# Why "expanding space"?

by Peter Watkins
Tags: expanding space
P: 294
 Quote by Ich But no need to calculate, just imagine the process to get a different view on cosmological redshift.
No, I don't want to just "imagine the process" Bunn & Hogg use, because I'm convinced it's wrong. It is wrong to use accumulated SR Doppler shifts to calculate cosmological redshift, because SR Doppler shift includes an element of SR time dilation which has no place in the light path between a clock-synchronized emitter and observer in the FRW model. It is also wrong to use accumulated non-SR classical Doppler shifts for the same purpose, because this approach cannot be demonstrated to yield even a remotely correct numerical result.
 Quote by Ich Classical Doppler shift depends on absolute speed, or your speed through the medium. Here, for each state of motion, you are supposed to be at rest wrt "the medium".
If one is at rest wrt the medium in each adjacent observer's small local space along a light path, then each such observer must measure 0 classical Doppler shift. An infinite accumulation of 0's is 0, clearly an invalid result. That approach is a dead end.
 Quote by Ich Take an empty spacetime (age: 13.7 GY), as there exists an relatively easy unambiguous answer. Flat speed now: 0.999969 cosmological speed now: 255 Flat distance now: 13.69958 GLY cosmological distance now: 3493.5 GLY If you're going to stop the expansion now (cosmological time), the flat distance (now the only sensible one) would be 10^110 GLY.
I don't follow what your definition of "cosmological speed" and "cosmological distance" are in this context. Your empty spacetime has no gravity, so no two observers should be able to have recession velocities > c with respect to each other, regardless of how you parse it.
 Quote by Ich Just have a look at the Davis&Lineweaver paper, a famous one, a good one, and often cited. Section 4.2: horribly uninformed and wrong. These are supposed to be educational papers.
What specifically is your problem with the cited passage?
P: 1,883
 No, I don't want to ...etc.
You're going in circles. Comoving observers still are not synchronized, time dilatation still can be ignored, and the approch with classical redshifts still yields correct results, as sketched in #37.
 If one is at rest wrt the medium in each adjacent observer's small local space along a light path, then each such observer must measure 0 classical Doppler shift. An infinite accumulation of 0's is 0, clearly an invalid result. That approach is a dead end.
That was hard work, quoting out of context to make sure you can deliberately misunderstand the rest. Or did you really not understand what I was saying? If so, I apologize.
 I don't follow what your definition of "cosmological speed" and "cosmological distance" are in this context.
Today, I can't follow neither, because I forgot the logarithm yesterday. The (hopefully) correct values are:
Flat speed now: 0.999969
cosmological speed now: 5.545
Flat distance now: 13.69958 GLY
cosmological distance now: 76 GLY
If you're going to stop the expansion now (cosmological time), the flat distance (now the only sensible one) would be 1754 GLY.
 Your empty spacetime has no gravity, so no two observers should be able to have recession velocities > c with respect to each other, regardless of how you parse it.
You're getting close. Go to Ned Wrights calculator and enter zero density everywhere. See what happens, and click on "comoving radial distance" for an explanation.
 What specifically is your problem with the cited passage?
Davis & Lineweaver fail to correctly state the SR-only case. That's obviously the empty universe with H~1/t (not H=const.), and it's excluded by SN data alone by 3 standard deviations, not 28 or whatever they claim.
P: 1,883
 Quote by Old Smuggler The issue we are discussing is whether or not spectral shifts observed between comoving observers in any FRW model can reasonably be interpreted as due to motion in flat space-time for small distances/times.
Exactly.
 Quote by Ich This specific "model" is nothing but a differet coordinate representation of a specific (the empty) FRW solution. Nothing wrong with it.
Yes, for the empty FRW model that representation is equivalent - but it gives you the false impression
that you can use such an representation approximately for all FRW models to correctly decide the issue we are discussing.
Actually, I can use the model to first order precision. I'll explain the procedure later.
 However, if you try to represent any FRW model as Minkowski space-time foliated by flat hypersurfaces, the world lines of the particles representing the expansion will not in general coincide with the comoving observers' world lines.
This is a point where I brought confusion into the discussion by referring to "flat" approximations. The approximation I have in mind neglects second order terms and thus describes a flat spacetime. But that does not mean that it uses a foliation where space is flat, it simply is insensitive to curvature.
 All I am saying is that any spectral shift can reasonably be interpreted as a Doppler shift in flat space-time only if this shift is also present in the tangent space-time. That is, take the 4-velocity of the emitter and parallel transport it along the null curve to a nearby receiver. Calculate the spectral shift. Do the same procedure in the tangent space-time. If the spectral shifts coincide to the relevant accuracy, the shift can reasonably be interpreted as a Doppler shift in flat space-time. If not, it cannot.
I agree.
 Again an illustrating example is a FRW model with flat space sections. What you really do here, is to transform the space-time curvature of the FRW model into a velocity field in Minkowski space-time.
No, I transform explicitly time-dependent coordinates to coordinates with the standard minkowski interval. That has nothing to do with spacetime curvature. It's just a coordinate transformation accompanied by a linearization. In the new coordinates, comoving observers have a definite velocity. If this had to do with curvature, the linearization you eliminate these velocities.
Specificly, here's the procedure:
I start with the FRW metric ds² = dt² - a(t)²dr². Depending on the details of the spacetime and the transformation I use, the other two space dimensions deviate from flat space in second order. that doesn't bother me, I'm after first order effects only.
Now, at a specific epoch t0, I can linearize the funktion a(t) by setting a(t)=const. * (t-t0'), where
$$(t_0-t_0')=1/H_0=a/ \dot a$$
and the constant ensures that a(t0)=1.
Now that a(t) is linear, I can get rid of it by the same transformations that bring the empty FRW coordinates to Minkowski coordinates, i.e.
$$t_{FRW} = \sqrt{t_{mink}^2 - x^2}$$
$$r = 1/H_0 \tanh^{-1}(x/t_{mink})$$
In these standard coordinates, comoving observers have the claimed velocities. That works because these velocities are proportional to $$\dot a$$ and independent of $$\ddot a$$. They are not a curvature effect.
P: 87
 Quote by Ich Specificly, here's the procedure: I start with the FRW metric ds² = dt² - a(t)²dr². Depending on the details of the spacetime and the transformation I use, the other two space dimensions deviate from flat space in second order. that doesn't bother me, I'm after first order effects only. Now, at a specific epoch t0, I can linearize the funktion a(t) by setting a(t)=const. * (t-t0'), where $$(t_0-t_0')=1/H_0=a/ \dot a$$ and the constant ensures that a(t0)=1. Now that a(t) is linear, I can get rid of it by the same transformations that bring the empty FRW coordinates to Minkowski coordinates, i.e. $$t_{FRW} = \sqrt{t_{mink}^2 - x^2}$$ $$r = 1/H_0 \tanh^{-1}(x/t_{mink})$$ In these standard coordinates, comoving observers have the claimed velocities. That works because these velocities are proportional to $$\dot a$$ and independent of $$\ddot a$$. They are not a curvature effect.
You have assumed that the effects of curvature must be in $$\ddot a$$ or higher order terms
and that $$\dot a$$ is always independent of curvature effects. That assumption is quite wrong.
The point is that the affine connection is curved in general. This means that the Riemann tensor is non-zero, but also that there are curvature effects via non-zero connection coefficients.
For a flat connection, the non-zero connection coefficients can be eliminated via a suitable coordinate
transformation. That cannot be done (globally) in a curved manifold.

To illustrate this point; for a flat FRW model the time-dependent connection coefficients are proportional to $$\dot a$$. The only difference between the line elements of Minkowski space-time on the one hand and of the flat FRW model on the other (using standard coordinates), is the presence of a time-dependent scale factor. Yet the latter line element yields non-zero connection coefficients proportional to $$\dot a$$. The coordinate systems used are the same, so the non-zero connection coefficients cannot be blamed on a coordinate effect in flat space-time. The only reasonable explanation is that the non-zero connection coefficients (and thus $$\dot a$$) comes from curvature. This means that your assumption that $$\dot a$$ is independent of curvature effects is incorrect.

If you do not agree with this, we should agree to disagree.

I see that I wrote somewhere that the interpretation of spectral shifts in a non-empty, open FRW
model can always be interpreted as motion in flat space-time for small times/distances. In light of
my subsequent posts this view would be wrong - for a non-empty open FRW model the spectral shift
should be interpreted as a mix of curvature effects and velocity in flat space-time. Also some
comments on the initial-value problem for open FRW models were a bit misleading. Otherwise,
what I have written in this discussion should be reasonably correct (except some minor nitpicks).

Anyway, since it is now quite clear at what points we disagree, we should round off this discussion.

By the way, it's All April Fool's Day today. Do you consider yourself fooled?
P: 1,717
 Quote by Ich Davis & Lineweaver fail to correctly state the SR-only case. That's obviously the empty universe with H~1/t (not H=const.), and it's excluded by SN data alone by 3 standard deviations, not 28 or whatever they claim.
Say what? The SR only case is that redshifts are a function of velocity using the Doppler shift. A major difference between SR models of the redshift and expansion models is that an assumption of homogeneity has different implications for observations from deep space.

I suspect you are mixing up definitions of H. In conventional GR based cosmology, H can be defined as (da/dt)/a where a is the scale factor (wrt to present) and t is a proper time co-ordinate. H~1/t is the nice simple solution you get for an empty universe using GR.

In SR, you can't define H in terms of a scale factor, because SR doesn't have expansion of space. However, you can assume that particles with recession velocity v are at a distance proportional to v. That is, we all started out close together and have been moving apart at constant velocities. The particle with redshift z has recession velocity [(1+z)^2-1]/[(1+z)^2+1].c, and it is at distance given by recession velocity by some constant time T by its recession velocity, on the assumption that everything started out from close together; that is, v is proportional to distance. H can be defined as the relationship between distance and v.

Now of course, under this assumption, the value of "H" for an observer at different times is proportional to 1/t. We can't test that, because we can't take observations billions of years apart in time.

However -- and THIS is the key point you seem to be missing -- H is defined here as a common feature of all observations, no matter how distant they may be. In GR the function H is a function of proper time, and so when you look into deep space you are seeing things when H was larger. Given information about time between events in deep space (SN data, for example) you can put strong constraints on the development of the scale factor over time. That is, there is a function from z to age, and from age to the scale factor, and from that to a value for H which was in play at the time the photon left whatever we are observing.

But in the SR model, H is a description of the observation, and it is identical for every particle we observe. When we look at distant particles, we are looking back in time, but the H is the same for all those particles. THAT's what is meant by constant H, I am pretty sure.

Davis and Lineweaver is excellent as an educational tutorial, helping to clear up all kinds of common popular misconceptions. It's perfectly normal to think they've done something wrong; and this is precisely because they tackle popular and entrenched misconceptions. If you think that they have made a mistake, you are probably in a good position to be learning something.

Cheers -- Sylas
P: 1,883
 You have assumed that the effects of curvature must be in$$\ddot a$$ or higher order terms and that $$\dot a$$is always independent of curvature effects. That assumption is quite wrong.
No. I approximated the scalar function a(t) by its tangent at the point of interest, therefore there are no first order deviations. That's not an assumption, that's basic calculus. All deviations are of second order in cosmological time, therefore at most second order also in private time and private space.
 The point is that the affine connection is curved in general. This means that the Riemann tensor is non-zero, but also that there are curvature effects via non-zero connection coefficients.
That's not a point, as these are second order contribuions.
 For a flat connection, the non-zero connection coefficients can be eliminated via a suitable coordinate transformation. That cannot be done (globally) in a curved manifold.
So what? It can be done locally, and that's what we are talking about. More to the point, I actually showed how it is done locally, so unless you're objecting to specific points in the transformation, there's no use telling me that curved space is not globally flat. I know this.
But you should know also that, in suitable coordinates, spacetime can be, locally and to first order, approximated by flat minkowski spacetime with zero connection coefficients (to be sure: first order). You simply have to find the correct local tranformation, and then show that lines of constant r have the appropriate velocity in these coordinates. That's what I have done, maybe you should try also.
 The only difference between the line elements of Minkowski space-time on the one hand and of the flat FRW model on the other (using standard coordinates), is the presence of a time-dependent scale factor. Yet the latter line element yields non-zero connection coefficients proportional to LaTeX Code: \\dot a . The coordinate systems used are the same, so the non-zero connection coefficients cannot be blamed on a coordinate effect in flat space-time.
Wow, the line element is different, but the coordinates are the same. Now that's interesting.
And what does "flat FRW model" mean? The empty one? One with flat space?
 If you do not agree with this, we should agree to disagree.
Yeah, I also get more and more the impression that this discussions makes no sense.

 I see that I wrote somewhere that the interpretation of spectral shifts in a non-empty, open FRW model can always be interpreted as motion in flat space-time for small times/distances. In light of my subsequent posts this view would be wrong
In light of your previous writings, which were consistent with basic premises of GR (equivalence principle), observations (we actually see nearby galaxies receding), and the paper we're discussing, you should re-think this statment:
 what I have written in this discussion should be reasonably correct (except some minor nitpicks).
because now you're struggling with all three points.
 By the way, it's All April Fool's Day today. Do you consider yourself fooled?
I admit, I had the impression already back in March. Someone knowledgeable denying either the existence of local inertial frames or Hubble's law (you see, it's a first order effect in the local frame), maybe there's some spacetime fooliation going on.
Anyway, it was fun.

cheers
Ich
P: 1,883
 Quote by sylas I suspect you are mixing up definitions of H. In conventional GR based cosmology, H can be defined as (da/dt)/a where a is the scale factor (wrt to present) and t is a proper time co-ordinate. H~1/t is the nice simple solution you get for an empty universe using GR.
With all respect, I suspect that you're missing a crucial point: The empty expanding universe is a valid FRW solution, with $$\Omega_{\Lambda} = \Omega_M = 0$$, and it's only 3 sigma away from LCDM. Empty spacetime is flat. SR can handle a flat spacetime, you simply have to use a different set of coordinates. Predicted observations, such as redshift of test particles, are independent of the choice of coordinates.
Maybe you want to read what Ned Wright has to say, or you want to convince yourself.
$$ds^2=dT^2-T^2dr^2$$
and apply the transformations
$$T = \sqrt{t^2 - x^2}$$
$$r = T_0 \tanh^{-1}(x/t)$$
You'll get
$$ds^2=dt^2-dx^2$$
and you can perform the necessary calculations (redshift, luminositiy distance, angular size distance...) purely in SR.
 It's perfectly normal to think they've done something wrong; and this is precisely because they tackle popular and entrenched misconceptions. If you think that they have made a mistake, you are probably in a good position to be learning something.
Ha, that's what I'm telling crackpots all along.
Pease understand that I'm not trying to sell a pet theory of mine. Davis&Lineweavers' analysis contradicts textbook wisdom, you can convince yourself if you're familiar with th idea of a metric, you can read what other authorities in the field have to say. Or you can take the fact that even Old Smuggler, who disagrees generally with everything I say, agrees with me as evidence with the status of a mathematical proof.
Really, I'm not doing original research here, that chapter is simply wrong.
P: 1,717
 Quote by Ich With all respect, I suspect that you're missing a crucial point: The empty expanding universe is a valid FRW solution, with $$\Omega_{\Lambda} = \Omega_M = 0$$, and it's only 3 sigma away from LCDM. Empty spacetime is flat. SR can handle a flat spacetime, you simply have to use a different set of coordinates. Predicted observations, such as redshift of test particles, are independent of the choice of coordinates.
That is not what is meant by an SR model. I do know about the FRW solutions.

The SR model described in Davis and Lineweaver is the model obtained by taking redshift as due to motions in a simple non-expanding space, and calculated as Doppler shift.

That's DIFFERENT from the FRW solution with an empty universe.

There's no error in the Davis and Lineweaver paper on this point, because they are quite clear on what they mean by SR model. It's not just taking an FRW solution and applying SR. It's taking redshift as being a Doppler shift in non-expanding space.

The luminosity distance with z arising from Doppler shifts for particles receding with at uniform velocity from a common origin event is different from that in the empty FRW model.

Cheers -- Sylas
 Sci Advisor P: 1,250 I hope I can contribute here. I think you (sylas and ich) are both basically right. The empty FRW universe is indeed only 'ruled out' at 3 sigma, but as sylas suggests this is not the model D&L mean by saying 'SR model', they are referring to a particular assumption, valid at low redshift, that gives a bogus result at high redshift. The point that leads to disagreement is actually a bit subtle. In post #61 ich makes a conformal tranformation between the empty FRW metric and a Minkowski like metric. This is all well and good, however this is only valid radially. If you put the angular terms back into the first line you will see that your transformation does not return a fully conformally Minkowski metric. This means that you cannot use this to determine either the angular diameter or luminosities distances. You need to do a more complex fully conformal transformation in order to do this. Some technical details of this can be found here. I*think* that the error in the SR model the D&L discuss is that if you work through the details, you can see that that way we define distance in the SR model violates simultaneity, which is why it is okay for small distances but gets worse and worse the further you go. So yes, a *correct* SR model is identical to an empty FRW universe and to work out the relationship between the FRW co-ordinates and the co-ordinates of this model you need to do the fully conformal transfomation, but D&L are talking about a model that, due to the misidentification of the meaning of co-ordinates, is only a low redshift approximation. I hope that helps!
P: 1,883
 The SR model described in Davis and Lineweaver is the model obtained by taking redshift as due to motions in a simple non-expanding space, and calculated as Doppler shift.
No, the SR model they use is, frankly, BS. Read this:
 Quote by D&L However, since SR does not provide a technique for incorporating acceleration into our calculations for the expansion of the Universe, the best we can do is assume that the recession velocity, and thus Hubble’s constant, are approximately the same at the time of emission as they are now6.
A non accelerating universe has $$\dot a = const.$$, thus H=1/t. Constant H is the de Sitter universe, nothing to do with SR, and the extreme case of an accelerating universe.
 There's no error in the Davis and Lineweaver paper on this point, because they are quite clear on what they mean by SR model. It's not just taking an FRW solution and applying SR. It's taking redshift as being a Doppler shift in non-expanding space.
Please try to understand my point: in the empty universe, the only difference between "expanding space" and "constant velocity in non-expanding space" is a coordinate transformation. That's nothing more than just taking an FRW solution and applying SR.
 The luminosity distance with z arising from Doppler shifts for particles receding with at uniform velocity from a common origin event is different from that in the empty FRW model.
No, if "uniform velocity" means what it should for particles receding from a common origin.
Now this is your claim, please back it up with calculations. You are probably in a good position to be learning something.
P: 1,250
 Quote by Ich A non accelerating universe has $$\dot a = const.$$, thus H=1/t. Constant H is the de Sitter universe, nothing to do with SR, and the extreme case of an accelerating universe.
Hmm, good point. I guess the best we can say then is that D&L introduce a very bad model and then demonstrate that it doesn't fit the data. I'm not sure that they intended it to be a 'correct' model in the sense of it correctly using relativity (SR and GR are of course identical if the universe is empty), I think they were trying to show that a mugs 'SR' model doesn't work, but maybe it was a bit too muggy.

I really think the points of agreement are much more than those of disagreement here, stemming from maybe some loose terminology. I think we can all agree that the 23 sigma model from D&L is not a 'correct' SR model. The disagreement appears to be just how incorrect it is, yes?
P: 1,883
 Quote by Wallace In post #61 ich makes a conformal tranformation between the empty FRW metric and a Minkowski like metric.
Stop, no! I just match coordinates locally to first order, and drop all the higher order terms. It's neither valid radially nor in the transverse directions, if you're looking at higher orders.
What I'm doing here is an exact coordinate transformation. The angular directions (hyperbolic to flat space) transform correctly, no need to bend the laws of physics. We're talking about a flat spacetime in both cases.
 This means that you cannot use this to determine either the angular diameter or luminosities distances.
Of course you can. The hyperbolic space in FRW coordinates stems solely from the definition of the radial coordinate as being measured by comoving observers. If you "fix" that, everything is ok again.
 but D&L are talking about a model that, due to the misidentification of the meaning of co-ordinates, is only a low redshift approximation.
Yes, they talk about the wrong model and therefore come to wrong conclusions. I think this is most clearly seen in the passage I quoted before, where they identify "not accelerated" with "constant H", which is bogus.
P: 1,250
Alright, I don't want to introduce additional disagreement. As you say, minkowski space and an empty FRW metric are both flat space-times (they have a vanishing Ricci scalar). You can transform between these two co-ordinate systems, without being forced to be vaild only to a given order, via a fully conformal transformation.
 Of course you can. The hyperbolic space in FRW coordinates stems solely from the definition of the radial coordinate as being measured by comoving observers. If you "fix" that, everything is ok again.
Right, this 'fixing' is exactly what the transformation does.

I get what you are saying, any co-ordinate transformation is exact, so if your original space-time is flat the transformed one is as well. Just pointing out that the one you suggest doesn't work, on it's own to relate FRW co-moving co-ordinates to their Minkowski counterparts. Clearly you agree with this point, it just wasn't clear to me what you were demonstrating with it original, but now I see.
P: 1,883
 I guess the best we can say then is that D&L introduce a very bad model and then demonstrate that it doesn't fit the data.
Yes, it's a strawman.
 I'm not sure that they intended it to be a 'correct' model
They say "the best we can do", so I'd say that they simply didn't know better. I'm convinced that this section would look quite different if they'd write it today.
 The disagreement appears to be just how incorrect it is, yes?
Of course. D&L claim incorrectly that the "Doppler/SR interpretation" is ruled out by 23 sigma by SNIa observations alone, I (we) say it's ruled out by ~3 sigma. Taking other observations into account, I think we're rather getting back to 23 sigma.
P: 87
For the benefit of readers who may possibly fall for the misunderstandings Ich seems to be promoting,
I will contribute with one last post in this discussion.
 Quote by Ich No. I approximated the scalar function a(t) by its tangent at the point of interest, therefore there are no first order deviations. That's not an assumption, that's basic calculus. All deviations are of second order in cosmological time, therefore at most second order also in private time and private space.
Your "linearisation procedure" is, with a suitable choice of constants, equivalent to expanding a(t) as a truncated Taylor series around some arbitrary time t_0. That is, you set a(t) = a(t_0) + $$\dot a$$(t_0)(t-t_0) and neglect higher order terms. But then you assume that no space-time curvature effects are included since the series is terminated after the linear term. This is not necessarily true since curvature effects
may be included into $$\dot a$$(t_0) (as is the case, in general). This simple misunderstanding may be appropriately called "the Bunn/Hogg fallacy", and you have endorsed it.
 Quote by Ich So what? It can be done locally, and that's what we are talking about. More to the point, I actually showed how it is done locally, so unless you're objecting to specific points in the transformation, there's no use telling me that curved space is not globally flat. I know this.
The problem is that you do not do what you think you do. Take again the FRW model with flat space
sections. The non-zero connection coefficients are proportional to $$\dot a$$, as usual. But here, since
we cannot perform any relevant coordinate transformation in order to change the connection coefficients
(the coordinates already have the standard form), the correct flat space-time approximation is to neglect $$\dot a$$ altogether. On the other hand, for a non-empty, open FRW model where
the line element is expressed in comoving coordinates, a coordinate transformation to standard
coordinates will change the connection coefficients, but not get rid of them altogether. What is left
should be due to curvature and must be neglected in the correct flat space-time approximation. It
is only for the empty FRW model a coordinate transformation from comoving to standard coordinates
can completely get rid of all the connection coefficients.

On the other hand, approximating a(t) as a Taylor series to first order the way you do, is effectively to include all the crucial effects of the connection coefficients (expressed in comoving coordinates) at the time t_0, since the relevant connection coefficients expressed in comoving coordinates are always proportional to $$\dot a$$. After making a local transformation to standard coordinates, the resulting non-zero velocity field is then just an expression of the fact that the connection coefficients (expressed in comoving coordinates) at the time t_0, are proportional to $$\dot a$$(t_0). You have absolutely no guarantee that these connection coefficients do not include some effects of curvature so that this procedure yields the correct flat space-time approximation for the issue we were discussing. In fact, it fails.
 Quote by Ich Wow, the line element is different, but the coordinates are the same. Now that's interesting.
You think so? Of course you can keep the coordinate system and change the metric as long as
the coordinate system covers the relevant part of the manifold. That is basic differential geometry.
You should try to learn it some time.
 Quote by Ich And what does "flat FRW model" mean? The empty one? One with flat space?
I have consistently used "flat FRW model' to mean the FRW model with flat space sections.

That concludes all I have to say in this discussion. You are on your own now. Good luck.
P: 1,717
 Quote by Ich No, if "uniform velocity" means what it should for particles receding from a common origin. Now this is your claim, please back it up with calculations. You are probably in a good position to be learning something.
There's no question about that! I'm sure most of you guys here know more than I do about GR, and metrics and tensors. I learn a lot by trying to work through these kinds of problems.

In any case, I'll go away and try my own analysis, and report back.

Cheers -- Sylas
P: 1,717
 Quote by sylas In any case, I'll go away and try my own analysis, and report back. Cheers -- Sylas
OK; I have now done this more thoroughly for myself as you suggest. You're right; and I was wrong. In fact, the luminosity distance in the SR case is the same as obtained in the FRW model with an empty universe, and the SR model used in section 4.2 of Davis and Lineweaver has no sensible correspondence to anything. It is, as you point out, nonsense.

I'm not an expert in GR; I can solve the differential equations for scale factor and energy density which are used in the FRW models; but I can't derive the equations themselves. In any case, I didn't need any of that, because the issue is simply the SR model.

The SR model corresponds to a realistic situation that could, in principle, be set up and tested right now, and SR is the appropriate way to analyze it.

Take a large collection of particles, and at a point in time, have them all start moving at constant velocity from a common point. (An explosion in space.) After elapsed time t, an observer on one of the particles makes observations of all the others.

Consider a signal received by one exploding particle from another, and compare with the signal from another equivalent particle at the same distance, but with no velocity difference. The signal received from the moving particle is weaker by a characteristic amount. The factors to consider are
• Redshift. Each photon arrives with less energy, by a factor (1+z).
• Time between photons. The time between successive photons is increased by precisely the same factor as the distance between wave crests. Think of a radiator sending out pulses of radiation, according to an onboard clock. The individual photons are redshifted. The frequency at which pulses of radiation arrives is reduced also, by the same factor. This reduces the energy by another (1+z).
• Angular size of the radiating surface. This is unchanged. There is no Lorentz contraction perpendicular to the direction of motion, so the stationary particle and the moving particle subtend the same angle at the same distance.
Hence, the signal received from the moving particle is weaker than a signal from the stationary particle at the same distance, by a factor $(1+z)^2$. Equivalently, the angular distance is less than the luminosity distance by this factor.

But that is precisely the relation for all the FRW models, empty or otherwise. Davis and Lineweaver, in their section 4.2, used a factor of (1+z) for the so-called SR model, which can only be seen as an error. There are still differences in comparing z with the apparent magnitude across the different FRW solutions, but the ratio of angular distance and luminosity distance is the same for everything.

Using Ned's formulae for the empty universe, I get the angular distance as follows:
$$D_A = \frac{c}{H_0}(1-(1+z)^{-2})/2$$

Using Lorentz transformations for the SR model I have described here, and using H0 as the inverse of time since the explosion, which makes sense, I get the same thing. Hence the SR model gives the same relation between z and luminosity distance as the empty FRW solution.

Thanks very much. I have learned something indeed.

Cheers -- Sylas