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As I understand it, one theory of the multiverse is that there are an infinite number of universes separated by small distances in other (than our observable 3/4) dimensions and that gravity is weak because it permeates these other dimensions.

But if there are an infinite number of these universes then surely the gravitational force we would feel would be infinite?

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Perhaps you can link to the sources of your information, so we can better explain them to you. – Jeremy Jul 2 '14 at 21:04
Do multiverse theories say that there must be an infinite number of universes? – HDE 226868 Aug 6 '14 at 18:26
Some do, certainly – adrianmcmenamin Aug 9 '14 at 19:47
Right, but not necessarily all. What specific theories say that there must be an infinite number of universes? – HDE 226868 Aug 9 '14 at 20:25
Well, eternal inflation is one, infinite brane cosmologies are another – adrianmcmenamin Aug 10 '14 at 18:06
up vote 3 down vote accepted

An exact answer would require a rather specific and mathematical formulation of the multiverse in consideration.

As a simple approximating example, suppose we have a countably infinite number of (observable) universes of the same mass $M$. Suppose the dimension of the full multiverse is one higher than each individual universe, and suppose the universes are all separated by the same minimum distance $\epsilon>0$ from each other. In a 2-D picture, this would just look like a bunch of parallel lines all separated by the same distance.

Pick your home universe and put an observer. Another universe of distance $n\epsilon$ away (meaning they're $n$ universes up or down from you in the 2-D picture) exerts a gravitational force on the observer in its direction approximately proportional to $\displaystyle{\frac{M}{n^2\epsilon^2}}.$ With the right units, we can just say "approximately".

The maximum gravitational force occurs when the observer is at the bottom (or top) of the picture, and all over universes are above it (or below): universes on each side will pull in opposing directions and so lead to cancellations. So the net gravitational force from the other universes (in the right units) is at most $$\sum_{n=1}^\infty \frac{M}{n^2\epsilon^2} = \frac{M}{\epsilon^2}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2 M}{6\epsilon^2}<\infty.$$

If the observer was "in the middle"—infinitely many universes above and below, with the distribution identical in either direction— the net gravitational force from the other universes is exactly 0.

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Thanks. Very clear answer. – adrianmcmenamin Jun 26 '15 at 9:33

The answer to your question for brane cosmology is simply that gravity follows the inverse-square law. We don't feel the gravitational pull of, say, Mars because it's so far away. Any other universes are so unimaginably (all right, that's hyperbole) far away in comparison that any gravitational effects by the matter in them would be very, very, very, tiny.

A similar argument might be to say that an infinite series always goes to infinity because you keep adding on terms. But this is often not the case, as with the series $$\sum_{n=0}^{\infty}2^{-n}=2$$ It is infinite, but it converges.

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Uh, call me crazy, but I think $1^{-n}=1$, so that sum diverges. Perhaps you meant $\sum_{n=0}^{\infty} 2^{-n}=2$? – Warrick Jun 25 '15 at 5:11
@Warrick I appear to be the crazy one. Thank you. – HDE 226868 Jun 25 '15 at 20:32
The series argument is quite convincing. – Florin Andrei Jun 26 '15 at 0:25

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