I have often heard it said (by professional cosmologists) that if the universe is infinite, then there necessarily exist infinitely many copies of me repeated throughout.

The reasoning seems to be that any finite volume of space can contain at most finitely many fundamental particles existing in finitely many configurations. From here it is implied that all possible (permissible) configurations must both exist and occur infinitely often.

This does not sound like a sound argument to me, so I have to believe that I have misunderstood the argument.

Is this unsettling implication true in an infinite universe, and if so how is it justified.

  • $\begingroup$ @HDE226868 For example, why could not a finite number of configurations repeat infinitely often, allowing my configuration to be unique. $\endgroup$ – user57876 Aug 26 '14 at 21:32
  • 1
    $\begingroup$ It just brings home the old fact that no scientist (not even cosmologists), is qualified to talk about issues completely outside of his or her field. The people who are saying these things in public are not likely to write them down in a peer reviewed paper, which they know wouldn't pass review, because they are simply not true. It does sound great on tv, though. $\endgroup$ – CuriousOne Aug 26 '14 at 21:38
  • 1
    $\begingroup$ @HDE226868: I would suggest to take a look at the mathematical literature about "infinity". There is more than one type of infinity. So what's the cardinality of the visible universe, what's the cardinality of an infinite (unseen and thus unphysical) universe and what is the cardinality of the number of possible future states of the universe? Neither of these questions has a scientific answer, so it's completely nonsensical to draw a conclusion that there are infinitely many copies of a person (which is not even a definable physical property of the universe). $\endgroup$ – CuriousOne Aug 26 '14 at 22:03
  • 2
    $\begingroup$ @HDE226868: I will claim my right to be a humorless physicist now. :-) $\endgroup$ – CuriousOne Aug 26 '14 at 22:37
  • 1
    $\begingroup$ @CuriousOne I will claim my right to be a humorless physics enthusiast. :-) $\endgroup$ – HDE 226868 Aug 26 '14 at 22:41

The argument is sound given a few oft-omitted (but not too unreasonable) assumptions. Here is one way it can be formulated.

Consider a volume $V$. Suppose it has a (possibly infinite) set of possible configurations; call this set of states $S$. Suppose we are interested in a particular configuration, $c \in S$, to within a certain tolerance. Let $C \subseteq S$ be the set of all states close enough to $c$ to count for our purposes.

Assumption 1: There is a probability measure $\mu$ on $S$ corresponding to the probability of $V$ manifesting in a particular state in some selection process I'm intentionally being vague about.

Assumption 2: $\mu(C) > 0$. That is, there is a strictly positive chance of a "randomly" (again, being vague) selected state matching our desired configuration.

Assumption 3: There exists a "horizon distance" $d$ such that if two volumes are more than $d$ apart, their states are entirely independent.

Assumption 4: The universe is infinite.

Assumption 5: The universe is homogeneous. In particular, $S$ and $\mu$ are the same for any $V$ chosen.

(1) means we can meaningfully talk about the chances a state $s_i$ manifests in a volume $V_i$. (2) and (5) tell us that for any $V_i$, $P(s_i \in C)$ is a constant, positive number. If we define the indicator variable $$ \chi_i = \begin{cases} 1, & s_i \in C \\ 0, & s_i \not\in C, \end{cases}$$ then the expectation of indicator, $\langle \chi_i \rangle$, is this same positive number. (3) and (4) together mean there are infinitely many uncorrelated volumes $V_i$ in the universe (in addition to possibly many other volumes correlated with these) to choose from. If $I$ indexes finitely many mutually uncorrelated $V_i$, then we can see $\langle \sum_{i\in I} \chi_i \rangle$ can be made arbitrarily large by augmenting $I$.

Of course, no one is asserting the claim "there are infinitely many copies of you in existence" as fact, because these assumptions can always be questioned, some in important ways.

(2) seems justified based on your existence, and (5) is a common assumption about the universe.1 But (1) really calls for more than a little bit of philosophy, and (3) and (5) together worried enough cosmologists in another context that they came up with inflation to essentially rid themselves of (3). And of course (4) is certainly not known, and strict scientific positivism would say that sentence isn't even deserving of being called science, for it is fundamentally untestable.

1 I find the history of 20th century cosmology interesting, in that homogeneity was assumed/hoped for before it was validated observationally. After all, most of the universe doesn't look like Earth, or the Solar System, or the Milky Way, or the Local Group, etc. Only with really big galaxy surveys did we see an end to the hierarchical structure.

  • $\begingroup$ This is a most interesting analysis that you have given here, as well as being persuasive. The sum over the chi(i) formulation is really clever. Thank-you for taking the time to provide such a excellent answer. $\endgroup$ – user57876 Aug 27 '14 at 2:24
  • $\begingroup$ My biggest objection on reading these is assumption 1. For something to qualify as a probability measure, the measure must never be negative and integrating/summing over the entire sample space must yield exactly one. If the sample space itself is not finite, this means the probability distribution must have local maxima and must tend toward zero at infinity. These arguments implicitly assume a constant, nonzero value over all space. That's not a probability measure because the integral over all space diverges rather than converges to unity. $\endgroup$ – David Hammen Oct 8 '14 at 16:41
  • $\begingroup$ @DavidHammen By using the index set $I$, one can avoid dealing with an infinite universe, merely an arbitrarily large finite one. The configuration space $S$ might be infinite, but I don't need a uniform distribution on it, just for $C$ to have positive measure. Or looking at $S$ modded out by "close-enough" indistinguishability, then $C$ can very well be one element of a finite set. $\endgroup$ – user10851 Oct 8 '14 at 17:48

The argument rests on the assumed validity of Ergodic theory (see http://en.wikipedia.org/wiki/Ergodic_theory). Quoting it "A central concern of ergodic theory is the behavior of a dynamical system when it is allowed to run for a long time. The first result in this direction is the Poincaré recurrence theorem, which claims that almost all points in any subset of the phase space eventually revisit the set". Thus, unless you can assume that a physical mechanisms is violating the ergodic assumption, then if the universe is infinite, there necessarily exist infinitely many copies of it repeated throughout. Actually, there is an extra assumption (by symmetry?) to get this last assertion: that the ergodic property also holds in a collection of separated systems at a single time, not only in a single system across time. The reason is that if the recurring states across time are random, then you get the same result if instead you have an infinite random collection of "disconnected" regions in the universe (the distribution of states across time should be no different than the distribution of many samples at a given time).

  • $\begingroup$ Yes, that is very helpful. You are starting to convince me that it may not be as far fetched as I had believe. And I do seem to recall that Poincare's theorem was mentioned on some occasions. Thanks. $\endgroup$ – user57876 Aug 27 '14 at 2:12
  • 2
    $\begingroup$ No, this is wrong. The Poincare recurrence theorem doesn't apply here, because it assumes a fixed volume, whereas the volume of the universe is expanding. $\endgroup$ – user4552 Nov 11 '14 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy