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If galaxies beyond the cosmological event horizon move faster than light is then that motion a combination of their KE and space expansion? Their KE alone isn't enough for them to move faster than light. But is this finite speed of an object simmilar to the motion of vibrations inside a material and that in that case the second reason of a so fast motion, expanding space could be simmilar to a evenly growing material?

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  • $\begingroup$ There's no limit to the speed of distant objects in curved spacetimes. That's strictly a local thing. $\endgroup$
    – D. Halsey
    Jun 5, 2022 at 17:04
  • $\begingroup$ I think the question's a good one, because that sub-molecular wobble seems essential for the widely-assumed homogeneity & isotropy of the universe (local or otherwise) to prevail (even if it's prevalent only on astronomical scales). $\endgroup$
    – Edouard
    Jun 12, 2022 at 16:32
  • $\begingroup$ The latest Sci Am article by L&D actually describes the sub-molecular "wobble" as a tiny reduction in that compression of tiny objects (including all of those comprising the entirety of larger objects, like the earth) which would otherwise result from gravity. Davis has explicitly stated (as quoted in a footnote of a paper by Chowdorowski) that spatial expansion is "not a force or drag" carrying astronomical bodies with it. $\endgroup$
    – Edouard
    Jun 12, 2022 at 18:00
  • $\begingroup$ The SA issue was the March 2005 one. $\endgroup$
    – Edouard
    Jun 12, 2022 at 18:07
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    $\begingroup$ see Susskind expaining this subject in his Cosmology lecture, part 2 at t=1h30m: youtube.com/watch?v=938_TNP4aUs&t=5429s $\endgroup$
    – Yukterez
    Jul 9, 2022 at 10:19

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No galaxy moves faster than light. In cosmology there is a quantity called the recession speed which can exceed 299,792,458 m/s. But that isn't the speed of light when you're talking about recession speeds. Recession speeds are not measured in the special way that speed must be measured in order for the statement "the speed of light is always 299,792,458 m/s" to be true.

It's not true (as Buzz's answer claims) that light emitted by an object whose recession speed is larger than 299,792,458 m/s will never reach us. If there were no cosmological constant, all light would eventually reach us, no matter how large the emitting object's recession speed. With the measured cosmological constant, there is a cutoff distance beyond which emitted light will never reach us, but the recession speed corresponding to the cutoff isn't 299,792,458 m/s. (In the current era it's roughly 340,000,000 m/s.)

Space doesn't expand, at least not in the way that many people seem to think. Say Alice and Bob are both at the bottom of a symmetric crater like this one:

and climb out in different directions. The distance between them, which we'll measure along a line of equal altitude (not through the bottom of the crater), starts small but increases as they ascend. Would you say that they are stationary relative to each other, but the dirt between them is expanding? You could say it, if you carefully defined "stationary" and "expanding" to make it true. But it's a bit silly. The dirt isn't doing anything. The distance increases because Alice and Bob are moving away from each other, by any reasonable definition.

If Alice and/or Bob climbs diagonally instead of straight up, how does that influence the rate of change of the distance between them? Well, it is what it is. They end up where they end up, and you can choose how you want to measure the distance between those locations.

When they climb diagonally, is part of the change of distance between them due to the dirt expanding, and part of it due to their diagonal motion? No, unless you insist on being silly.

The large-scale shape of spacetime is not too different from the shape of that crater. The cosmological time is like the altitude, and the recessional speed is like the change of distance between Alice and Bob as a function of altitude.

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  • $\begingroup$ "If there were no cosmological constant, all light would eventually reach us" - I don't think this is true in the FLRW model. Essentially you are saying that all universe is observable, if there is no dark energy. This may be true in reality, but not according to the current cosmology. $\endgroup$
    – safesphere
    Jul 9, 2022 at 5:46
  • $\begingroup$ @safesphere I really meant in a flat (or more generally open) model with zero cosmological constant, there is no event horizon. In a recollapsing universe there is one. The recession speed of the edge of the observable universe is around $3c$, but that's different from the event horizon. (The distance to the edge of the observable universe is $\int_0^\text{now} dt/a(t)$, and to the event horizon is $\int_\text{now}^\infty dt/a(t)$.) $\endgroup$
    – benrg
    Jul 9, 2022 at 6:13
  • $\begingroup$ Thanks for the clarification on the speed, but the whole FLRW universe is observable, if flat and without a cosmological constant? Can you provide an online reference? $\endgroup$
    – safesphere
    Jul 9, 2022 at 6:22
  • $\begingroup$ @safesphere I spent a few minutes searching the web and didn't find someone saying that exact thing, but you can find people saying that the distance to the event horizon is $\int_{\text{now}}^{\text{end of time}} dt/a(t)$, e.g. Wikipedia or this page. Then you can just observe that if $a''(t)\le 0$ then $a$ is bounded above by $At+B$ for some $A,B$ (e.g. take any tangent line), so the integral is bounded below by $\int dt/(At+B)$, which diverges. $\endgroup$
    – benrg
    Jul 9, 2022 at 6:47
  • $\begingroup$ So while using the wrong metric, FLRW still correctly predicts that all universe is observable since there is no “dark energy” and the universe does not accelerate. Thanks for your help +1 $\endgroup$
    – safesphere
    Jul 9, 2022 at 16:28
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The motion is primarily due to expansion. The KE is almost always irrelevant.

Imagine two objects with equal mass in a common circular orbit about their midpoint M. Assume that M is exactly at the point at which the expansion of the distance between the observers point O and the midpoint M is at the speed of light. Assume the orientation of the orbit is such that part of the line between O and M passes through the orbit as its diameter. Assume at this critical time one of the two objects has its motion close to towards the observer, and the other object has its motion close to away from the observer. Assume both objects are sending out high temperature light waves.

The object moving towards the observer moves at slightly less than speed of light towards the observer, and the light will travel at the speed of light towards the observer, but when it is seen when the wave reaches the observer, its wave length will be very much lenthened. The light waves from the other object will never reach the observer.

However, it will be a rather short time (how much needs to be calculated) after which the speed of both objects will be greater than the speed of light away from the observer.

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  • $\begingroup$ The D&L piece I cited in my earlier comment does have the light from some very far-distant objects approaching the observer only after the expansion of "the" Hubble sphere (or, I guess, the expansion of the observer's Hubble sphere, in multiverse models) encompasses it. $\endgroup$
    – Edouard
    Jun 12, 2022 at 16:24
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    $\begingroup$ Hi @Edouard - Your statement "the light from some very far-distant objects approaching the observer only after the expansion of 'the' Hubble sphere..." is confusing. Please define "the Hubble sphere". I note your quotes around "the" so I am guessing you have a particular sphere in mind. I am not aware of any particular Hubble sphere. The Hubble constant (which is a time changing variable) is a value for inverse time, that is 1 divided by a specific time. $\endgroup$
    – Buzz
    Jun 14, 2022 at 17:43
  • $\begingroup$ I agree: Actually, I should've said that "the Hubble sphere recedes" (which is to say that nearly all of its surface recedes further away from us, whenever we're watching it, as well as during the intervals between any such observations). For your purposes, I think the best report of Davis' and Lineweaver's work is at arxiv.org/abs/astro-ph/0310808.abs. The relevant pages are p.3, p.8-9, and p.11, with p.3 & 11 containing diagrams that show the evolution of the Hubble Sphere, and p.8-9 describing the processes involved. $\endgroup$
    – Edouard
    Jun 16, 2022 at 4:31
  • $\begingroup$ I'd left parts of the OP's question about my 1st comment on his answer without a response, so I should add that the Hubble sphere is, to my understanding, the region within which the redshift of light from some of the stars either beyond its surface, or that are presently, or were at one time, located within the ball bounded by that surface, can be observed by us. The implied possibility that there might be more than one Hubble sphere would apply only in multiverse models, such as Aguirre & Gratton's "Steady-state eternal inflation" of 2003. $\endgroup$
    – Edouard
    Jun 16, 2022 at 5:27
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    $\begingroup$ Hi @Edouard - The reference you posted (...abs) does not work for me. I am also still in the dark about your concept "Hubble sphere". Your description "the Hubble sphere recedes" remains confusing. What ever the sphere is, I can make no sense of "nearly all of its surface recedes". If the receding is about the expansion of the universe, then it would be necessary for ALL of the surface to recede. $\endgroup$
    – Buzz
    Jun 16, 2022 at 14:44

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