CMB

[David Levy]

But this does not seem to address your misconception about clusters at all.

 

Yes it is.

Yes to the first two.

 

Thanks

However, do you agree that if the expansion is not uniform then the CMB can't be isotropic?

Edited January 13, 2016 by David Levy

[Mordred]

No your looking at the term homogeneous and isotropic wrong. Those terms apply for an extremely large scale. Roughly 100 Mpc. Yes locally to a cluster filament there is an inhomogeneous and anisotropic condition. However expansion is scaled at 100Mpc for a homogeneous and isotropic condition. The part your missing is the measurement scale to get a homogeneous and isotropic condition

 

Take this for example look at the waves on a lake with no current. At close scales their is no uniformity. However move far enough away from the surface the lake waves will be measurable with uniformity.

This is a standard technique in modelling complex systems.

 

At 100Mpc there is no discernable preferred direction. The universe as a whole is far far larger than the LSS filaments. Measure at a large enough scale you will see uniformity in its web like patterns and won't see a preferred direction.

 

I should note a large scale structure is considered a local measurement not a global

When you look at a CMB image the temperature variations between hot and cold spots is roughly 1/1000 degree of a Kelvin. That's extremely homogeneous and isotropic. So close to zero variation it's a near perfect blackbody

http://wwwmpa.mpa-garching.mpg.de/galform/virgo/millennium/seqB_019a_half.jpg

Here is an example at 500Mpc/h.

 

You can easily see 500 Mpc is a small portion of the universe. Large scale structures at this scale are miniscule.

 

The observable Universe is roughly 28 billion parsecs in diameter. You wouldn't be able to discern one filament from another at this scale. A LSS anisotropy would be so miniscule it would be negligible.

Edited January 13, 2016 by Mordred

[swansont]

Yes it is.

I'm afraid you don't get to decide that. Whether or not you have a misconception is judged by others.

 

However, do you agree that if the expansion is not uniform then the CMB can't be isotropic?

Moot, since the CMB has been measured to be uniform to a very high degree.

[Mordred]

From Wikipedia on the temperature variation. 2.72548±0.00057 K. Extremely uniform.

 

https://en.m.wikipedia.org/wiki/Cosmic_microwave_background

Edited January 13, 2016 by Mordred

[David Levy]

No your looking at the term homogeneous and isotropic wrong. Those terms apply for an extremely large scale. Roughly 100 Mpc. Yes locally to a cluster filament there is an inhomogeneous and anisotropic condition. However expansion is scaled at 100Mpc for a homogeneous and isotropic condition. The part your missing is the measurement scale to get a homogeneous and isotropic condition

 

Take this for example look at the waves on a lake with no current. At close scales their is no uniformity. However move far enough away from the surface the lake waves will be measurable with uniformity.

This is a standard technique in modelling complex systems.

 

At 100Mpc there is no discernable preferred direction. The universe as a whole is far far larger than the LSS filaments. Measure at a large enough scale you will see uniformity in its web like patterns and won't see a preferred direction.

 

I should note a large scale structure is considered a local measurement not a global

When you look at a CMB image the temperature variations between hot and cold spots is roughly 1/1000 degree of a Kelvin. That's extremely homogeneous and isotropic. So close to zero variation it's a near perfect blackbody

http://wwwmpa.mpa-garching.mpg.de/galform/virgo/millennium/seqB_019a_half.jpg

Here is an example at 500Mpc/h.

 

You can easily see 500 Mpc is a small portion of the universe. Large scale structures at this scale are miniscule.

 

The observable Universe is roughly 28 billion parsecs in diameter. You wouldn't be able to discern one filament from another at this scale. A LSS anisotropy would be so miniscule it would be negligible.

Thanks

We must distinguish between evidence and theory.

The CMB is isotropic and near perfect blackbody.

This is EVIDENCE by definition. We all should agree about it!

 

However, the expansion is a theory.

This theory should give an explanation for the measured CMB.

Somehow, I still see a paradox as follow:

 

You claim that in large scale the expansion is homogeneous and isotropic:

 

However expansion is scaled at 100Mpc for a homogeneous and isotropic condition. The part your missing is the measurement scale to get a homogeneous and isotropic condition

 

However, in the same token you confirm that in small/local scale the cluster is inhomogeneous and anisotropic:

 

Yes locally to a cluster filament there is an inhomogeneous and anisotropic condition.

 

So, if it was just inhomogeneous and anisotropic in our local cluster - then yes, we could claim clearly that this local cluster has no real effect on large scale.

 

Hence, your example about the waves on a lake could be perfect.

 

Never the less, the inhomogeneous and anisotropic is not just a unique condition in our local cluster.

In reality – any cluster in the whole universe is inhomogeneous and anisotropic by definition.

 

Therefore, at any direction that we look, we should see infinite (almost) inhomogeneous and anisotropic clusters.

 

So, with regards to your example -

It is not just a local wave of anisotropic which should be neglected in large scale.

It is an aggregated wave that should be considered as the biggest wave - ever.

 

Therefore, I think that we should use a tsunami wave as an example.

 

With regards to our universe:

It is clear that at any direction that we look, we see all the clusters in that direction.

 

As all the clusters are inhomogeneous and anisotropic, how could it be that the total sum of those infinite (almost) inhomogeneous and anisotropic clusters is homogeneous and isotropic?

 

Also, those clusters set musk on the expansion process at their locations.

So, how could it be that the expansion could be isotropic under this condition?

Edited January 13, 2016 by David Levy

[swansont]

However, in the same token you confirm that in small/local scale the cluster is inhomogeneous and anisotropic:[/size]

Because clusters aren't subject to the expansion.

[imatfaal]

Clusters are homogeneous throughout the universe therefore their overall net effect would be the same throughout the universe. And even saying that; for the time that the CMB photons have been travelling before reaching up, 13.7ish billion years, the time spent around centres of mass will be vanishingly small - any change would be a tiny, statistically unrecognizable error rather than a paradox.

 

So clusters don't really matter in the wide scheme of things and would not change the isotropic nature anyway even if the effect was bigger

[Mordred]

Imatfaal essentially answered the question. This is confirmed by the uniformity in the CMB temperature. That temperature uniformity means the mass density is as uniform.

 

This is confirmed via the temperature and metalicity maps by Planck and WMAP. The axis of error on the 2012 maps was largely calibration in not filtering out completely out local redshift influences. The 2015 was better filtered and the axis of evil is far less pronounced. Coincidentally Matt Roose "Introductory to Cosmology" has a good section on how Earths movement will cause this effect on redshift datamapping.

Edited January 13, 2016 by Mordred

[Carrock]

Natural consequence of expansion. When you increase a volume with the same number of particles. (Aside from any phase changes) the temperature will lower. This is already well known.

 

pV=nRt.

 

The temperature variations is inversely proportional to the scale factor

This may be a little misleading. Classical physics does not claim that increasing volume violates energy conservation, as you seem to imply.

 

Assume a sealed container of gas.

When you slowly expand the container the gas does work on the container walls. Its temperature goes down because some energy has been removed from the gas in the form of work etc. Energy has not been destroyed.

 

Classically if you release some gas into vacuum, the gas is not in equilibrium and the volume containing all the gas increases indefinitely. The total energy of the gas does not decrease as the volume increases.

 

 

 

Black body radiation refers to radiation from a body in thermal equilibrium.

 

As the CMBR is not in thermal equilibrium in the sense required for equilibrium thermodynamics to be valid (which it has not been at least since ~400,000 years after the B.B.), the stretching of wavelength is due to the spatial expansion of the universe and nothing to do with near equilibrium thermodynamics.

 

Indeed this creation of space and wavelength stretching is in violation of classical physics.

 

As it happens, the stretched CMBR has the same spectrum as radiation from a 2.7K black body, with a few imperfections. This is pretty consistent with the CMBR being radiation from a black body at 2.7k but there is a massive amount of evidence inconsistent with this view.

 

 

It might be less confusing to think of the CMBR having a colour temperature of 2.7K much like a cool running LED lamp might be described as having a colour temperature of 3,000K.

ie colour temperature is a description of a spectrum, not the temperature of the source.

[MigL]

The universe does not have 'container walls' nor is it expanding into a vacuum, Carrock.

That's why Mordred's analysis works.

 

The CMB is a 'snapshot' of the uniformity of temperatures 300 mil yrs after the Big Bang.

The minute variations of that time have developed into the features we see today ( galaxies , clusters, voids, etc. )

 

The photons that make up that 'snapshot' from 13.4 bil yrs ago have become 'stretched' in proportion with universal expansion, but they are trivially affected by the large scale structures of the present universe.

[David Levy]

Clusters are homogeneous throughout the universe therefore their overall net effect would be the same throughout the universe.

Would you kindly prove this statement?

 

Unfortunately, I couldn't find any homogeneous cluster.

 

 

Never the less, I have found the following explanation about open star clusters:

 

http://adsabs.harvard.edu/abs/2012AstL...38..506G

 

"Based on published data, we have compiled a catalogue of fundamental astrophysical parameters for 593 open clusters of the Galaxy. In particular, the catalogue provides the Galactic orbital elements for 500 clusters, the masses, central concentrations, and ellipticities for 424 clusters, the metallicities for 264 clusters, and the relative magnesium abundances for 56 clusters. We describe the sources of initial data and estimate the errors in the investigated parameters. The selection effects are discussed. The chemical and kinematical properties of the open clusters and field thin-disk stars are shown to differ. We provide evidence for the heterogeneity of the population of open star clusters."

 

Hence, if an open star cluster is unhomogeneous by definition, how could it be that a giant cluster which includes diffrent types of glaxies at diffrent locations and many (millions?) of unhomogeneous open star clusters could be considered as homogeneous?

 

would you kindly offer even one cluster in the Universe which is confirmed as homogeneous cluster?

[Mordred]

Carrock the ideal gas laws are fundamental in Cosmology applications. Energy is conserved, pressure in the FLRW metric is the direct result of particle to particle interactions. This is termed an adiabatic gas in thermodynamic terms.

 

The temperature is Inversely proportional to the scale factor. Though you can use either Gibbs law or the Bose-Einstein and Fermi-Dirac equations. (Which will also allow one to calculate the particle contribution to the temperature of each particle species.

 

Here is the related equations of state.

 

https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology)

 

you can see from this that particles with higher kinetic energy exert greater pressure.

 

The temperature vs scale factor is

 

[latex]a\propto \frac{1}{T}[/latex]

In case your not familiar with the scale factor.

 

https://en.m.wikipedia.org/wiki/Scale_factor_(cosmology)

These articles each has decent coverage chapter 3 and 4 of the last link in particular

 

http://arxiv.org/pdf/hep-ph/0004188v1.pdf:"ASTROPHYSICS AND COSMOLOGY"- A compilation of cosmology by Juan Garcıa-Bellido

http://arxiv.org/abs/astro-ph/0409426An overview of Cosmology Julien Lesgourgues

http://arxiv.org/pdf/hep-th/0503203.pdf"Particle Physics and Inflationary Cosmology" by Andrei Linde

http://www.wiese.itp.unibe.ch/lectures/universe.pdf:"Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis

 

 

@David your ignoring the scale of measurement. Until you can wrap your head around this detail you will continue to go astray pointlessly.

 

Single large scale structures are miniscule compared to the size of the universe. So your last post is meaningless. It doesn't matter how inhomogeneous a cluster is. The scale that homogeneous and isotropy occurs at is far larger than a mere cluster.

[swansont]

The chemical and kinematical properties of the open clusters and field thin-disk stars are shown to differ. We provide evidence for the heterogeneity of the population of open star clusters."

 

Hence, if an open star cluster is unhomogeneous by definition, how could it be that a giant cluster which includes diffrent types of glaxies at diffrent locations and many (millions?) of unhomogeneous open star clusters could be considered as homogeneous?

 

The heterogeneity is quite clearly referring to the chemical and kinematic properties. There is no reference as to their spatial distribution. So your conclusion that they are "unhomogeneous (sic) by definition" is erroneous. You can't just latch onto a key word in an explanation. You have to parse the whole thing, too.

[David Levy]

 

The heterogeneity is quite clearly referring to the chemical and kinematic properties. There is no reference as to their spatial distribution. So your conclusion that they are "unhomogeneous (sic) by definition" is erroneous. You can't just latch onto a key word in an explanation. You have to parse the whole thing, too.

 

 

O.K

 

So, you claim that the spatial distribution of the open star clusters is homogeneous.

 

Spatial = of or relating to space, "the spatial distribution of population",

Homogeneous = of the same kind, Uniform, remaining the same in all cases and at all times,

​In other words, the density of population/matter in an open star cluster is Uniform/homogeneous.

 

However, if we place two homogeneous open star clusters with the same spatial distribution and the same size at an empty ball shape, could it be that the spatial distribution of the ball will also be considered as homogeneous?

The answer is - as long as those two open stars cluster keep their shape, then the ball itself can't be homogeneous, as in some arias (at the open stars clusters) the spatial distribution will be different from the empty arias in that ball shape.

 

In the same token -

If we place 10,000 homogeneous open star clusters around a galaxy, does it mean that the galaxy is homogeneous?

If that galaxy has a massive black hole at the center, how could it be that its spatial distribution is homogeneous?

Even if we find two homogeneous galaxies and place them in a cluster, does it mean that the spatial distribution of the cluster is homogeneous?

For example – Let's look at our cluster with about 54 galaxies (including the Milky Way and Andromeda galaxies). Could it be that the spatial distribution of our cluster is homogeneous?

 

Please look again at the following message from Mordred:

 

Yes locally to a cluster filament there is an inhomogeneous and anisotropic condition...

 

At close scales their is no uniformity.

 

 

Therefore, do you agree that all the clusters in the Universe are not homogeneous?

[Strange]

It has nothing to do with the homogeneity or otherwise of individual galaxies and galaxy clusters and superclusters. These are all different sizes, and varying distances apart. Rather like the atmosphere contains many different size atoms, molecules, grains of dust, droplets of water and even insects. But, on a large enough scale, the atmosphere can be treated as a homogeneous fluid and those components are irrelevant.

 

Similarly, on a large enough scale, the universe is roughly the same everywhere.

https://www.e-education.psu.edu/astro801/content/l10_p6.html

 

[swansont]

O.K

 

So, you claim that the spatial distribution of the open star clusters is homogeneous.

 

Spatial = of or relating to space, "the spatial distribution of population",

Homogeneous = of the same kind, Uniform, remaining the same in all cases and at all times,

​In other words, the density of population/matter in an open star cluster is Uniform/homogeneous.[/size]

 

No. We are discussing expansion. The density of population/matter within a star cluster is irrelevant. What imatfaal said is that the distribution of these clusters is homogeneous throughout the universe. What Mordred and I have said (in somewhat different terms) is that none of this matters. The clusters themselves are present on a scale that is much smaller than the resolution of any discussion about expansion cares about.

 

We are discussing the expansion of space. Clusters are not space. The clusters are gravitationally attracted to the point that they don't notice the expansion. Measurements of them tell you nothing about the expansion. The scientists that study the phenomenon know this. Expansion can only be measured using objects which do not have such gravitational influence.

 

Put a tiny piece of paper on a balloon, and blow it up. That paper holds together, but the balloon expands. You wouldn't measure the expansion of the balloon by looking at the individual piece of paper. You could look at another piece of paper that is not attached to the first one — that will tell you what the expansion is.

 

However, if we place two homogeneous open star clusters with the same spatial distribution and the same size at an empty ball shape, could it be that the spatial distribution of the ball will also be considered as homogeneous?

The answer is - as long as those two open stars cluster keep their shape, then the ball itself can't be homogeneous, as in some arias (at the open stars clusters) the spatial distribution will be different from the empty arias in that ball shape.

 

In the same token -

If we place 10,000 homogeneous open star clusters around a galaxy, does it mean that the galaxy is homogeneous?

If that galaxy has a massive black hole at the center, how could it be that its spatial distribution is homogeneous?

Even if we find two homogeneous galaxies and place them in a cluster, does it mean that the spatial distribution of the cluster is homogeneous?

For example – Let's look at our cluster with about 54 galaxies (including the Milky Way and Andromeda galaxies). Could it be that the spatial distribution of our cluster is homogeneous?

 

Please look again at the following message from Mordred:

 

 

Therefore, do you agree that all the clusters in the Universe are not homogeneous?

 

The question is meaningless without saying at what scale you are looking for homogeneity.

[Carrock]

The universe does not have 'container walls' nor is it expanding into a vacuum, Carrock.

That's why Mordred's analysis works.

I wasn't criticising Mordred's results, simply his use of part of a classical theory to prove results inconsistent with that theory.

I suspect we ascribe different meanings to the phrase 'Mordred's analysis works.'

 

Carrock the ideal gas laws are fundamental in Cosmology applications. Energy is conserved, pressure in the FLRW metric is the direct result of particle to particle interactions. This is termed an adiabatic gas in thermodynamic terms.

Some gas laws are used in cosmology. I didn't see pV=nRt being used in your offsite references as a well known explanation of the cooling of the universe. Quote please.

'Energy is conserved' : can you demonstrate that your definition of energy conservation is the same as that used in pV=nRt?

 

See eg. http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

Irritating isn't it when someone posts bare links because they're too lazy to quote or can't/won't present a possibly refutable argument?

 

'pressure in the FLRW metric is the direct result of particle to particle interactions. Reference and quote please, not just another set of offsite links.

 

If you believe the above, presumably you think there was a sudden drop in pressure when the CMBR decoupled from matter, or the CMBR is still in equilibrium ie its photons are still interacting.

Which is it?

 

'This is termed an adiabatic gas in thermodynamic terms.' CMBR cooling probably is adiabatic.

 

The temperature is Inversely proportional to the scale factor. Though you can use either Gibbs law or the Bose-Einstein and Fermi-Dirac equations. (Which will also allow one to calculate the particle contribution to the temperature of each particle species.

 

Here is the related equations of state.

 

https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology)

 

you can see from this that particles with higher kinetic energy exert greater pressure.

From your reference https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology):

In cosmology, the equation of state of a perfect fluid is characterized by a dimensionless number [latex]w[/latex], equal to the ratio of its pressure [latex]p[/latex] to its energy density [latex]\rho[/latex]:

[latex]w = \frac{p}{\rho}[/latex]

It is closely related to the thermodynamic equation of state and ideal gas law [i.e. pV=nRt].

As both equations agree 'that particles with higher kinetic energy exert greater pressure' and they are closely related I presume you are saying that [latex]w = \frac{p}{\rho}[/latex] and pV=nRt can be used interchangably in cosmology.

 

 

 

The temperature vs scale factor is

 

[latex]a\propto \frac{1}{T}[/latex]

In case your not familiar with the scale factor.

 

https://en.m.wikipedia.org/wiki/Scale_factor_(cosmology)

I'm familiar with more than one scale factor so I don't know which you're referring to.

Your scale factor is not referred to in your reference; perhaps you're confusing the symbol t(time) with T(temperature). Reading and trying to understand your references might help you avoid such errors.

 

I'm guessing your scale factor is the first referred to here.

According to the dependence of temperature to the energy of the particles, the temperature of relativistic fluids such as photons will fall as [latex]T\propto \frac{1}{a}[/latex]. For the non-relativistic fluids such as dust and cold dark matter, temperature relaxes with a faster rate of [latex]T\propto \frac{1}{a^2}[/latex]. This approach showed that the cooling of the cosmic fluid can not be explained with the classical concepts from the thermodynamics and cooling directly results from the expansion of universe which makes the thermal peculiar velocity of particles reduce during the stretch of the scales.

I expect you can easily prove/refute the above using pV=nRt but here I demonstrate my close mindedness by saying your analysis will be incorrect.

[Mordred]

Ok lets go through this. First I posted pv=nRt. It's a basic (very basic thermodynamic equation). It Doesn't get used much in Cosmology papers on arxiv simply because we use another form. I tend to use pv=nRt as many that post on this forum are weak in thermodynamic laws. Let alone mathematical Cosmology.(lol this post will likely cause difficulty) (The formula shows the relevant relations without too much complexity)

 

The more common form describes an adiabatic expansion. (Via an adiabatic and isentropic fluid) Now if the universe is adiabatic then the first law of thermodynamics apply. Rather than posting a bunch of latex. Here you go...(The more common form first equation on link)

 

https://en.m.wikipedia.org/wiki/Thermodynamics_of_the_universe

 

First law of thermodynamics means

 

https://en.m.wikipedia.org/wiki/First_law_of_thermodynamics

 

Now I did an example in another thread for another poster.

 

First take the first law of thermodynamics.

 

[latex]dU=dW=dQ[/latex]

 

U is internal energy W =work.

 

As we dont need heat transfer Q we write this as DW=Fdr=pdV

 

Which leads to dv=-pdF.

 

[latex]V=r^3[/latex]

 

Through several steps we will end up with.

 

[latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex]

 

We will use the last formula for both radiation and matter.

 

Assuming density of matter

 

[latex]\rho=\frac{M}{4/3\pi r^3}[/latex]

[latex]\rho=\frac{dp}{dr}[/latex]

[latex]\dot{r}=3\rho \frac{\dot{r}}{r}[/latex]

 

Using the above equation the pressure due to matter gives an Eos of Pressure=0. Which makes sense as matter doesn't exert a lot of kinetic energy/momentum.

 

For radiation we will need some further formulas. Visualize a wavelength as a vibration on a string.

 

[latex]L=\frac{N\lambda}{2}[/latex]

 

As we're dealing with relativistic particles

 

[latex]c=f\lambda=\frac{2L}{N}[/latex]

[latex]U=\hbar w=hf[/latex]

 

[latex]U=\frac{Nhc}{2}\frac{1}{L}\propto V^{-\frac{1}{3}}[/latex]

 

Using the formula above.

 

[latex]p=1/3\rho[/latex] for ultra relativistic radiation.

 

Those are examples of how the first law of thermodynamics fit within the equations of state. There is more intensive formulas involved. In particular the Bose-Einstein statistics and Fermi-Dirac statistics but the above serves as a good approximation.

The above clearly demonstrates how kinetic energy is involved in pressure. By the way the universe doesn't have container walls. So ask yourself why it would exert pressure ? The answer I have you but you seemed to deny it "particle to particle interactions"

 

I already linked you a paper covering the Fermi-Dirac statistics and Bose-Einsten statistics.

 

 

 

primary relations that are involved from the above

 

[latex]E=(\rho c^2+p)R^3[/latex]

Expansion is adiabatic if there is no net flow or outflow of energy. So that

[latex]\frac{de}{dt}=\frac{d}{dt}(\rho c^2+p)R^3=0[/latex]

 

(The last formula demonstrates the first law)

Let p be proportional to [latex]\rho c^2[/latex]

This leads to the relation

[latex]p=w\rho c^2[/latex]

 

The acceleration equation is

[latex]\frac{\ddot{a}}{a}=-\frac{4\pi G\rho}{3c^2}(\rho c^2+3p)[/latex]

This leads to

[latex]H^2=\frac{\dot{a}}{a}=\frac{8\pi G\rho}{3c^2}-\frac{kc^2p}{R_c^2a^2}[/latex]

where k is the curvature constant.

now the curvature constant can have three main configurations 1,0-1.

You've probably know about the stress energy tensor but this set of relation is handy to know.

[latex]T^{\mu\nu}=(\rho+p)U^{\mu}U^{\nu}+p\eta^{\mu\nu}[/latex]

Which correlate the stress energy tensor to energy density/pressure in Minkowskii metric form.

 

some articles more intensive in the Einstein field equations state that when

 

[latex]T_{\mu\nu}=0[/latex] energy is conserved. ( often the conservation statements are in the math, not the written portion)

 

The FLRW metric to distance formula is.

[latex]d{s^2}=-{c^2}d{t^2}+a{t^2}d{r^2}+{S,k}{r^2}d\Omega^2[/latex]

[latex]S\kappa r= \begin{cases} R sin r/R & k=+1\\ r &k=0\\ R sin r/R &k=-1 \end {cases}[/latex]

 

note the usage of the scale factor. In the last set of equations. These are standard formulas. Not alternative models.

 

now that I showed some energy density to pressure relationships.

 

Let's step back to pv=nRt.

 

As I stated its a simple formula. But n is number of moles so how can we use this ? Well let's say I want to calculate the number density of a particle ie photons or neutrinos. Via pv=nRt (in a far more complex form.). Well I can use these formulas.

 

[latex]q=\frac{N}{V}+\ge+n_q[/latex]

 

for boson particles Bose_Eintein statistics is

[latex]n_i(\varepsilon_i) = \frac{g_i}{e^{(\varepsilon_i-\mu)/kT}-1}[/latex]

 

for fermions you use the fermi-dirac statistics

[latex]\bar{n}_i = \frac{1}{e^{(\epsilon _i-\mu) / k T} 1}[/latex]

 

the De-Broglie wavelength is

[latex]\frac{V}{N\Lambda^3} \le 1 \[/latex]

 

 

Note the mention of the second law of thermodynamics on this link in regards to Bose-Einstein distribution/statistics

 

https://en.m.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics

 

at any given temperature it is possible using Those equations to calculated the number density of any particle at any given temperature including the CMB.

 

The steps in that chapter 3 section I mentioned. (Though a good book on statistical mathematics shows the above better) By the way you can confirm all of the above in "Modern Cosmology" by Scott Dodelson.

 

Muchanov's "Fundamentals of Cosmology". Also steps into this area.

 

Probably the simplest to follow though is Matt Roose "Introductory to Cosmology"

 

Those links I provided earlier I chose as they match the textbooks used to teach Cosmology. There is no alternative theory in them. The link by Liddle is a free full length textbook. He admits some of it is out of date. In particular he's using SO(5) instead of SO(10). However it's still useful in teaching the basics.

 

 

PS I'd like to know what the author of the paper you posted means by physical boundary. Kind of a too brief a paper for my taste but he states the universe as adiabatic.

 

http://arxiv.org/pdf/physics/0603087v1.pdf

 

 

Here is an interesting tidbit to blow your mind if it's not already. The equation of state for dark energy aka the cosmological constant is identical to an incompressable liquid.

 

W=-1.

 

 

Imagine that

(Hint positive energy density leading to a negative vacuum. https://en.m.wikipedia.org/wiki/Cosmological_constant ) figure that one out lol

(Hint vectors in modelling,but what do I know, PS the clues are provided this thread. Double PS a negative energy/fensity

Edited January 16, 2016 by Mordred

[David Levy]

It has nothing to do with the homogeneity or otherwise of individual galaxies and galaxy clusters and superclusters. These are all different sizes, and varying distances apart. Rather like the atmosphere contains many different size atoms, molecules, grains of dust, droplets of water and even insects. But, on a large enough scale, the atmosphere can be treated as a homogeneous fluid and those components are irrelevant.

 

Similarly, on a large enough scale, the universe is roughly the same everywhere.

https://www.e-education.psu.edu/astro801/content/l10_p6.html

 

 

O.K.

 

"on a large enough scale, the atmosphere can be treated as a homogeneous fluid and those components are irrelevant."

 

Let's agree that this diagram represents homogeneous Universe.

I assume that the universe is surrounded by an empty space at a temp of 0K.

 

However, what about the heat transfer from the universe to that empty space?

In the following explanation about the CMB this heat transfer isn't part of the calculation.

http://www.phy.duke.edu/~kolena/cmb.htm

 

"At some point about 400,000 years after the Bang, the universe had cooled to the point where the matter became neutral, at which point the universe's matter also became transparent to the radiation. (Completely ionized matter can absorb any wavelength radiation; neutral matter can only absorb the relatively few wavelengths that carry the exact energy that match energy differences between electron energy levels.) The temperature at which this transition from ionized to neutral (called the "moment of decoupling") occurred was roughly 3000 K."

It indeed had the blackbody spectral shape predicted, but the peak in the microwave spectrum indicated a temperature of 2.726 K. Although this temperature is clearly insufficient to ionize hydrogen, the entire spectrum has been redshifted from that at the moment of decoupling (when the temperature was 3000 K) by the expansion of the universe. As space expands, the wavelengths of the CMB expand by the same factor. Wien's blackbody law says that the wavelength peak of the CMB spectrum is inversely proportional to the temperature of the CMB. Therefore, the drop in the CMB temperature by a factor of 1100 (= 3000 K/2.73 K) indicates an expansion of the universe by a factor of 1100 from the moment of decoupling until now."

 

Hence, about 400,000 years after the B.B the Universe temp was 3000K. Today, due to the expansion factor of 1100, the temp had been reduced by the same factor to 2.73K.

Not even one word about Heat transfer!

https://en.wikipedia.org/wiki/Heat_transfer

 

"Heat transfer always occurs from a region of high temperature to another region of lower temperature."

"Heat transfer changes the internal energy of both systems involved according to the First Law of Thermodynamics"

"Thermal equilibrium is reached when all involved bodies and the surroundings reach the same temperature."

 

So, let me ask you the following question about heat transfer:

 

-Assuming that the Universe is homogeneous and its temp is 3000K.

-It is surrounded by an empty space at a temp of 0K.

-There is no expansion. (Please ignore it completely)

 

What is the expected temp of the universe (due to heat transfer only) after about 13 Billion years?

Edited January 16, 2016 by David Levy

[Mordred]

It's not part of the calculation as our universe (Observable) is surrounded by the Unobserved portion which is the same as our observable portion. So there isn't a temperature difference. We don't know the true size of the entire universe it could be infinite or finite.

 

That being said Unruh radiation is the heat transfer to the Unobserved portion. However that's due to not being able to measure the particle that's moved beyond our observational portion. Not due to temperature difference.

[David Levy]

It's not part of the calculation as our universe (Observable) is surrounded by the Unobserved portion which is the same as our observable portion. So there isn't a temperature difference. We don't know the true size of the entire universe it could be infinite or finite.

 

That being said Unruh radiation is the heat transfer to the Unobserved portion. However that's due to not being able to measure the particle that's moved beyond our observational portion. Not due to temperature difference.

Wow

What do you mean by: "surrounded by the Unobserved portion which is the same as our observable portion." ?

So do you mean that when our universe temp was 3000K, then this surrounded aria was also 3000K?

As the temp of our universe had been reduced due to the expansion, does it mean that also the surrounded aria temp had been reduces to the same temp?

Why? how could it be? Is it some sort of matter/anti-matter?

It is possible that this "unobserved portion" is also expanding at the same rate as our universe?

Could it be that we are located at the "unobserved portion" of the Universe?

Edited January 16, 2016 by David Levy

[Strange]

I assume that the universe is surrounded by an empty space at a temp of 0K.

 

Why would you assume that? Perhaps you should learn a little bit about modern cosmology. Or even just ask queations, rather than making silly assumptions.

As the temp of our universe had been reduced due to the expansion, does it mean that also the surrounded aria temp had been reduces to the same temp?

Why? how could it be?

 

As it expands it gets cooler.

 

It is possible that this "unobserved portion" is also expanding at the same rate as our universe?

Yes, it is a working assumption that the rest of the universe behaves largely the same as that which we can see. (Because we have no evidence that tells us otherwise).

Could it be that we are located at the "unobserved portion" of the Universe?

 

By definition, we are in the observable universe. Because that is the part of the universe that we can observe - i.e. a sphere of about 90 billion light years diameter around the Earth.

[Mordred]

The one mention section you posted David on the behaviour of a higher energy-density/pressure/temperature is accurate. Any higher state in the aforementioned three will naturally develop into a lower value.

 

Key note 'based upon heat or work transfer to amnther outside thermodynamic state

 

Here is the problem of development of this principle into a "cosmological constant".

 

Lets take your example first. Strict higher temperatures moving to a zero temperature state. (If you study thermodynamics this leads to a preference direction. {Hot to cold}=

 

Measurements show no sign of the above action. (CMB temperature)

 

Now the question comes into play.... can our uniform measurement of the CMB agree with this action. My knowledge (after 30 years study days no).

 

If you noted the conversation between Carrock and others during this thread. The subject came up. Thus far this thread no evidence or supported papers have shown up on anything other than a homogeneous and isotropic fluid in Cosmology terms been presented). Any suppositions presented have lacked following supported material.

Edited January 17, 2016 by Mordred

[swansont]

 

So do you mean that when our universe temp was 3000K, then this surrounded aria was also 3000K?

 

There is no surrounding area. We're talking about the universe. That's all there is. There's no "outside"

[David Levy]

Thanks Mordred

 

I do appreciate your patience & direct answers.

With regards to the following:

Lets take your example first. Strict higher temperatures moving to a zero temperature state. (If you study thermodynamics this leads to a preference direction. {Hot to cold}=

Measurements show no sign of the above action. (CMB temperature)

Now the question comes into play.... can our uniform measurement of the CMB agree with this action. My knowledge (after 30 years study days no).

 

So, if I understand you correctly -

You claim that technically, if we use my example, then we should see a heat transfer from Hot (Universe) to cold (outside the Universe)

However, Based on CMB, there is no sign for that heat transaction.

Therefore, the science estimates that there is no heat transfer outside the universe.

Hence, Swansont claims that there is no "outside":

 

 

 

There is no surrounding area. We're talking about the universe. That's all there is. There's no "outside"

 

Did I get it correctly?

 

I also would like to ask the following question:

Based on our best knowledge, what was the size of the Universe when its temperature was 3000K (400,000 years after the B.B.)?

Edited January 17, 2016 by David Levy