How does the gravity well change as space expands? If we assume that the Earth's gravitational field curves flat space to create a gravity well then how does the gravity well change as space expands also is the change in gravity well measurable.
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1$\begingroup$ I have no idea, but my best bet is that it does not change. Or the change is minute, not measurable. $\endgroup$– babouCommented Aug 20, 2014 at 12:55
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$\begingroup$ The gravity well is just an analogy. The gravitational potential generated by a stationary mass does not change with time, and does not care about space expansion. I don't understand the question. $\endgroup$– ACuriousMind ♦Commented Aug 20, 2014 at 13:21
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$\begingroup$ ACuriousMind - I know its an analogy but its close description to reality My questions is weather the gravity well becomes wider or shrinks as space expands. $\endgroup$– WhyMECommented Aug 20, 2014 at 13:29
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$\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/2110/2451 , physics.stackexchange.com/q/61320/2451 , physics.stackexchange.com/q/123061/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Aug 20, 2014 at 13:47
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2 Answers
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For a stationary massive object like Earth or the Sun, the gravity well does not change from the expansion of space. The gravity well at one time is the same as the gravity well at future times for the same mass. Any minute force imposed by the expansion of space does not constitute a change in the gravity well itself. The well is not stretched or compressed. It remains constant in proper distance scales and not comoving distance scales.
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$\begingroup$ Does that mean that the big stretch (or big rip, I do not remember the name) is not to be feared. Not that I worry for my kids. $\endgroup$– babouCommented Aug 20, 2014 at 14:23
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$\begingroup$ @babou No, the space between objects is expanding. The big rip refers to this. Just like how we don't say that a magnet changes the gravity well but still lifts a paper clip off the ground, the Big Rip wouldn't change the gravity well, it would simply offer a "force" greater than gravity that would rip apart structures $\endgroup$– JimCommented Aug 20, 2014 at 14:29
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1$\begingroup$ Not true. For example in a de Sitter background the Schwarzschild metric becomes the de Sitter-Schwarzschild metric. $\endgroup$ Commented Aug 20, 2014 at 15:08
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$\begingroup$ @JohnRennie Ok, I over-simplified. And it's true that mixing de Sitter and Schwarzschild results in a space where the event horizon is moved outward from the normal position (or missing entirely). But that isn't the gravity well that has changed, that is the curvature and expansion/contraction of spacetime that changes the position of event horizons. If you claim it is a stretching of the gravity well, then one has to also explain the inward shift of the cosmological event horizon for that metric $\endgroup$– JimCommented Aug 20, 2014 at 17:20
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If we assume that general relativity is the correct theory for this case, and there are currently no indications, that I am aware of, that it isn't, then the expansion of the universe adds a small modifying term to gravity wells. I doubt that it is measurable at the scale of the solar system. The current best estimate for the Hubble constant is 67.80±0.77km/s/Mpc. The size of the solar system (using the heliopause as a border) is about 0.001pc, so the total influence of the expansion would amount to approx. 1e-3pc*68km/s/Mpc=68um/s at the border of the solar system. I seriously doubt, that we can separate that from the gravitational noise of the solar system and the diverse effects of solar wind, light pressure, outgassing etc. that act on spacecraft that may be used to measure this effect.
Having said that, the effects at the edge of our Milky Way are much larger. Now we are talking about 30kpc*68km/s/Mpc=2.04km/s. While the errors bars are much larger (probably mostly due to dark matter distributions), it may be possible to find populations of objects, which show considerable effects on their orbits around the center of the Milky Way. But that's just the gut feeling of a physicist who is not an astronomer, so my intuition into the systematic errors that would undermine such a measurement could be way off.
Update: For those who would like to know what all of this means for orbits of bodies, I believe this paper may contain the required analysis:
https://arxiv.org/abs/1009.6117
"Analytical solution of the geodesic equation in Kerr-(anti) de Sitter space-times", Eva Hackmann, Claus L¨ammerzahl, Valeria Kagramanova and Jutta Kunz
answered Aug 20, 2014 at 13:32
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$\begingroup$ Actually, I think the 68μm/s figure you give is very large. Considering the lifetime of our planet estimated at $4 10^9$ years, that makes $1.26 10^{17}$ seconds. Multiplying by 68μm/s expansion speed for the solar system, that gives a size change of $8.6 10^{12}$ meters, i.e approximately $3 10^{-4}$ parsec, i.e. 30% of the solar system current size (hoping I made no error). This would have had a significant impact on the mechanics and evolution of the solar system, which I never heard of. - - - - Very large/small numbers can be deceptive because exponentials are deceptive. $\endgroup$– babouCommented Aug 20, 2014 at 14:16
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$\begingroup$ The 68um/s is the amount by which the movement of a body in the solar system would be modified in total in comparison to a solar system with perfectly Newtonian gravity, it's not an acceleration term that accumulates over time, so, fortunately for us, this large gain in time does not apply. $\endgroup$ Commented Aug 20, 2014 at 14:23
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$\begingroup$ Back of the envelope I did. But no envelope will help me if I do not understand the physics. $\endgroup$– babouCommented Aug 22, 2014 at 10:00
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$\begingroup$ And some people can do all the math in the world, and they still won't understand the phenomenology. The Hubble constant is ONLY measurable, because of the broken symmetry of the mass distribution in the universe, which defines a special "rest" system in cosmological coordinates. Take all the mass away, and you can't pin coordinates A and B into the metric. So that means, if I am already measuring the Hubble constant in a broken symmetry, I am allowed to argue the same way for "bound" particles. And then it's clear, that the Hubble constant modifies the escape velocity of bound particles. $\endgroup$ Commented Aug 22, 2014 at 18:51