Timeline for Can the Hubble constant be measured locally?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2021 at 5:01 | comment | added | benrg | @JerrySchirmer Rereading the answer I see I suggested in the second paragraph that all matter clumps. I'll fix that. | |
| Jan 19, 2021 at 3:53 | comment | added | benrg | @JerrySchirmer All I'm saying is that there's nothing special about cosmology. Approximating the widget distribution as visible widgets plus uniform widgets at the average density of visible+invisible widgets over a larger area is exactly as accurate as it would be in a non-cosmological context. I think the local dark matter density is probably orders of magnitude higher than the local dark energy density, so there probably is a local dark matter force but it's nothing like the ä/a force that Cooperstock et al calculated. | |
| Jan 18, 2021 at 23:09 | comment | added | Zo the Relativist | But the universe is mostly dark matter and dark energy, neither of which clumps. Why is "the solar system is visible matter, with a perturbative background of diffuse dark matter and dark energy" not valid? In particular, since dark energy is, as far as we can tell, a uniform $\Lambda$ with no structure? | |
| Dec 24, 2020 at 19:13 | comment | added | tparker | Nitpick: your second-to-last sentence in correct. The shell theorem doesn't hold in general relativity in the presence of a spherically symmetric matter distribution; see link.springer.com/article/10.1007/s10714-017-2267-y. And Birkhoff's theorem doesn't apply in the presence of matter fields. | |
| Dec 24, 2020 at 17:42 | history | bounty ended | knzhou | ||
| Dec 17, 2020 at 21:11 | comment | added | The_Sympathizer | Yes, which is taking advantage of sources of information present nearby to infer things about distant regions. So as I was saying, you can only distinguish it if you do so take advantage, and in that regard the measurement cannot be purely "local". | |
| Dec 17, 2020 at 21:08 | comment | added | benrg | @The_Sympathizer You could track the locations of objects outside the box by their gravitational fields, since gravity can't be shielded. In principle you could even measure the cosmic graviton background and fit it to a cosmological model. Galileo's ship experiment fails for the same reason: you can distinguish different speeds by the dipole of the graviton background. I'm basically assuming "Galileo's ship rules", and disallowing that like optical astronomy is disallowed. | |
| Dec 17, 2020 at 20:59 | comment | added | The_Sympathizer | I think I get it now. The trick is that you need to know the distribution of the other mass in order to filter out how much of what you're seeing is due to the scale expansion and how much is due to the action of the other mass, for the same general reason you cannot know individual terms of a sum from the value of the sum alone. And if you measure the other matter, you are no longer conducting a "local" measurement or, to put it another way, a "local" measurement is only possible if you have been informed in advance of other aspects of the universe in question that are not local. | |
| Dec 17, 2020 at 20:57 | comment | added | benrg | @The_Sympathizer Perturbation theory works. The behavior of your two particles depends on the initial conditions. If you give them just the right masses and initial velocities and there are no other masses nearby then the distance between them will stay proportional to the scale factor. With other initial values it won't. You can't know whether they're following the scale factor unless you measured it beforehand and chose the initial conditions on that basis. You can measure dark energy locally but you can't know how it relates to the scale factor without looking outside the box. | |
| Dec 17, 2020 at 20:48 | comment | added | The_Sympathizer | Also, I see this: "It represents the dark energy, which is present locally, starting to dominate. As you go to larger radii, the total contained dark energy goes up roughly as r3, and r∗ is the radius at which the repulsive force from that equals the attractive force of the central mass." - why at small radii through is there mathematically exactly zero repulsive force (which is what you need if you want to say it is flat impossible to measure the scale factor there, no?)? | |
| Dec 17, 2020 at 20:30 | comment | added | The_Sympathizer | E.g. if we assume two particles of extremely, extremely tiny mass so their gravitational attraction is negligible compared to the FLRW expansion, why would they not expand at all? You say it would have to be "entirely" uninfluenced by other matter - but what is the mathematical reasoning for this exact zero? I won't say it can't be so, but I want to see more justification for it. | |
| Dec 17, 2020 at 20:28 | comment | added | The_Sympathizer | can or cannot usefully be applied in a particular instance. Why can't it be here? | |
| Dec 17, 2020 at 20:28 | comment | added | The_Sympathizer | This is an interesting answer, but I still don't quite get why it works out. General relativity is at least linear under a small perturbation, isn't it? So given two weak influences, why can't one treat the result as a superposition of their individual contributing effects? You even seem to use that superposition process in supposing it can be decomposed as a "bunch of Schwarzschild metrics" (presumably integrating it with infinitesimal mass over the whole bulk mass distribution). "Nature" may not "do perturbation theory", but that is irrelevant to whether that perturbation theory | |
| Dec 17, 2020 at 19:33 | history | answered | benrg | CC BY-SA 4.0 |