# Does the expansion of the universe affect gravitational forces?

1. Mar 7, 2015

### m_robertson

I'm a complete and utter layman here, so please forgive any misunderstandings, but I'm just curious about something and wonder if people could say some words on the subject.

So as we understand it, the universe is constantly expanding, but expanding at an accelerated rate, so wouldn't this have some effect on the way in which bodies of matter can gravitate towards each other later in the universes life? If we take that into consideration, wouldn't it then become more difficult for black holes to form later in the universes life since the expansion of space is pulling matter further and further apart and thus consequentially weakening those gravitational interactions? Wouldn't that then imply that it would be impossible for all particles of matter to be eliminated due to black hole radiation since the gravitational forces between residual particles would be so weak that black holes simply couldn't be formed? If we took a relativistic view, couldn't we then say that the distribution of space in the universe is in fact static and that gravity is simply getting weaker as a force over time? I know observations tell us that the expansion of the universe is accelerating, and gravitational forces seem to remain constant (G), but if we considered that gravity as a force was weakening over time, wouldn't that then prevent Einstein's model of a static universe from collapsing and allow it to remain static?

A lot to ask, a lot to take in, and probably a lot of educational mistakes, but the thought just crossed my mind and was tempted to ask. I'd like to further my layman's knowledge and perhaps understand it a little better, and just gain a little enlightenment. Thanks for your time.

2. Mar 7, 2015

### Staff: Mentor

Expansion in general (whether it's accelerating or not) has this effect; bodies that are not already gravitationally bound get further apart, so it gets harder for them to attract each other.

Not necessarily. Black holes in the universe now are forming in systems that are already gravitationally bound, like galaxies. The expansion of the universe does not affect how difficult it is for black holes to form and gain mass in such systems.

"Weakening" only in the sense that the distance is greater (between bodies not gravitationally bound). The interaction between bodies a given distance apart (such as a million light-years) does not change.

Impossible? No. That's much too strong a claim.

What does "distribution of space" mean? The observed fact is that galaxies and galaxy clusters are moving apart. What other observable do you propose to use to tell us what the "distribution of space" is?

Also, we can measure the force of gravity directly, by putting known masses at a known distance apart and measuring the force between them. This tells us that gravity is not getting weaker over time. You appear to recognize this in your very next sentence, so I'm confused about exactly what hypothesis you are advancing here; gravity can't both be constant and getting weaker. Which is it?

I don't know because I don't know what "gravity as a force weakening" means. There is no "force of gravity" in Einstein's equations; there is just spacetime curvature. When we talk about the "force of gravity" in GR, we are talking about an informal way of describing particular effects of spacetime curvature, not about anything that's in the Einstein Field Equation. So what would have to change in the equations to correspond to "gravity as a force weakening"?

3. Mar 7, 2015

### m_robertson

Thanks, I realize I was making a lot of assumptions, you clarified a lot of things for me there. By "distribution of space" I meant the entire stretch of space from both the observable and unobservable universe considered as a single body. Obviously thanks to Hubble we know that the universe is not only expanding but also accelerating, but I was speaking merely from a thought experiment perspective out of interest, as opposed to trying to make a scientific claim. Instead of imagining the expansion of space accelerating and the gravitational attraction between body a and body b getting weaker, you could look at it from an alternative perspective, that gravity as a "force" is getting weaker causing degeneration of the attraction between body a and body b, instead of the acceleration of space but in a static universe. Obviously as you and I pointed out G is a constant and these ideas go against both experiment and observation, so it kind of scraps my idea, but entertain me for a second and put some of those things aside. I was just curious as to whether if we considered G a "force" of attraction like electro-magnetism, but gave that "force" a property that caused it to get weaker over time, whether it would prevent Einstein's static universe from collapsing. What kind of effects would that have on the universe we've observed and measured?

Again I realize I'm going against the grain here, I'm just asking this out of mere curiosity.

4. Mar 7, 2015

### Staff: Mentor

That would be in contradiction to observations, because apples on earth still fall down at the same speed as they did in the past. Measurements of the orbit of moon (its distance is known with centimeter-precision) are precise enough to rule out relevant variations in the gravitational constant.

You can make up all sorts of hypothetical universes, but they would look completely different. Without expansion of space, the universe would still be in a very compact very hot state, and no structures could form at all.

5. Mar 7, 2015

### guywithdoubts

There is something I've never really wondered much about (and should have) and I guess this is the right thread to do it. What does it exactly mean for two objects to be gravitationally bound? Does it imply some sort of orbiting? I understand the whole "objects getting farther appart" thing, I would just like to know whether there is a very specific definition for gravitationally bound systems or not.

6. Mar 7, 2015

### Staff: Mentor

Orbiting or a future collision, yes.

7. Mar 7, 2015

### Staff: Mentor

Again, I don't know, because I don't know how to model this in the framework of GR. Gravity is not a "force" in GR, so there's nothing in the math that corresponds to "how strong the force is".

8. Mar 7, 2015

### m_robertson

Is this something we've measured on the Planck scale?
In regard to the apples, is this something we've measured on the Planck scale? What sort of scaling do we use to measure the falling of the apples, and over what time period? Is it possible for us to make these measurements to determine whether the apples aren't falling at slower speeds at increments of Planck length and time? Again not trying to throw scientific claims around, just asking general questions in relevance to the thread subject.

9. Mar 7, 2015

### Staff: Mentor

Of course not. The Planck scale is 20 orders of magnitude smaller than the smallest scale we can access in experiments, which is about $10^{-15}$ meters (about the size of an atomic nucleus). And experiments at that scale can't measure gravity, because it is so much weaker than the other three interactions that those interactions dominate by many orders of magnitude. I'm not sure what the smallest scale is on which we've measured gravitational effects, but I expect it's something like micrometers ($10^{-6}$ meters).

However, we don't need to make measurements of gravity at the Planck scale to know that its strength isn't changing over time; in fact, in order to measure that, you want to make measurements over larger scales of distance and time, not smaller. Since measurements on larger scales show that the strength of gravity is not changing, even if it does fluctuate at the Planck scale, the fluctuations must average out to no change on larger scales.

10. Mar 7, 2015

### m_robertson

Can you give me an example of such longer scale experiments or observations?

Last edited: Mar 7, 2015
11. Mar 8, 2015

### timmdeeg

Is it true that gravity produces tidal forces due to local acceleration in GR? Then gravity isn't a force but is the cause of a measurable force.

12. Mar 8, 2015

### Staff: Mentor

The orbit of the moon, as mentioned in post #4. A relative change of the strength of the gravitational constant (or, in Planck units, of the masses of earth and moon) has to be smaller than ~2*10-12 per year (smaller than 3% if extrapolated linearly to the age of the universe) and there are limits on the second time-derivative as well.
Reference: Variations of the gravitational constant from lunar laser ranging data

There are also supernova measurements that cover a much longer timescale, but they are a bit model-dependent. Constraining a Possible Variation of G with Type Ia Supernovae

13. Mar 8, 2015

### Staff: Mentor

No. Tidal gravity causes relative acceleration of nearby freely falling objects, but that isn't a force; the objects are freely falling, feeling no force. If there is any force present, it is due to non-gravitational forces that prevent an object or its parts from traveling on freely falling worldlines. It's the same as the force you feel as weight when standing on the surface of the Earth: the force isn't due to gravity, it's due to the Earth pushing on you, keeping you from following a freely falling worldline.

14. Mar 8, 2015

### slatts

M Robertson said,
By distribution of space I meant the entire stretch of space from both the observable and unobservable universe considered as a single body....

I think the impression that there's any limit on the extent of space comes from imprecision or over-dramatization in the use of language, mostly in popularizations of science. One or two of the originators of the BGV Theorem (requiring an origin for inflation) have made clear that inflation does not imply that space did not previously exist. The curvature of space in relativity is, in the most consistent descriptions of it I've been able (due to limited mathematical abilities) to follow, a result of aspects of its relation to time that are only appreciable on very large scales of speed, distance, or duration. So I don't see that there's any distribution of it that can be considered.

15. Mar 8, 2015

### timmdeeg

Ah, I see, here I had a severe misconception. Thanks!

16. Mar 11, 2015

### yogi

It means that if the gravitational pull between two masses results in a greater acceleration toward each other than the acceleration of expanding space urges them to separate, the two are gravitationally bound.

17. Mar 11, 2015

### Staff: Mentor

There are a number of different ways to describe what it means; yogi gave one. His description, however, has the drawback that it uses "acceleration", which, as he's using it, is a coordinate-dependent concept.

Here's a way of describing it that doesn't have that problem: two objects are gravitationally bound if there is no non-gravitational interaction between them (i.e., no electric or magnetic fields, etc.), and if you would have to add energy to at least one of the objects to separate them. (This is sometimes described as the potential energy of the system composed of the two objects being negative; however, "potential energy" as a concept isn't always well-defined, while the way I stated it is always well-defined.)

Not necessarily. You are gravitationally bound to the Earth, but you're not orbiting it.

18. Mar 12, 2015

### Staff: Mentor

I think that needs some rephrasing, as you always have electric or magnetic fields somewhere - even if they are negligible like for astronomical objects.

Two objects are gravitationally bound if, switching off every no non-gravitational interaction between them (i.e., electric or magnetic fields, etc.), you would have to add energy to at least one of the objects to separate them.

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