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- Thread starter Sagar_C
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The reason for this is because of the tidal forces. You know Earth is an oblong shape because of this, well now, all mass exerts a gravitational force, and the tidal bulges on the Earth exert a gravitational pull on the Moon. Because the Earth rotates faster (once every 24 hours) than the Moon orbits (once every 27.3 days) the bulge tries to "speed up" the Moon, and pull it ahead in its orbit. The Moon is also pulling back on the tidal bulge of the Earth, slowing the Earth's rotation. Tidal friction, caused by the movement of the tidal bulge around the Earth, takes energy out of the Earth and puts it into the Moon's orbit, making the Moon's orbit bigger.

The Earth's rotation is slowing down because of this. One hundred years from now, the day will be 2 milliseconds longer than it is now.

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In fact, galaxies as a whole are gravitationally bound i.e. a galaxy will not rip apart due to Hubble's Law - only galaxies relative to each other will accelerate away.

Thanks. But this logic confuses me because galaxies are gravitationally bound in a galaxy cluster and then what?

- #4

Bandersnatch

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https://www.physicsforums.com/showthread.php?t=261161

It's kinda huge, but even the first few pages can help unerstand the current cosmological models better.

This pdf in particular might be of interest to you:

http://www.mso.anu.edu.au/~charley/papers/LineweaverDavisSciAm.pdf

It deals with the balloon analogy and some common misconceptions about the expanding universe, including the one in question.

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This pdf in particular might be of interest to you:

http://www.mso.anu.edu.au/~charley/papers/LineweaverDavisSciAm.pdf

It deals with the balloon analogy and some common misconceptions about the expanding universe, including the one in question.

Thanks. The pdf is just right for a layman like me.

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Drakkith

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Thanks. But this logic confuses me because galaxies are gravitationally bound in a galaxy cluster and then what?

Galaxy superclusters are the next organizational level. It is at this point that expansion starts to push things apart if the supercluster isn't massive enough to hold on to all of it's galaxies or galaxy clusters.

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- #8

Drakkith

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Nope, the Moon is gravitationally bound to the Earth, the Sun, and the rest of the Milky Way and does NOT experience any recession due to the expansion rate of the universe. The key lies in the fact that under GR, the specific math model that we use to determine the expansion of the universe says that the universe is homogenous at all levels. Obviously it is not, as we have clumps of matter in the form of Stars, Planets, etc, and also have wide tracts of nearly empty space. The universe does expand, but only on the largest scales where we can model it as being approximately homogenous. Locally the math doesn't work the same and we get zero expansion.

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Does this mean that the radius of the electron orbiting the center increases?

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Does this mean that the radius of the electron orbiting the center increases?

is answered by

The universe does expand, but only on the largest scales where we can model it as being approximately homogenous. Locally the math doesn't work the same and we get zero expansion.

If i go 4 billion years in the future, and i look at a hydrogen atom, it will still be the same.

I would like to know what TV program that was, just curiosity.

- #11

mfb

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The Big Rip (as it is called) is one possible scenario for the future. There are other options, and at the moment nothing indicates that the Big Rip will happen.I've heard from tv program that finally the atoms will rip apart too.

No - unless you look at the last 10Does this mean that the radius of the electron orbiting the center increases?

- #12

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FAQ: Does everything expand equally because of cosmological expansion?

No. If everything expanded by the same percentage per year, then all our rulers and other distance-measuring devices would expand, and we wouldn't be able to detect any expansion at all. Actually, general relativity predicts that cosmological expansion has very little effect on objects that are small and strongly bound. Expansion is too small an effect to detect at any scale below that of distant galaxies.

Cooperstock et al. have estimated the effect for systems of interest such as the solar system. For example, the predicted general-relativistic effect on the radius of the earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus; if the earth's orbit had expanded according to the cosmological scaling function [itex]a(t)[/itex], the increase would have been millions of kilometers.

To see why the solar-system effect is so small, let's consider how it can depend on [itex]a(t)[/itex]. The Milne universe is just flat spacetime described in silly coordinates, and it has [itex]\dot{a}\ne 0[/itex], i.e., a nonvanishing value of [itex]H_o[/itex]. This shows that we should not expect any expansion of the solar system due to [itex]\dot{a}\ne 0[/itex]. The lowest-order effect requires [itex]\ddot{a}\ne 0[/itex]. Since a rescaling like [itex]a(t)\rightarrow 2a(t)[/itex] has no physical meaning, we can guess that the effect is proportional to [itex]\ddot{a}/a[/itex]. Based on units, we expect that multiplying this by the size of the solar system might give an estimate of the anomalous acceleration with which the solar system expands, and this is indeed the result of Cooperstock's rigorous calculation. The fractional rate of anomalous acceleration [itex]\ddot{r}/r[/itex] is about [itex]H_o^2\sim 10^{-35}\ s^{-2}[/itex]. The result for [itex]\ddot{r}/r[/itex] is valid for [itex]r\ll 1/H_o[/itex] and is independent of r, so it can be applied to similar systems with circular orbits at other scales, such as the earth-moon system or a pair of galaxies in circular orbits about their common center of mass. It can't be applied to systems bound by non-gravitational forces, such as atoms and nuclei.

A nice way of discussing atoms, nuclei, photons, and solar systems all on the same footing is to note that in geometrized units, the units of mass and length are the same. Therefore the existence of any fundamental massive particle sets a universal length scale, one that will be known to any intelligent species anywhere in the universe. Since photons are massless, they can't be used to set a universal scale in this way; a photon has a certain mass-energy, but that mass-energy can take on any value. Similarly, a solar system sets a length scale, but not a universal one; the radius of a planet's orbit can take on any value. A universe without massive fundamental particles would be a universe without distance measurement. It would obey the laws of conformal geometry, in which angles and light-cones were the only measures. This is the reason that atoms and nuclei, which are made of massive fundamental particles, do not expand.

Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

No. If everything expanded by the same percentage per year, then all our rulers and other distance-measuring devices would expand, and we wouldn't be able to detect any expansion at all. Actually, general relativity predicts that cosmological expansion has very little effect on objects that are small and strongly bound. Expansion is too small an effect to detect at any scale below that of distant galaxies.

Cooperstock et al. have estimated the effect for systems of interest such as the solar system. For example, the predicted general-relativistic effect on the radius of the earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus; if the earth's orbit had expanded according to the cosmological scaling function [itex]a(t)[/itex], the increase would have been millions of kilometers.

To see why the solar-system effect is so small, let's consider how it can depend on [itex]a(t)[/itex]. The Milne universe is just flat spacetime described in silly coordinates, and it has [itex]\dot{a}\ne 0[/itex], i.e., a nonvanishing value of [itex]H_o[/itex]. This shows that we should not expect any expansion of the solar system due to [itex]\dot{a}\ne 0[/itex]. The lowest-order effect requires [itex]\ddot{a}\ne 0[/itex]. Since a rescaling like [itex]a(t)\rightarrow 2a(t)[/itex] has no physical meaning, we can guess that the effect is proportional to [itex]\ddot{a}/a[/itex]. Based on units, we expect that multiplying this by the size of the solar system might give an estimate of the anomalous acceleration with which the solar system expands, and this is indeed the result of Cooperstock's rigorous calculation. The fractional rate of anomalous acceleration [itex]\ddot{r}/r[/itex] is about [itex]H_o^2\sim 10^{-35}\ s^{-2}[/itex]. The result for [itex]\ddot{r}/r[/itex] is valid for [itex]r\ll 1/H_o[/itex] and is independent of r, so it can be applied to similar systems with circular orbits at other scales, such as the earth-moon system or a pair of galaxies in circular orbits about their common center of mass. It can't be applied to systems bound by non-gravitational forces, such as atoms and nuclei.

A nice way of discussing atoms, nuclei, photons, and solar systems all on the same footing is to note that in geometrized units, the units of mass and length are the same. Therefore the existence of any fundamental massive particle sets a universal length scale, one that will be known to any intelligent species anywhere in the universe. Since photons are massless, they can't be used to set a universal scale in this way; a photon has a certain mass-energy, but that mass-energy can take on any value. Similarly, a solar system sets a length scale, but not a universal one; the radius of a planet's orbit can take on any value. A universe without massive fundamental particles would be a universe without distance measurement. It would obey the laws of conformal geometry, in which angles and light-cones were the only measures. This is the reason that atoms and nuclei, which are made of massive fundamental particles, do not expand.

Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

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Nope, the Moon is gravitationally bound to the Earth, the Sun, and the rest of the Milky Way and does NOT experience any recession due to the expansion rate of the universe. The key lies in the fact that under GR, the specific math model that we use to determine the expansion of the universe says that the universe is homogenous at all levels. Obviously it is not, as we have clumps of matter in the form of Stars, Planets, etc, and also have wide tracts of nearly empty space. The universe does expand, but only on the largest scales where we can model it as being approximately homogenous. Locally the math doesn't work the same and we get zero expansion.

The effect on the earth-moon system is not zero, just 20 orders of magnitude too small to measure. Homogeneity doesn't explain why there is almost no local expansion. Homogeneity is just an assumption about initial conditions, but the smallness of local expansion requires an explanation in terms of the dynamical laws (the Einstein field equations) governing the expansion.

- #14

Drakkith

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The effect on the earth-moon system is not zero, just 20 orders of magnitude too small to measure. Homogeneity doesn't explain why there is almost no local expansion. Homogeneity is just an assumption about initial conditions, but the smallness of local expansion requires an explanation in terms of the dynamical laws (the Einstein field equations) governing the expansion.

I swear we came to the conclusion that there was no expansion locally, even a tiny tiny fraction of a little bit, here on PF a while back.

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So just like gravity (which is a weak force) doesn't squash you together when it pulls you towards the centre of the Earth, vacuum energy won't rip the molecules in your body apart.

One theory is however though is that vacuum energy may increase as space itself increases, eventually overcoming gravity, and eventually overcoming the strong and weak nuclear forces. Such a scenario however is only plausible in a universe with an open geometry.

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Nope, the Moon is gravitationally bound to the Earth, the Sun, and the rest of the Milky Way and does NOT experience any recession due to the expansion rate of the universe. The key lies in the fact that under GR, the specific math model that we use to determine the expansion of the universe says that the universe is homogenous at all levels. Obviously it is not, as we have clumps of matter in the form of Stars, Planets, etc, and also have wide tracts of nearly empty space. The universe does expand, but only on the largest scales where we can model it as being approximately homogenous. Locally the math doesn't work the same and we get zero expansion.

It seems to me that the underlying physics should be the same at all scales. If space expands then the the space between my toes expands just like the space between super clusters. The difference is that my toes are held in place by chemical bonds which have a preferred length. If you pull my toes apart or squeeze them together they spring back to the same separation. (experimentally verified just now):tongue:

I get that gravity warps space-time while expansion tries to stretch it and that the net affect in some situations may be zero, but in my mind that does not negate the existance of the components which sum to create that zero affect.

FAQ: Does everything expand equally because of cosmological expansion?

No. If everything expanded by the same percentage per year, then all our rulers and other distance-measuring devices would expand, and we wouldn't be able to detect any expansion at all. Actually, general relativity predicts that cosmological expansion has very little effect on objects that are small and strongly bound. Expansion is too small an effect to detect at any scale below that of distant galaxies.

Cooperstock et al. have estimated the effect for systems of interest such as the solar system. For example, the predicted general-relativistic effect on the radius of the earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus; if the earth's orbit had expanded according to the cosmological scaling function [itex]a(t)[/itex], the increase would have been millions of kilometers.

To see why the solar-system effect is so small, let's consider how it can depend on [itex]a(t)[/itex]. The Milne universe is just flat spacetime described in silly coordinates, and it has [itex]\dot{a}\ne 0[/itex], i.e., a nonvanishing value of [itex]H_o[/itex]. This shows that we should not expect any expansion of the solar system due to [itex]\dot{a}\ne 0[/itex]. The lowest-order effect requires [itex]\ddot{a}\ne 0[/itex]. Since a rescaling like [itex]a(t)\rightarrow 2a(t)[/itex] has no physical meaning, we can guess that the effect is proportional to [itex]\ddot{a}/a[/itex]. Based on units, we expect that multiplying this by the size of the solar system might give an estimate of the anomalous acceleration with which the solar system expands, and this is indeed the result of Cooperstock's rigorous calculation. The fractional rate of anomalous acceleration [itex]\ddot{r}/r[/itex] is about [itex]H_o^2\sim 10^{-35}\ s^{-2}[/itex]. The result for [itex]\ddot{r}/r[/itex] is valid for [itex]r\ll 1/H_o[/itex] and is independent of r, so it can be applied to similar systems with circular orbits at other scales, such as the earth-moon system or a pair of galaxies in circular orbits about their common center of mass. It can't be applied to systems bound by non-gravitational forces, such as atoms and nuclei.

A nice way of discussing atoms, nuclei, photons, and solar systems all on the same footing is to note that in geometrized units, the units of mass and length are the same. Therefore the existence of any fundamental massive particle sets a universal length scale, one that will be known to any intelligent species anywhere in the universe. Since photons are massless, they can't be used to set a universal scale in this way; a photon has a certain mass-energy, but that mass-energy can take on any value. Similarly, a solar system sets a length scale, but not a universal one; the radius of a planet's orbit can take on any value. A universe without massive fundamental particles would be a universe without distance measurement. It would obey the laws of conformal geometry, in which angles and light-cones were the only measures. This is the reason that atoms and nuclei, which are made of massive fundamental particles, do not expand.

Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

I didn't follow most of that. I was under the impression that expansion was just another factor in what is or is not a stable orbit. For example in the earth's orbit that you mentioned, I would think that expansion along with momentum would pull the earth away from the sun while gravity pulls it toward and the forces balance out. In other words, if expansion stopped, the earth would, over many years, drop to a lower orbit. Is this incorrect? If so, why? My knowledge of math is limited to algebra and the most basic calculus. Is it possible to dumb it down to that level? :-)

I think I get the fact that solid objects are unaffected by expansion because the chemical bonds which hold them together have preferred length. Try to stretch them and electrostatic forces pull them together, try to shorten them and pauli exclusion trys to push them apart, there is a certain length that has the lowest energy. Did I get that about right?

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This is a good question. This is indeed what happens, for example, due to the small outward acceleration of the earth due to the 1/r^2 force of the sun's light pressure. The result is exactly like dialing down the strength of gravity a little.I didn't follow most of that. I was under the impression that expansion was just another factor in what is or is not a stable orbit. For example in the earth's orbit that you mentioned, I would think that expansion along with momentum would pull the earth away from the sun while gravity pulls it toward and the forces balance out. In other words, if expansion stopped, the earth would, over many years, drop to a lower orbit. Is this incorrect? If so, why? My knowledge of math is limited to algebra and the most basic calculus. Is it possible to dumb it down to that level? :-)

If you look at p. 7 of the Cooperstock paper, there are a couple of things to note about eq. 4.2. It has an anomalous acceleration term [itex](\ddot{a}/a)r[/itex]in it due to cosmological expansion. It's proportional to r, not 1/r^2, which makes sense because you expect a universal expansion to separate things in proportion to their current distance (like raisins in rising bread dough). Also, its sign depend on [itex]\ddot{a}[/itex], which is the rate of acceleration of the expansion, not on [itex]\dot{a}[/itex], the rate of expansion. There are clear physical reasons why it has to be this way (see #11), but it isn't at all what you'd expect from the raisin-bread picture. In particular, you can get an *inward* anomalous acceleration if the universe is expanding but decelerating (which is not the case in our present universe, but has been the case in its past). In 4.8, they take a unverse that is (counter-factually) matter-dominated with [itex]\ddot{a}<0[/itex]. The paper dates to 1998, which was the same year as the first evidence for cosmological acceleration, so I guess it's natural that they picked a model with this sign for [itex]\ddot{a}[/itex].

As to the reason why we get an ongoing trend in the orbital radius rather than just a new equilibrium, that seems to be hidden in the grotty celestial mechanics of eq 4.3-4.10. This kind of stuff is often counterintuitive, e.g., if you have a rocket in a circular orbit and you thrust in the forward direction, the effect is to reduce its velocity.

Similarly, it's counterintuitive that an inward anomalous acceleration leads to an increasing trend in orbital radius, while an outward one leads to a decrease.

So this is an interesting and surprising fact -- circular orbits are currently *shrinking*, not expanding, due to cosmological expansion!

I think this is probably misleading. It makes the gravitational and electrical cases sound analogous, but they aren't, because you get a trend in orbital radius in one case but not the other. It makes it sound like the effect would be proportional to [itex]\dot{a}[/itex], when actually it has to go like [itex]\ddot{a}[/itex].I think I get the fact that solid objects are unaffected by expansion because the chemical bonds which hold them together have preferred length. Try to stretch them and electrostatic forces pull them together, try to shorten them and pauli exclusion trys to push them apart, there is a certain length that has the lowest energy. Did I get that about right?

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The thing is, I have never seen how we could have this "spherical" map of all the galaxies in the universe, while it's said that the universe has no center. For me it was controversial.

However, it's

Even though I believe people have mentioned this like over 9000 times, I did realize that by now, amazing! I did not grasp that logical statement, maybe because someone did not mention the thing I just said, but still, thanks!

And if you wonder why I write this is because my head figured it out while reading this forum thread, thanks, this was a huge problem in my head to visualize!

Best Regards

Robin Andersson

- #19

mfb

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Right. Our observable universe is just a small part of the whole universe. And we are in the center of our observable universe.However, it'sspace that expands, and the the map of the galaxies is just light, of the observable universe. And it's the observable universe that is spherical, right? Not the entire universe itself! Space can be flat happily by itself while the observable universe is spherical, right???

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Gah! I still didn't get this quite right. According to the Cooperstock paper, the sign of the anomalous radial force acting on an object in a circular orbit depends on the sign of [itex]\ddot{a}[/itex], but the sign of the secular trend in the orbital radius depends on the sign of [itex](d/dt)(\ddot{a}/a)[/itex]. I'll start a separate thread on this.

New thread: https://www.physicsforums.com/showthread.php?p=4061088

New thread: https://www.physicsforums.com/showthread.php?p=4061088

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