become 'dark'. Do we call this radiation pressure?

-S

Maybe. What bothers me about this is that no one appears to care. So either the explanation is so obvious (where the photon energy goes) that they (physics community) don't bother to explain it, or the magnitude of energy is so small that it is deemed unimportant (I have a hard time believing that since background radiation fills the entire universe), or no one cares.

It seems like a lot of energy to just ignore.

Hi Roger, as you noted, I tried to 'illuminate' the issue on my previous reply (to S). As far as the balloon analogy is concerned, there is no problem. In hyperspace, the photons have been 'lifted' to a higher energy plane - remember they lost kinetic energy in order to keep angular momentum constant, but they gained potential energy. That's why I originally blundered in saying "the 'lost energy' went into the expansion". Point is, they were 'lifted higher' by the balloon, not going there on their own accord.

It is somewhat equivalent to a particle orbiting a massive object in an elliptical orbit, continuously trading kinetic energy for potential energy (and visa-versa), while keeping the total mechanical energy constant. We know that photos cannot really orbit like that - even near black holes there is only one orbit, and that is circular and unstable. In the balloon analogy, we just force the photon's hand, so to speak. And so does the universe, apparently.

-J

Hi Jorrie,

In reply to what you wrote above, to what potential energy are you referring? From the standpoint of within the universe, the photons are losing energy but there is no potential energy gain as far as I can seeunless you are referring to gravitational? Can you clarify?

Also, I was wondering, have there been solutions done for an inhomogeneous universe? I ask not because I believe the universe to be inhomogeneous, but rather because I think it could elucidate local spatial effects. Local space is certainly inhomogeneous.

Roger

Hi Roger. The potential energy is most definitely gravitational here. Weird as it may seem, photons that are spread out more thinly (less radiation density) possess more potential energy than more dense counterparts. This is because radiation gravitates and it takes work to 'move the photons apart', so to speak. A static 'radiation only' cosmos would have collapsed under its own gravity and hence convert potential energy into shorter wavelength photons again. Perhaps it would disappear into a singularity, perhaps not...

There are various approximations for inhomogeneous universes. Local inhomogeneities are difficult to bring into the model, mostly done through perturbation theory. AFAIK, it is not possible to seamlessly match local solutions to Einstein's filed equations (e.g. the Schwarzschild solution) to the global solution, like Friedman's. I'm not too well clued up on inhomogeneous solutions though.

-J

Jorrie,

Thanks and Thanks. You're gravitational potential energy explanation seems reasonable. The energy of a photon is proportional to 1/λ (wavelength is just a distance) and the gravitational potential energy between photons is proportional to 1/r.

It feels like cosmic expansion is akin in many ways to a timelike gravitational well, which now that I think about it, is exactly what it is. A space like gravitational well means that as you move away from the well (out of the gravitational potential), the photon is redshifted because it's gravitational potential has increased.

Think about cosmic inflation in a similar way, except instead of moving through space, your moving through time. As time increases you are further and further outside of the gravitational well in the sense that your potential energy has increased, you have redshifted, hell, even the curvature of space has decreased (or is it increasing, I can't seem to get this straight).

How do you feel about such an interpretation Jorrie? It really seems right to me, I hope I'm explaining it well.

Roger, yes, " ...expansion is akin in many ways to a timelike gravitational well ..."

I'm not too sure what you mean by "timelike" and "spacelike" gravitational wells, but in a sense, the cosmos is expanding out of a gravitational well. There is one proviso though: the center of that well sits in a hyperspace dimension, never in a spatial dimension. The gravitational potential of the homogeneous cosmos is the same everywhere in space. It is only when 'lumpiness' is introduced that you have some local gravitational wells, but these are not part of the LCDM model. It may also be valid to view the gravitational well as sitting in a time dimension, but I am a bit wary of that.

The curvature issue seems to still cause some trouble! Remember that the cosmic balloon does not simulate cosmic curvature at all. The balloon's geometric curvature changes in the wrong sense, but it does not enter into any of the equations used, so it is irrelevant. It is perfectly valid to simulate a flat cosmos (with no curvature at all) by means of a balloon of arbitrary radius - all the calculations work perfectly.

-J

You wrote:"I'm not too sure what you mean by "timelike" and "spacelike" gravitational wells, but in a sense, the cosmos is expanding out of a gravitational well."

I guess what I was trying to say is not that the cosmos is expanding out of a gravitational well, but rather that it is a gravitational well and as we move forward in time we essentially are moving out of the well. Take a look at the diagram below:

I'm sorry the diagram is so messy. Basically the y-axis is the energy density (in the universe) and the x-axis is time. As we go forward in time the density decreases (as matter and radiation get spread out). Compare this with the gravitational well of a planet and it is very similar, except instead of Time on the x-axis, you would have space (radius r). I hope that makes what I'm trying to say clearer.

For the second part you wrote:"The curvature issue seems to still cause some trouble! Remember that the cosmic balloon does not simulate cosmic curvature at all."

I'm sorry, I wasn't thinking of the cosmic balloon when I was asking, I just couldn't remember which it did (I'm still unclear). In my mind, if big bang was a singularity then it seems like curvature should have decreased since then, but I have the nagging feeling you said that it actually has been increasing, which is why I'm confused.

To further clarify my point. Imagine we are a photon in space some distance from a black hole. We would expect as we increase our r (distance) from the black hole, that potential energy with respect to that black hole would increase, at the same time we would expect that our wavelength would be redshifted (because we are climbing out of it's gravity well)

I'm saying that this cosmological system is behaving the same way, except that instead of moving through space, the photon is moving through time. (I mean the photon is moving through space as well, but the redshift from cosmic expansion depends only on time). It's as though the Photon feels the gravity from the original singularity through time.

The cosmological constant thus would be a sort of measure against escape velocity of this singularity from back in time. Higher than 0 and the universe expands forever (light redshifts forever as it climbs higher and higher out of the gravity well). Lower than 1 and the universe eventually crashes back together in a big crunch. At 0 it doing neither, but such a position would be unstable.

Hi again Roger, you wrote: "We would expect as we increase our r (distance) from the black hole, that potential energy with respect to that black hole would increase, at the same time we would expect that our wavelength would be redshifted (because we are climbing out of it's gravity well)"

Qualitatively, yes. Quantitatively, no. The redshift around a black hole scales with √[1-2GM / rc²], while the cosmological redshift scales straight with 1/a, which is equivalent to 1/R for the balloon. So, be careful in not pushing this "analogy" to far - you may get seriously tangled!

"It's as though the Photon feels the gravity from the original singularity through time."

I suppose you can say that, but it's simpler to just consider that all the matter/radiation/vacuum energy of the universe set up a present time gravitational field and that photons and matter particles react to that.

-J

BTW, I've finished the particle 'redshift' (momentum decay) thread and has scheduled it to launch in a few hours from now. I've found the de Broglie wave interpretation very interesting (and have included it), but more difficult to work with than straight relativity.

Jorrie,

I really appreciating your indulgences on this subject. I want you to know that when you respond I think carefully about what you said, sometimes long after I've responded to you. I feel as though you really are helping me get an understanding of this subject. I'm ordering this book to read up some more.

As for you're caveat about redshift near a black hole, I agree. Lets say the analogy is more accurate the further you get from the black hole (where relativistic effects are less pronounced).

One side question that occurred to me when I was thinking about my timelike singularity (I should probably come up with a different name for this approximation, I'm open to suggestions).

Is there a rotating homogeneous universe cosmic model that you are familiar with? I'm just thinking that it would be an interesting twist (sorry about the pun).

Roger

Hi Roger, your diagram is just the wrong way round! Density starts out high and decreases as time goes on.

What you describe is akin to the so-called "kinematics model" of the universe. Jon and myself did quite a bit of work on that in this thread. It is useful to some level, but has a few serious problems. (i) It cannot handle the v > c recession velocities of the real cosmos. (ii) It does not agree with the accepted fact that time runs at the same rate everywhere in the homogeneous cosmos.

You must not equate the physical curvature of the balloon with cosmological curvature - they have nothing to do with each other. Cosmic curvature is an energy balance issue - if kinetic energy of expansion equals potential energy, the curvature is zero. If slightly off zero, it diverges farther from zero as time goes on. I have defined this reasonably well in that long thread on the design of the cosmic balloon (in a reply to S).

-J

Hi Jorrie,

You Wrote:"Hi Roger, your diagram is just the wrong way round! Density starts out high and decreases as time goes on."

You're right of course, sometimes I draw what I'm thinking rather than what I'm saying. What I put in that diagram was how energy density effected space, but I labelled it as energy level. Please disregard that diagram and give me a do over. I really fouled up what I was trying to say with that diagram, let me try again.

Here is a 3D representation of a black hole.

Let's call the position of the Singularity zero. As you move to the right from the singularity you climb out of it (lets not worry for the moment about if that's possible, or start above the event horizon if you can't take the impossibility of what I'm saying).

What I'm trying to say is when you're moving to the right, rather than moving through space, imagine you're moving through time. Each movement to the right is a movement further in time.

I'm saying this because cosmic redshift of radiation behaves exactly like the redshift experienced by a photon climbing out of a gravity well would. This is not the kinematic model you discussed before. This is a hyperspace model. It's just that it's a singularity in time rather than in space.

You Wrote:"You must not equate the physical curvature of the balloon with cosmological curvature"

I never think of the balloon ever when I think about this stuff. Ever.

You Wrote:"Cosmic curvature is an energy balance issue - if kinetic energy of expansion equals potential energy, the curvature is zero."

Lets just drop this. One of the difficulties of communicating by posts is that it's harder for me to assure you that I understand what you're saying. When I say I forgot which way the curvature went, I meant it literally, I just forgot. I remember now, curvature increases as we move forward in time (not smoothly I know).

Hi Roger, OK, I understand what you are trying here.

BTW: "I never think of the balloon ever when I think about this stuff. Ever." I'm afraid you do not realize what you miss. It's the only no-nonsense way (apart from serious mathematics) that I know of to make sense out of a difficult subject. I still have to find a general case that I cannot explain rather simply using the balloon!

Your "timelike singularity" probably just means that there was a spacetime singularity at some past time. I agree that you should find a better term for your idea. In any case, it does not look much different from the kinematics model, possibly suffering from the same issues that I listed previously.

Your diagram resembles Schwarzschild spacetime geometry. Yes, photons do redshift as they climb out of the gravitational well, but this is gravitational redshift, quite different (technically) from cosmological redshift. Inhomogeneities tend to look like Schwarzschild, but Friedman does not work with inhomogeneities. Good luck if you want to pursue this avenue!

"Is there a rotating homogeneous universe cosmic model that you are familiar with? I'm just thinking that it would be an interesting twist (sorry about the pun)."

No, I have never heard of such a model that works and observations seem to rule it out (we do not see any 'twist').

-J

Hi S, you wrote: "Maybe you've discovered the Dark Energy. As the light (photons) get red-shifted they become 'dark'. Do we call this radiation pressure?"

I think my "The 'lost' energy goes into expansion, I guess" was maybe a bit misleading.

Photons do not create pressure that can expand the universe - any 'radiation pressure' would have been randomly distributed in all directions and cancel out. Photons simply act as gravitating energy that tries to collapse the universe.

For the first 50 thousand years or so, radiation energy was the dominant energy doing the 'braking' on expansion. Then the radiation energy density dropped lower than the particle energy density and normal matter (light and dark) took over as the 'braking force' on the expansion.

Dark energy only took over some 7 billion years later, both as a 'brake' and as an 'accelerator', with the accelerator winning, it seems. By that time, radiation energy density only made up some .02% of the total energy density, so it made a negligible impression - even less today.

So, where did the 'lost energy' go? OK, maybe still into expansion, but it was never used to cause expansion. Inflation caused the initial expansion (we think) and then the expansion coasted, 'thinning out' all the other energy forms (reducing densities), while keeping the overall total cosmic energy constant.

-J

Energy is conserved in a single reference frame but is not invariant between reference frames. The 'lost' energy is an artifact of changing reference frames and is not, in fact, 'lost' at all.

I might also mention that the General Theory of Relativity does not describe a global COE law. The Cosmos, as a single entity, does not obey COE laws. In reference to cosmologically red-shifted photons, Peebles¹ writes:

"Where does the lost energy go? ... The resolution of this apparent paradox is that while energy conservation is a good local concept ... and can be defined more generally in the special case of an isolated system in asymptotically flat space, there is not a general global energy conservation law in general relativity theory."

_{1. Principles of Physical Cosmology, 1995, p. 139}

Hi e, I wrote:

"Yes, the photons lose both linear momentum and energy as they travel. The 'lost' energy goes into expansion, I guess".

I based this on the believe that the cosmos is at critical density, Ω=1 (as it may well be, from observation); hence its total energy content (including expansion energy) remains constant - exactly zero in Newton's terms or E=mc² in Einstein's terms. I do however realize that this is a quite slippery concept on the global scale and I may have it wrong.

-J

Jorrie, concerning your link in footnote 1: I like it. Now if only we could convince marcus in the physics forum that space expands. Then maybe we could all get on the same page. Like that's gonna happen. This is another example of the illogicalness that's pervading cosmology these days.

You may convince Marcus, but I cannot hold that Space expands because for me Space is the void into which the Universe may be expanding.

Further, all discussion about redshift still appears to be predicated on the Universe expanding away from we observers in a more or less uniform way, which would be acceptable if our place was somewhere near (!) the centre of the Universe, as we were in the old days at the centre of the solar system.

But how do we know that we are not right out on the edge of the Universe, relatively speaking.? As we peer further and further, do we see the event horizon at a uniform distance away from us or is it an illusion that this is the outer remanants of the Big Bang expansion?

Hi Duikerbok.

Cosmologists refer to the "void into which the Universe may be expanding" as hyperspace.

We are precisely in the center of our observable universe. We have no idea how much more universe there is, so it does not really help to speculate on our position relative to the whole. If the cosmos happens to be closed (hyper-spherical), it has no center, just like Earth's surface has no center. It it happens to be infinite in size, it also does not have a center.

You asked: "But how do we know that we are not right out on the edge of the Universe, relatively speaking?"

Right in the edge, no. This much we do know, since we see an equal amount of space in all directions. What we see at the limits of observability are remnants of the BB, yes, but we see it as it was some 13.6 billion years ago, only 400 thousand years after the BB.

Jorrie. Thanks for putting me straight. I've gone back to the original blog which I hadn't taken in, after thinking about the void that was left at the heart of the original universe.

I had always taken it as given that the BB created a sphericalish universe which was occupied right out to the expanding surface, as would be expected by the analogy with an explosion back here. Perhaps I'm too much of a Hoylian to have considered a void, but at least we might account for some missing matter.

What is the void called if the outer void is Hyperspace?

Hi duikerbok.

"What is the void called if the outer void is Hyperspace?"

By void you mean the space we inhabit? Well, just space, or the vacuum...

If you mean outside of hyperspace, well, more hyperspace. Maybe in other dimensions...

-J

Hello Jorrie.

I think I must be missing something here. As I read the blog we inhabit the universe that is the expanded shell (equivalent to the balloon fabric).

Inside this shell is more void (or hypo-space).(?) Outside the shell is hyper-space.(?)

Hi Duikerbok, sorry, been away for a a while...

Inside and outside the 'shell' there is hypothetical hyper-space, just a fictitious dimension so that we can visualize things. We cannot observe those spaces, but we can detect that our normal three dimensional space is not 'flat' in the presence of matter. So, to me it is just convenient to ignore one of our normal 3-space dimensions and use that as a fictitious 4th dimension, where space can curve in.

-J

A hyperspacial 'exterior' is a useful artifice (and certainly a lot more comfortable to visualize than non-existence, which I've had no luck visualizing at all), but our dear reader must not assume it has a physical reality. It's a crutch, albeit a useful one.

"... Space is the void into which the Universe may be expanding."

What, then, distinguishes your 'Space' from the space which is already a component of the Universe in which we live? If everyday space and the 'Space' - your 'Space' - into which the Universe is expanding are fundamentally the same kind of space, then any distinction between the two is necessarily an artificial one. Either yours is a different kind of space entirely, or your statement makes no sense whatsoever. Do you see the problem?

Concerning the "center" of the Universe. We are at the exact center of the Universe; we and everything else in the Universe. All of it. Why might this be so? (Hint: into what did the Big Bang explode?)

Wikipedia uses H for the Hubble parameter instead of h. Which one is used more? So the paper should have used H₀ instead of h and should have called it the Hubble constant instead of the Hubble parameter? That's what I am getting. It's all quite frustrating.

I got the following from this source. What do you think of it?

1. Redshift with Distance

It is a remarkable coincidence that there are actually two ways that you get a redshift with distance (and strangely only the big bang explanation was considered which is the most complicated explanation, the most simple explanation was never considered).

Fig.1 - In Big Bang cosmology, the universe is all that exists, thus to prevent it gravitationally collapsing an expanding universe was proposed. The discovery of the redshift with distance seemed to confirm this, where the redshift is assumed to be a Doppler effect of receding motion due to an expanding universe.
The problems with the Big Bang theory are obvious!

What is a 'Big Bang' and how does it create Space, Time, Matter and Motion?

What is outside the expanding universe?

Fig.2 - In WSM Cosmology, the observable universe is just a finite spherical region of infinite eternal space. We can only see and interact with other matter in this region. Thus there is no need for an expanding universe, as other matter around our observable universe prevents it from collapsing. This is the equivalent of Einstein's Cosmological / Antigravity constant, but it is just normal gravity of matter outside our observable universe within infinite space.
So why the redshift with distance?
Because as we look at matter farther away from us, we find that we share less overlap of a common finite spherical observable universe. And this means that there is less energy exchange, which equates to a redshift with distance.

See The Cosmological Redshift Explained by the Intersection of Hubble Spheres
This article shows that each wave center 'particle' is the center of its finite spherical observable universe (Hubble Sphere) within infinite Space. As two wave center 'particles' move apart there is less overlap of common Hubble spheres / observable universes, thus less wave interactions with increasing distance, thus less energy exchange which then provides a simple sensible explanation of the redshift with distance.

-S

Hi S,

"Wikipedia uses H for the Hubble parameter instead of h. Which one is used more?"

I don't know which Wiki-article you are referring to, but if that's what it says, it's simply wrong. H is called the 'time varying Hubble constant', also sometimes explicitly denoted by H(t). I do realize that it is sometimes erroneously called the 'Hubble parameter', but the cosmological convention is that a parameter is a dimensionless entity, which H is not. I use the Peebles 1993 conventions. It is also clear that the paper that you referenced last time uses h in the standard way.

I'll respond to the rest of your reply later...

-J

Hi Jorrie,

I think I could have found worse quotes than the ones you posted. I couldn't make sense of a sphere around a distant particle (like it could observe us). And it said their explanation was the simple one! Ha, yours is far more simple. It has to be if I can understand it! One for the trash can, I think.

-S

Hi Jorrie, ok as we discussed on the tethered galaxy blog, here are some thoughts about the cosmological redshift. Some of your observations on this subject are similar to mine, but I think there's one final conceptual bridge to cross in order to appreciate how simple and elegant the cosmological redshift is.

1. First, I wholeheartedly agree that the 'expansion of the spatial hypersphere' cannot itself cause travelling EM waves to stretch. (To simplify the description of the basic principle, at this stage I will ignore the acceleration of expansion caused by vacuum energy). The primary lesson the tethered galaxy exercise teaches is that the cosmic expansion itself does not act as a 'force' and is incapable of causing the proper distance between any two things to separate from each other unless they began with a proper velocity away from each other in the initial conditions (like galaxies receding in the Hubble flow began). The hypersphere acts like a flywheel's momentum (1st derivative of distance) rather than like an accelerator (2nd derivative of distance). It should be obvious (I can explain further) that successive wave crests or successive photons do not have an initial velocity away from each other upon emission. Although the cosmic gravity field arguably accelerates the each already-emitted wave crest or photon toward the observer, the same gravity field decelerates the recession velocity (HD) of the emitter by at least the same amount by the time the next wave or photon is emitted. So absent any initial mutual proper velocity, a physical stretching apart of wave crests in this scenario would flatly violate the tethered galaxy principle.

This conclusion allows us to avoid treading on thin ice dealing with tricky quantum mechanical issues such as, (1) the fact that a photon has no defined size to be stretched (as you mentioned), (2) the question of whether a single photon is emitted 'all at once' or 'wave-by-wave' over a short period of time (the latter seemingly being required for a proper velocity differential to occur), or (3) whether a wave crest is physically capable of possessing and retaining expansionary momentum even if succeeding crests were emitted with an initial proper velocity differential relative to earlier crests.

2. As we've discussed, the momentum of all particles (as measured locally in their then-current comoving Hubble frame) decays at the rate of 1/a in an expanding universe, whether they are massive particles or photons. As Peacock says (p.87): ... [I]t is momentum that gets redshifted; particle de Broglie wavelengths thus scale with the expansion, a result that is independent of whether their rest mass is non-zero." This loss of local momentum is itself the cause of the cosmological redshift, and not merely an effect of it.

3. A photon's loss of momentum does not occur once at the time of reception. Rather, an incremental portion of it occurs at each infinitesimal local Hubble frame crossing along its worldline. This reflects the requirement that the photon's peculiar velocity must remain exactly at c in each local frame.

Each successive local Hubble velocity (HD) that a photon passes through is incrementally faster relative to the emitter. If the photon were observed in each succeeding local frame, its proper velocity away from the emitter must increase in order to keep the same peculiar velocity of c. In exchange for this 'velocity boost' (as Tamara Davis refers to it) energy conservation requires the photon's momentum to be incrementally lower in each such succeeding frame.

As Peacock says (p.87): "Perhaps more illuminating, however, is to realize that, although the redshift cannot be thought of as a global Doppler shift, it is correct to think of the effect as an accumulation of the infinitesimal Doppler shifts caused by photons passing between fundamental observers separated by a small distance:

δz / (1 + z) = H_zδl_z / c (Eq. 3.67)

4. For some reason I don't understand, Peacock fails to connect the very last dot in this picture to provide a complete description of the cosmological redshift. The critical point is that the 'Doppler shifts' he refers to are not relativistic, they are classical Doppler shifts. The reason for this is that in FRW coordinates all fundamental observers share a common cosmological proper time. The cosmological redshift they observe from other galaxies at rest in the distant Hubble flow cannot contain any element of time dilation from the emitter, because their clocks run at the same rate, and their atoms emit light at the same wavelengths as the emitter.

The formula for classical Doppler shift with a stationary observer and moving source is (1 + v/c). Using this, I came up with a simple integration formula for the cosmological redshift:

λ_o / λ_e = _e∫^o 1 + H^_t δt / c

This equation simply accumulates multiplicatively the changes in local Hubble velocity (HD) relative to the observer as experienced by the photon over its worldline. This integration of accumulated infinitesimal redshifts yields a finite z, quite different than what a 'global' calculation of velocity differential yields. I believe this formula is equivalent to Peacock's 3.67. Note that δl = δt for light travel in units of c, but it's more straightforward to calculate the latter.

I roughly integrated this formula to z=1023 and got an answer within 10% of the correct answer. I also tried a version that included infinitesimal SR time dilation, and the error increased to 40%.

Why didn't Peacock acknowledge that his reference to "accumulated Doppler shifts" meant classical Doppler shifts? I don't know, but there seems to be a general reluctance to accept that classical Doppler shift can ever be applied to light. One need not look at it that way; instead one could conceptually just calculate the infinitesimal SR Doppler redshifts, and then divide each one by the SR time dilation to reflect the undilated time relationship between FRW fundamental observers.

5. I have great respect for Davis and Lineweaver, but it seems clear to me that they are wrong to claim that infinitesimal SR Doppler shifts can be extended to large distances by integration, as claimed in Appendix A-1 of the Davis thesis paper, which I assume is derived from a more limited claim by Peacock at his p. 87. An examination of their math reveals that what they are calculating is really representative of redshifts resulting from the peculiar velocity of the emitter, not those caused by a cosmological recession velocity. Of course, a peculiar velocity of the emitter (or observer) does cause an SR Doppler shift even in FRW coordinates, but that redshift should be multiplied by the cosmological redshift to obtain the total redshift (as Davis and Lineweaver demonstrate elsewhere in their papers).

After all, if infinitesimal SR Doppler shifts could actually be integrated to yield the correct cosmological redshift, D&L and many other authors would have demonstrated that fact long ago using numerical integration examples. Yet no one has, and they never will, because that calculation does not yield the correct result except at small, nonrelativistic Hubble velocities (which happens only because the erroneous SR time dilation component is vanishingly small at those distances).

6. I will write more later about the linear expansion of a train of cosmologically redshifted photons (e.g. a light pulse), but first let's discuss what's on the table.

Thanks, Jon

Another way to treat relativistic Doppler in cosmology, is by considering it like in Peacock's "A diatribe on expanding space":

"If we think of the observer as lying at the centre of a sphere of radius r, with the emitting galaxy on the edge, then the sense of the gravitational shift will be a blueshift: the radial acceleration at radius r is a = GM(< r)/r² = 4πGρr/3, so the potential is thus ..."

His eq. (16) is an (almost) exact treatment of relativistic Doppler and redshift together, but he then wrote: "To second order, it is exactly correct to think of the cosmological redshift as a combination of Doppler and gravitational redshifts (see Bondi 1947 and problem 3.4 of 'Cosmological Physics')."

-J

There is a fundamental problem with Peacock's formula, in that the equation 'blows up' as the Hubble radius is approached, as with all solutions that embed a 'global' Doppler equation. It's difficult to believe that he has actually used it for calculation of real redshifts at any supra-relativistic Hubble velocities. How can it be the correct equation if it can't actually be used to make calculations? In any event, I don't recall this equation has been widely embraced or tested by other authors. (This is a not a peer reviewed paper.) I have a copy of the seminal 1947 Bondi article he cites (describing the L-T-B model), but I can't figure out his math well enough to make practical use of it. It does refer to a velocity shift and an Einstein shift combining to produce the redshift.

The cosmic gravity field most definitely plays a role in the cosmological redshift, but I'm struggling at the moment to put it into my own words. I thought I had it cancelled out, but now it is squirming around on me. In any event, it can't act as a modifier to the classical Doppler shift formula, because the latter seems to be correct on its own without any further modification factors.

Also, I seriously doubt that an integration done with infinitesimal SR Doppler shifts alone will yield a correct answer regardless of the number of integration steps.

Jon

I'm not sure about the form of the Peacock equation that you reference, but I think an escape velocity expansion in a relativistic "interior solution" (not inside black hole) can handle super-luminal recession speeds just fine. In the frame of the center of an infinite, frictionless gas cloud, the coordinate speed of light at infinity should be infinite, but I'll have to look at what the form of the equation is...

You said: "Also, I seriously doubt that an integration done with infinitesimal SR Doppler shifts alone will yield a correct answer regardless of the number of integration steps."

There are two ways to look at the Newtonian Doppler redshift. You can have a moving emitter or a moving receiver (or both) and the results are different. If we choose the inertial reference frame of the source as reference, then the Newtonian Doppler transformation at the receiver end is given by a factor 1/(1-H_t dt). When I use this in a simulation, it gives a larger error than the relativistic Doppler. Essentially, both Newtonian ways are wrong, but they converge at an infinite number of steps.

-J

Hi Jorrie,

Both classical Doppler equations are 'wrong' if you are in an inertial frame. But you're not, this is FRW coordinates and there is gravity involved. It is perfectly correct for an FRW fundamental observer to consider themselves to be 'stationary' while all other galaxies are receding. So the only 'correct' answer in FRW coordinates is to use the incremental (1 + v/c) equation for stationary observer, moving emitter.

The relativistic Doppler shift is clearly wrong because there is NO time dilation between the emitter and observer in FRW coordinates. Look at the time part of the RW line equation: it's entirely linear. How can you explain that if there were time dilation involved?

Jon

Hi Jon, you wrote: "So the only 'correct' answer in FRW coordinates is to use the incremental (1 + v/c) equation for stationary observer, moving emitter."

You may be correct that this gives the "smallest wrong", but I still believe that the whole Doppler shift interpretation of cosmic redshift is fundamentally flawed. Doppler shift only works when inertial frames can be defined - so if you can't, don't use Doppler...

-J

Hi Jorrie,

I don't follow your point. It is well established in GR theory that there are a multitude of infinitesimal inertial frames all along the worldline of a photon in FRW coordinates.

When a Doppler shift of light occurs within a local infinitesimal inertial frame, it is SR Doppler shift. But when it occurs across two adjacent infinitesimal inertial frames, the FRW metric specifies that there is no time dilation, so it is classical Doppler shift.

This is straightforward.

Jon

Hi Jon.

"It is well established in GR theory that there are a multitude of infinitesimal inertial frames all along the worldline of a photon in FRW coordinates."

I agree.

"... But when it occurs across two adjacent infinitesimal inertial frames, the FRW metric specifies that there is no time dilation, so it is classical Doppler shift."

I cannot agree. Two "adjacent infinitesimal inertial frames" in FRW are in relative motion, so the Doppler shift must be relativistic.

I think this touches upon a deep truth: just like in the FRW metric, there is no absolute meaning to pure SR time dilation. When two inertial frame are in relative motion, it is not possible to tell which one's "clocks tick slower" - they are completely equivalent, so their clocks must be ticking at the same rate. It is just an observational issue.

As I view it, the FRW metric does not have gravitational time dilation, but it does sport velocity time dilation, i.e.:

dτ²/dt² = 1- v²/c², where v = a(t) dΣ / dt

(derived from Wikipedia's FLRW, the very first equation)

-J

Hi Jorrie,

Of course in my opinion your derivation of velocity time dilation across adjacent inertial frames in FRW coordinates is wrong. You have artificially inserted a factor of gamma which is not part of the metric. The FRW metric is the complete description of what happens non-locally in FRW coordinates; it does not need or accept any relativistic "add-on" factors. (Distinguish this from the relativistic momentum of peculiar velocity, which is meaningful only locally, not across adjacent inertial frames).

One way to think more deeply about the FRW metric is to compare the (almost) empty FRW model with the SR Milne model. In the 'X' spatial coordinate of the Milne model, there is indeed time dilation as between comovers at different velocities. That is illustrated by the hyperbolic curve of constant t in Tamara Davis' Fig 4.1. When the Milne coordinates are transformed to FRW coordinates, however, the line of constant t becomes flat. This is because FRW inherently adopts the common proper time of all comovers as its time coordinate. As Tamara explains on p.62, FRW chooses common proper time in order to achieve homogeneity in the Milne universe, which otherwise would be inhomogeneous. FRW coordinates (specifically the RW line element) achieve a homogeneous distribution of galaxies along a surface of constant t by transforming away time dilation as between comovers.

FRW's transformation to eliminate time dilation has the additional effect of introducing negative, hyperbolic spatial curvature. This spatial curvature is simply a byproduct of applying the Lorentz transformation (which in effect causes a reverse Lorentz-'expansion' of the flat space of the Milne model). This is the change from the SR 'X' coordinate (which measures proper length only in the observer's own frame) to the FRW 'D' proper distance coordinate (which measures proper length in all fundamental observers' frames)

In an FRW universe at critical density, however, because sufficient gravity is present (which introduces an element of counteracting positive spatial curvature, analogous to Schwarzschild coordinates), the same Lorentz transformation does not result in any 'net' spatial curvature, and the FRW metric remains spatially flat. But just as with the (almost) empty version of FRW, a single time coordinate is common to all interactions between comovers, and it underlies and affects all observations they make of each other. This is dictated by the FRW metric.

Jon

Hi again Jon.

"Of course in my opinion your derivation of velocity time dilation across adjacent inertial frames in FRW coordinates is wrong."

Easy to say, but can you show me where my derivation is wrong?

For completeness, here is the full derivation, starting with Wikipedia's FLRW's generic metric: -c²dτ² = - c²dt² + a(t)² dΣ²

Then: dτ²/dt² = 1 - a(t)² dΣ²/(cdt)² = 1 - a(t)² v²/c²,

if I take v = dΣ/dt. This is standard SR time dilation, just with the a(t) modifier for expansion. No "added in" relativistic time dilation factors - they are already inside the FLRW metric. This is to be expected, since the FLRW metric is a solution to Einstein's field equations.

What this tells me is that if you and me are adjacent fundamental observers, some distance apart, we will each observe the others clock to run slow by this derived factor. When I make any measurements in my local inertial frame, I can give you the results and you can use the above time dilation, including a(t), to determine the values in your own local inertial frame. You can obviously also make the measurements in your own frame and get the same results. This includes cosmological redshift.

The same is true in the opposite direction - when you give me your measurement results. This still means we are both observing the same 'cosmological time' - as I tried to show before, SR clocks in relative motion do not run at different rates in any absolute sense. It is only when you bring in gravity or acceleration that absolute differences can occur.

-J

Hi Jon, an afterthought.

When the steps are infinitesimal, then the relativistic and Newtonian approaches must obviously converge on the same value for redshift. To my mind, that still does not make the Newtonian approach the correct one. Contrary to what you have once said, I believe (and have shown above) that it is the Newtonian one that is in error and only comes close because of the non-relativistic velocity steps.

To argue otherwise is decidedly un-relativistic!

-J

Hi Jorrie. Sorry about the long delay in responding. I was attending to a death in the family.

First, we will not resolve our disagreement about the cosmological redshift by resorting to superficial claims that one approach is better because it is more 'relativisitic' than the other. It is also incorrect to characterize my description as 'Newtonian.' I am simply applying the Robertson-Walker line element as it is intended to be applied in GR.

Your derivation of FRW time dilation goes wrong when you insert the factor v = dΣ/dt into the FRW metric. You characterize a(t) as a mere 'modifier' to SR time dilation, but that is incorrect; a(t) is the sole and entire time-dynamic element of the FRW metric. As the Wikipedia article you cite explicitly says: "dΣ does not depend on t — all of the time dependence is in the function a(t), known as the 'scale factor'." If, on the contrary, time dΣ contained an SR time dilation element as you propose, then dΣ would change over time as the Hubble rate (v) changes. Which is incorrect.

The standard SR time dilation factor applies only within inertial frames, e.g. within any single infinitesimal local FRW frame. SR cannot be applied in the normal way across non-local comoving FRW frames, even if there is no gravity or acceleration. (Of course it can be applied within a single global Minkowski frame, but that is an entirely different coordiate system from FRW, which puts each fundamental observer in a separate comoving frame, each of which is fully identical and interchangeable, and each of which precisely agrees on all mutual observations). In fact the fundamental characteristic of FRW coordinates is that they specifically transform away the SR Lorentz factor as between comoving coordinates, by adopting a surface of constant time (as distinguished from Lorentz-dilated time). That's why the Milne 'X' coordinate is not the same as the FRW 'D' (proper distance) coordinate even when gravity is vanishingly small.

From my experience, the only way to come to grips with the difference between SR Minkowski coordinates and FRW 'fundamental-observer' coordinates is to closely examine the transformation between Milne and vanishingly empty FRW, as described in my previous post. Peacock walks through the transformation on p.88, and as I said Tamara Davis' graph is helpful. (Ned Wright has a similar graph in his tutorial).

Jon

Hi Jon.

Happy New Year to you and family. I'm sorry to hear about death in the family over the festive season, but I guess that's life.

OK, so would you say that I could take dΣ as just a constant spatial interval and divide a(t) by cdt, e.g., getting expansion rate H(t)? With appropriate units chosen,

dτ²/dt² = 1 - a(t)²/(cdt)² dΣ² = 1 - H(t)² dΣ²

This is still the same time dilation and hence changes nothing in the infinitesimal redshift calculations. You still have a proper time and a coordinate time. Using either method, I get the relativistic Doppler slightly closer to the proper cosmological redshift (1/a) than the Newtonian, but both are wrong! This causes me to still believe the whole idea of accumulating infinitesimal Doppler shifts as seriously flawed.

BTW, your "The standard SR time dilation factor applies only within inertial frames, e.g. within any single infinitesimal local FRW frame" is not quite true. SR time dilation is how two different inertial frames measure each others time intervals, in a reciprocal fashion. It does not mean that any frame's clocks run at a different rate, so it is compatible with FRW frame, where all inertial frames clocks tick at the same rate.

-J

Hi again, Jon.

Further to my #65, I note that Bunn & Hogg, in their paper The kinematic origin of the cosmological redshift (2008), where they strongly push the Doppler shift interpretation for the redshift, use the special relativistic versions in their Eqs. (6) and (7).

I'll have to check if they have a trick that I have missed (giving me divergent answers when integrating over large distances).

-J

Hi Jorrie,

You confuse me when you say that the classical Doppler calculation has a larger error than the SR Doppler. I thought you said earlier that was only true if you calculated classical Doppler using a stationary emitter and moving observer. My response to that was that the redshift measured by the observer should be calulated using the classical Doppler formula for a moving emitter and stationary observer, (1 + v/c). Any FRW fundamental observer is entitled to treat themselves as being stationary in the metric, and besides from the perspective Bunn & Hogg take, the cosmic gravitational action is toward the observer, not away from it, and so the emitter should be treated as having been in freefall toward the observer. (This gravitational action toward the observer is of course separate from the emitter's expansionary momentum away from the observer).

Assuming you calculate it the way I suggest, what are the respective errors you obtain with the classical and SR Doppler formulas?

I think the quasi-integration formula I gave is only an approximation when calculated in the simplest way (which is how I did it in my own small spreadsheet) because the Hubble rate used for each integration period is not exactly aligned with the corresponding dt value. Unless one goes through extra gyrations of averaging, the Hubble rate will correspond to either the beginning or the end of the integration period, and will not precisely correspond to the average Hubble rate over that period.

I read the Bunn & Hogg paper when it was first published, but the more I read it, the more I agree with their major point, which is that cosmological redshift can be alternatively depicted as a Doppler shift or as a gravitational shift. I have 3 quibbles with the paper:

1. As I have maintained, the 'correct' calculation should use classical Doppler shift rather than SR Doppler, due to the absence of time dilation as between FRW fundamental observers. As I read it, B&H offer no specific rationale for using SR Doppler in their equations, and resort to claiming that any two adjacent observers can be treated as approximately residing in the same local inertial frame. That may be ok as an approximation, but as B&H admit, they are tolerating a margin of error in the range of (r / R_c)². In my opinion that error arises specifically because of their assumption that two adjacent observers reside in the same local inertial frame. If on the contrary one explicitly accepts that such observers are in separate nonlocal, frames which are moving non-inertially with respect to each other, then one is compelled to account for the fact that there is no time dilation between FRW fundamental comovers. Thereby avoiding their margin of error.

2. In describing the alternative gravitational redshift explanation, they ought to mention that gravitational redshift and Doppler redshift are not directly interchangeable under the equivalence principle unless the observer's freefall is at exactly the Newtonian escape velocity. At any other rate of freefall, the gravitational spatial curvature and SR Lorentz (length) contraction will not exactly offset each other. Of course one needn't worry about this point if one assumes a homogeneous spatially flat FRW model.

3. They also feel oblliged to repeat the stuffy caveat that relativity purism frowns on measuring distances and velocities at large distances:

"Talking about v_rel is precisely the sort of thing that the enlightened cosmologist described in Sec. I refuses to do because, in a curved spacetime, there is not a unique way to define the relative velocity of objects at widely separated spacetime events. Determining the velocity of one object relative to another involves comparing the two objects' velocity four-vectors."

In fact, their description of parallel transport of 4-velocity seems perfectly appriorate to me when one is calculating the relative distance and velocity of two isolated observers in a homogeneous FRW space. (I think there approach is the same as mine except for their error in using SR Doppler). I find it annoying that cosmologists so frequently say or imply that one can't 'truly' solve these problems. The basic mechanics of the FRW metric are not all that complicated, and in my opinion all of these caveats just help to maintain the mystical aura around GR and rationalize away the failure after so many years to clearly articulate some basic mechanical aspects about how the cosmic expansion operates, how the cosmological redshift works, and about superluminal recession velocities. [End of editorial.]

You said: "OK, so would you say that I could take dΣ as just a constant spatial interval and divide a(t) by cdt, e.g., getting expansion rate H(t)?"

I don't follow how you get a(t) / cdt = H(t). We start with H(t) = [dot]a(t) / a(t). How do you derive the next step, a(t) / cdt = [dot]a(t) / a(t)?

Jon

Hi Jon, you wrote: "You confuse me when you say that the classical Doppler calculation has a larger error than the SR Doppler. I thought you said earlier that was only true if you calculated classical Doppler using a stationary emitter and moving observer."

Different round-off errors during lengthy integrations make that result somewhat dubious. Doing an integration for true cosmological redshift z from 0 to 4.0, using 4000 steps of dz=0.001, indeed produces a relativistic error that is larger than the Newtonian error, but both are way-out. Numerically, the values I got are: z_cosmo = 4.000, z_rel = 4.783, z_Newton =4.781, i.e., both errors of more than 15%.

I must still try and work it out analytically, but it indicates to me that the use of Doppler shift over large distances is a flawed way of calculating cosmological redshift. I feel likewise about treating it as a gravitational redshift. Why use it, when the straight cosmological redshift is so easy to work with?

You wrote: "I don't follow how you get a(t) / cdt = H(t). "

Yes sorry Jon, my last step in post #65 was wrong; should have been:

dτ²/dt² = 1 - a(t)²/(cdt)² dΣ², = 1 - (dΣ dot-a(t)/c)², which is a velocity time dilation in the appropriate units, because dot-a(t) is the normalized expansion rate. H(t) = dot-a(t)/a(t) is only a proper velocity when a(t) = 1.

-J

I don't understand why you're getting such a large error rate. I have 100 step series multiplication in my spreadsheet out to z=1023, and the classical Doppler equation 1+H(t)dt yields a redshift of 1146. That's less than 10% error.

By the way, technically speaking Bunn & Hogg don't actually accumulate redshifts through a multiplicative series or integration. Instead they do a parallel transport of the emitter's 'global' recession velocity to the observer, which of course involves the same multiplicative series as an accumulation of redshifts would.

On further thought I'm more skeptical of B&H's claim that cosmological redshift can be intepreted as gravitational redshift through a change in coordinates. As I understand it, the gravitational redshift arises from relative differences in clock rates between the emitter and observer, not from the photon gaining energy as a result of 'gravitational pull'. But B&H do not propose that the relative clock rates of the emitter and observer are different in the FRW model, so the basis for gravitational redshift seems to be lacking. Perhaps gravitational redshift should simply be disregarded in calculating the cosmological redshift.

Jon

Hi Jon.

Ah, I see what you were doing. But, (1+H₁dt)(1+H₂dt)....(1+H_tdt) does not represent a Doppler shift (I've shortened H(t) to just H_t). Since H_t = da/(a_tdt), it means that H_tdt = da/a_t, which is not a velocity and hence your calculation cannot be called a Doppler shift. In fact, it can be shown that (1+da/a₁)(1+da/a₂)....(1+da/a_t) ~ 1/a_t, which is just the normal cosmological redshift ratio, because z = 1/a_t - 1.

I have tried B&H's equations and they give identical results to the algorithm I'm using for redshift, based on the difference in Hubble velocity between steps in the integration.

-J

Jorrie, I think you're misinterpreting the equation. The 'Hubble rate' H_t alone is not a velocity, but H_tdt is. Recall that dt = dx for a photon's worldline in units of c=1. As I mentioned early on, dt is just an easier-to-use stand-in for dx in this equation. H_tdx is the 'Hubble velocity' differential, the relative proper recession velocity as between any two spatial points the photon passes along its worldline. This is the incremental change in local velocity which dictates a corresponding increment of Doppler shift.

In this analysis we are not trying to measure how much cosmic expansion per se occurs. Rather, we are interested specifically in the movement of the photon, and the increments of change it experiences in the velocity of each successive comoving galaxy it passes along its worldline, relative to the immediately preceding galaxy it passed. Accomodating those incremental velocity changes causes the photon to accumulate locally-measured losses of momentum. (In the same general way a non-relativistic massive particle loses locally-measured momentum at 1/a.)

Because of the duality of dt = dx, I too felt that the mathematical equivalence of a multiplicative series of (1+H_tdt) to the 'standard' cosmological redshift formula (as you point out) may prove that my formula is exactly correct!

Jon

Sorry Jon, but I cannot agree with your method and then calling it a Doppler shift. IMO, it is a corruption of the meaning of Doppler shift.

We are not working here with dt=dx, because we are not computing Doppler for photon velocities, but for emitter velocities (V_recession), very much smaller than c, over very small distances at earlier and earlier times.

You wrote: "H_tdx is the 'Hubble velocity' differential, the relative proper recession velocity as between any two spatial points the photon passes along its worldline. This is the incremental change in local velocity which dictates a corresponding increment of Doppler shift."

I checked this for z=1 with dz=0.001 and found a 'Hubble velocity' differential H_tdx of 0.00017, while H_tdt produces 0.0005. This discrepancy grows larger for more distant galaxies. I still think, as I showed last time, that what you are calculating is just an approximate cosmic redshift due to expansion alone. You can just as well use the 'canonically correct' value of z+1 = 1/a.

-J

Note: For anyone else that may still be following (or may stumble upon) this discussion, I want to categorically state that IMO, the Doppler shift interpretation of cosmic redshift is fundamentally flawed. The modern meaning of Doppler shift is defined for objects not moving with the Hubble flow (i.e., peculiar velocities in cosmology), not objects moving "with the Hubble flow" (for which cosmological redshift is defined). The real redshift of a distant object is in fact a combination of the two, but when we are talking cosmological redshift, we are not including Doppler shifts.

Hi Jorrie, I'm sorry you disagree, but I really don't see what's wrong with calling each local frame crossing by the photon a Doppler shift. One simply thinks of the photon as being emitted and received sequentially and repeatedly at each local frame crossing along its worldline. That's what Peacock says (p 87):

"One way of looking at this issue is to take the rigid point of view that 1+z tells us nothing more than how much the universe has expanded since the emission of the photons we now receive. Perhaps more illuminating, however, is to realize that, although the redshift cannot be thought of as a global Doppler shift, it is correct to think of the effect as an accumulation of infinitesimal Doppler shifts caused by photons passing between fundamental observers separated by small distances..."

Any photon traveling through an FRW universe would note that the infinitesimal segments of its worldline in front of it and behind at are all constantly in non-inertial motion relative to each other and relative to both the emitter and receiver. Therefore it makes no sense to try to calculate a global solution for the worldline as a whole using equations designed for inertial frames. One must limit oneself to calculating the next crossing to the next local frame, and then accumulating the redshift as one goes on.

As long as dx is measuring the photon's local movement along its worldline, as I described, it is impossible for dx and dt to not be equal. So the difference in results you calculate just demonstrates that you're not applying the logic Peacock and I are using. dt must equal dx, as measured locally and accumulated.

Therefore, the cosmological redshift must be caused by accumulated local Doppler shifts.

Why don't I just go with the simple 1+z calculation alone, as Peacock rhetorically asks? Because the expansion of the universe itself provides no specific cause or explanation for why the redshift occurs. We know almost to a certainty that the 'expansion of the hypersphere' itself cannot be the cause, as demonstrated by the tethered galaxy analysis. On the other hand, saying that the physical cause is unknown and unknowable is completely unsatisfactory.

Jon

Hi Jon, OK, I understand now what you are doing (I'm not sure that this was what Peacock intended, but I do not have his book with me at present).

You are working in redshift as a function of 'light-travel-distance', dx=dt, whereas I was using redshift against proper distance, dD = R dχ, where R is the radius and χ the 'development angle' on the hypersphere. It is true that at short distances, dx ≈ dD in the appropriate units, so I'm OK with it as an approximation.

In the end, using the factor (1+H_tdt) = (1+da/a) is mathematically guaranteed to produce a result that approximates 1/a, as I've shown before. As to whether this represents a Doppler shift, I'm not convinced. It is now pretty clear to me why a relativistic Doppler treatment produces larger errors - it is not a Doppler shift, because it depends only on the expansion factor (a), not on a relative velocity. IMO, treating a recession rate as a relative velocity, even over small distances, is still a fundamentally flawed concept.

-J

Hi Jorrie, ok now we are on the same wavelength, so to speak.

Yes I should have used the term 'light-travel-distance' to describe dx, that's a clearer way to put it. And of course dt is 'light-travel-time.'

And I agree that my formula is also guaranteed mathematically to produce a correct result, if precise 'average' values for H_t are used for each photon worldline segment.

If it walks like the duck and quacks like the duck, it seems to me it's the strongest candidate to be the duck. But I'm happy to consider a less flawed alternative, if one can be constructed. However, I don't see the flaw in this candidate.

Jon

One more explanation...

Remember it is the cosmic gravity that causes every infinitesimal local frame along the photon's worldline to accelerate non-inertially with respect to every other such local frame.

At any point in time along the photon's worldline, the present relative Hubble velocities of the local frames in the photon's past have no effect on the photon's ultimate redshift, and those of the local frames in the photon's future haven't been set yet to the values that will affect the photon's redshift. The same is true of the current relative velocity of the emitter frame (past) and observer frame (future). Thus there is no surface of constant time along which all of the agents of the photon's redshift have their effect simultaneously.

In those circumstances it is impossible for any global one-shot causal agent of the redshift to be defined. The effects must be accumulated sequentially over time.

It is certain that the photon must pass through every local frame at a peculiar velocity of exactly c. If the photon were able to adjust to the increasing relative Hubble velocities of the galaxies it passes without suffering a locally-measured momentum loss, it would be getting a free boost. Unlike a massive particle. But photons don't get a free boost, they get redshifted.

Jon

Hi Jon, you wrote: "And I agree that my formula is also guaranteed mathematically to produce a correct result, if precise 'average' values for H_t are used for each photon worldline segment."

This is not quite the case, firstly because the multiplicative (1+H_tdx) = (1+da/a_t) factor can only ever approach the correct (defined) value of 1/a, not reach it. Secondly, if you want to make it 'true Doppler', you have to do it relativistically (ask any cosmologist). In doing so, you deviate further from the 1/a target.

My objections to Doppler shift in the footnote of #75 still stand. If I have to choose a "best-buy" explanation for the cosmic redshift, I'll pick this one: Photons travel as waves, but are detected as particles. The traveling waves are stretched with the expanding fabric of space, hence the detected cosmic redshift. :)

A previous statement you made against this interpretation can similarly be frowned upon: "We know almost to a certainty that the 'expansion of the hypersphere' itself cannot be the cause, as demonstrated by the tethered galaxy analysis."

I do not recall any reason why that analysis shows anything like that. It clearly shows that 3-momenta of particles are reduced and that should include photons, which have no choice but to shed frequency.

Finally, perhaps it does not matter and I should stand by what I wrote in the OP of this thread: "I suppose in the end it is not too important which 'crutch' you use - stretching of wavelengths, conservation of angular momentum, or even Doppler shift, as long as it is accepted that the received to emitted photon wavelength ratio (λ/λ₀) is the same as the expansion ratio (R₀/R) since emission."

-J

Hi Jorrie;

1. All integrations or multiplicative series approach the correct value rather than hit it dead-on, but the error can become vanishingly small with a reasonable number of integration steps, so there is no reason to be concerned about it.

2. You are correct that cosmologists tend to think of relativistic Doppler as the 'only true' Doppler shift, but that is because the community as a whole has overlooked the role of classical Doppler shift in the cosmological redshift. Someday, maybe soon, they'll come around. Besides as I said before, you don't have to think of it as 'classical' Doppler, just think of it as relativistic Doppler with an adjustment to eliminate the time dilation portion because FRW coordinates already eliminate the time dilation as between fundamental comovers.

3. Your FN #75 says only: "The modern meaning of Doppler shift is defined for objects not moving with the Hubble flow (i.e., peculiar velocities in cosmology), not objects moving "with the Hubble flow" (for which cosmological redshift is defined)." That's your opinion, but it is contrary to mainstream current scholarly work such as Peacock (which I quoted to you ) and Bunn & Hogg's paper, among others, which refer explicitly to accumulated incremental Doppler redshifts along the photon's worldline. My impression is that the B&H paper has been well embraced by mainstream cosmologists.

4. I explained in my first post on this thread why the concept of travelling waves being stretched by the 'expanding fabric of space' (or hypersphere) seemed to be ruled out by the tethered galaxy analysis. B&H alludes to this point as well. Neither of those explanations is complete, and upon further review I'm less convinced of this point.

Here is a more complete explanation of my analysis.

In the normal exercise, the tethered galaxy analysis is performed using non-relativistic test galaxies, not relativisitic photons. In a gravitational, 0 Lambda scenario, the galaxy starts with no proper velocity away from the origin upon untethering, and we learn that the expanding hypershere does not act like a 'force', so it is incapable of imparting any outward recession velocity to the galaxy. The only force acting on the galaxy is the gravity of the 'Birkhoff's Theorem sphere' with its radius at the test galaxy, and this causes the galaxy to accelerate inward toward the origin.

A 'tethered photon' analysis behaves somewhat differently. (The photon itself is not tethered, it just gets emitted toward the origin from the test galaxy at the instant of untethering). Again, the expanding hypersphere does not act like a force, so at no time can it exert any outward recession velocity to the photon's travelling waves unless the waves had some outward velocity component in the initial conditions - say, at the time the second wavecrest is emitted. As in the normal scenario, the emitter galaxy does not experience any acceleration away from the origin between the emission times of the first and second wavecrests; so no outward "expansionary" velocity component is imparted to the second wavecrest relative to the first wavecrest.

The proper distance between the first and second wavecrests does increase over time, but not due to any outward acceleration vector acting on the second wavecrest (nor acting on the emitter prior to the emission of the second wavecrest.) Rather, the first wavecrest accelerates inward at a faster rate than the second wavecrest does, because the first wavecrest's proper distance from the emitter is greater than the second wavecrest's, so the former's Hubble velocity HD (relative to both the emitter and the origin) is greater than the second wavecrest's HD. Each separate wavecrest must have a peculiar velocity of exactly c in each infinitesimal local frame it passes through. So the wavecrests will be progressively physically 'stretched apart' during travel.

I suppose one can argue that the faster inward acceleration of the first wavecrest (relative to the second) is simply caused by always being located in a region where the hypersphere has a lower expansion rate than the region in which the second wavelength is then located. But it seems counterintuitive to rationalize that space expanding outward from the origin causes the wavecrests closer to the origin to stretch inward away from more distant wavecrests travelling inward along the same worldline. Describing this as the internal "stretching" of rubber-sheet substrate through which the wave travels doesn't really capture all of the dynamics. It seems clearer to describe this effect as being like the locomotive and caboose of a single long train which are on different stretches of track that happen to require different local operating speeds. The travelling wave's behavior is an accumulation of a series of disparate local events, governed only by its need to maintain a peculiar speed of c in each local frame. Nevertheless, I don't think I can say at this point that the expanding hypersphere approach is categorically ruled out.

In addition to the above, there of course is the additional inward acceleration vector caused by the cosmic 'cosmic tidal effect,' i.e. gravity's 'Birkhoff Theorem sphere' centered on the origin. The radius of that sphere, and therefore its gravitational potential, increases in direct proportion to the radius. Therefore the second wavecrest always experiences more inward gravitational acceleration than the first wavecrest. Nevertheless, the stretching apart of the wavecrests caused by the higher HD experienced by the first wave crest exceeds the counteracting gravitational effect, resulting in total net increase in distance between the crests that corresponds to the cosmological redshift.

By the way, the length of a train of photons (e.g. a burst of light) will increase radially during its travel by the same proportion as the wavelength increases, and due to the same cause. It's like a line of evenly-spaced joggers following each other at a constant peculiar velocity over a series of moving sidewalks which are moving in the opposite direction the joggers are running -- with each successive sidewalk having a lower 'negative' velocity than the prior one. Even if the speed of every such sidewalk is progressively simultaneously reduced over time by the same proportion of its current speed, the line of joggers will progressively stretch out.

Jon

Hi Jon, I suppose the different interpretations of cosmic redshift will always be arguable and fun, but in the end, no one is cast in concrete.

I still prefer to think of the observed (z_o) redshift like Prof. Peebles, rather than like Prof. Peacock. Principles of Physical Cosmology (Peebles) Eq. 5.48:

1+z_o = 1/a (1+v_p)/√[1-v_p²], where v_p is the radial component of the peculiar velocity. Now, this is using Doppler shift in the appropriate place!

Peebles do however say that at low recession speed the cosmic redshift is approximated by the first order Doppler shift, z ≈ H_tD_t, where D_t is the proper distance between nearby points. I still can't help thinking that since this is not an exact relationship, it tells us something against cosmic redshift as accumulated Doppler shift - although in the limit, the difference may become vanishingly small.

-J

Hi Jorrie, you said:

"Peebles do however say that at low recession speed the cosmic redshift is approximated by the first order Doppler shift, z ≈ H_tD_t, where D_t is the proper distance between nearby points. I still can't help thinking that since this is not an exact relationship, it tells us something against cosmic redshift as accumulated Doppler shift - although in the limit, the difference may become vanishingly small."

Clearly Peebles was on the wrong track when he chose D_t as the proper distance between points at a single instant in time, rather than as the light travel distance as Peacock did. Cosmological redshift is a function neither of the distance along any single surface of constant time, nor of the 'global' velocity between the emitter and receiver.

The fact that the accumulated multiplicative series (1+H_tD_t) along a photon's worldline equals the aggregate proportion by which the proper distance between the emitter and observer has expanded (a₀/a_t) during the photon's travel is not a coincidence, it is mathematically required.

As I explained in my last post, that's equally true regardless whether one prefers to attribute the recession velocities of galaxies to the 'expansion of the hypersphere' or to the less metaphysical kinematic movement of galaxies through space.

Jon

"The fact that the accumulated multiplicative series (1+H_tD_t) along a photon's worldline equals the aggregate proportion by which the proper distance between the emitter and observer has expanded (a₀/a_t) during the photon's travel is not a coincidence, it is mathematically required."

Sorry Jon, but I think you are not quite right on this one. I've shown before that:

(1+H_t1dt)(1+H_t2dt)....(1+H_tdt) ≈ 1/a_t, not an equality, just a first order approximation. So, if you are not satisfied with an approximation, cosmic redshift as Doppler shift doesn't quite work.

Further, H_t is defined in terms of comoving distance, which only approximates light-travel distance at short range. Add to this that Doppler shift is defined only relativistically, and your position of "mathematically required" is not too strong. (It may be publishable in an Engineering Journal, but I'm not too sure of a Science Journal. )

However, I'm happy with what you use as a very good approximation, but I think the caveat should be mentioned.

-J

Hi Jorrie,

Please explain exactly what you mean by "first order" when you describe my equation as a first order approximation.

Second I don't think the fact that H_t is defined in terms of comoving distance introduces any error into the equation, as long as the value of H_t is set at precisely the mean Hubble velocity over the worldline segment d_t. We are computing a ratio of H_t to H_t at slightly different points in time and space, so the error involved in using proper distance should wash out.

My equation conceptually captures the underlying physical cause of the redshift, so any inaccuracy must arise solely from the use of less than an infinite number of segments. Conceptually there is no reasonable alternative explanation for the cause of the redshift other than the accumulation of infinitesimal wavelength shifts. As I said, that is true even if one applies the 'expanding hypersphere' pardigm. The a₀/a_t solution is merely the summation of this series of discrete events, not a discrete spacetime event of its own. Its virtue is that it washes out any vanishingly tiny errors that arise from the fact that our computing tools are incapable of handling an infinite number of multiplicative segments.

Jon

Hi Jon,

The way Peebles used "first order" (AFAIK) is leaving out any non-linear terms, i.e. for Doppler shift, the Newtonian Doppler shift is first order and and a relativistic approximation is at least third order. With c=1:

1+dλ/λ = (1+v)/√(1-v²) ≈ (1+v)(1+v²/2) ≈ 1+v+v²/2 +v³/2 ≈ 1+v, if v << 1.

Similarly, (1+H_t1dt)(1+H_t2dt)....(1+H_tdt) ≈ 1/a_t has higher order terms in a_t at the end, which are neglected for small H_tdt.

-J

OK fine Jorrie, but Peebles does not recognize that there is no place for time dilation as between fundamental comovers in FRW coordinates, as I have explained many times. That's his oversight. If I am correct on that point, my equation is not a "first order approximation", it is actually correct.

Jon

Hi Jon,

"OK fine Jorrie, but Peebles does not recognize that there is no place for time dilation as between fundamental comovers in FRW coordinates, as I have explained many times."

I'm afraid you stand pretty alone here, because AFAIK, Peacock, Peebles, B&H, Lineweaver, etc. all use the relativistic Doppler shift. I've also tried to show you why the FLRW metric incorporates velocity time dilation. In #58 I wrote:

"For completeness, here is the full derivation, starting with Wikipedia's FLRW's generic metric: -c²dτ² = - c²dt² + a(t)² dΣ²

Then: dτ²/dt² = 1 - a(t)² dΣ²/(cdt)² = 1 - a(t)² v²/c²,

if I take v = dΣ/dt. This is standard SR time dilation, just with the a(t) modifier for expansion. No "added in" relativistic time dilation factors - they are already inside the FLRW metric. This is to be expected, since the FLRW metric is a solution to Einstein's field equations."

You commented on the fact that I used "v = dΣ/dt", which is not incorrect, IMO, because dΣ is a constant. It is probably better to consider the whole term a(t)² dΣ²/(cdt)² as the velocity (v²), but that changes nothing in the end. The end result is exactly the same as between two inertial frames in relative motion in Minkowski spacetime - completely reciprocal time dilation observations, but no "slow running clocks". Coordinate dependent observables, not absolute values...

IMO, your Doppler algorithm is not quite correct, but close enough for all practical purposes.

-J

Hi Jorrie,

After thinking through your derivation, I can't argue with it as far as it goes. With respect to radial distances (only) in a flat FRW model, obviously dx(t) = a(t)dr, where a(t) is time dependent and dr is not.

However, I think you may have omitted a step in calculating velocity. I do not think v = dx/dt. The dx term must be subjected to a Lorentz length expansion to calculate proper distance dl. Peacock includes this step at p. 88:

"Now, the time-dilation equation gives r in terms of t and t' as [eq omitted] and the radial increment of proper length is related to dr via length contraction (since dl is at constant t'): dl = dr/γ = ct'dω."

Peacock uses the term dl for proper length. I think that, when calculating radial proper recession velocity, the Lorentz transformation of the dx distance washes out the effect of time dilation in FRW coordinates, meaning that, in effect, all cosmological events are observed in constant cosmological time by all fundamental observers.

I'd appreciate your thoughts on this point.

Jon

Hi Jon.

Peacock uses somewhat confusing parameter names, as I summarized in the table in my eBook. If I had it right, he uses r as an angle, with comoving distance is R₀r, as indicated. Unfortunately, I have not listed proper distance, but l and dl seem to make some sense in terms of his other definitions. However, dl has quite a complex equation in standard hyper-spherical cosmology (e.g., an 'approaching' photon can have a positive dl and hence be receding from us in the early universe).

I don't have Peacock's book available atm, so I can't follow his derivation. Would it be possible to scan the page and post it as a .gif or .jpg? I'm a bit worried about his Lorentz contraction idea, but it may perhaps fall naturally out of his equations, as you said. As you know, I'm not all that comfortable to view the expansion as movement through space. I like to reserve that for peculiar movement only.

-J

Hi Jorrie,

Here, try this link to Peacock: Peacock p 88 excerpt Google Books.

The parameter names do get confusing because everyone uses something different. I used dl just to be consistent with Peacock's quote.

I think it makes the most sense to use dx to mean proper distance, and dr to mean non-proper distance.

By the way, the Lorentz length expansion of SR distance coordinates to generate conformal FRW coordinates in the Milne model is completely non-controversial. It is described and illustrated in the graphs of Tamara Davis and Sean Carroll which I have linked to previously on this thread. It makes good sense to me to apply this transformation before calculating proper radial velocity in FRW coordinates.

Also, I do not think the question about using classical vs SR Doppler shift to calculate accumulated cosmological redshift has anything directly to do with one's preference for the 'expanding hypersphere' or 'kinematic' paradigms. As I explained earlier, exactly the same underlying physical cause and mechanism I described for the redshift applies equally well to both paradigms.

Jon

Hi Jon.

Tx, but I'm afraid that I see nothing in the Peacock excerpt that changes my mind. Pages 88-89 deals with the Milne model (empty universe) only, which is not realistic. The derivations do not work for a matter-containing model, not even if it's flat.

I'm more interested in his page 87 comment that "particle de Broglie wavelengths thus scale with the expansion, a result independent of whether their rest mass is zero" (as I used in the next thread in this Blog: http://cr4.globalspec.com/blogentry/9443/Cosmic-Balloon-Application-III-Particle-Momentum-Decay), under "Particle Redshift?".

I still think Peacock's r refers to comoving distance, which is the same as the χ of Peebles, as shown on the right for a closed model. For an open model like Milne, the sin is just replaced by sinh.

The use of (even incremental) Doppler shift to describe cosmological redshift for a real universe can have quite confusing consequences. Say a distant object is moving directly towards us at very close to the local speed of light and we measure its redshift. It will be quite hard to model its observed redshift as incremental Doppler shifts.

-J

Nobody believes that the empty Milne model is a realistic representation of the actual universe. I reference it only for the same reason as Peacock -- to illustrate how Minkowski coordinates are mathematically transformed into FRW coordinates. The transformation involves a Lorentz length expansion which cancels out the effect of SR time dilation. End of story.

I agree that Peacock's dr refers to the comoving coordinate. In order to obtain a distance coordinate, one multiplies the comoving coordinate by a(t). Just as I did. If dr is stated in terms of FRW coordinates, then it is already a proper distance. But if it is stated in terms of Lorentz contracted Minkowski coordinates, then obviously it does not constitute a proper distance.

Your example of a distant object moving toward us at near the speed of light presents no difficulty at all. Presumably the motion toward us is peculiar motion (unless you're talking about the collapsing phase of an overdense model) of a massive object. The 'global' SR Doppler shift caused by this peculiar motion component is simply multiplied against the accumulated cosmological redshift caused by the recession velocity component (of a fundamental comover at the galaxy's location), yielding total redshift. Lineweaver & Davis demonstrate how this combination can cause an object moving away from us to have net blueshift, and vice versa.

The SR Doppler shift indisputably applies to the peculiar component of galaxy velocities. But it doesn't apply to the recession component of velocities of fundamental comovers.

As I've said repeatedly, your preference for stretchy space is all well and good, but that preference is not relevant to validating the mathematical details of how the cosmological redshift accumulates.

Jon

OK Jon, I think we can close this saga now - this will be my final reply on the merits of Doppler shift as cosmological redshift.

I've decided never to use a Doppler shift interpretation of cosmological redshift in any teaching of cosmology. To use a 'Lorentz expansion' to cancel a SR time dilation (or essentially a Lorentz contraction) smacks a little contrived. I will stick to: there is no Lorentz contraction in the first place, because the cosmo-redshift is not caused by relative motion through space. The incremental Newtonian Doppler interpretation of cosmo-redshift may approximate the 'real thing', but I won't go through the bother to try and defend it...

-J

Hallo Jon, interesting views!

I'll just comment where I have questions or reservations. You wrote:

"... As Peacock says (p.87): ... "It is momentum that gets redshifted; particle de Broglie wavelengths thus scale with the expansion, a result that is independent of whether their rest mass is non-zero." This loss of local momentum is itself the cause of the cosmological redshift, and not merely an effect of it."

I agree. I view an "angular momentum" relative to the hyper-center as constant for all particles, given by: L = γ m R V_pec = constant, where γ is the Lorentz factor and R the hyper-radius, proportional to expansion factor a.

You wrote: "The formula for classical Doppler shift with a stationary observer and moving source is (1 + v/c). Using this, I came up with a simple integration formula for the cosmological redshift:

λ_o / λ_e = _e∫^o 1 + H^_t δt / c

... This integration of accumulated infinitesimal redshifts yields a finite z, quite different than what a 'global' calculation of velocity differential yields. I believe this formula is equivalent to Peacock's 3.67. Note that dl = dt for light travel in units of c, but it's more straightforward to calculate the latter.

I roughly integrated this formula to z=1023 and got an answer within 10% of the correct answer. I also tried a version that included infinitesimal SR time dilation, and the error increased to 40%."

I think your errors are simply numerical integration ones, by virtue of too large a step size. When making step sizes infinitesimal, the classical and SR approach should give the same result.

You wrote: "After all, if infinitesimal SR Doppler shifts could actually be integrated to yield the correct cosmological redshift, D&L and many other authors would have demonstrated that fact long ago using numerical integration examples. Yet no one has, and they never will, because that calculation does not yield the correct result except at small, nonrelativistic Hubble velocities (which happens only because the erroneous SR time dilation component is vanishingly small at those distances)."

I don't think a numerical integration of a non-linear function can ever prove an analytical result. What D&L did was to employ the low-speed approximation of SR Doppler (which is exact as the steps approach zero length) to prove the result analytically.

Personally, I dislike the integrated infinitesimal Doppler shift idea. For me the best route is through the conservation of angular momentum, with resultant local momentum decay during any form of expansion...

Hi again Jon.

I do not quite understand your integration equation: λ_o / λ_e = _e∫^o 1 + H^_t δt / c

Did you mean: λ_o / λ_e = _e∫^o (1 + H_t δt / c) dt ?

Or is it more like: z = λ_o / λ_e - 1 = _e∫^o H_t dt / c ?

I'm trying to figure out if your method is not equivalent to mine...

-J

Well, calculus isn't my strong point, but with H in units of c it's just:

λ_o / λ_e = _e∫^o (1 + H^_t δt)

Jon

Hi Jon.

What you are doing is not quite an integral, but a multiplicative progression, like a power series.

1+z = γ₁(1 + H_t1 δt₁) γ₂(1 + H_t2 δt₂).....γ_n(1 + H_tn δt_n)

It does work (approximately) if n is very large (very small steps). What it means is putting up a local inertial frame for each intermediate receiver, which then becomes an emitter in turn. Each emitter's wavelength is immediately transformed to the inertial frame of the next receiver and hence red-shifted by a factor γ_n (1 + H_tn δt_n), where γ is the Lorentz factor. It then stays the same over the discrete time interval.

It is interesting that I also get your observation: for non-infinitesimal steps, the Newtonian version (with γ = 1, identically) gives errors slightly smaller than for the relativistic γ. Both approach the same result if one makes the number of steps huge. I think the second order nature of the relativistic calculation makes it more sensitive to the discrete, non-zero step size. Or, maybe, discrete, incremental Doppler shifts are just plainly wrong...

The above: "It [wavelength] then stays the same over the discrete time interval" is a bit of a disqualifier for me. I will stick to momentum decay, which is both simple and continuous. What's more, de Broglie wavelength redshift covers it both ways, mass-less or massive.

-J

Hi Jorrie, you said:

"The above: "It [wavelength] then stays the same over the discrete time interval" is a bit of a disqualifier for me."

Yes but any accumulation of incremental steps is going to involve infinitesimal local frames. Certainly that's what Peacock is referring to, even if he partly misses the point. I do not think it is reasonable to deny the mathematical propriety of accumulating infinitesimal steps. That's what calculus is all about, even if this particular equation isn't a pure integration.

You also said:

"I will stick to momentum decay, which is both simple and continuous. What's more, de Broglie wavelength redshift covers it both ways, mass-less or massive."

Well feel free to stick to what you're comfortable with. I certainly agree that the cosmological redshift is 'caused' by momentum decay. But that just begs the real question: what is the cause of the momentum decay??? I say it is the fact that the photons must maintain a peculiar velocity of c in each local frame they traverse, even if that means speeding up their velocity relative to the emitter. You say... well I don't know what you say causes the momentum decay. I'm looking for the local, physical cause of the momentum decay. What do you think it is?

Attributing it to deBroglie wavelength shift seems like ducking the question. Sure the wavelength shifts, but why? Are you sticking with the discredited theory that the 'expanding hypersphere of space' physically and locally causes wavelengths to expand, even though those wavelengths had zero mutual peculiar velocity at the time of emission? I thought you agreed that doesn't work.

Also, to what local cause do you attribute the 'preservation of angular momentum' in your balloon analogy? Is it just the expanding hypersphere, or something else specific?

Jon

Hi John, on your: "Also, to what local cause do you attribute the 'preservation of angular momentum' in your balloon analogy? Is it just the expanding hypersphere, or something else specific?"

I think the most plausible 'cause' comes from General Relativity's axiom "a photon's 4-Momentum is always conserved". In the balloon analogy that is equivalent to angular momentum conservation around the origin.

-J

OK Jorrie, I'd like to explore this point further with you.

Is a photon's 4-Momentum always conserved locally from the perspective of different FRW fundamental observers all along the photon's worldline (when they compare notes after the fact)? (I think not, because otherwise as the photon progressively redshifts at 1/a along its worldline, its local 3-Velocity would need to increase (which of course is impossible) in order to maintain a constant local 3-Momentum.)

Is a photon's 4-Momentum always conserved globally from the perspective of a single FRW fundamental observer (as compared to that same observer's perspective of the photon's non-local 4-Momentum at the time of emission)? (Perhaps. When considered from the emitter's frame, an outgoing photon moving across the Hubble flow gains non-local 3-Velocity, so it is not velocity-redshifted).

I also note that the 'angular' momentum of the balloon analogy derives from the 2-curvature of the balloon's surface. But that surface curvature is merely an artificial construct of the particular model, it does not represent any actual attribute of curvature in 3-space. So is the 'angular' momentum just a stand-in for normal linear 3-Momentum?

Jon

Hi Jon, I think 4-momenta are always conserved, no matter who observes it. It is a Lorentz invariant, just like the spacetime interval in SR.

On local (FRW) conservation, you wrote: "I think not, because otherwise as the photon progressively redshifts at 1/a along its worldline, its local 3-Velocity would need to increase..."

I do not follow. A photon's momentum is not a function of its 3-velocity, but only of its frequency.

You also wrote: "I also note that the 'angular' momentum of the balloon analogy derives from the 2-curvature of the balloon's surface."

The balloon surface represents 3-D space, but the radius does not represent spatial curvature at all. It simply represents the expansion factor a. The angular momentum around the hyper-center hence represents a 4-momentum - at least it plays the exact same role.

In the end, we must accept that the balloon analogy is not a physical representation of the cosmos, just a crutch. The conservation of angular momentum is such a nice, easily understandable concept to use as a crutch. And it seems to answer all the usual questions...

-J

Hi Jorrie, you said:

"A photon's momentum is not a function of its 3-velocity, but only of its frequency."

Again, that is true within a local inertial frame. But when an outgoing photon traverses the Hubble flow radially away from the emitter, as it crosses multiple infinitesimal inertial frames it must gain coordinate velocity in the emitter's rest frame. Its proper velocity away from the emitter becomes greater than c. Surely that velocity is an element of the photon's non-local momentum, as considered from the emitter's frame.

But as measured individually in each local frame a photon crosses, its peculiar velocity is always c, so proper velocity is not a 'variable' parameter of its momentum. As you say. In that case, the momentum decays at 1/a as the photon redshifts.

If a photon were capable of redshifting without any time dilation occurring between emitter and observer, and without any change in coordinate velocity (e.g., considered from the perspective of the emitter), then how could momentum ever be conserved in FRW coordinates?

Jon

Hi again Jon.

"If a photon were capable of redshifting without any time dilation occurring between emitter and observer, and without any change in coordinate velocity (e.g., considered from the perspective of the emitter), then how could momentum ever be conserved in FRW coordinates?"

IMO, as I've said in my previous post, the time dilation is there and the changing 3-momentum from frame to frame is just a function of frequency (p = h/λ) and λ is a function of time dilation/length contraction between frames.

What happens to non-local momentum (w.r.t. proper velocity) is not really an interesting question. I don't think there is any form of conservation law there.

-J

OK, Jorrie, but if you think only locally measured momentum is relevant, then your earlier statement that a photon's 4-Momentum is conserved is wrong. As I understand it, a photon the 4-Momentum is the same as its 3-Momentum, because the photon's local velocity never changes. We agree that the locally measured momentum of a photon decays at 1/a in FRW coordinates, which clearly is a departure from conservation in that context.

Jon

Hi Jon.

No, I understand a photon's invariant 4-momentum (if we consider just time and one space dimension): (P = |p|, h/λ) as quite different from its 3-momentum (p = h/λ), which is not conserved in an expanding cosmos.

By definition, 4-momentum is a 4-vector, which is invariant under Lorentz transformations. So, transforming from one infinitesimal frame to the next, 4-momentum must be conserved locally. AFAIK, photon velocity does not enter into photon momentum, except perhaps c as a constant for conversion of units.

-J

Jorrie,

It doesn't make any sense to say that a photon's 4-momentum remains the same in each subsequent local frame it passes through, as measured locally. Clearly the 4-momentum decays at the rate of 1/a as the photon's wavelength is progressively redshifted, as viewed by comoving observers in subseqent local frames. Otherwise CMB photons would have the same (locally observed) 4-momentum today as they did when emitted, despite their wavelength having increased by a factor of ~ 1100.

It's also common sense that as a photon pursues and then catches up with a distant observer with a high recession velocity, the photon's 4-momentum is going to be measured by that observer to be much lower than by the emitter.

I think in your balloon analysis the concept of 'conservation of angular momentum' is meaningful from the inertial frame perspective of the emitter. That's what your graph is portraying. In that sense I agree that 4-momentum is conserved.

But conservation of 4-momentum cannot apply as between different observers who are moving at different relative radial velocities, all of whom are observing the same photon.

Jon

Hi Jon, on the 4-momentum of photons, I believe this post by Wallace on PF summarizes it in a nutshell. The 3-momentum decays, but not the 4-momentum.

I found the same basic statement in the book General Relativity by Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby, page 121.

My balloon-analogy deals only with 3-momenta, where the 'angular momentum' is just a 'crutch' to make thing more palatable. I'm surely not claiming my angular momentum is equivalent to 4-momentum.

-J

Hi again Jorrie,

OK I've come around on 4-momentum after reading several sources which were less ambiguous than the two you cited. I now see that a photon's 4-momentum is indeed conserved. But the 4-momentum is never what is directly observed by any observer - that's where the locally-observed momentum translation due to the accumulated Doppler shift comes into play.

Photon 4-momentum is *analagous* to measuring a massive particle's 4-momentum in the particle's own proper time. Of course the proper time lapse for a massless photon's worldline is always zero, so the analogy is loose.

Jon

Hi Jorrie,

Here's another interesting aspect of the cosmological redshift. A photon clearly must maintain a local speed of exactly c in every local frame along its worldline. But if a particular photon isn't absorbed in a given local frame, how does the photon 'know' what speed it is traveling through that frame? (I don't mean to be anthropomorphic here about a photon's ability to sense or react.)

Presumably the local speed of light is governed by the inherent permittivity of the vacuum. But unless the nearby vacuum is literally "moving" along with a receding galaxy, permittivity does not explain how the photon's coordinate velocity changes in synch with changes in the local Hubble velocity (relative to the coordinate origin).

One could hypothesize that a photon can sense its local environment (e.g. the relative speeds of nearby galaxies), but that simple idea is unsatisfactory. After all, what would define the maximum distance at which a photon could sense the relative speed of a galaxy it passes, and what happens when multiple galaxies with different peculiar velocities are equally proximate to the photon?

The only reasonable answer I can see is that the photon's local velocity is dictated by the gravitational field (spacetime curvature) present in the local frame. At each local frame crossing, the photon changes its coordinate velocity (relative to a distant observer or emitter) so as to align its speed relative to the rest frame of the local gravitational field.

Note that the local gravitational field is a summation of the influences of both nearby and distant masses. In fact in a perfectly homogeneous universe, the local gravitational field would be dominated by the influence of the most distant masses within the local light cone, because the amount of mass increases proportional to the square of the distance, the same rate at which the gravitational field of any given mass decreases. This suggests that the photon's coordinate velocity is controlled by the smooth, homogeneous influence of distant masses, rather than by the lumpy, less homogeneous influence of nearby masses with their non-zero peculiar velocities. Sounds like Mach's Principle at work.

Contrast the situation in a Milne model where there is no gravity. There the photon doesn't need to ever change its coordinate velocity (relative to any single comoving coordinate origin, such as the emitter or observer) because there is a single global inertial frame.

I recognize that the 'expanding hypersphere' paradigm moots this inquiry by assuming that local regions become progressively more distant from each other without actually moving through local space, so the permittivity of the vacuum could be said to entirely control the coordinate velocity of light without help from gravity. So my inquiry probably applies only in the context of the kinematic paradigm.

Jon