# Infinite universe x Deceleration of expansion

1. Jan 22, 2012

### fbs7

I understand we don't know if the universe is finite or infinite, but I know that in the center of the Earth there is no gravity because the attraction from Earth's gravity is the same on all directions, so the final result is zero.

So, if the universe is infinite and matter is equally distributed, then the attraction of gravity from the universe on any given point should be the same in all directions... therefore the final result should be zero, so the effect of the universe's mass in any point in space should be zero.

That means the universe's mass should have no effect on slowing down its expansion. Ergo, the fact that the universe slowed down its expansion after the big bag should be evidence the universe is finite.

Of course this argument is wrong... but I can't find where. Any help, please?

2. Jan 22, 2012

### jewbinson

1. Bear in mind that the exact point of centre of gravity of the Earth is continuously changing. Also, it depends on, for example, what matter you actually consider to be part of the Earth (how far out w.r.t. the atmosphere, etc).

2. The resultant Force on a POINT PARTICLE only will be zero. This includes only the fundamental particles whose space occupies the exact spot at the centre of the Earth.

But matter is not equally distributed throughout the universe.

I have ignored the rest of your post for now because the rest of your argument follows from a false premise.

3. Jan 22, 2012

### phinds

I congratulate you on an intelligent statement of your question. I get very frustrated by folks who, unlike you, state what they feel is a clever argument and then ask how it is that therefore physics as we know it is so obviously wrong. I wish everyone had your thoughtfulness; my teeth would grind much less.

4. Jan 22, 2012

### fbs7

I mean in average - certainly there are local clumps, but I thought that in the large scale the universe looked the same whatever direction we look at. Something about a Cosmological Principle?

Or, asking the other way around, if I take a grain of sand and remove the local influences (Earth, Moon, Sun, galaxy, galaxy cluster)... will the gravity due to the rest of the universe be zero or non-zero?

I suspect my problem is that I know what acceleration/deceleration due to gravity is (I know it is a nice little vector, at least in common experience), but I only have a vague comprehension of what "expansion of the universe" is. I suspect that "deceleration of the expansion of the universe" may be a bit different than "deceleration due to Earth's gravity".

5. Jan 22, 2012

### chronon

You can argue in two ways

A) In a homogeneous universe, gravity will balance out, so overall will not have an effect

B) A homogeneous spherical shell of matter will have no effect on matter inside the shell. Hence, if we consider a sphere of matter, we think of the matter outside it as making up a series of spherical shells, and so we can ignore the matter outside it. The matter in the sphere will be pulled inwards. But this applies to any sphere, hence overall matter will be pulled together.

Isaac Newton claimed that (A) was the case, but it was never entirely convincing. (A) tends to make sense if you believe in absolute space, so that there's no reason for any matter to move from its original position. If you don't believe in absolute space then (B) looks much more convincing, and so following Einstein's theory of relativity has become the accepted viewpoint.

6. Jan 22, 2012

### Naty1

You likely mean that the average gravitational attraction from the earth is close to zero, assuming a homogeneous and perfectly round earth...neither is exactly true. Also there would be some gravitational affects from the sun and moon (and other stuff) that are usually not discussed in the scenario you are thinking about.

There are also other 'gravitational affects'...like time dilation that continue unabated even though the net gravitational force is close to zero. In other words the gradient of the gravitational potential (attractive 'gravity') is close to zero, but the gravitational potential [time dilation] remains.

Further, even if the universe is infinite, gravity from beyond the cosmological horizon [beyond the observable/visible universe] would not have yet reached us since light has not and both travel at 'c'.

PS: Why is 'deceleration of expansion' in your title??

Last edited: Jan 22, 2012
7. Jan 22, 2012

### Naty1

8. Jan 22, 2012

### DrStupid

This applies to the center of a finite homogeneous mass distribution but not to an infinite homogeneous mass distribution. The integration of the gravitational forces in an infinite classical universe results in an integration constant that can not be determined. Therefore the gravitational force acting on a particle in such a situation is simply unknown.

Within classical mechanics deformations of a mass distribution are caused by tidal forces and the tidal forces of an infinite homogeneous mass distribution slows down its expansion. Surprisingly for a classical flat universe the result is the same as for the Einstein-de Sitter model in GR.

9. Jan 22, 2012

### fbs7

That's because I read somewhere that during some time since the big bang the rate of expansion of the universe decreased, and then started to increase during the last billion years or so.

I read that increase in the expansion of the universe being referred the acceleration of universe... so I just guessed that the period through which the rate of expansion of the universe was decreasing, should be called deceleration.

I didn't really think much of it - my terminology is uneducated. Probably should have referred to as decrease of expansion.

Last edited: Jan 22, 2012
10. Jan 22, 2012

### fbs7

Wow, that's fantastic - I never thought an infinite, homogeneous and isotropic universe would get non-zero gravity. That's mind-boggling!!

Thankfully, if I understood this and the previous posts, the "inifinite" part does not apply, right, because a location is only affected by what's inside its observable radius (the 13.7 billion light-years thing)? Therefore, for the purpose of calculating the gravity due to the rest of the universe, we are indeed inside a very large, but finite, sphere, which in the large scale is homogenous, is that correct?

So, back to the main issue... the overall gravity due to distant bodies at the center of that mega-sphere should be zero - so it should be zero everywhere, given any point will be in the center of its own observable universe. So how can the mass of the universe have an effect in its expansion, give the gravity effects at large scale cancel themselves everywhere?

Last edited: Jan 22, 2012
11. Jan 22, 2012

### phinds

That all sounds perfectly reasonable to me. That's the way I think of it as well (but I'm not a physics major )

I inflated for a tiny fraction of a second then decelerated for about 6 or 8 billion years and then the mass became spread out enought so that "dark energy" took over and the expansion not only stopped decelerating, it started accelerating and has been doing that for 6 or 8 billion years.

Hm ... "give or take a billion" makes me sound like a politicion spending our military budget.

12. Jan 23, 2012

### Naty1

yes....It IS believed that matter dominated the early universe, but energy [described as 'negative pressure' below]is taking over. As the universe expands the constant energy density of new space adds energy and that seems to be speeding up expansion.

ok, here is a better explanation...the entire article is of general interest:

http://en.wikipedia.org/wiki/Big_Bang#Hubble.27s_law_and_the_expansion_of_space

The implications of all this are mostly beyond imagination!!!

13. Jan 23, 2012

### DrStupid

In GR gravitation is defined by the local stress-energy-tensor and in classical mechanics it is defined by the complete infinite mass distribution. I can't see the justification for a limitation to a finite mass distribution.

14. Jan 23, 2012

### Naty1

http://en.wikipedia.org/wiki/Metric_expansion_of_space

This is also interesting within the same article:

http://en.wikipedia.org/wiki/Metric_expansion_of_space#Local_perturbations

15. Jan 23, 2012

Thank you