We know that the universe is expanding, and that means everything is spreading apart. So does that mean in the future all the stars will dim and eventually disappear in our night sky because of the expansion? Just a curious thought that came to mind.
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$\begingroup$ According to expansion theory there is not really any expansion going on within galaxies such as our Milky Way. So even if other galaxies drift further away from us you would still be able to see all the stars in our galaxy. $\endgroup$– AgerhellCommented Oct 2, 2022 at 18:33
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Actually not everything is moving apart. On a scale of millions of light years gravity dominates and there isn't metric expansion of space. This is why (for example) the Andromeda galaxy is moving towards the Milky Way.
So nearby stars (only a few tens of light years away) are not affected by the expansion of spacetime at all.
In the very distant future (countless trillions of years, ie long after the sun has died) it is possible that the expansion of space accelerates to the point that objects on smaller and smaller scales move apart. This is sometimes called the Big Rip. It is strictly hypothetical, and observations tend to suggest it won't happen at all.
So the short answer is "no", stars won't get dimmer due to the expansion of space, because in locally, it's not expanding.
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$\begingroup$ Why is it not expanding seen locally? Is there something about solving the Einstein field equations that gives expansion basically between galaxies but not locally? $\endgroup$– AgerhellCommented Oct 2, 2022 at 18:41
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$\begingroup$ There is too much mass locally for it to expand. ProfRob could give chapter and verse. The expansion (as described in the Friedmann equations) assumes a homogeneous universe - which is true on average over large volumes, but not true locally. $\endgroup$– James KCommented Oct 2, 2022 at 18:53
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1$\begingroup$ @Agerhell GR with a cosmological constant behaves somewhat like a modified Newtonian gravity with a second term in the acceleration: $a=GM/r^2-\frac13 Λr$. If you equate that to $ω^2 r$ you get $ω=\sqrt{GM/r^3 - Λ/3}$. If $r$ is large enough, the quantity in the square root is negative and no circular orbit is possible. At smaller $r$, the second term just slightly changes the $ω$-$r$ relationship. $\endgroup$– benrgCommented Oct 5, 2022 at 22:15