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Hubble's law

by gptejms
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Tzemach
#55
May8-06, 09:42 AM
P: 80
Garth thank you for your help and the links, this has helped me sort out a couple of questions. I think your work is great, when I digest it all and do some more work I might have a few more clues.
gptejms
#56
May9-06, 11:02 AM
P: 343
Quote Quote by Garth
Yes, the behaviour of geodesics, described by R(t), are determined by the solution of the Einstein field equation. A gravitational field where all the components of the Riemannian are zero has a characteristic behaviour, just as when it has non-zero components.

What it is saying is that, given the universe is expanding, if there is no source of gravitation i.e. no gravitational forces, then the universe expands linearly, i.e. there is 'nothing to slow it down'. (when the cosmological constant [itex]\Lambda = 0[/itex])
'Taking the limit [tex] \rho [/tex] tending to zero(empty universe) starting with a universe with matter' is not the same as 'starting with non-gravitational bodies and then doing the calculations'.It could well be that the universe needs a trigger to start expanding and when that trigger is withdrawn([tex] \rho\rightarrow 0 [/tex]) it doesen't stop expanding:---it has an inertia of motion....or let me say inertia of expansion.

I googled for Milne cosmology today and found an interesting paper 'An interpretation of Milne Cosmology' by Alasdair Macleod.You'll see from the paper that Milne didn't like the idea of spacetime curvature.He tried to give an explanation of Hubble's law from SR.It needs a very special(and unlikely) initial condition for it to work.[tex] \rho [/tex] tending to zero is also called as Milne cosmology though it's a misnomer and not the original idea.

This author also argues that Milne and empty universe models are not mathematically equivalent.
Garth
#57
May9-06, 04:07 PM
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Quote Quote by gptejms
'Taking the limit [tex] \rho [/tex] tending to zero(empty universe) starting with a universe with matter' is not the same as 'starting with non-gravitational bodies and then doing the calculations'.It could well be that the universe needs a trigger to start expanding and when that trigger is withdrawn([tex] \rho\rightarrow 0 [/tex]) it doesen't stop expanding:---it has an inertia of motion....or let me say inertia of expansion.

I googled for Milne cosmology today and found an interesting paper 'An interpretation of Milne Cosmology' by Alasdair Macleod.You'll see from the paper that Milne didn't like the idea of spacetime curvature.He tried to give an explanation of Hubble's law from SR.It needs a very special(and unlikely) initial condition for it to work.[tex] \rho [/tex] tending to zero is also called as Milne cosmology though it's a misnomer and not the original idea.

This author also argues that Milne and empty universe models are not mathematically equivalent.
Milne had his own cosmological theory called "Kinematic Relativity" in which space-time is flat, do not confuse that with his name being attached to the empty GR curved space-time model, which is what we are talking about here.

As far as your first comment is concerned, you have to be clear about whether you are discussing SR or GR.

In SR if non-gravitating bodies are mutually receeding then the red shift observed is due to their motions through (Minkowski) flat space-time and is relativistic doppler shift.

In GR the bodies are normally assumed to be at rest in space-time (co-moving particles) and it is space-time itself that expands and carries the bodies with it. The red-shift observed is cosmological red shift as I have discussed above.

I have been considering GR cosmological red shift all along.

Garth
gptejms
#58
May10-06, 11:12 AM
P: 343
If you consider non-gravitating bodies(to start with),then of course you are in the SR domain(i.e. the original Milne cosmology) and you have to use the Doppler effect(as was done by Milne).What I argued in the posts above was that even in this case the Doppler effect may be traced back(as an alternative explanation) to the 'acceleration phase'.So even though we are not talking of cosmological red shift here, the red shift may be thought of as either due to the 'expansion'(where this is not cosmological expansion but 'expansion' of the universe due to bodies moving away from one another) or due to a gravitational effect(by equivalence principle the 'acceleration phase' may be thought of as a phase of gravity coming into play).So the 'two statements' that you have been claiming are equivalent(for the cosmological case) are equivalent even for this case.

Now coming to Milne cosmology in the (GR) sense i.e. [tex] \rho\rightarrow 0[/tex](i.e. the sense that you have been talking),let me first say that this is not a model for non-gravitating bodies moving away from one another.This is a model for(to start with) gravitating bodies moving away from one another for which gravity has become extremely weak or negligible at some stage---such a universe has an inertia of expansion and keeps expanding.At the stage when gravity is switched off,I do not see the equivalence of the 'two statements' holding any longer---now the only way to explain the red shift is the cosmological expansion;there is no cosmological gravitational field.
Garth
#59
May10-06, 12:28 PM
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You have to define your terms very carefully and state exactly how they are measured.

We were discussing the term "cosmological expansion", there is no such thing in SR, neither is there an "'acceleration phase'", these can only be discussed in GR.

As I have said, the surprising thing in GR is that the empty universe expands - linearly so for there is nothing to slow it down.

Test (infinitesimal mass) particles in such a universe are carried along with the linearly expanding space and exhibit mutual red shift, that is cosmological red shift, which depends on their position in space, rather than doppler red shift, which depends on their velocity through space.

To any observer the red shift of the other test particles may be attributed to the expansion of the universe or to the universe's (null) gravitational field, that is, the geometry of its space-time. They are two different ways of interpreting the same effect.

Perhaps where there is confusion is over the GR concept that co-moving objects in an expanding universe are not moving through space, but rather are been carried along by the expanding space in which they are embedded.

Garth
gptejms
#60
May10-06, 10:44 PM
P: 343
Quote Quote by Garth

We were discussing the term "cosmological expansion", there is no such thing in SR, neither is there an "'acceleration phase'", these can only be discussed in GR.
Agreed.The only thing I've been saying in the above posts is that 'the two statements' are equivalent even for the original Milne cosmology(in the sense that I've mentioned above(see post #58)).

As I have said, the surprising thing in GR is that the empty universe expands - linearly so for there is nothing to slow it down.

Test (infinitesimal mass) particles in such a universe are carried along with the linearly expanding space and exhibit mutual red shift, that is cosmological red shift, which depends on their position in space, rather than doppler red shift, which depends on their velocity through space.
Right.

To any observer the red shift of the other test particles may be attributed to the expansion of the universe or to the universe's (null) gravitational field, that is, the geometry of its space-time. They are two different ways of interpreting the same effect.
Here,there's a difference in our perspectives.If you look at my post no. 1,I say that there are two sources of red shift---cosmological expansion &. clocks getting faster due to decreasing (overall)gravitational field.You said these are one and the same thing.
Now coming to the [tex] \rho \rightarrow 0 [/tex] case,clocks ain't getting any faster due to a (null) gravitational field(!)---so how are the two statements equivalent?I hope I am able to convey my point.When I talk of a gravitational field, I am looking for clocks getting faster(or slower)--there is no such thing in the case of a (null) gravitational field.So the two statements are not equivalent 'in the sense' that I have been talking.

The way you are making them equivalent is an equivalence that exists per se---it does no value addition.It's not the same as the equivalence that I have been talking about.

Perhaps where there is confusion is over the GR concept that co-moving objects in an expanding universe are not moving through space, but rather are been carried along by the expanding space in which they are embedded.
There is no such confusion.
Garth
#61
May11-06, 02:11 AM
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Quote Quote by gptejms
There is no such confusion
BTW A Very Happy Birthday!

If there is no confusion and we agree that in cosmological red shift objects are treated as if they had no proper motion through space, only that of the expanding space in which they are embedded. Then the observed red shift can only come from the evolution of the scale factor with time.

That scale factor, R(t), and the curvature factor, [itex](1 - kr^2)^{-1/2}[/itex], determined by the average density, are the only descriptions of the "cosmological gravitational field" in the cosmological R-W metric.

Therefore in the Milne universe (where k = -1), the observed red shift is due to the scale factor, R(t) ~ t, i.e. the null gravitational field. (See d'Inverno "Introducing Einstein's Relativity" pages 324-5 for a derivation of the cosmological red shift)

I think it is confusing that this cosmological red shift is also called doppler shift when the objects are not moving through space.

Garth
gptejms
#62
May11-06, 01:16 PM
P: 343
Thanks, Garth!
ratfink
#63
May11-06, 03:19 PM
P: 58
Isn't this a bit strong
If there is no confusion and we agree that in cosmological red shift objects are treated as if they had no proper motion through space, only that of the expanding space in which they are embedded. Then the observed red shift can only come from the evolution of the scale factor with time.
(bolding etc is from me)
Just a thought, my understanding is that the Doppler interpretation is abandoned because if the redshift was due to proper motion then galaxies would distort as they approach the speed of light.
They don't ergo we say that it is space expanding carrying the galaxies along with it.
1) is this true?
2) Is there any test that we can carry out to differentiate between the two scenarios?
Ratfink typing with slapped wrists
Garth
#64
May11-06, 05:48 PM
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Hi ratfink.
If you think "Then the observed red shift can only come from the evolution of the scale factor with time." is a bit 'strong', then where else might it come from?

Note: we are in the field of GR, which is a theory that may not be the last word on the subject and might in future be adapted or modified, however, while we are understanding cosmology under the standard GR paradigm then we are accepting that cosmological red shift is due to the expansion of space itself rather than the motions of galaxies etc. through space.

As in GR we understand the universe to be expanding and the proper distance to these distant objects increasing with time then it is understandable that the observed red shift is described as Doppler shift. All I am saying in my posts above is that this is the same phenomenon as that described by the expression "evolution of the gravitational field".

Garth
gptejms
#65
May18-06, 02:01 AM
P: 343
I have a question on CMB.Due to the expansion of the universe there should be two sources of cooling of background radiation--1.)adiabatic expansion like that of any gas,2.)stretching of wavelengths due to cosmological expansion of space.Is the latter effect taken into account in calculations?
gptejms
#66
May19-06, 12:57 PM
P: 343
Okay,I see from Ned Wright's cosmology tutorial that the latter effect is indeed taken care of.Now my question is about the first effect--is that taken care of?
Garth
#67
May19-06, 01:51 PM
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Quote Quote by gptejms
Okay,I see from Ned Wright's cosmology tutorial that the latter effect is indeed taken care of.Now my question is about the first effect--is that taken care of?
There are two temperatures referred to in cosmology, the black body temperature of the CMB, which is at the present time equal to 2.760K, and the temperature of matter, whcih can be anything today above 2.760K. (There may be super-cooled gas around below this temperature in which case please let me have any links to published papers)

The temperature of the CMB is slowly decreasing because of the "stretching of wavelengths due to cosmological expansion of space". [itex]T \propto R(t)^{-1}[/itex].

When the universe was ionised the plasma within it was heated to the temperature of the then temperature of the CMB by photon-particle interactions.

Once the universe became transparent the radiation cooled adiabatically until other processes, such as it forming dense halos, changed the physical conditions, the matter temperature evolution then became a bit messy!

I think SpaceTiger is the expert in this regime!

Garth
hellfire
#68
May19-06, 03:01 PM
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Quote Quote by gptejms
I have a question on CMB.Due to the expansion of the universe there should be two sources of cooling of background radiation--1.)adiabatic expansion like that of any gas,2.)stretching of wavelengths due to cosmological expansion of space.Is the latter effect taken into account in calculations?
When a (relativistic) photon gas cools adiabatically, the dependence of its energy with volume is [itex]E \propto V^{-1/3}[/itex]. This is equivalent to [itex]E \propto 1/a[/itex] with a scale factor. Since [itex]E = h \nu[/itex] and [itex]\nu \propto 1/a[/itex] in an expanding space, this energy loss in an comoving (expanding) volume is due to the decrease of the frequency. As far as I know this accounts for all the energy loss and one has not to consider anything additionally.
gptejms
#69
May20-06, 05:02 AM
P: 343
I think both the effects mentioned in post #65 can be taken into account.Consider an ideal bose gas(background radiation) undergoing cosmological expansion.Internal energy of a bose gas per unit volume, [tex]U/V \sim T^4 [/tex],where T is the temperature.Now consider V increasing due to (cosmological) expansion.[tex]V\sim R^3[/tex],so it would seem that T goes as [tex]R^\frac{-3}{4}[/tex].

But since every photon in the background radiation is red shifted by an equal factor,the (total) internal energy U also goes down by the same factor i.e. U goes as 1/R.So one can see that the temperature T goes as 1/R rather than [tex]R^\frac{-3}{4}[/tex].

There could be loopholes in the argument(!) but because it's appealing I am reporting it as soon as I've thought of it.
gptejms
#70
May20-06, 02:29 PM
P: 343
The above argument is obviously wrong.Temperature can not depend on volume,so all the relation means is that for a gas at temperature T,larger volume means larger internal energy.In fact it sheds no light on how the temperature varies with increasing size of the universe.

Now assuming that temperature T~1/R (due to cosmological expansion),how does internal energy scale as a function of R?This is the only valid question to ask.Does the above relation answer this question?What is the answer?

Well, the answer is U ~ 1/R(!).See,there is a difference in what 'exactly' was said in the last post and this one.If in post #69 we were to 'if U goes as 1/R due to cosmological expansion then T goes as 1/R too' then it would be a correct statement.The present post says the reverse('if T goes as 1/R then U goes as 1/R).

So we have said nothing so far about cooling of the bose gas due to the effect of adiabatic expansion.Hope the cosmo tigers here have at least something to say on this.
hellfire
#71
May20-06, 03:23 PM
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If you apply the assumption of adiabatic expansion to a photon gas with [itex]P \propto u/3 = E/3V[/itex]...

[tex]dE + PdV = 0[/tex]

[tex]dE = - \frac{EdV}{3V}[/tex]

[tex]E \propto V^{-1/3}[/tex]

With [itex]V \propto R^3[/itex], this is:

[tex]E \propto R^{-1}[/tex]

For the internal energy per unit volume:

[tex]u \propto R^{-4}[/tex]

But I am confused. This follows merely from the assumption of adiabatic expansion, without taking into consideration the expansion of space. If a photon gas expands adiabatically in a piston of some characteristic lengt L (in static space), it will also increase its wavelengh, because [itex]E \propto L^{-1}[/itex] and [itex]E \propto \lambda^{-1}[/itex]. However, in an expanding space the same relation applies. Why?
Garth
#72
May20-06, 04:09 PM
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If the photon number remains constant there is no problem. Adiabatic expansion yields [itex]E \propto R^{-1}[/itex], which means the energy of each photon [itex]E_\nu \propto R^{-1}[/itex] that reveals itself as cosmological red shift.

The expansion of space dilution of each photon's energy is consistent with the time dilation cosmological red shift.

[itex]1 + z = R_0/R(t)[/itex]

Garth


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