2
$\begingroup$

As far as I understand, Hawking radiation is formed at the edge of a black hole, when a particle/anti particle pair is formed and one of the particles falls into the black hole before the particles has time to annihilate. This made me think of other situations where this could occur which led me to think about the expansion of the universe.

Since rate of expansion of the universe is increasing all the time due to dark energy, and at some point in the future the expansion will be so fast as to tear matter apart, wouldn't there be a point where particle/anti particle pairs that spontaneously form from quantum fluctuations be torn apart faster than they can annihilate, thus forming infinite amounts of new matter in the ever expanding universe?

Can anyone with a better understanding of the maths behind this explain what would actually happen?

$\endgroup$
1
$\begingroup$

The expansion of the universe won't necessarily get to the point where it rips everything apart. It depends on the equation of state of the dark energy. The phenomenon you describe is called the Big Rip, but if dark energy behaves like a cosmological constant (which is currently consistent with observations) there will be no Big Rip.

But the expansion caused by dark energy does form the equivalent of an event horizon. At some distance from it the expansion is so great that nothing travelling at the speed of light or less could reach us even given infinite time. This causes a horizon similar to the horizon round a black hole, and just like a black hole this horizon does emit Hawking radiation. The Wikipedia article on dark energy discusses this, and places the horizon at 16 billion light years so it will be a while before we see it :-)

The paper Expanding Confusion goes into the maths if you fancy the challenge.

Response to comment:

The virtual particle pair being pulled apart description of Hawking radiation is only a metaphor. Remember that the Feynmann diagrams showing virtual particles are just an illustration of one term in a calculation and are not intended to be taken literally. Depending on who you ask, and what you mean by exist, virtual particles don't really exist. See this article by Matt Strassler on virtual particles for an excellent, though inevitably difficult, explanation of virtual particles. See this article by John Baez for more on the nature of Hawking radiation.

The calculation showing that Hawking radiation exists is done by comparing what a quantum field looks like to an observer at infinity in the presence of a horizon. Hawking radiation is seen because observers near and far from the horizon will disagree about what constitiues a vacuum. I can't say any more about this because I don't understand it well enough. If you're interested this paper describes the calculation in layman's terms (for a small subset of extraordinarily well informed laymen).

So the question does a Big Rip pull apart virtual particles is probably meaningless, though the question could presumably be rephrased along the lines of how a Big Rip affects the apparent QFT vacuum. Even then I'm not sure this has an answer. The Hawking calculation is done by extrapolating to infinity and requires an asymptotically flat or effectively flat spacetime. How you'd do the calculation during the approach to the Big Rip I don't know.

If you were an observer in the closing stages of a Big Rip you would see yourself surrounded by a spherical horizon, emitting Hawking radiation, and the horizon would shrink in towards you at an exponentially increasing rate. I'd guess the Hawking radiation would become more intense as the horizon sweeps towards you but i don't know if it would tend to infinity or not.

$\endgroup$
  • $\begingroup$ But what would happen if the big rip scenario is the correct one? If that's what actually happens, then at some point every part of the universe must be expanding fast enough to separate the particle pairs as soon as they form? $\endgroup$ – Elias Mårtenson Oct 30 '13 at 13:16
  • $\begingroup$ @EliasMårtenson: I have edited my answer to respond to your comment, though I suspect you'll find my discussion disappointing. $\endgroup$ – John Rennie Oct 30 '13 at 16:21

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.