Relativity and Cosmology

Jorrie,

I assumed that you meant the direction of the change in curvature. I fear that this implies a growing universe with a reducing radius of curvature everywhere. I feel I too must have got hold of the wrong end of this stick?

Fyz

I hate that Jorrie sleeps. Now I have to wait till tomorrow to continue this discussion.

Fyz, yes this stick is very slippery! See what I wrote to Roger.

Fortunately, since Ω_tot is apparently so close to unity and hence R so large, it does not matter too much which end of the stick we get hold of.

-J

Hi Roger. Sorry, my wording: "... It means we can only use the balloon over a shortish timescale, because the balloon curvature actually goes in the wrong direction." was pretty sloppy.

As Fyz suggested in #141, I meant to say that the 'direction of change in curvature actually goes the wrong way'.

Curvature is a function of Ω_tot, with the radius of curvature: R=R_H/√(Ω_tot-1), where R_H=978/Ho (Gly) is the Hubble radius. It tends to infinity if Ω_tot = 1, the flat universe. If Ω_tot <> 1, it evolves away from 1 - it is essentially an unstable condition (Google critical density for more info). This means that if Ω_tot > 1, it grows larger and R shrinks. This is obviously the opposite of what the balloon radius and curvature does in an expanding balloon.

You are right that matter and Λ curve space in the same direction, just not that curvature evolves in the same sense as what balloon curvature does. This is a serious limitation of the balloon analogy, because the cosmic radius of curvature is not a physical radius of anything.

-J

You Wrote:"You are right that matter and Λ curve space in the same direction, just not that curvature evolves in the same sense as what balloon curvature does. This is a serious limitation of the balloon analogy, because the cosmic radius of curvature is not a physical radius of anything."

Ok, my confusion is gone, thanks.

So how about this. My viscous liquid on a sphere model creates adjustable expansion (through modifying vapor pressure) with constant curvature. This is not correct as we just discussed.

So what if instead of a sphere we take the same design except put it on a balloon that we can inflate? We would have to adjust the vapor pressure to compensate for the balloons inflation. The curvature of the fully inflated balloon would be that of space without matter or radiation (curvature just due to vacuum energy).

Early in the timeline we can inflate the balloon as the liquid condenses on it. This will result in to causes of expansion, but they can be added together and would appear as one net effect. The nice thing about it is because you've disentangled the expansion with the curvature, you can adjust the system so that it imitates space.

I think. (Thanks for staying up to answer my questions, this is fun. )

My model probably has serious flaws, have at it.

Also, as an aside unrelated to my last two posts, I also have a serious issue with the statement that matter isn't effected by expansion. I say this because if light is red-shifted, then so too should be de Broglie Waves. And since a redshift in a de Broglie wave corresponds to a reduction of momentum, I would argue that momentum of a particle traveling through space (lets say an electron) is not conserved but actually lost..

In other words, just as light loses energy as it is red-shifted by cosmological expansion, so does matter (de Broglie waves).

Roger

I'm not quite clear exactly whay you mean here.

I don't think you can mean that matter itself changes in an way, as that would result in all sorts of issues.

If you are referring purely to the momentum of matter, as opposed to its structure, I'm inclined to agree - but only when we are considering free "linear" movement. (What I'm trying to say is that I would expect no changes in the momentum of orbiting structures (bound electrons, for example)).

Regards

Fyz

Hi Fyz,

You Wrote:"I'm not quite clear exactly what you mean here."

I specifically was thinking of beta particles in extragalactic cosmic rays. In my mind that seemed the cleanest case since you've got no strong force and the electron is moving over vast distances (like light does).

You Wrote:"I don't think you can mean that matter itself changes in an way, as that would result in all sorts of issues."

That is not what I was trying to say in the previous post.

You Wrote:"If you are referring purely to the momentum of matter, as opposed to its structure, I'm inclined to agree - but only when we are considering free "linear" movement. (What I'm trying to say is that I would expect no changes in the momentum of orbiting structures (bound electrons, for example))"

Yes, this argument I was presenting in the post was meant to be specifically for free, moving electrons. Actually, as I said before, I was specifically thinking of cosmic rays.

Looks like you can't stay away.

Roger

My wife says my problems are not calculus, but algorithms.

Some of us have had some fun watching you and Roger go around.

When I have been most courageous I was most willing to make a fool of myself.

Is there an algorithm that spins my imperfect hotdog twisted balloon?

Of course post 156 cheered me up some because I can't figure out where the big bang came from unless the balloon is twisted into two halves.

When I think of microbes that asexually just split in half, there is progress partly because neither half is perfect.

I end up thinking perfect is imperfect as Bernard Shaw said about humanity and God.

"If God is perfect, then if God made man, I must be perfectly imperfect."

I myself have never had a perfectly round balloon.

Indulge me for a moment and think of the different images.

The balloon inside another balloon is is like an inner tube in a car tire.

Still the valve stem is there for either the tubeless tire, or the tire with a tube.

In the Wizard of Oz there was a guy behind the curtain turning knobs, and pressing on valves that was sort of a disappointment.

Frankly I don't expect the Universe to work with any particular input from me.

If I'm lucky I might get to play with some knobs, but a full explanation is not appropriate and I myself don't like the vernacular.

I like the Infinity Circle Balloon Model revolving on a string where it is accepted that though it may look like there is a perfect division, this is an impossibility, that makes all other possibilities a fact.

Dear Jorrie, Now as I understand it the cosmological principle is that from anywhere the Universe will look the same.

Then there is the Steady State Theory of the Universe.

In both cases the Universe has either according to one theory fluctuations in density, or local irregularities.

The goal is to make a Model that Looks like the Universe according to the cosmological principle and the Steady State Theory, plus all of cosmology.

In the Milne, and then Bondi and Gold work seems both say, it will look the same, but be different somewhere.

This is why I become more willing to defend my twisted balloon image, than less.

Hi Jorrie, sorry I haven't been around the past few days but I've been under the weather. I'm eager to jump back in but having reread all the posts, I'm unclear now as to what the current model is. The fascinating observable universe argument aside, I think we are still trying to build a model, right? The double skin balloon has been thrown out and I don't know what version of the single skin balloon we're promoting. Can you please refer me to the current model so I can attack it with vigor? (perhaps the original posting is up to date, in that case just let me know).

Hello Roger and welcome back. Hope the 'weather' has cleared.

The opening post is fairly up to date and we are trying to model the de Sitter (vacuum-only) model, using a balloon (and not just the Friedman equations). I've been down a few blind alleys and at present it seems like the only way forward is with the 'knobs' that you suggested.

Standardsguy (S) is looking at a program for us, starting with the Friedman equations and (hopefully) later with standard 'balloon dynamics'. I'm working on a representative (maybe simplified) set of equations for the latter, using a normal gas-filled balloon with a 'knob' on the pump.

This does not rule out your 'vapor model' - I just hope we will be able to write the equations of motion for it.

-J

Jorrie,

You do realize that in my model, space is the fluid sitting on top of the balloon, not the balloon itself, right?

I ask because I feel that all models where the skin of the balloon is space are flawed because of the curvature issue. That issue, to be more explicit, being that in such a model the rate of change of the curvature is directly proportional to the expansion of space.

In my model however, where space is a fluid sitting on the surface of the balloon (frictionlessly), only the curvature of the balloon, not the expansion, is conveyed to the fluid (space) sitting on the balloon. That way the balloon is only responsible for the curvature aspect of space.

Lets discuss so we can make a decision how to proceed.

Roger

Roger, yes I think I realize what your model says.

"In my model however, where space is a fluid sitting on the surface of the balloon (frictionlessly), only the curvature of the balloon, not the expansion, is conveyed to the fluid (space) sitting on the balloon. That way the balloon is only responsible for the curvature aspect of space."

One problem is that if spatial curvature is constant, the balloon plays no further role and you lose a lot of the explanetary powers of the balloon analogy, i.e., you are no better off than a flat model that is infinite and expands in all directions, which is very hard to visualize.

However, I do think we must keep your model as an alternative to the 'standard' balloon and see if it can provide the same (or better) insight. I will write up what I've got on the dynamics of the 'standard' balloon shortly.

-J

You Wrote: "One problem is that if spatial curvature is constant, the balloon plays no further role and you lose a lot of the explanetary powers of the balloon analogy, i.e., you are no better off than a flat model that is infinite and expands in all directions, which is very hard to visualize."

Hi Jorrie. That doesn't seem anything like what I'm talking about. Any liquid sitting on a balloon's surface will have the same curvature of the balloon. If you blow the balloon up, the curvature of the liquid sitting on it would decrease. If you let air out of the balloon, the curvature of the liquid sitting on the balloon increases. You get precisely the curvature of the balloon, without the expansion of the balloon (since the liquid sits frictionlessly on the balloon.

I don't see how that will have the problem you are stating above. I can't tell if you are misunderstanding me or if I'm missing your point.

Roger

Roger, the expanding balloon analogy describes a large amount of cosmological observations, e.g. the expansion law, the redshift, look-back time, etc. In order for your 'frictionless fluid' to succeed, it must describe the same things without making use of the balloon's expansion. This, you may find, is not easy.

Further, curvature is not considered a major contributor to the cosmic characteristics anymore (it was virtually constant over the life of the cosmos). Since this is all that the balloon part of your model deals with, you lose a great deal of the balloon's benefits and you are essentially no better off than describing a flat cosmos, expanding linearly.

That said, I'm still intrigued by the 'condensing vapor' idea, especially if it works by inserting molecules into the fluid, so that existing particles are pushed apart. You must just find the dynamical characteristics for such a process.

-J

You Wrote:"Roger, the expanding balloon analogy describes a large amount of cosmological observations, e.g. the expansion law, the redshift, look-back time, etc. In order for your 'frictionless fluid' to succeed, it must describe the same things without making use of the balloon's expansion. This, you may find, is not easy......"

Does that mean we shouldn't even try? You posted this so that we could develop a model that mirrors cosmology, that's what I'm trying to do. I refuse to be encumbered by what others have done, especially when we've established what they've done is wrong, as small an error as we'd like to portray that error of curvature to be (oh, don't worry about that error, it's a minor one? can you imagine if I was saying that about the model I was promoting?).

You Wrote:"Further, curvature is not considered a major contributor to the cosmic characteristics anymore (it was virtually constant over the life of the cosmos). Since this is all that the balloon part of your model deals with.....l"

I don't feel the balloon model should be sacrosanct, especially if it is not correct, which it appears not to be. Why are we making excuses for a model?

You Wrote:"That said, I'm still intrigued by the 'condensing vapor' idea, especially if it works by inserting molecules into the fluid, so that existing particles are pushed apart. You must just find the dynamical characteristics for such a process."

Yes, well that's the other half of this, isn't it? I'm not sure what you want though. If you want I can give the equations for surface tension (This was a mistake, I should have said Viscosity here) which I feel is analogous to gravity, and pressure (this is a mistake, I should have said force due to gravity (g)) which will cause the expansion in my model. The part I like best about the vapor idea is that it mirrors cosmic expansion in that anywhere where space exists, molecules can condense and cause expansion. I really feel it does an excellent job of mirroring space.

So please, tell me how we should proceed. Let's give this a model a chance. I'm not saying it will work, I just don't think we've found that out yet. Tell me what we need next to investigate this model.

I wrote: "The next step would be to model the 'design' with the special fluid and the pump and ensure that it gives the same curve."

One problem is that the "special (exotic) fluid" (with p = -ρc²) mimics the Friedman equations faithfully, but gives no further insight. Realizing this, I had a look at what a standard balloon with a pump and a 'pressure knob' would yield.

I found the relationship between balloon pressure p, radius R and elastic modulus E for an isotropic material (for small deformations):

p ~ 4 Do/Ro E (R - Ro)/Ro or

R ~ Ro + p Ro²/(4 E Do)

where Ro is the radius of the completely strain-free balloon, i.e., an extrapolation to p=0, and Do is the skin thickness. This says that a specific pressure p will result in a static and stable balloon radius R. In order to inflate (or deflate) the balloon, the pressure must change. This can be achieved by adding/withdrawing gas to/from the balloon.

If we assume that the balloon material retains its perfect elasticity over a reasonable range of balloon radii (at least enough range to illustrate the principles), one may 'build' the following system. A sensor measures the balloon radius (R) directly and also determines the rate of change (ΔR/dt). A pump system supplies gas to the balloon at a pressure that will keep ΔR/dt directly proportional to R. This is exactly what pure de Sitter expansion does.

Now this expansion rate might not sound too remarkable, until one realizes that it is an exponential inflation! The system will presumably have to turn up that 'pressure knob' at an interesting rate. Double R and the rate of expansion ΔR/dt must also double, requiring eight times the rate of gas flow than before. But, before we get too discouraged, it is good to consider the rate at which the balloon must inflate in order to simulate the real cosmic expansion closely.

Based on observed curvature limits, the R of our present hypersphere is at least 100 Gly and by using the present Hubble constant of ~74 km/s/Mpc (~0.074 Gy^-1), R increases by about 7.4 ly every year. Translate this to a balloon with one meter radius and the increase is some 7.4 / (100 x 10^-9) ~ 10^-10 m/year - yes, that's only 0.1 nm/year! Gas flow dynamics should not be a problem...

To make any change perceptible, we will have to speed up the simulation by a trillion times or so.

Comments?

-J

What model is this supposed to be describing? Surely not mine, right?

Any model where the balloon is the space isn't going to work in my opinion. The only model that will work is if you decouple the expansion from the curvature (as far as I can tell).

Hi Roger, yes, it's a standard balloon.

You wrote "Any model where the balloon is the space isn't going to work in my opinion."

When the surface of the balloon is our space, I see no reason why it will not work. All that is needed is the correct expansion profile. I have not found any cosmological principles that I cannot describe using the hyperspherical model. Granted, we ignore the curvature issue, but as I stated above, it is not really important anymore, mainly because of the immense minimum size of the total cosmos, making curvature unobservable.

The issue here is to find a credible mechanism that we can put in front of an engineer and say, hey, play with this model and convince yourself that you get the standard cosmic principles from it. This 'credibility' is one reason why I shy away from exotic things like my special fluid with p = -ρc².

-J

You Wrote:"Granted, we ignore the curvature issue, but as I stated above, it is not really important anymore, mainly because of the immense minimum size of the total cosmos, making curvature unobservable."

I don't agree with this statement. Surely in the early universe this was very important. I thought the point of this was to create a model, not just rubber stamp the existing one.

You Wrote:"This 'credibility' is one reason why I shy away from exotic things like my special fluid with p = -pc2

Ok, now we are getting somewhere. Why in the world do you think the above would require a "special" fluid in any way? All fluids have surface tension holding them together and pressure pushing outward.

Hi Roger, I'm afraid here is some miscommunication. (I'm referring to your #192)

If have written on this somewhere in this thread already: the curvature of the early universe (directly after inflation) was about 60 orders of magnitude less than it is today, not higher! If not precisely zero, curvature evolves away from zero. With curvature so small today, it was utterly negligible in the very early universe.

The curvature of the balloon has in fact nothing to do with the curvature of the universe - they evolve in different directions! This is an unfortunate misleading visualization of the balloon model, but it does nothing to invalidate the model.

"Why in the world do you think the above would require a "special" fluid in any way?"

I'm not sure what 'above' you refer to, but the usual cosmological model includes a special fluid with p = -ρc², which is what I referred to (as described in the opening post). No such fluid exists, except in the 'false vacuum' of space.

"All fluids have surface tension holding them together and pressure pushing outward."

OK, but that's positive pressure, not negative. Put enough of any standard fluid together and its pressure also creates gravity (over and above its gravitating mass). The pressure of the 'vacuum fluid' creates negative gravity, which creates the accelerating expansion.

That said, on an ordinary balloon's skin (non-cosmological scales), a fluid can expand as you attempted to show. The next step would be to show how this will work - it's not good enough to say that the thickness of the fluid remains constant and therefore other dimensions must expand. I see that you have started to put the idea into equations - I'll see if I can follow them and then comment further.

The ordinary balloon with the ordinary pump/compressor of #183 does a quite credible job of the de Sitter expansion. You will have to match that...

-J

You Wrote:"If have written on this somewhere in this thread already: the curvature of the early universe (directly after inflation) was about 60 orders of magnitude less than it is today, not higher! If not precisely zero, curvature evolves away from zero. With curvature so small today, it was utterly negligible in the very early universe."

I see. I had this backwards. So gravitation tries to curve space the opposite direction? If so then I think I understand.

I'm going to proceed with putting out the equations that would govern the liquid on a balloon model.

Roger, you wrote: "So gravitation tries to curve space the opposite direction? If so then I think I understand."

Not quite. See the definition I gave to S in #206. If you want to see the equations for the evolution of Ω, a nice set is given in these lecture notes (page 2), where Ω₀ = Ω_R + Ω_M + Ω_Λ, and Ω_R refers to radiation, not radius!

I haven't had the energy to work through your first equation set. I think I'll wait until you have them more sorted before I'll give serious consideration (they are a bit outside of my expertise).

Am I right in assuming that your fluid represents only vacuum, not matter or radiation in space?

-J

Jorrie,

I think you should stop calling the balloon system a model. It's an analogy to help people visualize what spatial expansion is. It doesn't correctly represent curvature, It's mass density is constant during expansion. There's other issues.

I've quit on my model, it's too much work and I obviously don't understand the underlying cosmology we are trying to model well enough. The initial assumption for these cosmological models is that space is a homogeneous fluid so it makes perfect sense that you would use a liquid to model it, not a balloon, but it's a lot of work.

Besides, I've become sidetracked by something else. Do you know what the predicted wavelength of the highest energy gravity waves are? I ask because I recently read in the APS newsletter that they will be monitoring the period of pulsars to look for deviations that could be caused by gravity waves passing between us and said pulsar.

http://www.newscientist.com/article/dn1797-pulsars-beat-could-reveal-gravity-waves.html

Roger

Hi Roger, no, I do not think it is wrong to call it a model - if we could make it to faithfully follow the ΛCDM model, it would perhaps fit the term.

The problem is that we both have problems in doing that - the only way I could make it work is to employ the exotic (p = -ρc²) fluid that cosmology uses anyway. When I have to revert to a balloon with pump/compressor, programmed to follow the Friedman equations, I would agree that it is just an analogy. However, once that part is done, the dynamics of particles on the surface may again fit the characteristics of the real universe and it may perhaps be called a model again...

The highest frequency gravitational waves that I know of is some 10 kHz, but I have no idea if they are the most powerful waves. I guess that depends on amplitude considerations and hence the distance of the source.

I think that the perturbation of pulsar frequencies would be indirect evidence of passing gravitational waves, yes. It is probably as indirect as the decay of binary pulsar orbital periods anyway. LIGO and LISA may provide more direct measurements.

-J

You Wrote:"The problem is that we both have problems in doing that - the only way I could make it work is to employ the exotic (p = -ρc²) fluid that cosmology uses anyway."

Then You Wrote:"When I have to revert to a balloon with pump/compressor, programmed to follow the Friedman equations, I would agree that it is just an analogy. However, once that part is done, the dynamics of particles on the surface may again fit the characteristics of the real universe and it may perhaps be called a model again..."

I assume on this "balloon model" matter sits on the surface of the balloon right? The depressions made on the surface of the balloon where matter sits represents the gravitational field of that matter?

Any balloon that expands, the skin gets thinner and less elastic, which means that as you inflate the balloon, the indentations caused by the matter would be less. In other words, in the balloon model gravitation weakens as the universe expands, not to mention that a thinner balloon skin means that the mass density and thus the energy density changes.

So I guess you're using a balloon whose skin doesn't get thinner as you expand it and remains the same elasticity. That seems like an unusual balloon.

You Wrote:"I think that the perturbation of pulsar frequencies would be indirect evidence of passing gravitational waves, yes. It is probably as indirect as the decay of binary pulsar orbital periods anyway. LIGO and LISA may provide more direct measurements."

Here's an article with more information on the subject.

http://meetings.aps.org/Meeting/APR09/Event/102932

You wrote: "I assume on this "balloon model" matter sits on the surface of the balloon right? The depressions made on the surface of the balloon where matter sits represents the gravitational field of that matter?"

No, in this 'programmed pump'-balloon analogy, particles of matter does not make depressions on the skin, just like the ΛCDM model does not allow for that. Matter is simply spread homogeneously in both cases and the skin could be considered massless, or its mass could be included in the surface mass.

The 'dynamics of particles' that I referred to is like what Jon and I discussed at length in this thread, i.e., moving particles that are only influenced by the expansion profile of the balloon. The properties of the skin play no role - not even its elasticity, because the pump system's feedback loop takes care of any changes (reply #183). The skin is just a surface containing the interior gas that blows up the balloon and for particles to move on freely.

-J

So how is this model anything like the cosmological model? The only think you seem to want to recreate is the expansion. All other details seem to be trivial.

Energy Density of space remains constant? Nope, the balloon skin becomes thinner as it expands.

Expansion accelerating? Nope, only if you force it with a pump, the balloon actually wants to decelerate expansion as it expands.

Curvature increasing as expansion increases? Nope, actually the balloon does the opposite.

You Wrote:"The skin is just a surface containing the interior gas that blows up the balloon and for particles to move on freely."

The balloon system isn't allowed to have a particle, as it doesn't express the gravity of the particles correctly and any particle would break the homogeneity argument anyway.

I'm sorry but I simply am not seeing how this is a great model.

Hi Roger,

The 'programmed-pump-balloon' thing is an analogy, not a model anymore, as I said above. I can make it a model by supplying it with a special fluid, though!

I must log out for other commitments - will come back to your other concerns later.

-J

No need to address the other concerns if you don't want. Let's discuss the special fluid you're talking about. Can you describe that model?

Hi again Roger, you requested: "Let's discuss the special fluid you're talking about. Can you describe that model?"

I have written quite a bit on it in the "Design" section of the opening post and in the replies referenced there (change proposals reply #117 and reply #149).

For convenience (in this long thread), here's an extract and more description.

"The basic design is very simple. A hypothetical balloon with a massless and infinitely stretchable skin is filled with a special fluid with an equation of state w = -1, meaning pressure p = -ρc². This pressure will balance the self-gravity of the fluid, keeping the balloon in a static, yet unstable equilibrium.

The skin is just a 'membrane' to separate the interior of the balloon from the vacuum outside. Ordinary material particles sit on the skin and can move without friction along the surface. Radiation moves along the surface at local velocity c.

A pump system with an infinite reservoir of the special (p = -ρc²) fluid monitors the pressure inside the balloon and adds or withdraws fluid to keep the pressure constant at a preset value at all times. If the balloon is made large enough (in empty space), the 'false vacuum' of space can be the 'infinite reservoir'."

If the "static, yet unstable equilibrium" is disturbed, the balloon will either expand or contract, like the ΛCDM model. With the correct ratio of particle:fluid mass-energy, the expansion curve can match the ΛCDM model for any epoch. If matter energy dominates, the rate of expansion will decrease and with fluid energy dominating, the expansion will accelerate.

The pump in this system is not programmed, but very simply set to keep the negative pressure inside the balloon constant (very simple if it was not for the fact that it is negative pressure inside and vacuum outside). This is one reason why I did not propagate the model further.

I think the 'programmable pump' with normal fluid is much more accessible to engineers. This thing we can build!

-J

I wrote: "I think the 'programmable pump' with normal fluid is much more accessible to engineers. This thing we can build!"

I think it is time to wrap up this unwieldy long thread and update the opening post to reflect "this thing we can build!" Reply #226 will remain as a record of the 'special fluid' idea and can be discussed further if so required. Reply #183 touched upon the 'programmable pump' design:

"If we assume that the balloon material retains its perfect elasticity over a reasonable range of balloon radii (at least enough range to illustrate the principles), one may 'build' the following system. A sensor measures the balloon radius (R) directly and also determines the rate of change (ΔR/dt). A pump system supplies gas to the balloon at a pressure that will keep ΔR/dt directly proportional to R."

To generalize, we must change the last sentence to: "A pump system supplies gas to the balloon at a pressure that will keep ΔR/dt = R H, where H is a function of R and the energy density makeup of the cosmos to be simulated. H is defined as follows:

H = Ho √[ (1-Ω)/R² + Ω_m/R³ +Ω_r/R⁴ + Ω_Λ ]

where Ho is a constant, Ω = Ω_m + Ω_r + Ω_Λ is the total energy density parameter, Ω_m the matter energy density parameter, Ω_r the radiation energy density parameter and Ω_Λ the vacuum energy density parameter, all as present measured values."

For clarity, in this model Ho have the units Gy^-1, relating to the usual Hubble constant H0 (in km/s/Mpc) as Ho = H0/978 Gy^-1. Energy density parameters are dimensionless, because they are fractions of the critical density: Ω = ρ/ρ_crit, where ρ_crit = 3Ho²/(8ΠG).

The expansion curve for this model is shown in the figure. Present (2009) values are: Ω_m= 0.258, Ω_r= 0.0000812, Ω_Λ= 0.742. A present hyper-radius Ro=100 Gly was used, because it is the minimum compatible with the data.

I am contemplating only a few more rounds of question/comments on this simulation before finalizing the opening post and closing down this thread.

-J

PS: I may perhaps open another thread called "Applications of the CR4 Cosmic Balloon", to further discuss the relative merits of models.

Jorrie,

I played around with the spreadsheet from your supplied link. Changing the omega values (m,r,v) to (0,0,1) gave me the de Sitter curve, right? Putting them at (0.5,0,2) gives a very interesting chart, and (0.4,0,2) looks like the Big Rip!?? I am amazed that the chart re-scales the time for each choice I make. Did you have to do anything special to make that happen? It looks like I am too far behind you to be of any use.

I also opened QuickBasic in WinXP and it only gives me a small window. I would have to boot up with a floppy or use an old computer, neither of which excites me. With summer "honey-do" projects on the horizon, I must "bow out", but hope to do a simulation at a later time. Your spreadsheets will show you what my simulation would have for a "standard" graph to compare the 'model' to. I take it that you're giving up on it anyway. Are you saying that a positive pressure inside the balloon and a lesser constant positive pressure outside won't give the proper expansion curve?

Thanks for the spreadsheet and for a stimulating thread.

regards,

-S

Hi S, I'm glad that you found the spreadsheet interesting.

Yes, (0,0,1) gives the de Sitter curve. The (0.5,0,2) case looks something like a 'rip', after a period of very small expansion rate. Sadly, (0.4,0,2) just crashes the spreadsheet, because such a universe contracts shortly after 7 Gy and the spreadsheet does not handle the contracting phase (√ of a negative value) - it was not designed for contracting phases - I run a from 0 to 1, so it cannot decrease. It is however interesting that a huge vacuum energy component (like 2 out of 2.4) causes a contraction after some time - strange beast, this cosmological constant!

BTW, normal vacuum energy is not able to cause a 'big rip', despite the appearance of the (0.5,0,2) case, which is just an exponential expansion. AFAIK, it requires a different form of dark energy (quintessence) to do a 'big rip', but that's outside of my knowledge base.

You asked: "Are you saying that a positive pressure inside the balloon and a lesser constant positive pressure outside won't give the proper expansion curve?"

It will not work with constant pressure differentials, but my pump system programmed to follow the Friedman equations will work, as I have shown in #228. One only needs a feedback loop that measures R and dR/dt, run that through the pump's microprocessor, feed it to the pump, changing the flow rate to satisfy the Friedman dR/dt requirement. Simple!

-J

Jorrie, if that spreadsheet of yours crashes with certain inputs, as you said to StandardsGuy, then your pump microprocessor will also crash for those inputs. What happens to your balloon when the crash occurs? Or how do you prevent it?

Hi Guest, yes you are right.

The input values (0.4, 0, 2) that S used are actually invalid for the conditions of the simulation, which requires that the Ωs all be today's values. With (0.4, 0, 2), the model can never reach 'today', i.e., R cannot reach Ro, the present value of R (which is the same as saying a cannot reach 1 in the Friedman equations).

However, it would be better to check and stop invalid inputs, or ensure that the processor program keeps on running and issues a warning. This is standard practice in control system programs - one cannot allow the actual system to 'crash'!

BTW, your comment has shown up a flaw in the equation anyway: I should not have used R as I did:

H = Ho √[ (1-Ω)/R² + Ω_m/R³ +Ω_r/R⁴ + Ω_Λ ]

should have been normalized to:

H = Ho √[ (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ ]

where a = R/Ro.

One way to overcome the 'crash' issue is to use:

H² = Ho² [ (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ ]

in the control loop, with a 'switch' to change inflation to deflation of the balloon, where the direction of the switch is: sign( (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ). The plot on the right has been done like this, where Ro is taken as 100 Gly.

-J

PS: Case (0.4, 0, 2) means Ω_m = 0.4, Ω_r = 0, Ω_Λ = 2.0 and Ω = Ω_m + Ω_r + Ω_Λ = 2.4 in this case.

OK, I can accept that for now.

What still puzzles me is how you would convert a calculation of H² to a pressure to apply to the balloon. Or are you going for a "bang-bang" system where a valve on the gas supply is either open or closed? I would rather expect a valve that can control the flow rate.

For my own slow understanding, let me take this step by step.

1. Continuously measure the radius R of the balloon and calculate (ΔR/dt)² = R²H² = R²Ho² ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ), with all the constants given and a = R/Ro, where Ro is just the starting radius of the balloon, I presume (it can surely not be 100 billion light years??)

2. Measure ΔR/dt, square the result and compare it with the calculation.

3. If squared result smaller than the calculation, increase gas flow rate.

4. If squared result larger than the calculation, decrease gas flow rate.

5. Repeat loop

Am I making any sense so far?

Hi again Guest. Not 'slow understanding', but rather a very good reply!

Some comments:

"... where Ro is just the starting radius of the balloon, I presume (it can surely not be 100 billion light years??)"

Ro is not necessarily the starting radius; it must be the radius associated with the constants used. You can decide on (say) 1 meter for Ro and then start the balloon at a much smaller radius R to see what will happen.

I suppose you must also add the 'switch' for inflation/deflation to step 1 of your loop.

1b. Calculate switch = (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ. If switch is positive, set gas flow direction for inflating the balloon. Else, set gas flow direction for deflating the balloon.

In a rapidly inflating balloon there may be all sorts of stability issues with the control loop, so I think one must limit the rate of gas flow, which will be very easy in practice. Cosmic expansion is actually very, very slow (%-wise), so we can decide how much we want to speed up the simulation for observability.

-J

Ah, OK - thanks.

Another puzzle to me is the deflation when your switch ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ) comes out negative. How can a vacuum energy dominated model (like your case 0.4, 0, 2.0) be deflating at any time? Vacuum energy is supposed to be acting like anti-gravity!

Hi Guest, you asked: "How can a vacuum energy dominated model (like your case 0.4, 0, 2.0) be deflating at any time? Vacuum energy is supposed to be acting like anti-gravity!"

This is a very important question and easy to misunderstand. Look at it this way:

When you look at the expansion/collapse graph in #233, you will notice that the collapse starts at R ~ 34, meaning a ~ 0.34. At that time the matter density term Ω_m/a³ ~ 9.84, while Ω_Λ = 2.0 (constant), for a total of 11.84. It is clear that matter dominates at that this time (t ~ 7.5 Gy) in the expansion equation:

(ΔR/dt)² = R²Ho² ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ)

At the same time the curvature term (1-Ω)/a² ~ -11.84, canceling the energy density terms, making the expansion rate near zero. If ΔR/dt stays ever so slightly positive, it eventually inflates faster and faster again. If ΔR/dt goes ever so slightly negative, it causes a deflation that happens faster and faster. Vacuum energy does not allow a stable static condition - it demands that space either expands or contracts near-exponentially.

I hope my brief explanations are adequate - if not, please ask!

-J

Jorrie, you said

"If ΔR/dt stays ever so slightly positive, it eventually inflates faster and faster again. If ΔR/dt goes ever so slightly negative, it causes a deflation that happens faster and faster. Vacuum energy does not allow a stable static condition - it demands that space either expands or contracts near-exponentially."

Your equation (ΔR/dt)² = R²Ho² ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ) does not show this behavior, so this part that still bugs me.

Hi again Guest, you wrote: "Your equation ... does not show this [diverging] behavior, so this part still bugs me."

In a way the equation (ΔR/dt)² = R²Ho² ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ) does show the instability, but I agree that it does not clearly show in which direction the radius will diverge.

Fortunately, there is another independent Friedman equation to fall back upon - the acceleration of expansion:

ΔR²/dt² = R Ho² (Ω_Λ - Ω_m/2a³)

If both the rate of expansion and the acceleration of expansion have the same sign, it is pretty clear into which direction the curve will diverge.

It also happens to be true that if the curve has a negative slope at any time, it will always diverge negatively, because the 'pull' of mutual gravity becomes stronger and the 'push' of vacuum energy becomes weaker. This is why the 'latching valve' idea can work. If vacuum energy starts to dominate before mutual gravity can cause a negative slope, it always diverges positively.

Interestingly, there is a theoretical quasi-equilibrium for the case (0.5, 0, 2.0), shown on the right. The expansion curve approaches the radius a=0.5 asymptotically, since both expansion rate and acceleration approach zero there. However, any perturbation will still let the radius diverge.

-J

Would it be workable if I update the steps in my post 234 like this, based on your comments? I followed your method of bolding the parameters, because it is clearer that way.

0. Set gas flow direction valve for inflating the balloon.

1. Continuously measure the radius R of the balloon and calculate (ΔR/dt)² = R²H² = R²Ho² ((1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ), with all the constants given and a = R/Ro, where Ro is a value that corresponds to the R for which the parameters are given.

1b. Calculate switch = (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ. If switch goes negative, even temporarily, change the gas flow direction direction valve for permanently deflating the balloon. (??)

2. Measure ΔR/dt, square the result and compare it with the calculation of ΔR/dt².

3. If squared result is smaller than the calculation, increase gas flow rate.

4. If squared result is larger than the calculation, decrease gas flow rate.

5. Repeat from 1 until timed out (or something like that?)

It is an interesting exercise, but I still cannot figure what you want to do with it. You said you want to show applications. (?)

Regards.

Hi Guest.

I think your algorithm is good. Step 2 has just got a bracket missing. Should be:

2. Measure ΔR/dt, square the result and compare it with the calculation of (ΔR/dt)².

If you don't mind, I'll modify it slightly and transfer it to the opening post as part of the 'design section'.

You wrote: "It is an interesting exercise, but I still cannot figure what you want to do with it. You said you want to show applications. (?)"

I have written a few possible 'applications' in my previous post (the new change note). I guess ultimately it is just a 'learning experience'...

-J

Jorrie, you are welcome to use my humble (not Hubble) contribution.

Anyway, thanks for an interesting thread.

Cheers for now.

I wrote: "1b. Calculate switch = (1-Ω)/a² + Ω_m/a³ +Ω_r/a⁴ + Ω_Λ. If switch is positive, set gas flow direction for inflating the balloon. Else, set gas flow direction for deflating the balloon."

This logic will not work, because the flow direction will switch backwards and forward. As soon as the switch goes negative and a starts to reduce, the switch will go positive again, etc.

One way out is to have a latching valve. It must normally be in the positive position (inflating balloon), but if the switch calculation ever goes negative, the valve must latch into the negative position (deflating balloon).

It is possible to design something that will positively determine the required direction of the valve, but that's probably an unnecessary complication.

-J

Hi Roger.

Firstly, despite a lot of effort, we are not able to come up with a model that is anything like the real cosmos and it is pretty obvious why: we do not understand the cosmos. We do not know what dark matter is and we do not know what dark energy is. All we know is the effects that they have on the 4% energy of particles, radiation and normal matter that we can observe. The ΛCDM 'model' is just a set of equations and parameters that fit observations to within achieved accuracy limits.

Back to your concerns:

"Energy Density of space remains constant? Nope, the balloon skin becomes thinner as it expands."

That's exactly why I tried the double skinned model, where the skins plus fluid in the cavity represent space with a constant energy density. It gave an accelerated expansion, but with the wrong curve, unless I 'cheat' by cleverly changing the density in the cavity (i.e., programming the pump to artificially get the correct expansion curve).

"Expansion accelerating? Nope, only if you force it with a pump, the balloon actually wants to decelerate expansion as it expands."

With the inevitable 'programmed pump', I can get the correct expansion curve easily - independent of skin properties (except that it must not tear).

"Curvature increasing as expansion increases? Nope, actually the balloon does the opposite."

I have given the modern definition of cosmic curvature and it is not related to the balloon's curvature. In fact the (1/R) curvature of the standard balloon analogy is only used to show why all objects fixed to the surface moves apart when the balloon expands. But, as you well know, it is not this curvature (or cosmic curvature for that matter) that causes expansion, so its just a visualization aid.

"The balloon system isn't allowed to have a particle, as it doesn't express the gravity of the particles correctly and any particle would break the homogeneity argument anyway."

Just like any galaxy breaks the homogeneity argument of the ΛCDM model anyway! Despite this, the large scale cosmos follows the ΛCDM model very closely, so no problem - make the balloon very large...

"I'm sorry but I simply am not seeing how this is a great model."

I was also hoping for more and was quite exited about your efforts when mine failed.

That said, if all my 'programmed-pump-balloon' analogy does is to faithfully reproduce the ΛCDM model's expansion curve, it will give engineers something that is more accessible than dark energy (etc.) to cling to. When they then place a particle on that balloon's surface and give it a push, they will learn quite a lot. Ask Jonmtkicco...

-J

For whatever reason, I couldn't see the article - but I think I remember the basics.
Some pulsars have been measured quasi-continuously with Allan variance of about 1e-16, which is probably the most precise physical measurement ever made by man. I think that we might expect to detect something as the gravitational wavefront passes us??
Pulsars have been watched with precision improving from about 3e-16 since about 2000. The question seems to be one of the following:
How much more precise do we need to get?
How much more continuously do we need to watch? or
How long we will have to wait?

You Wrote:"For whatever reason, I couldn't see the article".

I didn't think the APS News article (which I read the hard copy of) would be online. Then I realized how unlikely that would be (that it wasn't online) and searched again and found the actual article I read (the other link was just describing the technique).

Here it is:

http://www.aps.org/publications/apsnews/200906/pulsars.cfm

This thread has been interesting but has sent me on a tangent. Specifically I've been interested now in gravity waves and perhaps how they interact with de Broglie waves (though I haven't gotten to that last part yet, I'm still reading about gravity waves).

One cool thing I've been reading about is the idea of cosmic gravity-wave background.

http://www.aip.org/pnu/2007/split/809-1.html

What I'm trying to figure out is if gravity waves , along with changes in curvature, constract and expand the space they pass through.

"the curvature of the early universe (directly after inflation) was about 60 orders of magnitude less than it is today, not higher! If not precisely zero, curvature evolves away from zero. With curvature so small today, it was utterly negligible in the very early universe.

I am bewildered by this. If the universe was a sphere at the moment of the Big Bang (or after inflation), and it is a sphere now, then the curvature is the same, no?

The curvature of the balloon has in fact nothing to do with the curvature of the universe - they evolve in different directions!"

I thought we were trying to simulate the universe as we think it is. I'm wondering what the point is if you don't believe it matches reality. What is your impression of this and this?

Didn't have time to read both articles, but they are helpful.

Thank you very much.

I'll try to finish them both sometime tomorrow.

I shall be looking specifically for how my hotdog twisted balloon image is made stupid and useless by them.

Hi S, you wrote: "I am bewildered by this. If the universe was a sphere at the moment of the Big Bang (or after inflation), and it is a sphere now, then the curvature is the same, no?"

No, the curvature of a normal sphere's surface is inversely proportional to its radius (1/R). The curvature of the universe is a somewhat different thing, which at large radii has only some resemblance to the curvature of the hyper-surface of the hypersphere. That is why a balloon-like thing is sometimes used to explain cosmic curvature and it is useful as long as we do not consider changes in curvature, which work in the opposite sense.

The cosmic curvature has to so with the energy balance of the expansion. If energy of expansion is precisely equal to the (negative of the) potential energy of the cosmos, it is said to be flat (total energy zero and Ω = 1). If energy of expansion is smaller than the (negative of the) potential energy of the cosmos, it is said to be closed, with positive curvature (Ω > 1) and with zero vacuum energy (Λ = 0), it will eventually collapse. With Λ > 0, even a cosmos with Ω > 1 will expand forever.

"I thought we were trying to simulate the universe as we think it is. I'm wondering what the point is if you don't believe it matches reality."

As long as we have the right definition of cosmic curvature in mind, the balloon analogy works very close to perfection. I have used it from t = 2 seconds after the BB, up to very far into the future and it agrees with all results from the standard cosmic model.

I know the references that you quoted - very nice and correct visualizations, especially for the popular press. They are not very useful for the purposes that I have in mind, which is to show that the balloon can satisfy the Friedman equations numerically. Trying to do everything in 3D is possible, but will clutter the issue quite a bit, as I remarked to GK on his 'cosmic teardrop'. Yo have to go 'all the way', with semi-transparent surfaces and all, to get a good picture in 3D.

-J

"the curvature of a normal sphere's surface is inversely proportional to its radius (1/R)"

I had gone to Wikipedia for a definition, and got "curvature is the amount by which a geometric object deviates from being flat, or straight in the case of a line,". I was thinking angles. I saw the long line of formulas under the curve that I couldn't relate to but missed the simple one! Anyway, I'm still confused.

"the curvature of the early universe (directly after inflation) was about 60 orders of magnitude less than it is today, not higher! If not precisely zero, curvature evolves away from zero. With curvature so small today, it was utterly negligible in the very early universe.

So you are saying that the universe is 600X more curved now than right after inflation? An expanding universe is like a contracting balloon? What about the effects of gravity when all the mass is close in? Doesn't that curve space more?

"The cosmic curvature has to so with the energy balance of the expansion. If energy of expansion is precisely equal to the (negative of the) potential energy of the cosmos, it is said to be flat (total energy zero and Ω = 1). If energy of expansion is smaller than the (negative of the) potential energy of the cosmos, it is said to be closed, with positive curvature (Ω > 1)"

I don't think I will ever understand this. Maybe I'm a simpleton, but I can only conceive of two possible universes: 1. Spherical (3 or more dimensions) and finite, and 2. flat and infinite. That's just the way I am. Maybe in the next life I'll be smarter.

Hi S, do not be too despondent about not understanding this curvature issue - very few people do. I keep my sanity by thinking like this (rightly or wrongly):

If the energy of expansion equation is exactly balanced (to zero), then the inside angles of triangles built in space add up to 180° and we have flat space. If the total energy is less than zero, the angles add up to more and we have positive curvature and visa-versa for negative curvature.

In the very early universe, the curvature was so close to zero that we cannot even comprehend it, i.e., the energy equation was very, very close to being balanced. The fact that we picture it as a balloon with a very tiny radius (high curvature) is irrelevant because it is a different kind of curvature. It may mean that the balloon analogy is very misleading in the very early universe.

However, by the time that the normal ΛCDM model kicks in (t~1 second, because before that it becomes dicey), the total cosmos might already have been so large that the balloon analogy curvature is close enough to zero for all practical purposes.

Remember that if cosmic curvature is exactly zero, the ΛCDM model says that the cosmos started out infinitely large... Drat! Now I've opened another can of worms.

Fortunately we know that the model fails at very early times, so that's a way around the issue.

-J

Please don't go.

I think you are right to "not understand" it in the way it is generally presented.

I'm probably wrong, but to my mind the problem is resolved by considering that the 4D space is itself a Riemann surface - i.e. it is actually a 4D surface in a higher-dimensional space. This also seems in accord with current physics at the level of elementary particles, where we can see QM states dropping out when 6-dimensions are used (shough 11D string theory seems too all-inclusive to me - it would appear that you can model so many different behaviours that it becomes non-predictive - just a prejudice, of course).

That said, 4D with modifiers would remain a convenient tool.

I think the best way to start getting equations for the vapor condensation fluid model to see if it holds up under scrutiny is to start with a static non-expanding puddle on a flat surface and see what those equations look like.

Let's simplify our puddle of liquid to two forces. One force that wants to push the fluid outward and another force, characteristic of the fluid, trying to hold it together. When these two forces are equal, the puddle doesn't spread out or contract.

The outward force can be described as:

F_outward=mg=ρAh_equilibriumg

where ρ (rho) is the mass density, A is the area, which times h the height gives you the volume, which times ρ (rho) gives you the mass. g is the normal 9.8 m/s² acceleration.

F_inward=-F_outward for static puddles. So:

F_inward=-ρAh_equilibriumg

Leading to:

F_net=F_inward + F_outward = 0

So now lets see what happens when we "condense" some more fluid on our puddle. Lets assume that condensing the fluid increases the height of the puddle.

F_outward=ρAh_newg

F_inward=-ρAh_equilibriumg

F_net=F_inward + F_outward = ρAh_newg - ρAh_equilibriumg = (ρAg(h_new-h_equilibrium)

So, there is an outward force that is proportional to the difference between the new height of the puddle and the old equilibrium height. Also notice that if h_new is lower than h_equilibrium, then the force points inwards (is negative).

Pressure is defined as force per area.

So the pressures can be defined as the following:

P_inward = F_inward/A = -ρh_equilibriumg

P_outward = F_outward/A = ρh_newg

P_net = F_net/A = (ρg(h_new-h_equilibrium)

I'll stop at this point and open the above to criticism. Please take a look and let me know if I'm making any errors or there are any objections with the method or how I'm proceeding.

Ok, the liquid on the surface of a balloon model has undergone some slight changes and I wanted to post a comprehensive update.

1. A liquid puddle, which represents space, is on the surface of a very large frictionless balloon which itself is just a template. The balloon and liquid interact weakly. Enough so that the liquid puddle stays put, but not enough for the expansion or contraction of the balloon to effect the expansion or contraction of the liquid (I may be in trouble with that last statement, it may not be possible to completely decouple balloon expansion and liquid expansion as I thought earlier, though we can make them weakly coupled with this model).

2. The liquid is an ordinary liquid with viscosity and pressure. It will spread out over as much of the balloon as it can until it reaches equilibrium. This happens because our model is built on Earth and the liquid feels g acceleration. We can prevent equilibrium by creating an environment where condensation (or evaporation) is constantly occurring along the area of the liquid. This will cause the liquid to spread out more (or contract). This corresponds with the expansion we see in space.

3. As the curvature of balloon underneath increases, the equilibrium of the liquid is displaced (the same amount of liquid spreads out more). The further we get from a flat surface, the faster this acceleration becomes (here's your antigravity Jorrie, or at least part of it).

4. Two knobs are associated with this system. One for controlling the condensation (or evaporation) rate, the other for controlling the contraction (or expansion) rate of the balloon.

Hopefully equations are on the way. This is just a summary of the idea.

Consider this modification to the structure of the sphere: Instead of being inflated from increased pressure within,as a balloon, consider it as a bubble rising in a liquid, expanding as it nears the top.It would retain all of the original matter and energy that it started with, without the need for an additional inside "pressure".The inflation would be caused by a decrease in outside pressure.

How will this affect the outcome of the experiment?

Hi Guest. Interesting comment.

Yes, over a short period, a bubble rising in a fluid could work - it would mean choosing the densities inside and outside (as well as viscosity of the fluid) for a specific epoch, in order to get the correct expansion rate. Otherwise, some form of control system will have to maintain a specific expansion profile over time.

For the longer term, there might also be the problem that the bubble must be rising forever and do so at an increasing rate (at least so that the bubble size grows faster and faster). Once you have built all that into a control system, it will probably be more complex than simply adding gas to the interior of the balloon at a pre-programmed rate.

-J

A bubble does increase it's velocity and diameter as it rises.The pressure inside the bubble actually decreases as it rises and expands.The mass density will decrease, but total mass remains the same.Could our universe be actually a bubble,rising through a fluid, increasing velocity and diameter as it rises? No need for dark energy to increase diameter and speed.

What if the bubble strikes an obstruction that slows it ascent, such as a slanted shelf for instance?This effect could be used to get just about any expansion rate by varying the angle of the obstruction,without the nescessity of controlling other difficult parameters.

Observe closely a bubble of air in a container of viscous liquid.The old Prell shampoo commercial comes to mind.The tilt of the bottle can control rate of rise.

Hi Guest,

Yes, it may be possible to perform such a test. One other problem is that buoyancy only works when there is gravity and local gravity distorts some of the tests that we may want to do.

An analogues method would be to put a balloon in a vast pressure chamber in free space. Then lower the pressure in the chamber (programmed) so that the balloon can inflate according to the desired profile.

Personally, I still prefer the programmable pump that inflates the balloon in empty space according to any profile I choose, even form practically zero radius.

-J

The problem I see with the balloon analogy is you have to keep adding mass to the balloon to increase size.My understanding of the universe is it retains the same amount of mass as it expands.Also, as the pressure increases, the temperature must also increase.The balloon will also be deformed from ideal by the attachment point of the inflation hose.

For the bubble theory, imagine a gas dissolved in a liquid,like champagne, before the cork is popped.Now instead of a sudden release of pressure, release it in a controlled manner to vary the size of the bubble formed.The mass of the bubble will remain the same, as it expands.Will this scenario fit the current theory of the expanding universe, or do I have a bubble in my think tank?

Hi Guest.

The design that we eventually settled upon does not care about the mass inside the balloon - the adding of gas/fluid only serves to blow it up at a programmable rate. We have established that there is no (known) way of making a balloon that will automatically simulate the real universe, so we abandoned that idea. It is a 'brute force' method with a huge pump...

Your idea of controlling the pressure outside the balloon is good, but it boils down to the same thing. You will have to measure the expansion rate of the balloon and then control the outside pressure by 'pumping out' gas/fluid at the appropriate rate. One problem is be that you may end up with a vacuum outside and unable to control the pressure differential any further. That fluid with unlimited negative pressure is very elusive...

-J

Jorrie,

Thanks for being a good sounding board for my idea.I am getting a better "ping" from your balloon idea than from mine.I appreciate your time and patience.