Suppose spacetime is expanding at an ever accelerating rate now, and supposing it continues doing so in the future, won't our universe at some point in the far far distant future look exactly like it did during inflation, just before the Big Bang? That is incredibly cold and homogeneous?

If the Universe is flat as it seems, does this mean this accelerating expansion can effectively go on for ever?

Would this then create massive amounts of Unruh radiation?

  • $\begingroup$ I read conflicting stuff on this, but I used for the info profmattstrassler.com/articles-and-posts/… $\endgroup$ Feb 18, 2015 at 4:19
  • $\begingroup$ Hmm. Seems I need to go back and look at my inflationary cosmology a bit more. You may be interested in this post, though. $\endgroup$
    – Kyle Kanos
    Feb 18, 2015 at 4:34
  • $\begingroup$ What does "spacetime is expanding" mean? $\endgroup$
    – WillO
    Dec 31, 2021 at 15:00

1 Answer 1


Assuming no new discoveries concerning either the nature of matter in our universe or the behavior of gravity at large distance scales, then yes, the late universe will resemble the inflationary universe. Both will exhibit exponential expansion, although the expansion rates will be different. And presumably the later inflationary period will never end; the inflation is being caused by the omnipresent dark energy and not an inflaton rolling down a potential.

But no, this unending expansion would not lead to Unruh radiation per se, although perhaps this is just an issue of semantics. Unruh radiation is the phenomena that an accelerating observer (coupled to a quantum field) would see a thermal bath of radiation. Here the expansion of the universe is accelerating. So the two aren't quite the same. But there is a strong connection.

Let's take de Sitter space. de Sitter is the maximally symmetric spacetime with positive cosmological constant and it also exhibits exponential expansion, just like the early inflationary universe or the late-time fate of our universe. The notion of a particle is an observer-dependent notion in curved space, but a static observer would indeed see a bath of thermal radiation due to the cosmological horizon. And the existence of the horizon is directly related to the exponential expansion. So while I wouldn't call the radiation Unruh radiation, there will indeed be radiation from the accelerating expansion. In both the current case, and Unruh radiation, (and also Hawking radiation), the radiation is caused by a horizon. In the Unruh case the horizon is present because an observer straps on a jet pack and chooses to accelerate, in the black hole case it's because the observer is hovering (perhaps far away) from a black hole, and in the cosmological context it's due to the cosmological horizon caused by the expansion.

Here's a nice article by Hawking on the subject: http://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2738

Even if you can't access the article, there's quite a bit of content in the abstract.

EDIT: Perhaps I should add that this late-time radiation due to the cosmological horizon will be very cold.

  • $\begingroup$ Ok, so assume an observer is one Planck volume, there would come an acceleration where his event horizon would be the Planck volume. What would the temperature of the horizon be? If the horizon is just the Planck volume, is the temperature of the horizon then the temperature of the observer? $\endgroup$ Feb 19, 2015 at 11:24

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