I Hubble's law and conservation of energy

[olgerm]

2 bodies that have distance d between them are distancing from each other because Hubbles law. at time t=0 distance between them was d(0) and speed between them was 0.
If no force interacts with them then distance is increasing by rate $\frac{\partial d}{\partial t}=H_0*d$
Is it correct?

Their potential energy is increasing by rate $\frac{\partial E_{pot}}{\partial t}=\frac{(m_1*m_2*k_G-q_1*q_2*k_E)*H_0*e^{-t*H_0}}{d(0)}$Is that correct?
Where is that energy coming from? How is total energy conserved?

If a force interacts between the 2 bodies, that keeps distance same ($\frac{\partial d}{\partial t}=0$)
Is the energy in that scenario converting to some other form of energy?
I know that Hubbles law is very small in that scale, but if the 2 bodies are proton and electron in hydrogen atom, would hobbles law make this atom unstable?

$H_0$ is Hubble parameter.
$k_G$ is gravitational constant.
$k_E$ is Coulomb's constant

 

[Drakkith]

A good question, and one whose answer isn't exactly satisfactory. The issue is that energy conservation applies in GR at a local scale, not necessarily at a global (or universal?) scale.

See this article: http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

 

[kimbyd]

A good question, and one whose answer isn't exactly satisfactory. The issue is that energy conservation applies in GR at a local scale, not necessarily at a global (or universal?) scale.

See this article: http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

Another, complimentary take on the matter:
https://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

 

[olgerm]

energy conservation applies in GR at a local scale, not necessarily at a global (or universal?) scale.

So it is theoretically possible to make pepertual motion machine by exploiting Hubble's law?

I know that the effect is too small to measure it but what is theoretical prediction about how does expansion by Hubble's law affect hydrogen atom? If it emits electromagnetic radiation what frequency this radiation has?

 

[Drakkith]

So it is theoretically possible to make pepertual motion machine by exploiting Hubble's law?

Not that I know of. I believe the distance scales would be so large your would run into the problem that different parts of your machine are receding from each other at absurd velocities. Perhaps FTL velocities.

I know that the effect is too small to measure it but what is theoretical prediction about how does expansion by Hubble's law affect hydrogen atom? If it emits electromagnetic radiation what frequency this radiation has?

It doesn't affect a hydrogen atom. Atoms are bound together by strong electromagnetic and nuclear forces and are not subject to expansion's effects. Expansion only affects objects when the forces binding them together are very, very weak, which is why it still takes tens or hundreds of millions of light-years of separation for even gravity (the weakest force) to become too weak to hold objects together.

The emitted radiation, when viewed from a great distance, would be redshifted due to expansion, but that is not an effect on the atom itself, but a result of the light moving through expanding space.

 

[PeterDonis]

So it is theoretically possible to make pepertual motion machine by exploiting Hubble's law?

No. A perpetual motion machine would require violating local energy conservation.

In any case, "lack of global energy conservation" in GR is really not a good choice of terminology; a better way to describe it would be that "energy" is not a well-defined global concept in GR (except in a very special class of spacetimes, stationary spacetimes, which does not include FRW spacetimes, the ones in which Hubble's law appears).

 

[olgerm]

Expansion only affects objects when the forces binding them together are very, very weak, which is why it still takes tens or hundreds of millions of light-years of separation for even gravity (the weakest force) to become too weak to hold objects together.

Or the effect of expansion on bodies is just very small if force binding the bodies together is not very, very weak, so that altough the effect can not be measured, theoretical predictions about what would happen could still be made?

The emitted radiation, when viewed from a great distance, would be redshifted due to expansion, but that is not an effect on the atom itself, but a result of the light moving through expanding space.

I mean the effect on atom itself not on radiation that it has radiated.

 

[PeterDonis]

Expansion only affects objects when the forces binding them together are very, very weak

This could be misinterpreted. If two objects are flying apart at some speed, and the strength of forces between them is weak enough, they will never stop flying apart and become bound. That has nothing to do with "expansion"; it would be true even in flat spacetime. "Expansion" means the global spacetime geometry is not flat, but has a certain curved shape that can be sliced into spacelike 3-surfaces in a certain way with certain properties. But it doesn't mean there is some mysterious "expansion" force on objects over and above the fact that they're flying apart.

 

[Buzz Bloom]

But it doesn't mean there is some mysterious "expansion" force on objects over and above the fact that they're flying apart.

Hi Peter:

I am probably misinterpreting what the quote above means. It seems to me to imply that the following analysis is incorrect.

Imagine two objects of the same mass M moving in circular orbits about their center of mass, assume separated by at a distance D>>r_M, the radius of an event horizon, which implies that the Newtonian gravity effect on the two bodies is a very good approximation to GR (if we ignore the effect of gravitational waves causing orbital decay). Also assume that it is far in the future, and the density of matter has become negligible, so the Friedmann equation becomes (approximately):

h/h₀ = 1,​

and h becomes a constant h₀.

This implies:

a = e^h₀(t-t₀),​

da/dt = h₀ a,​

and the accelerations

d²a/dt = h₀² a .​

If the effect of the expanding universe is ignored, then the radial acceleration A_G acting on one body by the other is:

A_G = -GM/D² .​

This must equal the negative of the centrifugal acceleration

A_C = 2 v²/D .​

where v is the tangential velocity of each body moving along its circular orbit. This implies
v² = GM/2D .
Thus, no matter how far apart the two bodies are, there is a circular tangential velocity for a stable orbit.

Now let us consider the effect of the expanding universe. If the two objects are far enough apart, and the gravitational effects between them becomes insignificant compared to the effect of the expanding universe, then (approximately)

dD/dt = h₀D .​

The acceleration between the two bodies is

A_H = d²D/dt² = h₀ dD/dt = h₀² D.​

If all three accelerations act on the pair of bodies, there will be a value of D for which the tangential circular velocity is zero, and the total of the three accelerations is zero.

-GM/D² + 2 v²/D + h₀²D = 0​

v = 0 gives

-GM/D² + h₀²D = 0​

implying

D³ = GM/h₀² .​

This implies that two bodies, each of mass M and stationary at a distance between them of

D = (GM/h₀²)^1/3​

will remain stationary. This is because the gravitational attraction is exactly balanced by the expanding universe acceleration.

Now, if this above analysis is false, what does happen to these two objects with these initial conditions.
Do they begin to fly apart? If so, do they ever catch up with the “normal” speed of

v = H₀D?​

If not, do they fall towards each other? In that case, is the dynamics of the fall independent of the expanding universe? If D was slightly larger than (GM/H₀)^1/3, what would happen then?

Regards,
Buzz

 

[Bandersnatch]

Also assume that it is far in the future, and the density of matter has become negligible, so the Friedmann equation becomes (approximately):
h/h0 = 1,and h becomes a constant h0.

This assumes dark energy, causing accelerated expansion. The bit you quoted from Peter talks about expansion only, sans dark energy, where (using Newtonian approximation) objects are just flying apart inertially. While it's possible to attribute force-like effects to dark energy, it's not correct to attribute them to expansion alone.

 

[Buzz Bloom]

The bit you quoted from Peter talks about expansion only, where (using Newtonian approximation) objects are just flying apart inertially. While it's possible to attribute force-like effects to dark energy, it's not correct to attribute them to expansion alone.

Hi Bandersnatch:

Thank you for your reposnse.

I said at the beginning of my post that I may have misunderstood Peter's quote. Your explanation still leaves me confused about several points.
1. Are you saying that Peter's quote does not imply the analysis I posted is wrong?
2. Are you saying that Peter's quote assumes no expansion, and that the motions of the objects he is referring to is only due to local gravitational effects?
3. I did not mention "dark energy". I simply assumed a cosmological constant without attributing to it any physical interpretation. With this context, I assume that the expansion (with the assumption that mass density has become very small) is due to the cosmological constant, not the other way around. Why is it incorrect to talk about the second derivative of the scale factor? If the geometry is expanding, and any two geometric points in space with a distance D apart having an acceleration

d²a/dt² = h₀²D,​

why does this not affect objects occupying these two points?

Regards,
Buzz

 

[PeterDonis]

Now let us consider the effect of the expanding universe.

You're not considering the effects of an expanding universe. You're considering the effects of a cosmological constant. That's something different.

The presence of a cosmological constant of the magnitude we think exists in our actual universe does, in principle, produce a force that pushes objects apart. But the magnitude of the force depends directly on the distance. At small distances, where "small" here means "the size of a galaxy cluster or smaller", the force is too small to significantly affect bound systems (like galaxy clusters or smaller ones).

In any case, the force due to a cosmological constant is not due to "expansion". It would be present regardless of whether the universe as a whole was expanding, static, or contracting.

 

[PeterDonis]

I did not mention "dark energy". I simply assumed a cosmological constant

Don't quibble. Whether you call it "dark energy" or a "cosmological constant" does not matter to the dynamics. You wrote down the dynamics explicitly, and that's what the responses you got were based on, not what you called it. You did exactly right by writing down the explicit math; but you also need to realize that when you write down the explicit math, we will actually read it and will ignore the ordinary language you use, since the physics is in the math, not the ordinary language.

 

[PeterDonis]

With this context, I assume that the expansion (with the assumption that mass density has become very small) is due to the cosmological constant

It isn't. The cosmological constant is only one factor involved, and up until a few billion years ago, it wasn't even the largest one--the universe was matter dominated before that, not cosmological constant dominated. So the cosmological constant by itself cannot explain the expansion. It can only explain why the expansion is currently accelerating.

 

[PeterDonis]

why does this not affect objects occupying these two points?

Because the force due to the cosmological constant is too small to pull apart bound systems of the size of a galaxy cluster or smaller (because the force gets larger with distance, so it only gets large enough to matter once you reach a large enough distance). It will slightly change the orbital parameters of objects in bound systems (though even this effect is too small to matter for the bound systems we observe), but it won't keep them from being bound.

 

[Buzz Bloom]

The cosmological constant is only one factor involved, and up until a few billion years ago, it wasn't even the largest one--the universe was matter dominated before that, not cosmological constant dominated. So the cosmological constant by itself cannot explain the expansion. It can only explain why the expansion is currently accelerating.

It will slightly change the orbital parameters of objects in bound systems (though even this effect is too small to matter for the bound systems we observe), but it won't keep them from being bound.

Hi Peter:

Regarding the first quote, I accept as completely valid your criticism about how I phrased my interpretation of the math. I seem to have difficulty in expressing properly what the math means conceptually. The concept of the cause of expansion seems particularly difficult to pin down.

If the Friedmann equation correctly describes the dynamic behavior of the a universe model, I am tempted to attibute the cause to whatever it is in the physical reality that the four Ω parameters (which add up to 1) of a particular math model (based on the Friedmann equation) actually describes. The h₀ parameter I do not consider as contributing a descriptive cause since it seems more like a result. (It is the value of (1/a)×da(t)/dt for the value of t corresponding to a = 1.) Each of the four Ω parameters is defined as a mass density (or mass equivalent density or something that behaves mathematically like a math density) divided by the critical mass density

ρ_c = 3h²/8πG .​

Would you accept the combination of the four density ratios as the combined cause of expansion (and possible contraction depending on the values) of such a universe model?

In the scenario of my analysis, I explicitly assumed that the three Ω parameters other than Ω_Λ all very close to zero. In this particular scenario, whatever Ω_Λ represents physically would be the only significant cause.

I confess I am not able to choose values for the four Ω parameters which would be equivalent to Einstein's first1917 GR cosmological model (prior to Hubble's contribution) in which the universe has neither expansion nor contraction. The following describes this model, a finite hyperspherical geometry.

https://en.wikipedia.org/wiki/Static_universe​

The article gives the following relationship among R_E, the radius of curvature, Λ_E, the cosmological constant, and ρ, the mass density.

Aside from

Ω_r=0​

I am not comfortable in trying to establish values for the other three Ω parameters. However, it would seem that in this scenario, the role of Ω_Λ would be the "cause" of the stable non-expanding and non-collapsing universe.

Your second quote did not take into account the scenario in my post #9 (assuming three Ω parameters = 0, and Ω_Λ=1 in which I presented a formula for a distance D in which two stationary bodies of mass M separated by a distance D remain stationary.

D = (GM/H₀)^1/3​

At that distance D the repelling effect of Ω_Λ would exactly balance the gravitational attraction between the two masses.

Regards,
Buzz

 

[kimbyd]

Hi Peter:

Regarding the first quote, I accept as completely valid your criticism about how I phrased my interpretation of the math. I seem to have difficulty in expressing properly what the math means conceptually. The concept of the cause of expansion seems particularly difficult to pin down.

If the Friedmann equation correctly describes the dynamic behavior of the a universe model, I am tempted to attibute the cause to whatever it is in the physical reality that the four Ω parameters (which add up to 1) of a particular math model (based on the Friedmann equation) actually describes. The h₀ parameter I do not consider as contributing a descriptive cause since it seems more like a result. (It is the value of (1/a)×da(t)/dt for the value of t corresponding to a = 1.) Each of the four Ω parameters is defined as a mass density (or mass equivalent density or something that behaves mathematically like a math density) divided by the critical mass density

ρ_c = 3h²/8πG .​

Would you accept the combination of the four density ratios as the combined cause of expansion (and possible contraction depending on the values) of such a universe model?

In the scenario of my analysis, I explicitly assumed that

|1-Ω_Λ|<<1.​

In this particular scenario, whatever Ω_Λ represents physically would be the only significant cause.

I confess I am not able to choose values for the four Ω parameters which would be equivalent to Einstein's first1917 GR cosmological model (prior to Hubble's contribution) in which the universe has neither expansion nor contraction. The following describes this model, a finite hyperspherical geometry.

https://en.wikipedia.org/wiki/Static_universe​

The article gives the following relationship among R_E, the radius of curvature, Λ_E, the cosmological constant, and ρ, the mass density.
View attachment 245496
Aside from

Ω_r=0​

I am not comfortable in trying to establish values for the other three Ω parameters. However, it would seem that in this scenario, the role of Ω_Λ would be the "cause" of the stable non-expanding and non-collapsing universe.

Your second quote did not take into account the scenario in my post #9 (assuming three Ω parameters = 0, and Ω_Λ=1 in which I presented a formula for a distance D in which two stationary bodies of mass M separated by a distance D remain stationary.

D = (GM/H₀)^1/3​

At that distance D the repelling effect of Ω_Λ would exactly balance the gravitational attraction
between the two masses.

Regards,
Buzz

What are you getting at? The static universe has been roundly disproven. It doesn't work theoretically because it's unstable (any inhomogeneity, like a bunch of galaxy clusters, will cause parts of the universe to contract and others to expand forever). It doesn't work observationally because of the observation of the Cosmic Microwave Background (which only works if our universe was much hotter and denser in the distant past).

 

[Buzz Bloom]

What are you getting at? The static universe has been roundly disproven.

Hi kimbyd:

My discussion of the 1917 Einstein model was to demonstrate (suggesting a mathematically possible model based on the Friedmann equation) that the cosmological constant could theorectically have a different role than the one Peter mentioned in the first quote. I did not intent that this model would match any currently observed cosmological observations.

Regards,
Buzz

 

[PeterDonis]

The concept of the cause of expansion seems particularly difficult to pin down.

No, it isn't. It's simple: the cause of the expansion is the Big Bang. The universe is expanding because it started out expanding at the Big Bang.

More precisely, in an inflationary cosmology, at the end of inflation, the inflaton field was rapidly expanding (there are technicalities here about how a field in its vacuum state can be "expanding", but we would need an "A" level discussion to go into those; suffice it to say here that they don't change what I'm saying), and when all the huge energy density in the inflaton field was transferred to the Standard Model fields at the end of inflation (this "reheating" event is the Big Bang as that term is properly used), the Standard Model fields changed from their vacuum states to a state that can be described as a very hot, very dense, rapidly expanding plasma of quarks and leptons.

Then for about 10 billion years that expansion slowed down because the universe was radiation and then matter dominated; a few billion years ago it became dark energy dominated and the expansion started speeding up.

 

[kimbyd]

Hi kimbyd:

My discussion of the 1917 Einstein model was to demonstrate (suggesting a mathematically possible model based on the Friedmann equation) that the cosmological constant could theorectically have a different role than the one Peter mentioned in the first quote. I did not intent that this model would matched any currently observed cosmological observations.

Regards,
Buzz

Except the cosmological constant in that model has the exact same role as it does in today's universe. The only difference is one of magnitude.

 

[Buzz Bloom]

No, it isn't. It's simple: the cause of the expansion is the Big Bang. The universe is expanding because it started out expanding at the Big Bang.
More precisely, in an inflationary cosmology, at the end of inflation, the inflation field was rapidly expanding (there are technicalities here about how a field in its vacuum state can be "expanding", but we would need an "A" level discussion to go into those; suffice it to say here that they don't change what I'm saying), and when all the huge energy density in the inflation field was transferred to the Standard Model fields at the end of inflation (this "reheating" event is the Big Bang as that term is properly used), the Standard Model fields changed from their vacuum states to a state that can be described as a very hot, very dense, rapidly expanding plasma of quarks and leptons.

Hi Peter:

Thank you much for explaining the role of inflation in creating the "big bang"as inflation ended (or soon after). I would much appreciate your posting a reference with more details (even though I know I will have difficulty understanding it). I had previously thought (since inflation is not yet included in the Standard Model for the early stages of the universe) there was still some uncertainty among the cosmology community of it's actually having happened.

Does any augmented Standard Model including inflation discuss what was happening before inflation? Was the universe a vacuum then? If so, where did all the energy come from to create "the huge energy density in the inflation field"?

Regards,
Buzz

 

[Buzz Bloom]

Except the cosmological constant in that model has the exact same role as it does in today's universe. The only difference is one of magnitude.

Hi kimbyd:

I think you are saying that my quote below is a mistake.

However, it would seem that in this scenario, the role of Ω_Λ would be the "cause" of the stable non-expanding and non-collapsing universe.

If so, please explain in what way what I said in this quote is an error? Einstein reluctantly put the cosmological constant into his 1917 model because he had no reason to believe at that time that the universe was expanding. He included it specifically for the purpose of obtaining a non-expanding universe because his model without the cosmological constant did expand. Therefore, I think it is quite reasonable to say from Einstein's perspective that the role of the cosmological constant was specifically to avoid expanding rather than to make acceleration of expansion happen, as Peter described it.

So the cosmological constant by itself cannot explain the expansion. It can only explain why the expansion is currently accelerating.

Thus whatever in the real world the cosmological constant represents, say X, with what was the state of knowledge at that time, X was the cause of what Einstein thought to be a stable non-expanding universe.

Regards,
Buzz

 

[kimbyd]

Matter attracts itself, tending to slow if not reverse the expansion. A cosmological constant tends to push matter apart. That's the role the cosmological constant plays. Whether that results in a universe that collapses back on itself or expands forever still depends upon the precise densities of each as well as the rate of expansion.

 

[Buzz Bloom]

A cosmological constant tends to push matter apart. That's the role the cosmological constant plays.

Hi kimbyd:

I agree that pushing apart is the role of the current cosmological constant, and it has been so since it was found to exist as something equivalent to a non-zero constant mass density independent of the scale factor. However, the Friedmann equation can also model theoretical universes in which Ω_Λ is negative, and its role would be additional attraction of pairs of distant points which would have the effect of reducing expansion. Such models would include the Einstein 1917 model.

Regards,
Buzz

 

[PeterDonis]

I would much appreciate your posting a reference with more details (even though I know I will have difficulty understanding it).

A good recent textbook on cosmology is Andrew Liddle, An Introduction to Modern Cosmology.

I had previously thought (since inflation is not yet included in the Standard Model for the early stages of the universe) there was still some uncertainty among the cosmology community of it's actually having happened.

I think some form of inflation is the majority opinion at this point. But note that all I really described in my post was what happened at the end of inflation. In other words, the Big Bang event. We have good evidence of that--that is, of the hot, dense, rapidly expanding state that, in inflationary models, occurs just after "reheating". I just explained it in terms of the end of inflation because that's, as I said, the majority opinion (AFAIK) about what came before the Big Bang. But you don't actually need to answer that question in order to know that the Big Bang is the cause of the expansion since then.

Does any augmented Standard Model including inflation discuss what was happening before inflation?

There are various alternatives. I'm not familiar enough with current work in this area to know which, if any, seems to be gaining ground.

 

[PeterDonis]

I agree that pushing apart is the role of the current cosmological constant

And the cosmological constant in the Einstein static universe. @kimbyd is correct that, as far as the cosmological constant is concerned, it works the same in that model as in our current model of our actual universe.

the Friedmann equation can also model theoretical universes in which Ω_Λ is negative

Yes, but...

Such models would include the Einstein 1917 model.

...no. As above, the cosmological constant in the Einstein static universe is positive, not negative.

 

[Buzz Bloom]

As above, the cosmological constant in the Einstein static universe is positive, not negative.

Hi Peter:

After some effort, I cannot figure out how a model based on the Friedman equation (with four Ωs whose sum is 1) can represent the 1917 Einstein static universe model. I need to give this a lot more more thought. Any suggestions would be appreciated.

Regards,
Buzz

 

[George Jones]

After some effort, I cannot figure out how a model based on the Friedman equation (with four Ωs whose sum is 1) can represent the 1917 Einstein static universe model. I need to give this a lot more more thought. Any suggestions would be appreciated.

In FLRW universes, the four $\Omega$s always sum to one, even in FLRW universes that are not (spatially) flat. You have even stated as much.

I confess I was not previously familiar with this usage of Ω, but I now see at the bottom of the Wiki section:
Ω_0,k - 1-Ω₀.
So I yield, and I am now convinced I was previously mistaken.

This equation is true at all times, i.e., $\Omega_k + \Omega = 1$, where
$$\Omega = \Omega_r + \Omega_m + \Omega_\Lambda.$$

Also, as you have stated, in the Einstein static is not spatially flat; it has positive spatial curvature.

The following describes this model, a finite hyperspherical geometry.
https://en.wikipedia.org/wiki/Static_universe

Therefore, in an Einstein static universe,

$$1 < \Omega = \Omega_r + \Omega_m + \Omega_\Lambda,$$

and $\Omega_k < 0$.

See equations (4.7) and (4.8) of

https://www.ast.cam.ac.uk/~pettini/Intro Cosmology/Lecture04.pdf

 

[olgerm]

it doesn't mean there is some mysterious "expansion" force on objects over and above the fact that they're flying apart.

Since here was a dispute about this claim, that I mostly did not understand, is this claim true or not?
I do not know a lot about GR, but since some interactions(gravitational interaction or all interactions?) are described with curvature of spacetime instead of forces in GR, is it meant about interpreting obsevations by classical physics?
If we attached ropes to distant objects and measured forces between this these objects and interpreted results by classical physics: would we not find there any additional "expansion" force?
for example measured force between 2 bodies that have masses $m_1$ and $m_2$ and charges $q_1=\sqrt{\frac{m_1*m_1*k_G}{k_E}}$ ; $q_2=\sqrt{\frac{m_1*m_1*k_G}{k_E}}$ ane therefore $F_{gravitational}+F_{electric}=0$?
Is it also true if we take into account acceleration of expansion?

 

[Bandersnatch]

If we attached ropes to distant objects and measured forces between this these objects and interpreted results by classical physics: would we not find there any additional "expansion" force?
Is it also true if we take into account acceleration of expansion?

If you picked a galaxy that is receding solely due to expansion of space, and attached a rope to a point at your location that is fixed and comoving, then that galaxy would be stopped by the tension in rope. It would be accelerated out of its local comoving coordinates until it stops receding w/r to the fixed point, and the tension in the rope disappears. At which point it'd just stay there, at a constant distance from the observer at the fixed point. This is similar to what would happen if you attached a rope to some mundane object (a ball, say) moving in accordance with Newton's equation of motion, with some initial velocity but no acceleration.

If the expansion is decelerating, the rope first stops the object, which then begins to approach the fixed point. This is similar to what would happen if you attached the rope to a mundane object that was flying away with some initial velocity and acceleration acting in the opposite direction.

If the expansion is accelerating, the rope stops the object, but the acceleration keeps the tension in the rope. Again, as above with the mundane object, but now with acceleration acting in the direction of initial velocity.

This is the tethered galaxy problem, discussed e.g. here: https://arxiv.org/abs/astro-ph/0104349

The point about not associating expansion with a force, is the same point one would make about not associating initial velocity in Newtonian motion with a force.

 

[olgerm]

@Bandersnatch ,were these measurement results different if we just were in special locations from which other objects are moving away?

 

[Bandersnatch]

I don't understand the question.

 

[Buzz Bloom]

If you picked a galaxy that is receding solely due to expansion of space, and attached a rope to a point at your location that is fixed and comoving, then that galaxy would be stopped by the tension in rope. It would be accelerated out of its local comoving coordinates until it stops receding w/r to the fixed point, and the tension in the rope disappears.

Hi Bandersnatch:

The quote above seems to disagree with my analysis in post #9.

Now let us consider the effect of the expanding universe. If the two objects are far enough apart, and the gravitational effects between them becomes insignificant compared to the effect of the expanding universe, then (approximately)

dD/dt = h₀D .​

The acceleration between the two bodies is

A_H = d²D/dt² = h₀ dD/dt = h₀² D.​

If all three accelerations act on the pair of bodies, there will be a value of D for which the tangential circular velocity is zero, and the total of the three accelerations is zero.

-GM/D² + 2 v²/D + h₀²D = 0​

v = 0 gives

-GM/D² + h₀²D = 0​

implying

D³ = GM/h₀² .​

This implies that two bodies, each of mass M and stationary at a distance between them of

D = (GM/h₀²)^1/3​

will remain stationary. This is because the gravitational attraction is exactly balanced by the expanding universe acceleration.

With respect to your discussion of the rope in your quote, a tension would remain since the distant end would experience a force F creating this tension

F = M h₀² D ,​

where M is the mass of the distant galaxy and D is the distance from you.

If you do disagree with this, can you describe some evidence that my analysis is wrong or post a citation of a reference that has this evidence?

Regards,
Buzz

 

[Bandersnatch]

The quote above seems to disagree with my analysis in post #9.

I don't think it does. You assumed exponentially accelerated expansion (that's when h=const=h0), and that corresponds to the third case in my post.

 

[Buzz Bloom]

I don't think it does. You assumed exponentially accelerated expansion (that's when h=const=h0), and that corresponds to the third case in my post.

Hi Bandersnatch:

I do appreciate your response, but I am still somewhat confused.

Would you please post a quote with the text corresponding to your "third case".

My understanding of the the quote above, is that you agree with my analysis for a future time when Ω_m<<1, but may have reservations about the expansion causing acceleration during the current era when Ω_m~=0.31. Or do you agree that even for the present era, there would be expansion caused acceleration, although the formula for its relation to D would be more complicated.

Regards,
Buzz

 

[Buzz Bloom]

Hi @George Jones:

I appreciate your effort to help me resolve my confusion regarding putting the Einstein static universe equation into the Friedmann form. However, the items in your post are not what confuses me.

Einstein's model without a cosmological constant involved assumptions that the density of matter exceeded the critical density, and therefore

Ω_r = Ω_Λ = 0​

Ω_m > 1, and​

Ω_k = 1 - Ω_m < 0.​

This produces an expanding universe that reaches a maximum radius of curvature, and then the universe contracts. That is, for this model, the stationary universe is not stable. What I am unable to see is how the addition of the cosmological constant creates stability. That is, given

Ω_r = 0,​

Ω_m > 1,​

Ω_k< 0​

and

Ω_Λ = 1 - Ω_m - Ω_k ≠ 0,​

I see no way for this form of the Friedman equation model to be static and stable. Therefore, I am unable to undestand why Einstein added a cosmological constant to his 1917 model.

Regards,
Buzz

 

[timmdeeg]

@Buzz Bloom In Einstein's static universe matter density and $\Lambda$ cancel each other (see the 2. Friedmann equation) and thus the second derivative of the scalefactor is zero (which means static).

 

[George Jones]

Ω_Λ = 1 - Ω_m - Ω_k ≠ 0,
I see no way for this form of the Friedman equation model to be static and stable.

This can't be seen directly from this equation; the time-derivative of this equation also needs to be considered. Einstein concocted a situation in which the time-derivative (rate of change) of each individual term is zero.

@Buzz Bloom In Einstein's static universe matter density and $\Lambda$ cancel each other (see the 2. Friedmann equation) and thus the second derivative of the scalefactor is zero (which means static).

The second derivative equal to zero implies that the first derivative is constant. Then, a condition needs to be satisfied that forces this constant first derivative to be zero.

1) I don't know if this is legible.
2) if it is legible, I don't know if this is understandable.
3) $\Lambda$ has been treated as constant of integration.

 

[timmdeeg]

The second derivative equal to zero implies that the first derivative is constant. Then, a condition needs to be satisfied that forces this constant first derivative to be zero.

1) I don't know if this is legible.
2) if it is legible, I don't know if this is understandable.
3) $\Lambda$ has been treated as constant of integration.

I'm not sure I understand what you mean.
$\ddot{a}=0$ yields $\Lambda_{stat}=4{\pi}G\rho$; [$c=1$]
I agree this "implies that the first derivative is constant." Would you let me know your concern?

EDIT ah I see, yes your script is legible.

 

[Buzz Bloom]

In Einstein's static universe matter density and Λ cancel each other (see the 2. Friedmann equation) and thus the second derivative of the scalefactor is zero (which means static).

Hi timdeeg:

Thank you very much for clearing up much of my confusion.

I think by Friedmann #2 equation you are referring to

What puzzled me is how to put this equation onto the Friedmann form

I also do not know how to put the pressure term into this form.

I see that in the first equation above that the curvature term is missing, and I had (mistakenly?) thought Einstein's solution with Λ also had assumed a finite hyperspherical universe. Is it not possible that Einstein intended somethink like the following put into a form involving the second derivative of a?

.
I think if that I differentiate the second or third equation I might be able to put these results into a form like the first.

ADDED:
I did differentiate the second equation, and I got a cubic equation in R (the radius of curvature) which should result in a solution value for R corresponding to a finite stable stationary universe. I will post this result at a later time. (There should also be solutions for a flat infinite universe and for a non-flat infinite universe. I am not sure that the solutions for the infinite cases are meaningful.)

Thanks again for the help.

Regards,
Buzz

 

[Buzz Bloom]

1) I don't know if this is legible.
2) if it is legible, I don't know if this is understandable.

Hi George:

Thanks for trying to help me, but I think it does appear to be unreadable. I am sympathetic to not liking the posting of a lot of equations. I have from time to time used LaTeX a bit, and I still find it very awkward to use.

I will try to copy what I think I am reading onto a paper I can more easily read. Then I can decide if I can understand it.

Regards,
Buzz

 

[George Jones]

View attachment 245636
What puzzled me is how to put this equation onto the Friedmann form
View attachment 245635

This is because the two equations are independent.

I see that in the first equation above that the curvature term is missing, and I had (mistakenly?) thought Einstein's solution with Λ also had assumed a finite hyperspherical universe.

Einstein's static universe does have non-zero positive spatial curvature.

 

[olgerm]

Question in post #29 was whether in frame of reference of any location there is some force that causes objects to move away. For example is there any force in rope between 2 very distant bodies, that have masses $m_1$ and $m_2$ and charges $q_1=\sqrt{\frac{m_1*m_1*k_G}{k_E}}$ ; $q_2=\sqrt{\frac{m_1*m_1*k_G}{k_E}}$? speed between bodies at time 0 is 0 ($v(0)=0$). My intuition is that there would be force in case of accelerating expansion.

I don't understand the question.

As I understand the expansion means that in every location objects tend to move father away with speed on average $d*H_0$. d is distance to object.
My question in post #31 was: If universe would not be expanding and Earth would be in a special location from which other objects tend to move away with speed $d*H_0$ - would the force measuremetns with ropes attached to distant objects give the same result?

 

[timmdeeg]

Thank you very much for clearing up much of my confusion.

My pleasure, seldom enough that an amateur can contribute.

I think by Friedmann #2 equation you are referring to
View attachment 245636
What puzzled me is how to put this equation onto the Friedmann form
View attachment 245635

I also do not know how to put the pressure term into this form.

I see that in the first equation above that the curvature term is missing, and I had (mistakenly?) thought Einstein's solution with Λ also had assumed a finite hyperspherical universe. Is it not possible that Einstein intended somethink like the following put into a form involving the second derivative of a?
View attachment 245637.

This equation with $H=0$ (universe static) and $\Lambda_{stat}=4{\pi}G\rho$ leads to $k/a^2=4{\pi}G\rho$

 

[Buzz Bloom]

This equation with H=0 (universe static) and

Λ_stat=4πGρ​

leads to

k/a²=4πGρ​

Hi timmdeeg:

I note that you use "a" where I used "R".
I get a different solution assuming H=0. I also use k=1.
If the static universe has radius of curvature R, and mass M corresponding to mass density ρ,

ρ = M /(2π²R³) .​

Therefore

(4/3)GM/πc²R³ - 1/R² + Λc²/3 = 0 .​

Multiplying by 3R³ we get

4GM/πc² - 3R + Λc²R³ = 0.​

However, this equation establishes only that the universe is stationary. For it to also have the acceleration zero (d²R/dt²=0), then differentiating the equation above gives an additional equation.

ADDED

3(dR/dt) = 3Λc²R²(dR/dt), yielding​

Λ = 1 / c²R².​

Combining this with the undifferentiated equation yields

4GM/πc² - 3R + R = 0.​

Simplifying

R = 2GM/πc².​

Simplifying

Λ = (1/c²) (π²c⁴/4G²M²)​

= π²c²/4G²M².​

Regards,
Buzz

I think I have now fixed all the errors, but experience has taught me to not have a high confidence I am correct.
ADDED June 26
I have now fixed 2 another errors.
The first correction is about the 3D volume as the containing boundary of a 4D hypersphere. It is

V = 2π²/R³.​

Reference:

https://www.quora.com/What-is-the-formula-for-volume-of-4D-sphere?share=1​

The second fix is corecting that I misused the word "stable".

 

[George Jones]

However, this equation establishes only that the universe is stationary. For it to also be stable

I have not tried to analyze what you have, but it has been known for almost a century (since at least 1930) that he Einstein static universe is unstable in the following sense.

Consider an Einstein static universe whose cosmological scale factor is $a_s$. Hit the universe with a stick (nicking a phrase from Robert Geroch) such that its scale factor becomes slightly larger than $a_s$. Then, the scale factor, over time, will grow in an unbounded way. Hit the universe with a stick such that its scale factor becomes slightly smaller than $a_s$. Then, the scale factor, over time, will decrease to zero.

 

[timmdeeg]

I get a different solution assuming H=0. I also use k=1.
If the static universe has radius of curvature R, and mass M corresponding to mass density ρ,

ρ = 2 M /(π²R³) .​

Therefore

(16/3)GM/πc²R³ - 1/R² + Λc²/3 = 0 .​

​

Can you show how you derive the first equation? Why do you intend to express $\rho$ by $M$ and $R$?​

​

I think the requirement that $k$ is positive and $(\rho+3p)=0$ is much simpler.​

 

[Buzz Bloom]

Can you show how you derive the first equation? Why do you intend to express ρ by M and R? I think the requirement that k is positive and (ρ+3p)=0 is much easier.

Hi timmdeeg:

I am not sure what equation you are referring to by "the first equation", or what post it was in. Since you are asking about my deriving it, I am guessing it was

ρ = 2 M /(π²R³) .​

This is derived from

ρ = M/V, and​

V=2π²R³.​

I just noticed that I made an error in my post #45 which I have just corrected
The formula for volume V is from

https://www.quora.com/What-is-the-formula-for-volume-of-4D-sphere?share=1 .​

I replace ρ with its formula using M and R because this makes the time derivative less complicated. Since R and Λ are the parameters for which the values are to be calculated that correspond to the first and second derivatives being zero, I felt it easier to have only R be a function of time.

I do not understand why you think (ρ+3p)=0 is related to the problem of finding a universe model which is not expanding nor contracting, and also has the acceleration (d²R/dR²)=0.

I did make k=+1, which made the middle term on the RHS: -1/R². (ρ+3p)=0 has to be wrong since the Einstein universe model has mass, ρ>0, and Einstein also assumed matter was just dust so the pressure p=0. Someone educated me about this recently in another post, but I am not able to find it right now.

Regards,
Buzz

 

[timmdeeg]

I replace ρ with its formula using M and R because this makes the time derivative less complicated. Since R and Λ are the parameters for which the values are to be calculated that correspond to the first and second derivatives being zero, I felt it easier to have only R be a function of time.

Why? In Einstein's static universe $R$ isn't a function of time.

(ρ+3p)=0 has to be wrong since the Einstein universe model has mass, ρ>0, and Einstein also assumed matter was just dust so the pressure p=0.

Yes, the pressure of dust is negligible. What counts is the negative pressure of $\Lambda$. Finally one obtains $\rho_M-2\rho_{\Lambda}=0$.

 

[Buzz Bloom]

Why? In Einstein's static universe R isn't a function of time.
Yes, the pressure of dust is negligible. What counts is the negative pressure of Λ. Finally one obtains ρ_M−2ρ_Λ=0.

Hi timmdeeg:

I used "R" instead of "a" because R varies the same way that a varies, except while a=1 corresponds to now, we would need to use R=R₀ since the value of R now is not 1. We could also define R=R₀ to be the sought after value of R that satisfies the two equation I have at the end of the next paragraph.

Einstein's static universe is a special case of a finite universe with two parameters: a constant mass M (non-varying with respect to R and therefore also t) and a constant Λ. If ρ remains in the equation to be differentiated, then it also varies with R and t creating some extra complexity. The special case is one in which

dR/dt=0, and​

d²R/dt²=0.​

I do not understand what form of the Einstein/Friedmann equation you started with, and how you get from that to

ρ_M−2ρ_Λ=0.​

Th Einstein static model also includes a radius of curvature term, which is -1/R² in the equation. This is for an unbounded finite universe with with the geometry of a 3D boundary to a 4D hypersphere. Without both the -1/R² term and the Λ term, there is no solution satisfying both

dR/dt=0, and​

d²R/dt²=0.​

It is also possible to have a model with an M term (or a ρ term) and a -1/R² term without a Λ term. This model's equation can also be solved for a value of R such that

dR/dt=0,​

but it cannot also satisfy

d²R/dt²=0.​

Regards,
Buzz