If the universe is sufficiently large enough, there should be a black hole in every direction of the sky. So, could there be a game of tug-of-war going on between all of the black holes in the universe? It could explain the accelerating rate of expansion due to there being more black holes that have grown larger. The universe wouldn't actually be growing, it would be stretching.
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3$\begingroup$ There are a lot more stars than black holes and the stars are a lot bigger than the black holes. Can you see a star in every direction? Neither can the Hubble space telescope. Now, the gravity of a black hole is no larger than the gravity of the object that collapsed into it, so why would the fact that matter has changed its state from an plasma ball into a black hole matter? It's still the same matter. $\endgroup$– CuriousOneCommented Jul 27, 2016 at 0:54
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$\begingroup$ If there was more mass in one part of the universe, that should be observable by measuring the velocity of multiple galaxies. Measurements suggest the universe is relatively flat and uniform, so while your argument could work in theory, there should be evidence of it as gravity pulls on closer objects more than further objects, you'd see stuff moving in that direction, not spreading out evenly everywhere and spreading out evenly everywhere is basically what has been observed. $\endgroup$– userLTKCommented Jul 27, 2016 at 1:11
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1$\begingroup$ I am voting to close this question as off topic because it is not asking about any aspect of mainstream physics, it is asking for opinions on a new theory to explain the expansion of the universe. $\endgroup$– sammy gerbilCommented Jul 27, 2016 at 1:28
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1$\begingroup$ There is no mainstream answer, so the topic is forbidden? $\endgroup$– devhlCommented Jul 27, 2016 at 1:50
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$\begingroup$ Hi @devhl sadly such is the way this community works, I too dared ask something not straight out of a physics manual and got downvoted to bits. At least you got some sort of replies other than "don't ask this". $\endgroup$– David Andrei NedCommented Oct 20, 2016 at 14:38
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Outside of its event horizon the gravity from a black hole is exactly the same as the gravity from any object that isn't a black hole. So black holes just make up part of the overall stuff in the universe. There are no effects on the universe expansion that are a special result of black holes being present.
The gravity of the black holes certainly does affect the expansion, but black holes affect the expansion in the same way as all the other matter. The geometry of the universe is approximately described by the FLRW metric, and the effect of the matter (black holes and all) is neatly summarised by the Friedmann equations. If we ignore dark energy (and assume pressureless matter) then the acceleration of the expansion is given by the second Friedmann equation:
$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\rho $$
where $\rho$ is the average density of all the matter including the black holes. And since the density $\rho$ is positive that means $\ddot{a}\lt 0$ i.e. the expansion must be decelerating. Black holes cause the expansion to decelerate just like all the other forms of matter. The only way to get an acceleration is to include the dark energy $\Lambda$ in which case our equation becomes:
$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\rho + \frac{\Lambda c^2}{3} $$
and we get an acceleration when:
$$ \frac{4\pi G}{3}\rho < \frac{\Lambda c^2}{3} $$
answered Jul 27, 2016 at 9:12
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$\begingroup$ Good answer. I initially posed the question because I thought if they have a singularity then their reach could extend further than a finite mass could. Also, how does p get calculated considering infinite densities? Are we just assuming that the infinity exists only on paper? $\endgroup$– devhlCommented Jul 27, 2016 at 12:21
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$\begingroup$ @devhl: $\rho$ is the average density i.e. the mass of everything contained in a cubic whatever divided by the volume of a cubic whatever, where a whatever would be some length on the order of $10^8$ light years. $\endgroup$ Commented Jul 28, 2016 at 3:34
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$\begingroup$ Understood, but if there is a singularity how can we compute p? $\endgroup$– devhlCommented Jul 28, 2016 at 15:39
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$\begingroup$ @devhl: assuming you mean $\rho$ rather than $p$, the mass of a black hole is well defined and is known as the ADM mass. So the average density of all the black holes in some volume $V$ is just the sum of all the ADM masses divided by the volume $V$. $\endgroup$ Commented Jul 28, 2016 at 15:42