3
$\begingroup$

Wikipedia defines a multiverse as a hypothetical set of finite and infinite possible universes. Similarly, it defines the Universe is as "all of time and space and its contents". On the surface, these definitions seem to contradict each other, unless one assumes that other universes don't contain time and space.

I understand that without being technical or specific to a theory it's hard to describe a multiverse, but if we don't use the "all of time and space" definition from above, when do formal Physical theories reach the need to define a multiverse?

$\endgroup$
  • 1
    $\begingroup$ You are correct, the multiverse is neither philosophically nor scientifically a good idea. In science one should rather talk about one universe that may be separated into non-communicating local sub-universes. Like the aether the "multiverse" has its time in the limelight and it will disappear into the dustbin of scientific history soon enough. $\endgroup$ – CuriousOne Jan 30 '16 at 20:26
  • 1
    $\begingroup$ If there is a universe, then there is a multiverse ---- just take the universe and cross it with the unit interval. This leaves open the questions of whether the universe is part of a more interesting multiverse, and whether studying the properties of that multiverse might reveal things we might not otherwise have discovered about the universe. This would not be unprecedented. It's not uncommon, for example, to learn things about the properties of a surface $S$ by studying a family of surfaces that degenerates to $S$. $\endgroup$ – WillO Jan 30 '16 at 21:24
2
$\begingroup$

the cutting by universes is a way :

  • to introduce possible new physics for each of these universes without leaving the homogeneity and isotropy cosmological principles, the known constants and the known physics of "our" universe

  • to defer the infinity issue from our universe to a parent structure : the multiverse

Homogeneity and isotropy are the main reasons. The multiverse scheme you cited assumes different inflations / expansions and different Hubble constants. It was necessary to make a distinction.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.