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Cosmic Balloon Application V: Cosmic Tidal Force

Posted July 16, 2009 2:00 AM by Jorrie
Pathfinder Tags: Cosmic tidal force cosmology

In this final 'application' of the cosmic balloon, the effect of expansion on 'rigid bodies' will be investigated. Two buttons are separated by proper distance D on the surface of the cosmic balloon. We tether the two buttons to each other by means of a semi-rigid tether and then inflate the balloon.

Gravitational tidal forces[1] attempt to either compress or stretch objects in a gravitational field. It has to do with the different spacetime curvatures at different parts of an extended object. Cosmic tidal forces have similar effects on extended objects, but it has to do with the expansion dynamics and not directly with gravity.

Cosmic tidal forces

Fig. 1 (right) shows the cosmic situation - two galaxies (red buttons) on the surface of the balloon, with a tether keeping them at a constant proper distance (D) apart. To determine the cosmic tidal force (if any), we measure the force on the tether, using strain gauges or some other practical method, taking into account that the forces can be either stretching or compressing.

It is reasonable to expect that cosmic tidal forces will depend on the inflation profile of the balloon. As it happens, it feels intuitively correct (and it is in fact easy to show) that a constant inflation rate (dR/dt = constant) causes no tidal force on the tether (apart from a transient when we start the inflation). With the balloon in uniform expansion, the buttons will essentially move inertially across the surface of the balloon. This is equivalent to a phase of the actual universe, around half of the present age, when the expansion rate was essentially constant (see this post).

If we blow up the balloon with a constant gas flow rate, the expansion rate will slow down over time and free buttons would tend to move towards each other, because they would have had initial momentum towards each other. The semi-rigid tether will prevent that and we should measure a compression force on the tether. This is equivalent to a phase in the actual universe before 7 Gy age, when the expansion rate was decreasing under the dominant matter density.

If we control the gas flow rate so that the expansion rate of the balloon increases over time, equivalent to the now dominant vacuum energy phase, one would expect the buttons to separate. Again, the tether will prevent the separation and we should measure a stretching force in the tether. This is all rather intuitive, so let's try to quantify these forces on a cosmological scale.

Our Local Group

Rather than working with billions of light years (as usual), let's keep it 'local' and make the tether 10 million light years long. This is about the diameter of our Local Group of galaxies (Andromeda, the Milky Way, Triangulum and some 27 dwarf galaxies). It will be interesting to find the magnitude of the cosmological tidal force on this scale (ignoring the local gravitational effects at first). Fig. 2 (right) plots the accelerative tidal force exerted on the 10 million light years long tether. The timescale starts some 10 billion years ago and ends far into the future. For the relevant equations, see note [3].

For the first 7 Gy, the tidal force was negative (compressing) and then it became stretching, as expected. At the present time (~13.6 Gy), the stretching tidal force over 10 million light years has an order of magnitude (OOM) 30 femto-g,[2] or 300 femto-Newton per kg mass. This is small, really small. However, when compared to the gravitational acceleration at the edges of the Local Group, it may not be so negligible.

Gravitational acceleration

Let's get an idea of how much effect the cosmic expansion may have on the structure of our Local Group of galaxies. It has a total mass of ~ 1.3 × 1012 Sols, or M ~ 3 x 1042 kg[4] and a radius r ~ 5 x 1022 meter. If we include dark matter (at ratio 6:1), it puts the mass of the Local Group at M ~ 2 x 1043 kg. The rough gravitational acceleration of a particle just outside of the Local Group is: a ~ -5 x 10-13 m/s2, (-50 femto-g).[5]

If my calculations are correct,[6],[7] this is of the same OOM as the 30 femto-g tidal stretching force at that distance. It may mean that this very feeble tidal stretching force has some influence on the structure of things like galactic clusters - and even more so on super-clusters. It probably only means slightly larger orbits for the galaxies right at the edge of clusters. Even so, never again 'sneeze' at an acceleration of a few tens of femto-g!

I have made a table of our family of gravitationally bound structures, using my equations and data from ref [7]:

I have included dark matter in the masses of the structures, although it does not make a huge difference. The last three columns are: g_acc is gravitational acceleration of a particle at radius r from the gravitational center of the structure (in femto-g), t_acc is the tidal acceleration at that same place; r/r_crit is the ratio of r to r_crit, the 'critical radius' where the two forces precisely balance each other.[3] eq. (6)

It is clear that the Milky way is gravitationally bounded very well, our Local Group cluster is marginally bounded, the Virgo Cluster is bounded very well (it has a great density of galaxies) and the Virgo Supercluster is marginally unbounded. Above the size of superclusters, we get the 'filaments' and 'great walls', which are not gravitationally bounded at all.

[Edit: Also see reply to Roger below (effect of cosmic tidal forces on galaxy formation).]

Jorrie

Notes:

[1] Gravitational tidal forces are 'real', coordinate independent forces, measurable by means of strain gauges or accelerometers.

[2] It is convenient to work in nano-, pico- or femto-g, because my calculations output the acceleration in units of 1/Gy, which is roughly one nano-g. An acceleration of 1 ly/y2 is roughly one g (9.8 m/s2), because the radius of spacetime curvature at Earth's surface is roughly one light year. One nano-g is equivalent to a gravitational radius of curvature of one Gly. (Can you think why this must be so?)

[3] Some relevant cosmological equations (read with the equations of the previous Blog):

The acceleration of expansion:

d2a/dt2 = a H02Λ - Ωm/(2a3)) ---------------(1)

(From Peebles 1993, eq. 13.3, p 312, where one must read H0 as implying the usual H0/978 Gy-1, in order to be compatible with our units convention. The factor Ωm/(2a3) - ΩΛ has historically been called the deceleration parameter q. It has the present value q ~ -0.6 (negative, because the deceleration is negative, i.e., it's an acceleration ).

The tidal acceleration over a fixed distance D follows exactly the same profile, but scaled to D. Any change in a causes the same % change in the proper distance between two free particles that were momentarily at rest relative to each other. Hence:

d2D/dt2 = D H02Λ - Ωm/(2a3)) ---------------(2)

At the present time, with a=1, it means

d2D/dt2 = D H02Λ - Ωm/2) ---------------(3)

Assuming the present values: H0 = 72/978 Gy-1, ΩΛ = 74% and Ωm = 26% of the critical density, this works out to a present proper tidal acceleration of:

d2D/dt2 = 0.00334 D Gy-1, ---------------(4)

or about 3.3 femto-g per Mly of proper distance (about 10 femto-g per Mpc).

If we work with radius r instead of with diameter D, we can find the critical radius of spherical structures (r_crit) where tidal and gravitational forces have the same magnitude. From equation (3) we can write:

GM/r_crit2 = r_crit H'02Λ - Ωm/2) ---------------(5)

where H0 must be expressed as inverse seconds (s-1) in order to keep the (SI) units (m/s2) of the two sides compatible, i.e., H'0 ~ H0 /( 3 x 1016) s-1. (Coming from 1 Gy ~ 3 x 1016 seconds, together with the usual H0/978 conversion to Gy-1). Hence:

r_crit = [GM / (H'02Λ - Ωm/2))]1/3 meters ---------------(6)

We can 'cosmologize' it to (say) Mly: divide by 1022, roughly the meters in a million light years.

[4] http://en.wikipedia.org/wiki/Local_Group

[5] The gravitational acceleration is obtained as: a ~ (-6.67 x 10-11 ) (2 x 1043) / (5 x 1022)2 ~ -5 x 10-13 m/s2, or a ~ -50 femto-g. This is not strictly correct for a very non-homogeneous mass concentration like the Local Group, but the OOM should be about right.

[6] The only applicable reference that I managed to find so far is a quite technical paper by Gregory S. Adkins et. al (2006): 'Cosmological perturbations on local systems'. Their conclusion is that matter dominated phases do not perturb the orbits of local systems, but for the vacuum energy dominated phase they say: "However, the cosmological repulsion becomes more important for more extended clusters, and would make a contribution comparable to that of the gravitational term for clusters that are only marginally bound."

Unfortunately, they do not give quantified results (at least not in a form that I can understand).

[7] I have now found a very accessible paper: "On the influence of the global cosmological expansion on the local dynamics in the Solar System" by Matteo Carrera and Domenico Giulini (2006), which seems to confirm my approach. They use a different, more general approach than myself, but the results seem to be compatible.

-J

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#1

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/16/2009 3:23 PM

Jorrie,

Interesting post.

Even at Constant Expansion, wouldn't the Hubble Flow would create a tension on the tether? A tension is a force and you are speaking of tidal force which is the first derivative of a force, so you may only be focusing on tidal force in this post. Can you clear this up for me?

Roger

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#2
In reply to #1

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/16/2009 11:30 PM

Hi Roger. Thanks.

No, a constant expansion rate causes no cosmic tidal forces - just like a hypothetical 'uniform gravitational field' that causes no gravitational tidal forces. As you can see from the graph (Fig. 2), decelerating expansion causes a compressive force on the tether and accelerating expansion a stretching force. Logically, where they change over there should be zero tidal effects.

Look at eq. (3) - I have now numbered them - and you will see that when ΩΛ = Ωm/2, there is no tidal acceleration. This result of mine agrees with Matteo Carrera and Domenico Giulini (2006), referred to in the PS of the OP.

Another way of looking at it is: for a constant expansion rate, there is no net force working in the hyper-spatial or spatial directions - those two red arrows in my Fig. 1 are like constant, linear vectors in hyperspace. If the expansion rate changes, those red vectors are no longer linear and that creates an acceleration between them, experienced as tidal forces by the tether.

-J

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#3
In reply to #2

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 1:05 AM

Jorrie,

I agree that a constant expansion would produce no tidal force. My question was does constant expansion cause tension in the tether? It seems like at even constant expansion the two buttons would spread out if it weren't for the tether, so I thought the there must be tension on the tether. Am I thinking of this wrong?

Roger

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#5
In reply to #3

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 10:56 AM

Roger, you wrote: "I agree that a constant expansion would produce no tidal force. My question was does constant expansion cause tension in the tether? It seems like at even constant expansion the two buttons would spread out if it weren't for the tether, so I thought the there must be tension on the tether."

But tension in the tether would mean the presence of tidal forces, which you agree are not there!

As we have established before (see Fig 1 of this thread), a constant expansion rate results in zero proper velocity for the 'tethered-then-untethered' galaxy. Zero proper velocity means zero tension in the tether. There will only be tension in the tether while the expansion rate changes.

-J

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#6
In reply to #5

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 11:35 AM

Hi Jorrie,

I'm confused. A tidal force is the gradient of a force. In this case the cosmic tidal force is coming from changes in the expansion rate.

However, even without changes changes in the expansion rate, a force is still present that is pushing the buttons (galaxies) apart. That's why you have to tether them in the first place, right? Even at a constant expansion rate, if the galaxies weren't tethered, they would move away from each other, right? The tether prevents that and keeps them at same proper distance. In doing so it would result in a tension on the tether I would think, since tension is simply a force, not a gradient of a force.

However, I may be confused. Proper Distance as opposed to regular distance may be the source of my confusion. I think I understand proper distance, but I may be missing a subtlety.

Roger

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#7
In reply to #6

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 12:55 PM

Hi Roger, you wrote: "... even without changes changes in the expansion rate, a force is still present that is pushing the buttons (galaxies) apart. "

Huh? The only time there is something resembling 'a force pushing buttons apart' is if vacuum energy dominates and the expansion rate increases. When matter dominates, there is something resembling a 'force trying pull the buttons together' and the expansion rate decreases. If these two fictitious forces balance (like at 7 Gy age) there is no effective force and the expansion rate remains constant. Eqs. (1) and (3) of the OP say it all!

In the case of constant rate of expansion, the tether only serves to bring the buttons to zero proper velocity relative to each other; thereafter, it plays no role and it may be cut with no effect on the buttons, like in Fig. 1 of this thread. I think the most difficult thing to get one's head around there is that the peculiar velocity decays over time, while the proper velocity remains zero relative to each other.

"However, I may be confused. Proper Distance as opposed to regular distance may be the source of my confusion. I think I understand proper distance, but I may be missing a subtlety."

Proper distance is 'regular distance', as measured across the surface of the balloon with a regular tape measure. I think the subtlety that you may be missing has to do with the fact that increasing expansion rate will try to move the tethered buttons apart (stretching tidal force), while decreasing expansion rate will try to move them together (compressing tidal force). It is then fairly obvious that constant expansion rate will do neither - no tidal force.

To convince yourself, look again at my segment of cosmic balloon in Fig. 1 above. In the case of uniform expansion, why would a free particle, once moving along a red vector, not keep on moving in that direction? Only changes in the balloon expansion rate can cause a change in the particle's direction of movement.

-J

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#8
In reply to #7

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 1:27 PM

Jorrie,

What you seem to be saying is that once two things have been tethered, as long as the the cosmic expansion remains constant, the proper distance will never change between those two objects, even if you untether them. That seems wrong to me, but I'm probably am just being dense.

Roger

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#9
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Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 2:05 PM

Roger, your: "... as long as the the cosmic expansion remains constant, the proper distance will never change between those two objects, even if you untether them."

That's it! Obviously, this is not a very realistic case, but as we have seen, around 7 Gy, our universe resembled such a state.

Do not feel too perturbed about your intuition being wrong - it took me a long time to understand this correctly. When you asked the question, I even had a few moments of doubt again! Some cosmologists disagree on it to some extent, but I suspect it is simply a semantics issue. After all, they all use the same model and math and the math is very clear on this issue. I am actually quite amazed at how well the cosmic balloon analogy 'explains' all of this.

-J

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#10
In reply to #9

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 3:44 PM

Jorrie,

But that doesn't square with your balloon analogy. What you're saying is if you inflate a balloon with tethered buttons on it, then cut the tether, they won't separate? With the balloon analogy, it seems to me they would separate. Is the balloon analogy wrong in this case?

(Disregard this comment, I think I figured it out in my next comment)

Roger

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#11
In reply to #10

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 4:01 PM

Jorrie,

Hang on, I think I understand. Basically what you're saying is that the tether gives the galaxy a velocity equal and opposite to the hubble flow and as long as the hubble flow is constant, the galaxy stays in place, even untethered, like a guy running on a treadmill. Is that right?

Essentially we're talking about Newton's First Law.

Roger

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#13
In reply to #11

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 9:33 PM

Hi Roger.

Yes, basically you are right. Just remember that for a constant expansion rate, the Hubble flow decreases over time until it tends to zero as time tends to infinity.[1] It's like a treadmill set to move slower and slower and so does the runner relative to the treadmill (her 'peculiar velocity').

Remember that as long as there is expansion, peculiar velocities decay; hence, Newton's first law can lead you completely astray! It would demand that her peculiar velocity remains constant and that she might fall over the front rail of the treadmill as it slows down. Actually, the treadmill is a very bad example, because as we all know from the 'huffing and puffing', it takes a lot of work to run on it. Frictionless buttons on the balloon perform no work. As we have discussed at length before, it is angular momentum[2] relative to the balloon center that is conserved, not linear momentum, which always decays.

-J

[1] Fig. 2 and Eq. 7 of the tethered galaxy Blog.

[2] Fig. 1 and Eq. 1 of the momentum decay Blog.

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#16
In reply to #13

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/18/2009 10:49 AM

Jorrie,

I understand. It's clear now to me that the hubble flow and the momentum of the galaxies (buttons on your balloon model) decrease at the same rate, so that as long as expansion rate is constant, tethered galaxies don't separate (and there is no tension on the tether).

It is also clear to me that were the expansion rate to decelerate, as in the earlier "matter dominated" epoch in the universe, that there would be a compressive force on the tether. I also get now that if the expansion rate were to accelerate, that the tether would experience tension as the galaxies tried to seperate.

In essence, the peculiar velocity and the hubble flow as well as the deceleration of peculiar velocity and deceleration of the hubble flow are in balance in a constant expansion, but are thrown out of balance when expansion accelerates or decelerates, leading to tension or compression forces on the tether.

I have been assuming that gravity acts as a tether (and is what the tether model is supposed to represent to a certain degree). What's interesting about gravity given the information above, is that as a "tether", it "ecourages" compression (in epochs where the expansion rate is slowing) and it "discourages" expansion (in epochs where the expansion rate is increasing).

Is my understanding ok at this point, because I have an interesting tangent for us but I don't want to proceed if I'm getting something wrong here?

Roger

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#17
In reply to #16

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/18/2009 6:27 PM

I think it's worth while to examine what kind of tether gravity is to further examine the consequences of this tidal force. For instance, gravity, as far as I know, in no way resists the "compression" force of a decelerating cosmic expansion, yet does resist the "tension" force of an accelerating cosmic expansion. So gravity wouldn't qualify as a "rigid" tether.

Jorrie, I don't think you ever said gravity was a "rigid tether", I believe you mentioned a "rigid tether" only as a way of just examining effects. If I'm wrong on that, let me know.

To get back to my point, an interesting consequence of this asymmetry may be found in local systems which are tethered gravitationally(like galaxies in galaxy clusters). Periods of deceleration of cosmic expansion result I think in higher kinematic periods for those clusters. I say this for two reasons:

1. The galaxies in the clusters feel the "compressive force" and are pushed toward each other. Now I know the acceleration due to this force is small, but over billions of years this may add up. Plus, this effect is amplified by the gravitational potentials of the galaxies. Accelerating something into a potential has a much larger effect than accelerating something into just space.

2. A decelerating cosmic expansion results in a slower redshifting of bragg waves. As a result, free particle deceleration is lower.

So, what I'm saying is, the asymmetry of the gravitational tether results in increased kinematics of galaxies in galactic clusters during periods of cosmic expansion deceleration and conversely, a reduction of kinematics during periods of cosmic expansion acceleration.

This may be obvious or wrong so feel free to jump in and correct where needed.

It feels like an entropy argument, but I'm not sure how to phrase it.

Roger

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#18
In reply to #17

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/18/2009 11:07 PM

Hi Roger, you wrote: "I don't think you ever said gravity was a "rigid tether", I believe you mentioned a "rigid tether" only as a way of just examining effects. If I'm wrong on that, let me know."

I've used the term 'semi-rigid tether' and you are right - only as a way of measuring or examining the cosmic tidal effect. Gravity is a weird form of 'tether', with an inverse square law of elasticity against length.

Another issue that must not be lost out of sight is the 'dual personality' of gravity in cosmology: it behaves differently in the homogeneous FRLW solution and in the inhomogeneous (local Schwarzschild) solution. In the balloon analogy, the FLRW solution only applies to the 5th (hyper-spatial) dimension and the Schwarzschild solution only to the 4D spacetime dimensions. The homogeneous balloon is precisely spherical, with no 'depressions' on the surface. The inhomogeneous solution represents 'depressions' or 'gravity wells' on the surface. It is only in these gravity wells that the 'gravitational tether action' can be postulated.

Nevertheless, I do not support the idea of 'gravity as a tether', because it is open to all sorts of confusions. In this post I have separated them completely. First, the tether just served to measure the strain (or rather acceleration) of cosmic tidal force in a homogeneous solution. Second, the result is compared to the normal gravitational acceleration at a given distance from the center of an inhomogeneity. This was in order to decide what 'polarity' the combined effect should have. This is where the 'critical radius' comes in, where they have the same magnitude, but not necessarily canceling out (they may be in the same direction).

In clusters, the galaxies are generally in orbit around the center of mass of the cluster. Orbits depend on the inverse square law of gravity. Cosmic tidal forces are directly proportional to distance from the center (not even inverse, like gravity). This should cause perturbations of the orbits of galaxies, probably making them precess and so on. Unfortunately, the effects are too small to be measured, I think.

"2. A decelerating cosmic expansion results in a slower redshifting of bragg waves. As a result, free particle deceleration is lower."

I don't think Bragg waves and cosmological redshift have anything to do with each other - can you explain?

The fact that normal gravity and cosmic tidal forces worked in the same direction in the early universe probably had an effect on early structure formation, but I have virtually no knowledge on that front. I understand both the Schwarzschild and the FLRW solutions to a reasonable extent, but I have no idea how to combine them.

-J

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#19
In reply to #18

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 1:44 AM

Hi Jorrie,

Regarding the "bragg wave" comment, I meant "de Broglie wave". Just a careless mistake on my part. Slowing momentum is equivalent to redshifting the de Broglie wave since:

In the equation above you can see that the slowing momentum causes the wavelength to get longer(redshift). This redshift is equal to the redshift experienced by photons. Rather than redshift you could say that photons are losing momentum, right? It's all the same thing.

The above is the momentum of a photon, you see what I mean.

I think you already know all this and were just thrown by my sloppiness in calling it a Bragg wave by accident, which has nothing to do with anything we're talking about. I'm sorry about that.

Regarding Gravity and Tethering

Although I understand your reluctance to call gravity a tether, I think we can say at least qualitatively gravity has played an important role "tethering" galaxies (among other things) together, else our "local group" wouldn't be nearly as local.

I guess the way I'm looking at it is that the cosmic tidal force that you calculated is probably a good 1st order approximation and could be applied as a perturbation (apply perturbation theory to the solutions) to the Schwarzchild solution. If this were done, I think it would show that during epochs of decreasing cosmic expansion, galaxies in a local cluster would experience a small constant force towards the center of mass. This would have little effect when the galaxies were far from the center of mass, but I think a significant effect when their orbits brought them close (1/r2 dependence). Since we are talking about GY, I think there would be time for some serious perturbing. Essentially this amounts to a decrease in the overall potential energy of the galactic cluster with the corresponding increase in kinetic energy of the cluster (basically all the galaxies are feeling a constant force towards the center of the galaxy cluster gravity well for billions of years (though a very very small force)).

On a side note

You probably could answer this question pretty quickly. I know if you make the gravitational constant half as strong G/2 instead of G, then the force of gravity from a massive object is half as strong(at least for newton). What I don't know is what that does to the cosmic expansion rate and cosmic tidal force. Can you tell me? To be clear, I don't think G is changing, its just an undergrad physics trick to better understand things(vary the variables and constants and see what happens).

Roger

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#20
In reply to #19

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 5:13 AM

Hi Roger, you wrote:

"If this were done, I think it would show that during epochs of decreasing cosmic expansion, galaxies in a local cluster would experience a small constant force towards the center of mass. This would have little effect when the galaxies were far from the center of mass, but I think a significant effect when their orbits brought them close (1/r2 dependence)."

The "small constant force towards the center of mass" would be more pronounced the farther the galactic orbit is from the center of mass - that's the way the cosmological tidal force works (see eq. 4 above).

On the 'half-G': I guess that a lot of cosmological constants, like the critical mass and so on, would have had different values and we would have had a very different picture than what we have today. I think it's impossible to say if inflation would have produced a slower or faster expansion initially, which means the tidal force could have been smaller or larger than it is today.

-J

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#21
In reply to #20

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 10:58 AM

Hi Jorrie,

You Wrote:"The "small constant force towards the center of mass" would be more pronounced the farther the galactic orbit is from the center of mass - that's the way the cosmological tidal force works (see eq. 4 above)."

d2D/dt2 = 0.00334 D Gy-1, ---------------(4)

In the equation above, D is proper distance, right, not distance? Assuming that's the case, what I see in the equation above is a proper acceleration that doesn't depend on distance from the center of mass. It looks like a constant acceleration. What am I missing here?

Roger

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#24
In reply to #21

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 2:32 PM

Hi Roger, you asked: "In the equation above, D is proper distance, right, not distance? "

I wrote in #7 above: "Proper distance is 'regular distance', as measured across the surface of the balloon with a regular tape measure." I fail to see how you can view proper acceleration as not dependent on distance - eq. 4 says it is directly proportional to the length of the tether! Note that eq. (4) is only applicable to the present time. Even for a constant D, tidal acceleration does vary over time, as per eq. (2).

-J

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#25
In reply to #24

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 4:30 PM

Hi Jorrie,

You Wrote:"I fail to see how you can view proper acceleration as not dependent on distance - eq. 4 says it is directly proportional to the length of the tether! Note that eq. (4) is only applicable to the present time. Even for a constant D, tidal acceleration does vary over time, as per eq. (2)."

I didn't see the D in the equation. I really don't know how I missed it, sorry about that. So it depdends on the length of the tether linearly. For us that means that the force drops off linearly the closer the galaxy gets to the center of mass.

Still, I think in terms of gravitational potential energy, the change is the same whether you are far away from the center of mass or close (in terms of percentage).

For instance:

If galaxy A is x distance from the center of mass and galaxy B is x/2 distance from the center of mass then galaxy A will experience FA-tidal whereas galaxy be will experience FA-tidal/2. These tidal forces accelerate galaxy A and galaxy B by aT and aT/2 respectively.

Since:

r= ri -1/2at2 or r-ri =-1/2at =Δr

so

Δra = 2Δrb

so for the same amount of time, Galaxy A moves twice as much into the potential as Galaxy B.

So lets say that Galaxy A moves from 1 to 1/2 distance while Galaxy B moves from 1/2 to 1/4 distance (any unit distance). The gravitational potential energy has a 1/r dependence meaning that the gravitational potential energy for both galaxy A and B has been cut in half for both with a corresponding increase in kinetic energy for both.

In other words, regardless of the distance of a galaxy in a local cluster from the cluster's center of mass, the gravitational potential energy of the galaxy with respect to the center of mass is reduced by the same percent by the cosmic tidal force.

This is certainly different from what I said earlier which was that the effect would be more pronouned closer to the center of mass. My mistake was that I was missing the linear dependence of the cosmic tidal force. Now that I've included it the change in potential is as I've stated above.

I think.

Roger

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#26
In reply to #25

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 1:07 AM

Hi Roger.

Yes, I suppose you can view the effect of tidal forces on the potential energy as you did. Just be careful in how you interpret it.

Like normal gravitational tidal forces, cosmic tidal forces do not generally contribute to present day potential energy changes. It may hold over millions/billions of years of structure evolution, but surely not in the present day dynamics of the structures. It is a bit like an extended object free-falling in a normal gravitational field. We do not include any potential energy caused by tidal stretching/squeezing in its mechanical energy balance (PE+KE=constant).

-J

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#27
In reply to #26

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 2:08 AM

Jorrie,

I Agree. My interest was purely with respect to galaxy formation and activity over billions of years. Increased kinematics means increased interactions among galaxies and increased star formation (from the increased gravitational tidal effects of galaxies more closely interacting).
Since we had a prolonged period of cosmic expansion deceleration in an earlier epoch, I'm wondering if galaxies at the end of that epoch tended to be more active (higher magnitude) than expected with normal models.
So I looked around and found the following:

http://www.sciencenews.org/view/generic/id/42419/title/Heavyweight_galaxies_in_the_young_universe
http://www.thaindian.com/newsportal/health/giant-stellar-factory-identified-in-early-galaxy_100151414.html

I think there may be something to this Jorrie. I think it's at least worth looking into further.

I knew to look back 6-7 million years for the effect because that was the end of billions of years of this effect taking place. That's when the cosmic expansion went from deceleration to acceleration.
What do you think?

Roger

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#28
In reply to #27

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 3:10 AM

Hi Roger, yes, very interesting.

It is only the Livescience article that properly addresses the issue of dark (vacuum) energy (probably the only 'popular writer' that understood what the scientists said).

One must remember that the cosmic tidal force is only one of the contributors to structure formation and I suspect it is not even specifically mentioned in the paper (or press release). I also suspect that it is automatically included in any structure simulation based on the (perturbed) Friedman equations.

The fact that the cosmic expansion of the ΛDCM model had to be much slower in the early epochs is probably the major contributor to faster/heavier structure formation (without any consideration of the deceleration and resultant negative tidal forces).

I read this indirectly in the article, but I'm not too sure of it and I also realize that it is a rather tricky inference. It will perhaps need a lot of research/discussion. One will need a full simulation and then do a sensitivity analysis - that's surely outside of my scope...

What I do understand is how the very fast initial expansion rate of a matter+radiation only cosmos (with vacuum energy permanently zero) would have hampered structure formation. Can you remember why a 'matter+radiation only cosmos' just had to expand very much faster initially?

-J

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#29
In reply to #28

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 3:28 AM

Hi Jorrie,

You Wrote:"What I do understand is how the very fast initial expansion rate of a matter+radiation only cosmos (with vacuum energy permanently zero) would have hampered structure formation. Can you remember why a 'matter+radiation only cosmos' just had to expand very much faster initially?"

I think the answer to your question is that a nonzero vacuum energy, because of mass - energy equivalence, interacts gravitationally with itself. Essentially the expansion has to overcome a vacuum gravitational potential which slows it down(that may be a bad way to put it, hopefully you understand what I'm trying to say here). Am I in the ballpark?

Roger

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#31
In reply to #29

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 5:15 AM

Hi Roger, no, the reason is much more mundane, yet subtle!

Given that there was deceleration (matter dominated) followed by acceleration (vacuum dominated), it had to start out slower in order for us to observe the expansion rate that we have today. If there was only deceleration (matter only), it had to start out much faster for it to be 72 km/s/Mpc today.

Maybe that's what you implied, but is is not clear to me that you have it right.

-J

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#32
In reply to #31

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 11:50 AM

Hi Jorrie,

I think I understand your logic (that wasn't what I was trying to say in my response, I think my response was just wrong, I remembered self gravity but forgot the negative pressure).

You Wrote:"Given that there was deceleration (matter dominated) followed by acceleration (vacuum dominated), it had to start out slower in order for us to observe the expansion rate that we have today."

Couldn't you also say that instead of starting out slower, the deceleration was stronger than we think (in that earlier epoch), or that the acceleration was weaker than we thought (in the later epoch)?

You Wrote:"If there was only deceleration (matter only), it had to start out much faster for it to be 72 km/s/Mpc today."

I agree, except again, couldn't you also say that the deceleration is much less than we thought (thus same initial speed)?

Or does the cosmic redshift pretty reliably tell us the acceleration and deceleration rates in those epochs, and so my suggestions above are effectively ruled out leaving what you suggested as the only possibilities?

Roger

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#33
In reply to #32

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 2:18 PM

Hi Roger, you asked:

"Or does the cosmic redshift pretty reliably tell us the acceleration and deceleration rates in those epochs, and so my suggestions above are effectively ruled out leaving what you suggested as the only possibilities?"

AFAIK, since we indirectly measure the Hubble constant and the deceleration and acceleration over time, there can only be a limited range of initial expansion rates matching the data inside different cosmic models. In essence, cosmologists tweak the parameter values of the various models to determine which model and which parameter set fit observations best. This determines the most probably Hubble constant and energy densities (matter, radiation and vacuum) and that sets the most likely expansion rate for any epoch, within rather tight limits.

So I would say that what I suggested are the most likely possibilities.

-J

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#34
In reply to #33

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 8:09 PM

Jorrie,

I guess that makes sense and agree what you suggested are the most likely culprits.

Roger

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#30
In reply to #28

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/20/2009 4:07 AM

Hi Jorrie,

I found the following article that talks about the filament and void structure being related to a cosmic tidal force. Unfortunately, from what I can tell, they are talking about something different from us when they speak of cosmic tidal force (theirs is a matter based tidal force). I'm posting the link here because it's interesting read and thought you might enjoy it, even if it's off topic a little.

http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.2480v1.pdf

I'm having a hard time finding articles on the cosmic tidal force we're talking about. Do they perhaps call it something different in the literature? Maybe I'm just tired. Anyway I suspect you're right that they include it in their models. The deceleration of the cosmic expansion was very large in the early early universe (<1GY), I have to imagine it's role was large back then.

Could you maybe plot the cosmic tidal force's strength over time? I'd be interested to see how much larger it was the first few 100 million years after the big bang.

I'm gonna sign out now. Thanks for the fun discussion.

Roger

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#22
In reply to #19

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 11:23 AM

Hi Roger,

From Fitzpatrick's intermediate QM course he says:

"Now, we saw, in Sect. 3.2, that a plane-wave propagates at the so-called phasevelocity, vp = ω/k . (3.36)

However, according to the above dispersion relation, a particle plane-wave propagates at vp = p/2m . (3.37)

Note, from Eq. (3.28), that this is only half of the classical particle velocity"

Does this change your picture of momentum in de Broglie waves?

-S

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#23
In reply to #22

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/19/2009 11:54 AM

Hi StandardsGuy,

I don't think it does. The key is that you're speaking of Phase Velocity, but it's Group Velocity that matters. I recommend the following link (I didn't want to post the entire derivation here).

If that link doesn't change your mind let me know and we can discuss it further.

Roger

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#4

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 3:39 AM

Nice presentation Jorrie. I've, also, taken a quick look (due to lack of time) at the 4 previous presentations concerning the cosmic ballon. They deserve a second -more detailed- reading though.

I wonder if we should consider this tidal, "antigravitational" forces (due to accelerated expansion - vacuum energy) as a 5th new kind of force (beside the 4 other well-known, fundamental forces i.e. electromagnetic, strong nuclear, weak nuclear and gravitational force).

(Taking the occasion of your post about the redshift, I think I have to make an introductory post about the whole wave and quanta issue... It is a very interesting and complex concept and it is, always, worth-while to discuss it...)

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#14
In reply to #4

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 9:39 PM

Hi George, thanks!

"I wonder if we should consider this tidal, forces (due to accelerated expansion - vacuum energy) as a 5th new kind of force ..."

No, I don't think so. Firstly, it is not "antigravitational", because the tidal forces are present without any form of dark energy - just straight old gravity from matter density gives a decelerated expansion and tidal compressive forces on the tether.

I would love to see a new thread on "wave and quanta issue"!

-J

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#35
In reply to #14

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/21/2009 6:30 AM

I don't support that we should consider this kind of "force" as a 5th fundamental force either.

But I made the following thought: Noone can claim that a tidal force due to gravity (e.g. on earth from the moon or on the whole universe from its total mass) is a new kind of force. Its primary cause is the gravity itself and this is the true, fundamental force. But, in the case of the accelerated expansion of the universe, the situation is -in a way- different. These tidal forces are not related to gravity. There is a new "reason" which produces these forces. This reason is -probably- the vacuum energy. This -always increased- vacuum energy is considered as an internal "presure" of the space itshelf. Furthermore, its result is these tidal forces which push everything -wherever it is- away from everything else. So, we have a new "push" from a new "cause".

(That's why I said this -just as a thought- in my previous post.)

Any comments???

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#36
In reply to #35

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/21/2009 3:37 PM

Hi George. You wrote: "There is a new "reason" which produces these forces. This reason is -probably- the vacuum energy."

Not for cosmic tidal forces - they are present whenever the expansion rate changes, even while it is slowing down due to matter domination. When vacuum energy density and matter energy density balances (as at 7 Gy age), there is no cosmic tidal force. Hence it is not specifically related to the vacuum energy.

That said, one may perhaps call the dark energy that accelerates expansion a 'new kind of force' ...

-J

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#38
In reply to #36

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/22/2009 10:45 AM

Yes, Jorrie. Your presentation was perfectly understood. Tidal forces are present whenever the expansion is either accelerated or decelerated.

There is always a 'fight' between two opposite 'tensions': an 'inside tension' produced by gravity and an 'outside tension' produced by sth else which is called dark energy. Btw the adjective "dark" has the meaning of the "unknown" (exactly as in the issue of the "dark matter"). Probably this dark energy is the vacuum energy (and that's why -in my previous post- I refer to 'vacuum energy' instead of 'dark energy'... if this bothers you, I'll stick with the 'dark energy'...)

The universe had 3 phases: 1) Inflation 2) Domination of gravity → Decelerated expansion 3) Domination of dark energy → Accelerated expansion (you could add a short phase between (2) & (3): Balance of gravity and dark energy → Constant expansion).

So, the total cause of these tidal forces is a combination of two components: the gravity and the dark energy. The gravity is a well known, fundamental force. But the dark energy is sth new. It is a kind of a "universal field" which affects everything in the universe. So -maybe- it could be considered as a "fundamental field" too (i.e. like the gravitational field).

In other words, -beside gravity- this new "field" plays its role to the production of these tidal forces. (Actually it plays a role on the expansion of the space itshelf.) That's why I had the thought that someone could consider it as another "fundamental field". (Although, I personally agree with you that we should not consider it this way.)

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#51
In reply to #38

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 2:32 PM

Hi George, you wrote:

"That's why I had the thought that someone could consider it as another "fundamental field"."

AFAIK, dark energy is not viewed as a field like gravity or electromagnetism. It is essentially the raw energy of empty space, with a negative pressure. Precisely what it is, nobody knows, but it has some characteristics that do not fit the description "field". An example is the fact that a cosmos that is static, due to dark energy balancing mass energy, can contract at an increasing rate, if slightly perturbed in that direction.

-J

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#12

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 8:22 PM

Hi Jorrie,

Good post as usual. In figure 1, it seems clear that the red arrows represent the distance between the buttons remaining constant with the passage of time (expansion of balloon). Also clear that the blue arrows indicate separation of the buttons if untethered at the start. What I am confused about is the words "Comoving untethered button" near each blue arrow. Are these labeling the arrows? Since the buttons would be moving apart in this case, is the term "comoving" appropriate? It doesn't fit my conception of comoving. Think about it and see if you meant to say it that way.

It's great that you put some values on this at this scale. So the Virgo super-cluster is slightly stretching apart by vacuum energy now?

-S

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#15
In reply to #12

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/17/2009 9:57 PM

Hi S, yes, your issue with the term "comoving" is very valid!

I've used the cosmological jargon, with comoving taken to mean 'moving with the Hubble flow' perhaps a little out of context, since it may here be mistaken as buttons comoving with each other. I guess "Untethered button, moving with Hubble flow" would have been better. I will update the figure in due course...

"So the Virgo super-cluster is slightly stretching apart by vacuum energy now?"

I believe so, yes, although it may be very marginal and only for those clusters at the edge of the structure. The peculiar velocities of the galaxies in the super-clusters are of the same order as the Hubble flow, so it is very difficult to measure this sort of thing for any structure. I've written a little more about this problem in chapter 16 (page 205) of Relativity 4 Engineers.

-J

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#37

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/21/2009 7:05 PM

Hi Jorrie,

I was thinking about an earlier discussion we had regarding the energy loss of cosmic redshifted photons and now of course we can say cosmic expansion induced momentum loss (from our deBroglie argument). Your response to my question as to where the lost energy goes is that while these things lose energy due to the redshifting, their gravitational potentials are increasing (as they are separated by the Hubble flow. That argument is sound and something I hadn't considered back then, therefore I shelved my objection of energy loss.

Recently however on a relaxing walk I came up with a thought experiment which seems to work around your explanation.

Imagine an expanding universe with only one photon with no other radiation or matter. As it expands it redshifts the photon and the total energy of the photon is gradually lost. Unfortunately there is no gravitational potential to explain away this lost energy. Since we know in our universe energy is conserved, how do we reconcile this?

After some thought I remembered that conservation of momentum actually would forbid a universe with a single photon (must exist in pairs). Similarly particles. In other words, the example I found where your argument broke down isn't valid as a possible configuration of the universe.

I know this is a bit convoluted, I just thought it was an interesting affirmation of your explanation.

Roger

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#39
In reply to #37

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/22/2009 4:07 PM

Hi Roger, you wrote:

"Unfortunately there is no gravitational potential to explain away this lost energy. Since we know in our universe energy is conserved, how do we reconcile this?"

and

"After some thought I remembered that conservation of momentum actually would forbid a universe with a single photon (must exist in pairs)."

Practically, your latter statement may be right, but your first statement is not really a problem for the hyper-spherical model. The single photon (if it could exist) must have energy that is uniformly spread over the total surface of the balloon - rather impossible, but a requirement of Friedman cosmology. The increasing radius of the expanding sphere would then convert the energy lost due to redshift to potential energy.

As it happens, the energy of the vacuum overshadows the energy of a single photon by far. In a way, the photon would create a fast-moving inhomogeneity of negligible magnitude, which will disappear entirely as time tends to infinity.

-J

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#40
In reply to #39

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/22/2009 9:32 PM

Hi Jorrie,

Maybe my line of thinking is wrong. Here's where I'm coming from. As far as I can tell, the cosmological model that exists today doesn't conserve energy. I say this because expansion is occuring (there is more space today than there was yesterday) and the energy density of space is constant. More space, same energy per space, this adds up to more energy. Where this energy comes from we doesn't matter for now, the point is the model doesn't account for it.

However, and this is a huge however, matter and radiation do conserve energy in their interactions. Also, it seems like any "forces" generated by cosmic expansion, be them real or pseudo, that act on radiation and matter, don't seem to lead to radiation and matter violating conservation of energy. In other words, if a photon is redshifted it has lost energy, however it also has a higher gravitational potential energy, the energy of the system is conserved.

My single photon thought experiment showed that the photon would be redshifted and yet wouldn't have gained gravitational potential energy thus violating conservation of energy. However this case is irrelevant because it violates conservation of momentum. It simply is a forbidden case. There is no doubt about it.

You Wrote:"The single photon (if it could exist) must have energy that is uniformly spread over the total surface of the balloon - rather impossible, but a requirement of Friedman cosmology."

I'm confused by this statement, please help me out. If we can talk about universes without matter or radiation, why can't we talk about a universe with say two photons? It's a calculation I'd like to do (intend to do), but if there is a fundamental reason why it doesn't make sense, I'd like to understand before I waste my time calculating energies.

Roger

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#41
In reply to #40

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/22/2009 11:00 PM

Hi Roger.

All I tried to indicate is that the Friedman (ΛCDM) model is only strictly correct for a homogeneous cosmos. If you want to use it to calculate for photons only, you need a myriad of them, 'evenly spread', approaching a uniform energy density.

In any practical cosmos, energy density is only approximately uniform and only on the large scale (the domain of the ΛCDM model), so we are only approximating anyway. I guess that you can use the method you suggested as an approximation, but I am not aware of any formal method for calculating the energy conservation of individual photons in an otherwise empty cosmos.

-J

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#42
In reply to #41

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/22/2009 11:35 PM

Hi Jorrie,

You Wrote:"All I tried to indicate is that the Friedman (ΛCDM) model is only strictly correct for a homogeneous cosmos. If you want to use it to calculate for photons only, you need a myriad of them, 'evenly spread', approaching a uniform energy density."


No, you're right. I'd like to compare the energy lost by a photon due to cosmic redshift with the energy gained by gravitational potential, but the question is; gravitational potential of what? I don't know if that makes any sense what I'm trying to say.

A better way to try for the energy balance would be a thermodynamic approach. Are you familiar with the idea of entropy in cosmology? I'd like to learn something of that subject if you have any personal opinions on it. I know nothing on the subject but it seems interesting.

Roger

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#46
In reply to #39

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/23/2009 8:03 PM

"The single photon (if it could exist) must have energy that is uniformly spread over the total surface of the balloon - rather impossible, but a requirement of Friedman cosmology. The increasing radius of the expanding sphere would then convert the energy lost due to redshift to potential energy."

In order to have redshift you have to have an observer. Let's take an object that clearly has mass, say a planet (only 1 in the universe). Will the planet change potential energy from the expansion as well? It seems that you need 2 planets which are moved apart by the expansion to change the potential energy (toward each other), which would go up. Am I grasping this correctly?

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#47
In reply to #46

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 2:19 AM

Hi S, you wrote: "In order to have redshift you have to have an observer."

Not really; in principle, redshift or stretching of wavelengths happens during expansion, irrespective of observers.

The cosmic potential energy of a single planet should be viewed relative to the center of the hypersphere, in which case there is no problem, or is there? Friedman would have demanded that this planet be ground to dust and then be spread evenly around the hyper-surface!

Two galaxies that move with the Hubble flow have cosmic potential energy in the hyper-space dimension. If they are far form each other, they do not have normal spatial potential energy relative to each other, because the entire hypersurface is at the same potential level. The galaxies do cause local inhomogeneities in this potential and close-by galaxies do influence each other because of it. The pure Friedman cosmology ignores such inhomogeneities, because it only consider the large scale, which is approximately homogeneous.

-J

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#48
In reply to #47

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 3:35 AM

Jorrie, any comment on my post #38??? Could the dark energy be considered -by someone- as a universal field???

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#50
In reply to #48

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 11:48 AM

Hi GK.

"Could the dark energy be considered -by someone- as a universal field?"

I think so, possibly the Higgs Field.

-S

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#53
In reply to #50

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 10:44 PM

Hi S.

"I think so, possibly the Higgs Field."

Agreed, from the particle physicist's POV. Personally, I do not understand how it can solve the cosmological dark energy problem, e.g., creating the negative pressure required by the Friedman cosmology.

-J

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#49
In reply to #47

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 11:36 AM

Hi Jorrie,

I have problems with this.

"The cosmic potential energy of a single planet should be viewed relative to the center of the hypersphere"

If the planet can not fall to the center of the hypersphere, then it has no potential energy in that "direction". That's why I used 2 planets in the last post. That's the only way I can see any potential energy existing. And, it seems to me, that's why Friedman demanded matter to be spread out.

-S

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#52
In reply to #49

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/24/2009 2:54 PM

Hi S, you wrote:

"If the planet can not fall to the center of the hypersphere, then it has no potential energy in that "direction"."

But the planet can! If the mass of the planet causes the hypersphere to contract, it will eventually end up at the center of the sphere. Granted, with a single (or even double) planet, the contraction of the hypersphere will not be spherically symmetrical, but still...

The potential energy of the Friedman solution works only in the hyperspatial direction. Potential energy in the normal spatial directions is the domain of Schwarzschild geometry (local inhomogeneities).

-J

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#54
In reply to #49

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/26/2009 1:10 AM

Hi S, further to my reply #52.

An interesting (if not quite rigorous) way of looking at the combination of the Friedman and Schwarzschild solutions is depicted on the right.

The black semi-circle represents the uniform density Friedman hypersphere and the blue curves represent a Schwarzschild perturbation caused by a huge mass concentration (highly exaggerated) in the center of the picture. Taken with the appropriate pinch of salt, the arrows indicate the interaction between cosmic expansion and the normal Schwarzschild gravity. The two vertical arrows indicate a 'critical distance' where cosmic expansion and normal gravity balance, at least temporarily. The inner and outer arrows respectively indicate the cases where either of the two 'forces' dominate at the time (loosely stated).

Now, of we take this to the limit of a black hole with mass M, the black hole's central singularity will perturb the skin of the balloon all the way to the center of the hypersphere, where the BB singularity sits (not shown in the figure). Since we are free to choose zero potential anywhere except at R=0, taking the 'Friedman potential' (PF = -GM/R) of the unperturbed surface as zero (by addition of a constant), the PF of the black hole singularity at R => 0 will become infinitely negative. Even in an expanding universe, it seems that a black hole singularity that forms after the BB will 'fall' all the way to the center of the Friedman balloon, losing all its potential energy in the process.

I realize that there may be any number of issues with such an interpretation, but it is interesting for as far as it holds. I think it is an approximate solution to the Einstein field equations, but I'm not sure what the limits of applicability are.

-J

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#55
In reply to #54

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/26/2009 10:27 AM

Hi Jorrie,

Interesting interpretation. As for post #52, I guess we don't really know the final outcome of the universe, though we like to think we do. So you may be right. Anyway, good thinking - keep it up!

-S

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#43

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/23/2009 12:34 AM

Ok, I'm going to try this fully aware it may be useless.

A photon has energy:

Mass equivalence says:

Masseffphoton = h/cλ

Gravitational Potential is equal to:

PE=-GM/r

For a photon, the gravitational potential it produces is:

PE=-Gh/cλr

So the potential energy gained for a change in distance Δr is

ΔPE=-Gh/cλΔr

So I need to equate the change in r with the change in λ which I will do in a later post (once I figure it out).

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#44
In reply to #43

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/23/2009 1:47 PM

Hi Roger.

I do spot one problem with your analysis. If the single photon is the only potential generating entity on an expanding hypersphere, the potential caused by that photon would remain constant (I have taken r in your P = -Gh/cλr to be the hyper-radius). The reason is that λ and r change as the inverse of each other, i.e. λr = constant.

If you are not talking about the hypersphere, you must have two photons that are moving relative to each other, but I'm not sure what that would mean in practice. I have never considered single/double photons as gravitational potential generating entities.

-J

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#45
In reply to #44

Re: Cosmic Balloon Application V: Cosmic Tidal Force

07/23/2009 5:26 PM

Hi Jorrie,

Thanks for the input. I'm not going to look at a single photon because it violates conservation of momentum(it would give the universe a nonzero momentum).

r in that equation isnt the hyper-radius, actually its the distance between the two photons. I'm sorry, I wasn't clear about that at all.

You Wrote:"If you are not talking about the hypersphere, you must have two photons that are moving relative to each other, but I'm not sure what that would mean in practice. I have never considered single/double photons as gravitational potential generating entities."

That's exactly what I'm trying to calculate. If you consider two photons without cosmic expansion, they still are climbing out of each others gravitational potentials as they move away from each other. So classically they should experience a redshift even without cosmic expansion.

On top of that there is also the cosmic redshift, which I think can be treated seperately from the classical redshift (since we know the classical redshift is conserved).

I will post soon some math on this. I'm not sure what it would mean in practice either, but I'm interested to see how the numbers compare.

Roger

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#56
In reply to #45

Re: Cosmic Balloon Application V: Cosmic Tidal Force

08/09/2009 12:18 AM

Hi Roger.

I have done some calcs on those earliest galaxies that you PM'd me about. They possibly formed in the epoch z ~ 19 to z ~ 9, or roughly 200 to 500 million years after the BB. At 200 My age, it was probably just a dark matter halo that formed (including some matter particles that would later form stars). If I take that halo as having the total mass of our MW galaxy (with dark matter included), it comes to some M ~ 3 x 1042 kg, with a halo radius of some 10 thousand light years, or r ~ 1020 meters (20% of the MW, according to the article).

A single particle orbiting just outside such a halo would experience a gravitational acceleration of order 10-9 g (one nano-g). The cosmic tidal acceleration during that epoch would have been inward, enhancing gravity, but its magnitude worked out to only ~ 10-13 g (100 femto-g), hence negligible compared to gravity.

A question that you asked is: would this tiny force, operating over millions of years not have compacted the galaxies? The paper that I referred you to (http://arxiv.org/abs/gr-qc/0602098) states that all that will happen is that the angular orbit velocity will be increased to compensate for the additional contracting force. The increase ratio is given as:

0 - ω)/ω = √[1 - sign(a''/a) (r/rcrit)3]

where a is the expansion factor, a'' the (negative) acceleration of expansion of that period, r the orbital radius and rcrit the radius where the gravitational and cosmic tidal forces are of equal magnitude for that mass.

Taking the worst case, at z~19 (when the dark matter is reckoned to have formed), a''/a ~ -112 Gy-1. The crucial value is however r/rcrit ~ 0.03, meaning that it is only at ~ 30 times the radius of the galactic halo that cosmic tidal forces equal gravity. This gives the utterly negligible order of magnitude: 0 - ω)/ω ~ 10-5.

Hence, any notion that cosmic tidal forces could have caused those stars to whiz around the galaxy at more than twice the velocity of the solar system's velocity is false. Likewise, cosmic tidal forces do not contribute to the compactness of early galaxies.

-J

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#57
In reply to #56

Re: Cosmic Balloon Application V: Cosmic Tidal Force

08/10/2009 12:53 AM

I'm sorry Jorrie, I'm not convinced.

When I have time (god I need time), I will calculate the central force lagrangian. I will treat the cosmic tidal force as a spring potential (equilibrium of the spring being at the center of mass) and see if I can solve to see how the orbits would be effected over billions of years. I'll post the derivation and results here.

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#58
In reply to #57

Re: Cosmic Balloon Application V: Cosmic Tidal Force

08/11/2009 12:57 AM

Hi Roger.

Before you spend time on it, consider the following:

  1. When a purely radial force (positive or negative) is applied to an orbit, there is a stable analytical circular solution that simply changes the orbital velocity to balance the changed radial 'gravitational force' with a change in centrifugal force. The radius needs no change in order to be stable.
  2. Practically, if a small constant radial force is applied to a normal, stable, circular orbit, the orbit will become elliptical. AFAIK, there is no analytical solution for such a case. Experience with general relativity (GR), which changes the force of gravity when compared to Newtonian gravity, convinces me that the orbit will be stable, i.e., not in-spiraling or out-spiraling.
  3. The radial force will now only be normal to the movement at apo- and periapsis and at an angle during the rest of the orbit, causing a precession of the ellipse. Let's take a negative radial force, working with gravity. During the 'in-falling' half of the orbit, the extra force will add to the orbital velocity and during the 'out-falling' phase, it will work against the orbital velocity, causing the precession. However, there is no net orbital energy transferred to the orbit; the in-falling and out-falling portions cancel out and orbital energy is conserved.
  4. The situation is well worked out for GR and there is no evidence of any in-spiraling our out-spiraling - other than due to tidal gravity and gravitational waves in extreme case.
  5. As a practical case, think of a solar sail. It does not work if the solar wind is made to impact the sail it at a normal angle - it needs to be tilted near 45 degrees relative to the radial.

-J

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