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Cosmic Balloon Application III: Particle Momentum Decay

Posted July 02, 2009 12:00 AM by Jorrie

We have seen how cosmic redshift can be understood as photons conserving angular momentum by 'shedding energy' in terms of frequency as the cosmic balloon expands. Is it reasonable to think that particles simply shed speed in order to also maintain a constant angular momentum? So it seems.

Intro

Fig. 1 (right) shows a single massive particle (pink arrow) being shot along the balloon surface at a time when R=25 units. As the balloon expands, the particle maintains its angular momentum relative to the balloon center by traveling more slowly along the balloon surface. This is visible as the shrinking of the angle between the arrow and the radial lines, representing static observers on the surface. Instead of continuously spiraling out, as a photon would have done, the particle's angle to a local radial line eventually approaches zero. This means that over time the particle's surface velocity (momentum) decays to zero.

This effect is qualitatively independent from the expansion profile, be it decelerating, constant or accelerating expansion, as long as there is expansion and not contraction. This is a rather surprising result of cosmic expansion, because it looks like a 'drag' or 'friction' that the cosmic balloon surface (space) exerts on particle movement. However, when there is no expansion (a static universe), the effect disappears, ruling out that it is a friction. What is more, should the cosmic balloon shrink, the particle will gain speed relative to balloon surface, as it has to in order to maintain a constant angular momentum.

Analytics

In Galilean dynamics, it is the kinetic energy (½mv2) and the momentum (mv) of a particle that decay due to an expanding universe. The velocity v is also known as the 'peculiar velocity' in order to distinguish it from the 'Hubble velocity', which is the apparent recession velocity (relative to us) of a distant object that is static relative to the balloon surface.

As the radius R of the balloon increases, the v of the particle decreases in order to keep angular momentum L=mRv constant - at least in Galilean (low speed) dynamics. In relativistic (high speed) dynamics, things are not much different, just with the special relativistic Lorentz factor gamma (γ) entering the equation.[1]

L = m R v γ = constant -------- (1)

where γ = (1-v2/c2)-0.5. It is relatively easy to plot the curve of Fig. 1 from this equation, together with the expansion equations of Friedman, of course. Depending on the speed of the particle and the expansion profile, the particle path may be virtually straight, as in Fig. 1, or it may be curved. In the end, it always ends up going more and more radially, meaning with less and less momentum along the surface.

Figure 3.5 above is from a doctoral thesis by Tamara M. Davis, under supervision of prof. Charles Lineweaver, titled "Fundamental Aspects of the Expansion of the Universe and Cosmic Horizons".[2] It shows the decay of the velocity of particles for various initial particle speeds, with light (v=c) the top curve. It is clear that the larger the kinetic (momentum) part of the total energy of a particle (see Eq. 2 below), the more it behaves like a photon.

Particle Redshift?

It is possible to analyze this case in terms of redshift of de Broglie waves.[3] Following the thoughts in the previous Blog entry ("The linear momentum of a photon is given by p = h, where h is the Planck constant") and Roger's suggestions on de Broglie waves, the following interpretation is very interesting.

The relativistic energy of a moving particle is given by

E = √[p2 + m2] c2 = √[(h)2 + m2] c2 -------- (2)

where p is the momentum, m is the rest mass and λ the de Broglie wavelength of the particle. It seems like a "de Broglie momentum" (h) is added vector-wise to the rest mass (m), giving the particle its wave-particle duality. Particle mass is presumed constant, so in order to keep angular momentum relative to the balloon center constant, the de Broglie wavelength of the moving particle must redshift more or less like for a photon.

Angular momentum of a particle moving on the balloon is given in special relativity by eq. (1) above as: L = m R v γ = constant.

But, according to ref. note [3]:

γ = h / (λ m v) -------- (3)

so we can also write:

L = R h / λ = constant -------- (4)

exactly as for a photon. If R is increasing, λ must be increasing by the same ratio in order to keep L constant. This is same as a velocity decrease in order to keep the relativistic angular momentum of the particle constant on the expanding balloon. The velocity equivalent to a specific λ can be extracted from Eq. 3, taking into account that γ is also a function of v.

Caution

One must always remember that the balloon analogy is just that, an analogy. It may help to get 'handles' on some of the puzzles of the cosmos, but it does not represent the real thing. As an example: the cosmos might be precisely flat, with no curvature, yet the behavior discussed here is apparently still present. How and why, I don't know.

Jorrie

Notes

[1] This is essentially the special relativistic form of Kepler's second law of planetary motion: "a planetary orbit sweeps out equal areas around the Sun in equal time intervals", where the equal time intervals are now measured in proper time of the moving particle. We do not need the general relativistic form here, because particles on the balloon feel zero net gravitational acceleration (the whole surface is at the same gravitational potential).

[2] http://arxiv.org/abs/astro-ph/0402278, a doctoral thesis by Tamara Davis, under supervision of Charles Lineweaver, page 50.

[3] http://en.wikipedia.org/wiki/De_Broglie_Wave

-J

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#1

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/02/2009 7:52 PM

But here we have mention in the noted literature of an empty universe.

To be part of the universe, and empty at the same time, this would imply to me more than one universe.

Of course the universe is defined as all existing things.

To have existing things, then we have nothing to balance.

There must be a nothing, as absolute as anything, for there to be anything.

Our model is only intended to look alright from afar.

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#2
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/02/2009 10:08 PM

Hi T, mentions of an 'empty universe' in the literature are in connection with a constant expansion rate. The mathematics are still the same, but the results (curves, etc) are easier to comprehend in certain case. This does not mean anyone proposes an empty universe! This thread is about particles (which could just as well have been galaxies) moving through in space, so it is very far from an 'empty universe'.

I think I may understand where your quest to have more than one universe comes from. In the BB, matter and antimatter should have been created in equal amounts. Yet, today, we basically find only matter occurring naturally and scientists do not know why. Maybe an equal amount of antimatter created another universe in another set of dimensions. We would never know, because there seems to be no interaction possible between such universes.

-J

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#3
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/02/2009 10:48 PM

I was just caused to continue due to my reading of the references.

Certainly as well, I am influenced by the report of a discovery of a great "Void".

Apparently this absolute nothing was predicted by String Theory.

Equality of anti-matter and matter is likely to be in flux.

What should be, and what is, could very well be both a reality.

Personally I feel that the mechanics of any other universe would essentially be the same. In prior discussions where I have postulated that the only difference I can imagine is that there may be a universe with a different constant for the speed of light, the essential mechanics are not altered.

Slower, and elements are more stable. Faster, and elements are more unstable.

This is the sort of thing I have learned from you and others in your discussions.

When I have studied the balloon representations of a finite universe, they touch in drawn representation as having characteristics of imagery that approaches the imagery that came to me in the word descriptions of the perfect balloon representation.

Where did the infinity circle come from?

When I study what the human mind can conceive, and even discover as facts, I am often awed and amazed.

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#4
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/06/2009 12:43 PM

Hi T, you wrote:

"When I have studied the balloon representations of a finite universe, they touch in drawn representation as having characteristics of imagery that approaches the imagery that came to me in the word descriptions of the perfect balloon representation.

Where did the infinity circle come from?"

I don't understand what you mean by an 'infinity circle". The circular representation of the balloon is always finite.

-J

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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/06/2009 10:58 PM

Tonight I spent some time looking for a good explanation of the infinity circle, and actually didn't find one.

If you have an old camera lens around it will have a focus setting for infinity that uses the image that has been driving me since I started reading your question so much as to post anything related to this balloon model.

This image coming from the lenses on cameras inspired my twisted hotdog balloon image, and continues to upset one universe perfection, and inspires me to be so bold as to suggest putting your two perfect balloons next to each other.

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#5

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/06/2009 3:43 PM

Jorrie,

I'm sorry I'm late to this interesting post. Thanks for putting this together.

I understand your center of the balloon angular momentum approach as you've explained it in this and previous posts, and it is interesting that it works. Is this your idea (analogy) or have you taken it from other sources (if you've already referenced the source in a past post, I'm sorry I missed it).

How approximately how much of a change in momentum does one get with the current rate of cosmic expansion?

I have other questions, but I want to read your references first.

Roger

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#8
In reply to #5

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/06/2009 11:16 PM

Hi Roger.

You asked: "I understand your center of the balloon angular momentum approach as you've explained it in this and previous posts, and it is interesting that it works. Is this your idea (analogy) or have you taken it from other sources ..."

Momentum decay of freely moving objects with the inverse of the expansion factor (i.e., p = p0/a) is today widely published. Tamara Davis does reference some more sources. I find this one the 'nicest': http://arxiv.org/pdf/0808.1552v1 by Hongbao Zhang.

Note that this is decay over billions of years and only for freely moving objects, so there are no implications for particles in CERN and other accelerators, AFAIK. You need 'expanding space inside the experiment' for it to be measurable, so it only really happens over cosmological distances and timescales.

-J

PS: My idea to 'couple' particle momentum decay to the 'cosmic balloon' analogy came only after your input on de Broglie waves, although it does not need the wave theory - particle (momentum) analogy works equally well. It is however extremely interesting that one can treat massive particles and photons on an equal basis as far as cosmic redshift is concerned.

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#9
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 12:48 AM

Thanks Jorrie,

I was wondering, in your model/analogy, the R is the radius from the center of the balloon to the surface, right? Is there anything this translates to in the real universe? I mean, for our universe, we don't know R (since R is just some radius of an imaginary model that can be used to simplify our understanding of redshift due to cosmic expansion)? There's no value we can assign for R for the real universe?

Also, what is the value of a in your equation P=p0/a. I mean the numerical value accepted today(a is the scale factor, right?).

Roger

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#10
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 4:56 AM

Hi Roger.

Yes, in the hyper-spherical model, R is the usual radius of the sphere. As far as translating it to the real universe, no, there is no rigorous way. The radius of curvature of the universe is either infinite or near it. A value that is presently thrown around, coming from current WMAP data, is 100 Gly minimum, but probably much larger, possibly infinite (flat universe).

For a flat or near-flat cosmos, none of the usual cosmological parameters and results care about the radius of curvature. We can choose any value equal to or larger than the Hubble radius (some 14 Gly) and all results will be the same. I normally use Rnow = 100 Gly for convenience.

The expansion parameter a = R/Rnow, so obviously a = 1 at present, with the evolution of a over time given by the Friedman solution.

-J

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#11
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 7:19 AM

If a=1, wouldn't that give no particle momentum decay and no photon redshift?

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#12
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 8:30 AM

Hi Roger, you asked: "If a=1, wouldn't that give no particle momentum decay and no photon redshift?"

Only if a=1 forever, meaning somehow expansion has stopped. Otherwise, a changes continuously and we observe photons that originated when a << 1.

-J

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#13
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 9:16 AM

I understand what you're saying. Here's the issue I'm concerned with:

Cosmic expansion is going on as we speak. It happens continuously. That means that particles are losing momentum as we speak. Let's not worry about bound particles for now (since everyone says they don't get affected and I don't want to debate it), but for free particles, right now at this very moment, they are slower than they were a moment ago. The difference is no doubt unbelievably small in their momentum from one second to the next, but it's there.

I'd like to know what the rate of decline in momentum (or just the rate of redshift) is at the given rate of expansion. Is this possible to know?

Roger

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#14
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 12:18 PM

Hi again Roger.

"I'd like to know what the rate of decline in momentum (or just the rate of redshift) is at the given rate of expansion. Is this possible to know?"

The present rate of change of a is indeed incredibly small. It changes by only about 10% in a billion years (~1 part in 1010 per year). The % change in momentum and wavelength are of the same order. Think about it: a has changed from 0 to 1 in 13 billion years. You can work it out per second, if you want, but it seems rather pointless.

Mathematically the present da/dt = H0/978 = 0.073/Gy. The 978 is just a conversion factor between km/s/Mpc to Gy-1.

-J

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#15
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 1:07 PM

So do astronomers not notice any redshift on galaxies a 100 million lightyears away? The way you are describing it it seems like anything under a billion lightyears away isn't really redshifted. Is that true?

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#16
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 2:06 PM

Roger, you wrote: "The way you are describing it it seems like anything under a billion lightyears away isn't really redshifted. Is that true?"

I do not understand how you get to this interpretation from what I wrote. The present rate of change of a or z or λ has little to do with the the actual redshift at any distance.

You can use the Cosmo-calculator on my website to verify the distances for various redshifts. A redshift of z ~ 0.0075 gives a distance of about 100 million light years and a recession speed of v ~ 0.0075c, valid for galaxies having no peculiar velocities. In practice, galaxies have ~0.001c peculiar velocities (on average) in semi-random directions, so there are some errors in redshifts for distances of 100 million light years. The Coma cluster, at some 330 million light years is about the closest galaxies with reasonably reliable redshift measurements (z ~ 0.025).

-J

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#17
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 2:36 PM

You Wrote:"I do not understand how you get to this interpretation from what I wrote. The present rate of change of a or z or λ has little to do with the the actual redshift at any distance."

I'm just trying to see how much of a reduction there would be in momentum for a very short timescale (like a second). I'm sure you are right that this would be an unbelievably small number. I just want to know what it is. The equation you give is a large timeframe approximation. See below:

P=P0/a

a=1 thus P=P0

That's not an exact solution, it's saying that the change is so small they are basically the same.

What I'm asking for is a solution that says "At the current accepted rate of expansion right now you lose x amount of momentum per second".

I'm sorry about the confusion. I realize the answer may be something like 10-100 kg·m/s/s of momentum is lost, or smaller, I know. I just want to know. Can you tell me that?

Roger

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#18
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 3:28 PM

Hi again Roger, you wrote: "That's not an exact solution, it's saying that the change is so small they are basically the same."

P=P0/a is a good solution for non-relativistic objects. P0 is the momentum at time t0 (the usual cosmological convention for the present, with a=1), but a changes with time (at the rate ~10% per billion years), so free object momentum changes at the same rate. With this you can easily calculate any time based rate that you desire. Obviously, we can have a=1 only for one instant of 'now', not for a succession of 'nows', which may or may not be the cause of your problem with it.

If your problem is with relativistic particles (v/c > 0.9), the calculations are obviously a bit more involved. Essentially, relativistic particle momentum decays less than slow particle momentum. If you really want that, I can help you, but only in our 'tomorrow'.

-J

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#19
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 3:49 PM

For instance, I get about 2.3 x 10-16 kgm/s/s reduction in momentum of an object that started with a momentum of 1 kgm/s at the current expansion rates. This seems waaaay too large. Think about it, there are 3.15 x 1016 seconds in a billion years. That means if you had a rock with a linear momentum 1 kgm/s, in 730 million years it would stop.

You may not like the idea of using a mass so large in the above example (large masses just have tiny wavelenths). Lets ignore that for now. Are you getting anything similiar with regards to momentum loss? Or did I make a mistake?

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#20
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/07/2009 10:29 PM

Hi Roger.

I have no idea how you got that result - you surely did not use the cosmic balloon!

At the present time, the rate of change is ~10% per billion years (as I've stated more than once), so how would that stop your rock in less than one billion years?

My cosmic balloon simulation yields that your rock's momentum will reduce to ~40% of its present momentum when the universe is double its present size, taking into account that the radius of the balloon will grow to more than double in that time, reducing the change rate to less then 5% per billion years by the year 27.4 billion. Even holding the ~10% per billion year decay rate constant for 13.7 billion years will reduce the momentum to ~ (1-0.1)13.7 ~ 0.24 of its present value, approaching zero asymptotically at infinity.

The size of the mass does not matter, because as I said before, we do not require the de Broglie waves to get to the correct result - straight momentum calculations work just as well and give the same answers.

-J

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#21
In reply to #20

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 7:30 AM

Your result sounds much more realistic to me. What I did was I took the Z from 100 MLY and divided it by 100 MY (converted to seconds) to get a Z per second. From that I calculated the change in momentum per second. I post the equations later so we can figure out where I went wrong (I have to go to work right now).

Also, just wanted to let you know my Cosmology book arrived today. I'm sorry to support the competition, but I figure the more explanations the better.

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#22
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 12:00 PM

Hi Roger.

It looks like your calculation produces dz/ds, where s is distance, instead of the dz/dt required for a rate of change. Since a = 1/(1+z), I find it easier to just use da/dt = H0/978 (from Friedman when a=1).

I hope you enjoy Weinberg's book - it's one of the most modern ones out there, but perhaps a bit technical, I've heard.

-J

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#23
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 2:18 PM

Hi Jorrie,

You've given me the additional information I needed to get this right (I hope). Can you check my derivation below and make sure the results are correct?

Derivation

1+z=λ/λ0

λ=h/P
λ0=h/P0

λ/λ0 = P0/P

1+z=P0/P

P=(1/1+z) P0

P= aP0

So.

dP/dt = da/dt P0
dP/dt = (H0/978) P0
dP/dt = ((74.2 km/s/Mpc))/978)P0
dp/dt = (.0759 Gy-1)P0

So it seems like there is roughly a 7.5% reduction in momentum every billion years? Is that correct? It still seems wrong but I don't know where I went wrong.

Roger

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#24
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Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 3:27 PM

Hi Roger.

You have it mostly right, but P = P0/a (not P = a P0, as you have it).

Despite this, for a~1, dP/dt ~ da/dt P0 = (H0/978) P0, as you have it, so the 7.5%/Gy is quite correct. I've used a slightly different value for H0 and then rounded it to ~10%/Gy. Remember that this is only valid for a~1, so you cannot use it for billions of years into the future. For that you need to integrate the Friedman equation numerically.

-J

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#25
In reply to #24

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 4:29 PM

You Wrote:"Despite this, for a~1, dP/dt ~ da/dt P0 = (H0/978) P0, as you have it, so the 7.5%/Gy is quite correct. I've used a slightly different value for H0 and then rounded it to ~10%/Gy. Remember that this is only valid for a~1, so you cannot use it for billions of years into the future. For that you need to integrate the Friedman equation numerically."

Thanks for checking my math (I agree with your correction of course) and pointing out the answer will be the same.

7.5%/Gy seems like a lot to me and I'm still worried it's wrong. It's quite remarkable if you think about it, that's a large amount of momentum lost and it isn't size dependent. It seems like it amounts to about an angstrom/second per second of lost speed (the loss of momentum is really loss of speed).

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#26
In reply to #25

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/08/2009 9:45 PM

I Wrote:" It seems like it amounts to about an angstrom/second per second of lost speed (the loss of momentum is really loss of speed)."

I should have been clearer here. What I was trying to say was:

There are 3.16 x 1016 seconds in a Gy

so

.075 Gy-1 = 2.3 x 10-18 sec-1

so

dP/dt= 2.3 x 10-18 sec-1 P0

since

P=mv and P0=mv0

then

dV/dt = 2.3x10-18 sec-1 V0

So the faster the particle is moving, the more it slows down per second, which makes sense I guess. All of this is based on the idea that the rest mass can't be changed due to expansion.

Anyway, based on the above, an object moving 107 m/s would lose about 0.2 Å/sec/sec off it's speed.

It's behaves like a force

Also intereseting, as you noted somewhere in the past, it acts like a frictional force (while expansion is occuring).

dp/dt = kPo
F=kPo

Please check what I've written here to see if it makes sense (and catch where I've inevitably gone wrong). Then I have an exciting idea that may be worthless or interesting.

If this behaves like a frictional force, perhaps we can discuss the idea of a spatial "normal force". It's a nice idea because the spatial "normal force" vector could flip during deflation and be zero during static universes meaning it might, if I'm thinking correctly, offer a new way to thoroughly think about this subject. Space having a frictional coefficient and a normal force.

geez, if I've gone off the deep end, feel free to tell me Jorrie.

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#27
In reply to #26

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 12:29 AM

Hi Roger.

You are perhaps pushing the 'non-relativistic particle' case too far here. Recall what I wrote in the main article (opening post): "In relativistic (high speed) dynamics, things are not much different, just with the special relativistic Lorentz factor gamma (γ) entering the equation.[1]

L = m R v γ = constant -------- (1)"

I think the simplest way to look at your problem is to use eq. 4 from the OP:

L = R h / λ = constant -------- (4)

Speed-wise, the faster a particle, the more it behaves like a photon (check Tamara's graph in the OP), simply because the de Broglie wave energy (same as kinetic energy) becomes more and more significant. The way I do sums on this is as follows: work out L at R0 and λ0; this remains constant for a constant mass, so it is easy to find λ for a new R.

You wrote: "Space having a frictional coefficient and a normal force."

No, you should rather not say this - it will confuse the living daylights out of everyone! The theoretical disproof lies in the fact that if you have no expansion, there is no 'friction' on moving particles in empty space - they simply carry on with a constant speed forever. It is only when a (R) changes over time that the change in linear momentum comes into play. Think 'balloon analogy' and the conservation of angular momentum.

In a relativistic sense, it is not a 'force' either - it is a natural consequence of expansion, full stop! Like in a gravitational field, the only detectable force on particles is a tidal force. This is present in accelerating/decelerating expansion and it may be the topic of another 'application of the cosmic balloon'.

-J

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#28
In reply to #27

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 7:22 AM

You Wrote:"In a relativistic sense, it is not a 'force' either - it is a natural consequence of expansion, full stop! Like in a gravitational field"

Hang on now, give it a chance. Basically expansion causes a dp/dt. That is the definition of a force.

F=dp/dt

Sure, that force changes when the expansion rate changes, and since the expansion rate is constantly changing, in truth we have a dF/dt (impulse I think). Never the less, if we didn't know about relativity or cosmology, yet had a sensitive experiment, what we would observe is a slowing of these objects as though they are being acted on by a force.

Gravity is defined as curvature of space, but it is also called a force. I don't see why this aspect of the geometry of the universe can't be expressed in such a way as well.

Give it another thought and let me know what you think.

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#29
In reply to #28

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 8:19 AM

Hi Roger, you wrote: "Hang on now, give it a chance. Basically expansion causes a dp/dt. That is the definition of a force."

My point is that if it was a 'proper force', I could have put an accelerometer on the object and read off a proper acceleration. In this case the proper acceleration remains zero forever. So, at best it can be a pseudo-force that is coordinate dependent. IMO, proper forces are not coordinate dependent.

"Gravity is defined as curvature of space, but it is also called a force."

I view the gravitational 'force' as a pseudo-force as well. The only proper force that you feel while sitting in your chair is the chair pushing you 'up' to prevent you from following a geodesic in spacetime. When you free-fall in space (following the geodesic), you do not measure any proper acceleration, only coordinate dependent 'pseudo-acceleration'.

-J

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#30
In reply to #29

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 10:07 AM

Jorrie,

You Wrote:"My point is that if it was a 'proper force', I could have put an accelerometer on the object and read off a proper acceleration. In this case the proper acceleration remains zero forever. So, at best it can be a pseudo-force that is coordinate dependent. IMO, proper forces are not coordinate dependent."

Jorrie, the dp/dt we are describing is a proper acceleration. As the universe expands the object experiences a physical deceleration, not a coordinate deceleration. That would make the force a proper force. An accelerometer would detect a change in acceleration.

You Wrote:"I view the gravitational 'force' as a pseudo-force as well."

In that context I'll agree to call this force a psuedo force (since it will be in good company with gravity).

My point is, if you had an electron flying through space for 100 years and you had an experiment that measured it's speed, you would oberve a deceleration in that electron as though there was a force working in the opposite motion of the electrons motion. Is that fair?

Rather than getting bogged down into the convention of whether it's a force or not, can we say it "behaves like a force" and examine that line of thinking?

Roger

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#31
In reply to #30

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 12:01 PM

Roger, no, your: "As the universe expands the object experiences a physical deceleration, not a coordinate deceleration. ... An accelerometer would detect a change in acceleration." is not correct. It is a coordinate acceleration only.

In an expanding universe, there is no single inertial frame that spans an extended distance. Place observers Jack and Jill at some distance (D) apart along the path of the particle, Jill first and then Jack, both at rest w.r.t. the CMB (i.e., with zero peculiar velocity). Due to the expansion, they are each in their own inertial frame, with proper velocity (v=H D) relative to each other. Each measure the velocity of the particle as it passes them. Jack will measure a lower velocity for the particle than what Jill measured, but neither will measure an change in velocity in their own respective inertial frames. Hence, a zero reading on the particle's accelerometer. The change in velocity between Jack and Jill's measurement is simply a coordinate change issue.

"Rather than getting bogged down into the convention of whether it's a force or not, can we say it "behaves like a force" and examine that line of thinking?"

I can live with that, provided that we are not led up the garden path by it, like that the 'force' can be measured by an accelerometer! Expanding space alone has no mechanism for exerting any proper force on any single particle, moving or stationary.

-J

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#32
In reply to #31

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 1:08 PM

Jorrie,

I must be misunderstanding what a redshift of a bragg wave means. In my understanding, the momentum, that is the physical momentum, of the free particle in question, is reduced. That means that object is physically moving slower. One could say that if exapansion continued long enough, that that free object (like a free electron moving with some linear momentum through space) would slow to almost a stop. Where am I wrong here?

Roger

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#33
In reply to #32

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 2:38 PM

Roger, OK, I think I see what the problem is. When you wrote:

"In my understanding, the momentum, that is the physical momentum, of the free particle in question, is reduced. That means that object is physically moving slower", it appears as if you ignore the fact that velocity and momentum are purely relative concepts, i.e., coordinate choice dependent. Redshifts of cosmological bodies are also coordinate dependent things. Astronomers in other galaxies observe other redshifts than what we do for the same astronomical objects.

What does "that free object (like a free electron moving with some linear momentum through space) would slow to almost a stop" mean anyway? In what frame of reference? In the balloon analogy, it means it 'comes almost to a stop' relative to the balloon surface, but that surface is made up of an infinite number of inertial frames. It seems to come to a stop only in one of the many - wherever it is at a particular time. In all other frames it may accelerate, e.g., in the frame of the origin in the 'tethered galaxy' thread.

'Decay of particle momentum' in cosmological terms means a loss of momentum as measured in successive inertial frames of co-moving (CMB-static) observers. Maybe I have inadvertently created the false impression that momentum decays in all inertial frames.

-J

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#34
In reply to #33

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 3:15 PM

You Wrote:"it appears as if you ignore the fact that velocity and momentum are purely relative concepts"

I just don't understand the argument here. Even at very low speeds (nonrelativistic) we would see this deceleration. You asked what we would observe in our reference frame, what we would observe is an object experiencing a deceleration. The argument you are making could be applied to any force as far as I can tell.

You Wrote:"would slow to almost a stop" mean anyway? In what frame of reference? In the balloon analogy, it means it 'comes almost to a stop' relative to the balloon surface, but that surface is made up of an infinite number of inertial frames. It seems to come to a stop only in one of the many - wherever it is at a particular time."

You have a point there. However your argument seems almost to suggest that there is no such thing as forces at all. All these reference frames you speak of, unless they were accelerating themselves, would detect a change in the speed of the electron over time.

Let's take a step back and go back to the basics.

Newton said anything in motion tends to stay in motion unless an external force is applied to it. What I'm saying, is from the Newtonian standpoint, this decrease in momentum can be described as a force because that's exactly how it acts.

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#35
In reply to #34

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 4:03 PM

Roger, I think we are talking in circles here. You wrote: "You asked what we would observe in our reference frame, what we would observe is an object experiencing a deceleration."

Any one spot on the expanding balloon surface could be our reference frame and if we put a particle in motion there, we will observe it to accelerate away from us, not decelerate! The only deceleration is relative to the balloon surface as it moves over it, but that it not our reference frame (as I said, it represents an infinite number of inertial reference frames).

"What I'm saying, is from the Newtonian standpoint, this decrease in momentum can be described as a force because that's exactly how it acts."

I think Newton would have agreed that if you and me were moving inertially relative to each other, then you could have measured a smaller momentum for the same particle than what I did. Does this 'decrease in momentum' mean that the particle had any forces working on it? Obviously not.

The same is true for various observers, static on the surface of the expanding balloon (observers comoving with the Hubble flow) - they are moving relative to each other and get lesser and lesser momentum values for the same inertial object as it passes each of them. If you agree that an accelerometer attached to this object does not register any force, then our difference is in semantics. If you do not agree, we have a serious conceptual difference.

-J

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#36
In reply to #35

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 5:18 PM

Jorrie,

You wrote:"The only deceleration is relative to the balloon surface as it moves over it, but that it not our reference frame"

I don't' think that's correct. Light that is redshifted due to cosmic expansion isn't only redshifted with respect the the balloon surface, that redshift is detectable in our reference frame. So too would be the change in momentum of a particle (a redshift of a particle wave).

You wrote:"If you do not agree, we have a serious conceptual difference."

I don't agree. Why is it we can observe cosmic redshift from our reference frame? Afterall, that redshift represents a reduction in the momentum of the light.

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#37
In reply to #36

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 7:57 PM

I will think on this more Jorrie. For you to be so insistent I feel like I need to give it some thought. I do tend to miss the subtleties of reference frames. Maybe a long think is in order on my part.

Lets give it a day of thought. In the meantime, if you have any interest in DNA, I've made a post recently, feel free to check it out.

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#38
In reply to #36

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/09/2009 10:29 PM

Hi Roger, look again at that first quote of yours in full context and things should become clearer.

I wrote: "Any one spot on the expanding balloon surface could be our reference frame and if we put a particle in motion there, we will observe it to accelerate away from us, not decelerate! The only deceleration is relative to the balloon surface as it moves over it, but that it not our reference frame ..."

You replied: "I don't' think that's correct. Light that is redshifted due to cosmic expansion isn't only redshifted with respect the the balloon surface, that redshift is detectable in our reference frame. So too would be the change in momentum of a particle (a redshift of a particle wave)"

To make the two 'experiments' (particle and photon redshifts) compatible, we should treat both as either moving away from us, or towards us. Sure, if some distant observer fires a particle at us at a certain momentum, we will eventually observe it as having lost speed (momentum) when it reaches us, just like for a photon wavelength. If we fire a particle or a photon towards the distant observer, she will get the same results. Both have lost momentum relative to CMB space[1] (the surface) en-route.

However, if we could track the particle that we have fired away from us, we will observe it to speed up in our reference frame - its proper velocity will increase (remember it will asymptotically join the Hubble flow), while its peculiar velocity (relative to CMB space) will decrease. See Fig. 3 of the next Blog. This is the crux of the whole matter. Think balloon analogy - you can save yourself a good number of headaches.

Finally, while you think it over, remember that I only asked if you agree that an accelerometer does not detect any force acting on an object moving freely in expanding space, despite the fact that it is being 'redshifted' as it goes (redshifted in relation to local CMB space). This is the same as saying it loses momentum in the CMB frame as it goes. But, it is still an inertially moving object, which by definition does not experience proper acceleration or proper forces.

-J

[1] 'CMB space' here means the CMB frame, i.e., the frame of an observer measuring the same average CMB temperature in all directions.

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#39
In reply to #38

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/10/2009 1:14 AM

Thanks Jorrie. Let me think on it a day. I hope it goes without saying that I enjoy these discussions (even though I just said it).

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#40
In reply to #38

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/14/2009 3:47 PM

Ok. I think it was a good thing I stepped away and came back. My general impression of the situation is the following. Let me start by saying that for simplicity I'm assuming a constant expansion rate d2R/dt2=0. I'm also assuming space is basically flat. Obviously this isn't exactly correct, but as you've told me plenty of times, these are good approximations. Based on those assumptions I'm noticing the following:

There is such a thing as an "expanding reference frame" that is analogous to a "rotating reference frame" such that

1. The expanding reference frame is an non-inertial reference frame.

2. The forces that are dependent on this expansion are fictitious, analogous to how centrifugal, coriolis, and euler forces are fictious forces of a rotating reference frame.

3. One example of a fictious force that exists in the expanding space reference frame has the form:

P = (1/a) P0

so

P= [H/(da/dt)]P0
dp/dt = H/a P0
F´= (1/a) H P0

This force disappears when there is no expansion H=0, just as the fictitious forces from a rotating reference frame disappear when there is no rotation.

One thing I'm worried about is that the Force seems to decline as the radius grows. Is that derivation correct Jorrie or did I make a mistake?

I'd like to get at this paper as it seems like it might be talking about a similiar idea, but I don't have access to it. Actually, I found a link and here it is. I don't know if it will be helpful or not for me. It's rather long but I'll give it a read when I have a chance.

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#41
In reply to #40

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/14/2009 9:12 PM

Hi Roger.

My problem with comparing expanding frames with rotating frames is that while it is correct that the centrifugal force is a 'fictitious force', the centripetal force is real, in the sense that it can be independently measured by means of an accelerometer at any single spot in the rotating frame. In cosmic expansion, there are no real forces that can be independently measured by an accelerometer at any single spot.

Your derivation seems correct to me, but I'm not sure of its usefulness as an analogy for cosmic purposes. I have not read the paper that you referenced (just the synopsis) and I suspect that it has to do with the gas flows of star formation, which is outside of the Friedman solution. Tell us what you find if you read it.

-J

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#42
In reply to #41

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/14/2009 11:22 PM

Hi Jorrie,

I have a real problem with your answer even though I know you didn't make this up. The term 'fictitious force' doesn't settle with me. This site talks about the merry-go-round. A person on the merry-go-round when spun builds up potential energy. When she lets go, that energy becomes kinetic energy and performs work. I would have to call it real. Did Sir Isaac Newton ever say "To every fictitious force there is an opposite and equal real force."? I don't think so. I think he would be rolling over in his grave if he heard this. What about Einstein? You know a lot about his life. Did he ever use the term fictitious force?

Here is a quote from the link above:

"It is important to note that the centrifugal force does not actually exist. We feel it, because we are in a non-inertial coordinate system. Nevertheless, it appears quite real to the object being rotated. This is because the object believes that it is in a non-accelerating situation, when in fact it is not"

Well, inanimate objects have consciousness, do they? I think I could adopt their belief systems quicker than I could adopt some physicist's systems.

-S

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#43
In reply to #42

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/15/2009 12:22 AM

Hi StandardsGuy,

Actually Jorrie is using an official term. Fictional force is a very specific thing that is unfortunately named. Here is a link describing it. My reaction was the same as yours as I didn't know the formal definition. When I first read what Jorrie had written, I objected and tried to argue the force was "real" which took us down a tangent. At a point later in this thread I took a step back and gave it a few days. I think it was a good thing I did.

Fictious Forces are consequences of noninertial reference frames. After rereading Jorrie's objections I altered my argument. I'm now suggesting that an expanding/contracting reference frame is a noninertial reference frame that gives rise to the ficticious force I was arguing earlier. Much in the same way that a rotating reference frame gives rise to a number of ficticious forces such as coriolous, centrifugal, and euler (if you click on that last link you'll see they refer to it as a fictious force).

Jorrie has reservations regarding this interpretation and I'm eager to hear them. I've read your interactions with him so I know I don't have to tell you he has a deep understanding of these issues. Personally I depend on his skepticism to keep me straight. In the case of the fictitious force comment, I think he did so.

If you're interested, there is a really cool paper on expanding/contracting reference frames found here. I'm sure the 10th time I read it I'll begin to understand it better.

As I said earlier, my initial reaction was the same as yours above, but Jorrie was absolutely right to call it a fictious force. My head hurts, but I love this stuff.

Roger

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#46
In reply to #43

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/15/2009 7:41 PM

Hi Roger,

Yes, I am aware (recently) that these forces have been named because of their origin. I can accept that, but when they say they don't exist (as the link in #42 states), that's when I have a problem. The centrifugal link (from Wikipedia) fortunately doesn't say that, however the first link in your post (also from Wikipedia) describes it: "... is an apparent force", which reinforces the 'doesn't exist' syndrome. This is an example of the illogicalness that I was referring to in a previous blog. Thanks for the cool paper link. I have already saved it from your previous reference, but don't know when I'll get time to read it. I have several new ones on my desktop.

-S

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#44
In reply to #42

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/15/2009 1:05 AM

Hi S.

I view a 'fictitious force' as a synonym for a 'coordinate dependent force', as against a 'real force', which is a synonym for a 'coordinate independent force' or a 'proper force'. The latter can be measured with a strain gauge, or an accelerometer (through F=ma, taking into account the mass of the thing being accelerated). The former cannot 'really' be measured.

As such, centripetal force is 'real' and centrifugal force is 'fictitious', just present to keep the object static in the rotating frame of reference (by balancing out the centripetal force). Similarly, what we normally consider as a 'gravitational force' towards Earth's center is 'fictitious'.

As I said before, when you sit in front of your computer, the only 'real force' that you can detect is the chair pushing against your behind. The reason why you do not accelerate towards the ceiling is that there is 'fictitious force' balancing out the real force - and that 'fictitious force' is only there because of a choice of coordinates (your room as the reference frame). Einstein abandoned gravity as a force and just called it geodesic movement in curved spacetime (that was before he talked to the Bohr...)

-J

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#45
In reply to #44

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/15/2009 10:37 AM

Hi Jorrie,

I read the gravity section of the fictitious forces entry in wikipedia and I think I finally understand the distinctions you're trying to draw.

It's true gravity is simply a manifestation of curvilinear space and if you were in a box in free fall in a gravitational field, you would feel no force.

In the same way, were you to build a box around a free particle traveling through space you also would not detect any force due to expansion since you, the box, and the particle would all slow at the same rate (due to this dp/dt dependence upon expansion that we've been talking about). One would not notice any decline in momentum on the electron since everything would be experiencing the same momentum loss.

As light or a matter wave travels through space, they are redshifted, just as if they were escaping a gravitational well. When escaping a gravitational well, it is movement through curved spacetime that causes this redshifting. In the case of cosmic redshift, it is traveling in expanding spacetime that creates the redshift. These redshifts behave similiarly and are analogous.

What I'm saying is that a particle or photon experiencing a cosmic redshift behave as though they are escaping a gravitational potential. I'm not saying they are escaping a gravitational potential, I'm just saying it's analogous.

So why did I bring this up? The point of an accelerometer is to distinguish a force independent of freefall. Even though the above is only an analogy, the point is that particles and photons are behaving as though they are experiencing a gravitational force. An accelerometer distinguishes between gravitational like forces and inertial forces. Thus an accelerometer wouldn't be able to distiguish the force I'm suggesting simply because this force is behaving like a gravitational (curved space) force. Just as you said.

Thus the froce should be considered fictitious.

I'm now ducking and covering.

Roger

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#47
In reply to #44

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/15/2009 8:09 PM

Hi Jorrie,

"I view a 'fictitious force' as a synonym for a 'coordinate dependent force', as against a 'real force', which is a synonym for a 'coordinate independent force' "

I like those definitions. I also like 'inertial force' for the former. Do we need a third category for cosmology?

"that 'fictitious force' is only there because of a choice of coordinates (your room as the reference frame). Einstein abandoned gravity as a force and just called it geodesic movement in curved spacetime"

I seem to recall a discussion between you and (G.K.?) where it was postulated that massive objects were "sucking in vacuum". Einstein's theory of gravity would almost make sense in such a scenario. Do you recall such a discussion?

"(that was before he talked to the Bohr...)"

Are you trying to be funny here, or was there a discussion? If so, what did Bohr say?

-S

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#48
In reply to #47

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/16/2009 1:34 AM

Hi S.

"I seem to recall a discussion between you and (G.K.?) where it was postulated that massive objects were "sucking in vacuum"."

I think it was with someone who wanted to know about falling into a black hole and seeing his feet...

It is just a choice of a different set of coordinates (free-fall coordinates), where you as in-falling subject are static in this 'vacuum', which is being "sucked in and stretched" in the process, causing tidal forces.

Since we believe that the vacuum contains energy, it cannot really be sucked into black holes (vacuum energy does not clump and collapse gravitationally). It is another coordinate dependent situation, which must be 'fictitious' by definition.

-J

PS: 'The Bohr' won the argument, using Einstein's own theory to refute his objection against quantum uncertainty...

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#49
In reply to #48

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/16/2009 8:35 PM

Hi Jorrie,

"I think it was with someone who wanted to know about falling into a black hole and seeing his feet..."

No, way before that, possibly a year or more ago, But you used the term "sucked in" in this post!

"(vacuum energy does not clump and collapse gravitationally)."

But you think it can be stretched! I was thinking the previous discussion was talking about the vacuum flowing into the earth, but could be mistaken. Let's suppose that Newton and Einstein met face to face and that Einstein convinced Newton that the vacuum was distorted around massive objects. I can imagine that Newton would say that the force of gravity is causing the distortion, but that Einstein would say that the distortion is causing the [force of] gravity. Is that how you see it? If so, what is causing the distortion? (What is the sucking force?)

-S

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#50
In reply to #49

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/16/2009 11:49 PM

Hi S.

"No, way before that, possibly a year or more ago, But you used the term "sucked in" in this post!"

No, I cant recall, but it is a valid choice of coordinates (called free-fall, or Finkelstein coordinates) where space is 'falling' into any object. Again, since it is a coordinate dependent concept, it is also a fictitious concept, just a mental aid in some discussions (like the 'feet inside the black hole').

Me: "(vacuum energy does not clump and collapse gravitationally)."

You: "But you think it can be stretched!"

Space can be 'stretched', but vacuum energy cannot - proof: vacuum energy density remains constant, despite space being stretched. More space, more vacuum energy - simple! This hints towards rather thinking of more space (vacuum) being created and not as space being stretched? The official term is in any case 'expanding space' and not 'stretched space'.

"I can imagine that Newton would say that the force of gravity is causing the distortion, but that Einstein would say that the distortion is causing the [force of] gravity."

I can imagine Newton getting very frustrated later - when Einstein explained Mercuries anomalous perihelion precession and Newton could not!

-J

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#6

Re: Cosmic Balloon Application III: Particle Momentum Decay

07/06/2009 4:01 PM

Jorrie,

So I've checked out the Tamara Davis Thesis. I checked her out here at her website and she seems on the level. That's cool because I was having a tough time finding anything on massive particle deceleration but I felt that it was logical that it must occur.

I'm going to calculate the change in momentum of a particle due to the cosmic expansion rate and post it. I will probably get it wrong (don't I always the first bunch of times....sigh), so check me when I do it.

Roger

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