Why isn't the pull of gravity neutral on large scales?

In summary: I'm struggling to understand how the universe would 'contract' due to gravity, which is the assumption used in the FRW equation as I understand it.I'm sorry, @phinds, I may have misunderstood your earlier post. Can we chalk it up to fatigue?I'm actually up way too late and should probably go to bed. o:)In summary, the universe appears to be expanding, but gravity may "cancel out" on larger scales.
  • #1
rede96
663
16
I'm only an interested layman so please forgive my lack of knowledge but assuming the universe is flat, infinite and the cosmological principle holds then why wouldn't the attractive force between galaxies cancel out?

If I think of a 2d analogy such as the surface of a sphere, and there were galaxies spread evenly on the surface, then geometrically speaking it would be impossible for everything to move towards each other (or away from each other) unless the size of the 3D sphere changed with time.

So does the fact that gravity and indeed dark energy produce the results observed mean we must live in a 4 dimensional spatial universe? Or is it correct to view space time as that extra spatial dimension?
 
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  • #2
rede96 said:
I'm only an interested layman so please forgive my lack of knowledge but assuming the universe is flat, infinite and the cosmological principle holds then why wouldn't the attractive force between galaxies cancel out?
What makes you think they don't?

If I think of a 2d analogy such as the surface of a sphere, and there were galaxies spread evenly on the surface, then geometrically speaking it would be impossible for everything to move towards each other (or away from each other) unless the size of the 3D sphere changed with time.
Yes, so? It DOES change with time. See the link in my signature.

So does the fact that gravity and indeed dark energy produce the results observed mean we must live in a 4 dimensional spatial universe?
Absolutely not. Why would it?
Or is it correct to view space time as that extra spatial dimension?
What extra dimension? None is needed.
 
  • #3
The key is that matter is not distributed uniformly on a local scale. If one piece of matter is even slightly closer to one of its neighbors than to the others, then it will feel a net force in that direction and will accelerate. And it turns out that such non-uniformity definitely existed in the early universe, as all matter existed as an extremely hot, extremely dense plasma.
 
  • #4
phinds said:
What makes you think they don't?

Gravity isn't canceled. Otherwise galaxies wouldn't form, let alone galaxy clusters and superclusters.
 
  • #5
phinds said:
What makes you think they don't.

Newtons theorem. I couldn't understand how it could ignore the effects of gravity outside the shell.
 
  • #6
Drakkith said:
The key is that matter is not distributed uniformly on a local scale. If one piece of matter is even slightly closer to one of its neighbors than to the others, then it will feel a net force in that direction and will accelerate. And it turns out that such non-uniformity definitely existed in the early universe, as all matter existed as an extremely hot, extremely dense plasma.

Yes I sort of understand that. But as mentioned above I still can grasp how Newtons theorem would work, as wouldn't it have to be based on matter not being uniformity spread, not an isotopic and homogeneous universe as used in the FRW equation?
 
  • #7
rede96 said:
Yes I sort of understand that. But as mentioned above I still can grasp how Newtons theorem would work, as wouldn't it have to be based on matter not being uniformity spread, not an isotopic and homogeneous universe as used in the FRW equation?

Maybe I'm misunderstanding what you're asking. Are you asking why galaxies are attracted to each other, how the universe can expand, or something else?
 
  • #8
Drakkith said:
Gravity isn't canceled. Otherwise galaxies wouldn't form, let alone galaxy clusters and superclusters.
Right. "locally" (on cosmological scales) things are lumpy, but overall they are smooth (the Cosmological Principle) and the Milky Way isn't being moved in anyone direction by anything other than the Local Group, right?
 
  • #9
phinds said:
Right. "locally" (on cosmological scales) things are lumpy, but overall they are smooth (the Cosmological Principle) and the Milky Way isn't being moved in anyone direction by anything other than the Local Group, right?

That's right. Our galaxy is under the influence of gravity from the local group, the virgo cluster and supercluster, and possible a larger structure, but once you get beyond that scale the universe has started to look very smooth and gravity starts to "cancel out".
 
  • #10
Drakkith said:
Maybe I'm misunderstanding what you're asking. Are you asking why galaxies are attracted to each other, how the universe can expand, or something else?

Yes, exactly. I can understand how locally things aren't uniform so there could be a strong attraction in a particular direction. That even makes sense on much larger scales, as I would imagine there would always be variation (as per quantum physics). But on larger scales, assuming the cosmological principle then I'm struggling to understand how the universe would 'contract' due to gravity, which is the assumption used in the FRW equation as I understand it.
 
  • #11
I'm sorry, @phinds, I may have misunderstood your earlier post. Can we chalk it up to fatigue? I'm actually up way too late and should probably go to bed. o:)

rede96 said:
But on larger scales, assuming the cosmological principle then I'm struggling to understand how the universe would 'contract' due to gravity, which is the assumption used in the FRW equation as I understand it.

Oh, I see what you're asking. Unfortunately I'm not an expert on General Relativity, so I can't really answer this.
 
  • #12
Drakkith said:
I'm sorry, @phinds, I may have misunderstood your earlier post. Can we chalk it up to fatigue? I'm actually up way too late and should probably go to bed. o:)
No problem.
 
  • #13
rede96 said:
But on larger scales, assuming the cosmological principle then I'm struggling to understand how the universe would 'contract' due to gravity, which is the assumption used in the FRW equation as I understand it.
I can't answer your question mathematically (it's a good one) but I can tell you that up until quite recently (maybe 15 years ago) it was believe that the universal expansion WOULD at the very least slow down and reach a steady state or more likely go into reverse and cause "the big crunch". BUT ... "dark energy" was discovered and "Einstein's Biggest Blunder" turned out to not be a blunder and the "Cosmological Constant" in his equations is given as one possible explanation for dark energy. So apparently the math does work out in the end.
 
  • #14
rede96 said:
Newtons theorem. I couldn't understand how it could ignore the effects of gravity outside the shell.
This looks like the crux of the problem. Is the issue with not understanding the derivation of the shell theorem, its conclusions intuitively, or its application to the derivation of Friedmann equations?
 
  • #15
rede96 said:
If I think of a 2d analogy such as the surface of a sphere, and there were galaxies spread evenly on the surface, then geometrically speaking it would be impossible for everything to move towards each other (or away from each other) unless the size of the 3D sphere changed with time.

So does the fact that gravity and indeed dark energy produce the results observed mean we must live in a 4 dimensional spatial universe? Or is it correct to view space time as that extra spatial dimension?
The first part is correct, the second doesn't follow. It is possible to fully describe an expanding space without having to embed it in a higher dimension. We tend to do that with an expanding 2-sphere as in your example, but it is just a visual aid, not a mathematical necessity. Furthermore, one needs to keep in mind that the universe might be actually flat and infinite, which makes the embedding in a higher dimension no longer useful (so it cannot be time either, as it'd only work in one special group of cases with positive curvature).
 
  • #16
@rede96 You have essentially restated Mach's principle.
 
  • #17
Bandersnatch said:
Is the issue with not understanding the derivation of the shell theorem, its conclusions intuitively,

Yes, that is certainly one area I am struggling to come to grips with. It's not so much the principle, as I accept that, even though I don't fully understand it. It is more how it works when applied simultaneously to all matter in the universe. For example if I choose myself to be in the 'center' and look at the gravitational effects on a star some arbitrary distance away, then I know the gravitational forces on that star is equivalent to all the matter within that shell. However there could be someone else on the other side of that star, an equal distance away, whose shell contains the same amount of matter as my shell. So which way does the star in the middle move?

Assuming a universe that was perfectly isotropic and homogeneous, as every point in the universe is equally valid, if I choose all points simultaneously then it would seem the gravitational effects would cancel out. Or more accurately there would be equal gravitational forces in all directions.
 
  • #18
rede96 said:
Yes, that is certainly one area I am struggling to come to grips with. It's not so much the principle, as I accept that, even though I don't fully understand it. It is more how it works when applied simultaneously to all matter in the universe. For example if I choose myself to be in the 'center' and look at the gravitational effects on a star some arbitrary distance away, then I know the gravitational forces on that star is equivalent to all the matter within that shell. However there could be someone else on the other side of that star, an equal distance away, whose shell contains the same amount of matter as my shell. So which way does the star in the middle move?

As stated already, these forces would cancel out when you're looking at the largest scales of the universe. But at local scales, such as within galaxies and galaxy clusters, the forces certainly don't cancel out and you have a net force on the star.
 
  • #19
Drakkith said:
As stated already, these forces would cancel out when you're looking at the largest scales of the universe. But at local scales, such as within galaxies and galaxy clusters, the forces certainly don't cancel out and you have a net force on the star.
Drakkith, the question is not about that, it's about the Newtonian derivation of the Friedmann equations, which has as a feature this apparent 'paradox' the OP is talking about.

I don't think I can give a good answer to the OP now, though. I'll think about it some more.
 
  • #20
Bandersnatch said:
Drakkith, the question is not about that, it's about the Newtonian derivation of the Friedmann equations, which has as a feature this apparent 'paradox' the OP is talking about.

My apologies. I just woke up. Rereading this thread a bit more, I see now this was already expanded on up above.
 
  • #21
Drakkith said:
My apologies. I just woke up.
Need some sleep ... just woke up ... excuses ... excuses ... :-p
 
  • #22
phinds said:
Need some sleep ... just woke up ... excuses ... excuses ... :-p

Indeed. I need to stop logging onto PF first thing in the morning and right before going to bed. o:)
 
  • #23
Drakkith said:
Indeed. I need to stop logging onto PF first thing in the morning and right before going to bed. o:)
Well, actually I do sympathize. I've certainly thrown in my 2 cents worth and had it go South for the same reason
 
  • #24
Chronos said:
@rede96 You have essentially restated Mach's principle

Thanks for the reply, I've actually not heard of Mach's principle, so had to look it up. From what I understand isn't it to do inertia? I was just concerned with gravitational forces to be honest. But I probably don't fully understand Mach's principle any way.
 
  • #25
Hadn't had an answer to this so thought I'd just bump the thread rather than post a new one.
 
  • #26
I'm not sure which question remained unanswered, perhaps this one:

rede96 said:
Yes, exactly. I can understand how locally things aren't uniform so there could be a strong attraction in a particular direction. That even makes sense on much larger scales, as I would imagine there would always be variation (as per quantum physics). But on larger scales, assuming the cosmological principle then I'm struggling to understand how the universe would 'contract' due to gravity, which is the assumption used in the FRW equation as I understand it.
This issue has nothing to do with local inhomogeneities or their neglection on large scales nor with any cancelling of forces. If there isn't enough repelling gravity, dark energy, or the cosmological constant resp. to keep the universe expanding forever then it will expand decelerated until it reaches its largest extension and then will contact. The similarity of throwing a stone upwards is obvious. That's how gravity works. The ball needs an initial speed, the universe an initial rate of expansion.

You can see that surprisingly simple from the second Friedmann equation (Wikipedia): unless ##\Lambda## is large enough in comparison to the matter density the second derivative of the scale factor is negative, which means deceleration.
 
  • #27
Perhaps I'll restate the question for the OP, as it seems that people keep missing it.

When one derives the Friedmann equations using Newtonian framework, one picks an arbitrary point in a homogeneous and isotropic distribution of matter as the origin of their coordinates, and writes the equation for acceleration of a particle P at distance R from the origin, which is then converted to mechanical energy equation.
With some manipulation we arrive at the first Friedmann equation (steps detailed here: http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf).
The derivation uses Gauss' law (or equivalently: shell theorem), to justify that all matter located farther than R has no bearing on the dynamics of P.

Since the choice of a sphere drawn in the homogeneous distribution of matter was arbitrary, we should be able to describe the dynamics of the same particle w/r to a different point, and get exactly the same equation. So, if P was e.g. moving away and decelerating w/r to origin O, it does the same w/r to origin O'.

However, since this deceleration is governed by Newtonian gravity, what stops us from saying that point P has no force acting on it, since for any arbitrarily chosen origin we can choose another one, exactly opposite the first and at the same distance, w/r to which P is pulled with equal but opposite force.

Taking into account that in Newtonian derivation we are dealing with particles moving on a static spatial background (rather than embedded in expanding space as in GR), we have a particle that experiences 0 net force.

So what is the justification for saying that it does move w/r to any chosen point (i.e. the sphere of matter is self-gravitating)?
I.e., what gives us the right to write the first step in the derivation, where particle P is accelerated towards the arbitrary origin point?
I was thinking about an answer, but am not quite happy with my reasoning. The best I could come up with was that we get 0 acceleration only with a change of reference frame. As long as you stick with describing the dynamics within one frame, there's no problem. Which would mean that the 0 acceleration result is an error coming from trying to use two reference frames at once.

There's also the possibility that this is just one of the limitations of Newtonian derivation, and that one needs to follow the GR derivation to get rid of it in a satisfactory manner (after all, the Friedmann equations were only arrived at once GR was available). Not being conversant with GR I can't be certain that's the case, though. Furthermore, while browsing a couple papers detailing differences between the two derivations these past few days, this issue was never brought up.

If anyone can fill in the blanks, that'd be great.
 
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  • #28
Bandersnatch said:
since this deceleration is governed by Newtonian gravity, what stops us from saying that point P has no force acting on it, since for any arbitrarily chosen origin we can choose another one, exactly opposite the first and at the same distance, w/r to which P is pulled with equal but opposite force.

Because if you switch origins, the original origin is now outside radius ##R## from the new origin, which means its effects cancel out (because everything outside radius ##R## from whatever your chosen origin is cancels out by the shell theorem). Similarly, with the original choice of origin, the "new" origin (the one you are contemplating switching to) is outside radius ##R## and its effects cancel out.

Bandersnatch said:
Taking into account that in Newtonian derivation we are dealing with particles moving on a static spatial background

Which is really the underlying issue here. The Newtonian derivation, IMO, does not actually explain why the entire universe decelerates if it is matter dominated; it's just a heuristic that can help to make the idea plausible, by showing how it makes sense as long as we confine ourselves to a finite region around some chosen origin. But it cannot be extended to a description of the entire universe like the one GR gives using standard FRW coordinates; those coordinates simply do not satisfy the assumption of a static spatial background (because our actual universe doesn't have one), so there's no way for a Newtonian model to ever arrive at them.

Bandersnatch said:
Which would mean that the 0 acceleration result is an error coming from trying to use two reference frames at once.

That's part of it, yes, as my first comment in this post implies. But the other part is the limitation described in my second comment above.
 
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  • #29
Bandersnatch said:
Perhaps I'll restate the question for the OP, as it seems that people keep missing it.

Thank you for that, it's very much appreciated.

PeterDonis said:
Which is really the underlying issue here. The Newtonian derivation, IMO, does not actually explain why the entire universe decelerates if it is matter dominated; it's just a heuristic that can help to make the idea plausible, by showing how it makes sense as long as we confine ourselves to a finite region around some chosen origin.

Thank you for your reply Peter. So can the Friedmann equations be derived without using Newtonian gravity?

PeterDonis said:
But it cannot be extended to a description of the entire universe like the one GR gives using standard FRW coordinates; those coordinates simply do not satisfy the assumption of a static spatial background (because our actual universe doesn't have one), so there's no way for a Newtonian model to ever arrive at them.

So if I imagine 3 shells A, B and C arbitrarily spaced in a isotropic and homogeneous universe so that no shell overlaps with another. As I understand it, observers at the center of A, B and C would see any matter within their shell being pulled towards them.

But from what I understand about how the universe is modeled, in a matter dominated universe (e.g. no dark energy) then any observer would see all matter moving in towards them. Which seems to contradicts what observes in the 3 shells would see.

So what is the correct way to think about this using the example I gave?
 
  • #31
rede96 said:
can the Friedmann equations be derived without using Newtonian gravity?

Of course. They are derived from the Einstein Field Equation. That's how they were originally derived, and that's the actual justification for them. The Newtonian derivation, as I said, is just heuristic.

rede96 said:
if I imagine 3 shells A, B and C arbitrarily spaced in a isotropic and homogeneous universe so that no shell overlaps with another. As I understand it, observers at the center of A, B and C would see any matter within their shell being pulled towards them.

Observers at the center of A, B, and C would see any matter within the radius a, b, c of their shell being pulled towards them, yes. But, and this is the crucial point, that statement is true regardless of the radius of the shell. In other words, observer A could just as well pick a shell that was large enough to include observer B (or C or both), and then he would observe B (or C or both) being pulled towards him. Similarly for B and C.

rede96 said:
from what I understand about how the universe is modeled, in a matter dominated universe (e.g. no dark energy) then any observer would see all matter moving in towards them. Which seems to contradicts what observes in the 3 shells would see.

No, it doesn't. See above. The only issue that arises, with the Newtonian derivation, is that allowing the shell to be arbitrarily large, while still maintaining a constant density of matter everywhere, is not really possible in Newtonian physics; Newtonian physics would require the matter density to go to zero at some finite shell radius (otherwise the gravitational potential energy would not be bounded).
 
  • #32
Bandersnatch said:

Thanks for that. I don't quite get the Math, but reading the conclusions helped.

Bandersnatch said:
But it doesn't. It's exactly the same statement. After all, the centres of any two spheres are accelerated towards the third

Ah of course, got that now thanks.

PeterDonis said:
Of course. They are derived from the Einstein Field Equation. That's how they were originally derived, and that's the actual justification for them. The Newtonian derivation, as I said, is just heuristic.

Ok great. I'll see if I can find any cosmology lectures on that. Cheers.

PeterDonis said:
Observers at the center of A, B, and C would see any matter within the radius a, b, c of their shell being pulled towards them, yes. But, and this is the crucial point, that statement is true regardless of the radius of the shell.

I get that now thanks, but being honest I still struggle with understanding the physics behind the how. I think this is where I confuse myself in my visualization.

I imagined the situation to be analogous to a long string of tennis balls, each connected to each other by lengths of elastic. If I move all the tennis balls away from each other equally, then obviously there will be a force created from the elastic between each tennis ball which wants to contract the system. If the ends were not bound and I let go all the tennis balls, then this could happen freely. However if I had an infinity long string of tennis balls then I thought for any smaller group of balls, as there was an infinite amount of force either side in both directions, then it would cancel out and be as if the ends were bound. Hence the system could not contract, even though the force remained.

However I am guessing that an infinite string of tennis balls as mentioned above would still collapse on itself? Or is the analogy just not applicable?
 
  • #34
rede96 said:
Or is the analogy just not applicable?

I would say not, because the actual GR model of what is going on does not assume a static background space, which your analogy does. Your analogy does, IMO, raise a similar issue with the Newtonian derivation to the one that I mentioned before--that you run into problems if you have an infinite expanse of matter.
 
  • #35
PeterDonis said:
See pg. 223 of Sean Carroll's lecture notes:

Ok, great. Thank you.
 

FAQ: Why isn't the pull of gravity neutral on large scales?

Why is the pull of gravity not neutral on large scales?

The pull of gravity is not neutral on large scales because of the distribution of mass in the universe. Gravity is a force that is directly proportional to the mass of an object, so the more mass an object has, the stronger its gravitational pull. On a large scale, such as the scale of the entire universe, there are massive objects like galaxies and clusters of galaxies that have a significant gravitational influence on each other, making the overall pull of gravity non-neutral.

How does the distribution of matter affect the pull of gravity on a large scale?

The distribution of matter plays a crucial role in determining the pull of gravity on a large scale. As mentioned before, the more mass an object has, the stronger its gravitational pull. On a large scale, the distribution of matter is not uniform, and there are regions with higher concentrations of matter, such as galaxies and galaxy clusters, that have a stronger gravitational pull than regions with lower matter concentrations.

Why is the pull of gravity stronger in some regions of the universe compared to others?

The pull of gravity is stronger in some regions of the universe because of the uneven distribution of matter. As gravity is directly proportional to mass, regions with a higher concentration of matter, such as galaxies and galaxy clusters, have a stronger gravitational pull than regions with lower matter concentrations. This uneven distribution of matter is a result of the formation and evolution of structures in the universe.

Can the pull of gravity on a large scale ever be neutral?

No, the pull of gravity on a large scale can never be neutral. As long as there is matter present in the universe, there will always be a gravitational pull between objects. Even if the distribution of matter were to become more uniform, there would still be a gravitational pull between all objects, albeit a weaker one.

How does the concept of dark matter relate to the pull of gravity on a large scale?

Dark matter is a hypothetical form of matter that is thought to make up a significant portion of the total matter in the universe. It does not emit or absorb light, making it invisible to traditional telescopes. However, its presence is inferred from its gravitational effects on visible matter. The distribution of dark matter is thought to play a crucial role in the formation of structures in the universe and, therefore, affects the pull of gravity on a large scale.

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