# Particle creation in an accelerating Universe?

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1. Mar 27, 2012

### johne1618

Hi,

I am interested to hear what people think of the following argument that in an accelerating Universe virtual particles that are separated by the Hubble radius $R(t)$ or more cannot annihilate and thus become real particles. Thus, by the uncertainty principle, particle pairs with energy $E \approx h c / R(t)$ are continually being created in an accelerating Universe giving it an intrinsic temperature that is inversely proportional to its Hubble radius.

Let us assume that, at cosmological time $t_s$, a particle-antiparticle pair pops into existence separated by a distance $d_s$.

The particle pair will annihilate in the time it takes a light signal to travel between them.

If the proper separation distance expands faster or at the same rate as the light signal proper path distance then the particle pair will never annihilate.

Let us consider the situation at the "present" cosmological time $t_p$.

Assume that the scale factor is given by $a(t)$ where $a(t_p) = 1$.

The proper separation distance of the particles at time $t_p$ is given by

$\Large D(t_p) = \frac{a(t_p)\ d_s}{a(t_s)}$.

Now if a light signal is emitted from one particle at time $t_s$, the proper distance it will have travelled by time $t_p$ is given by

$\Large P(t_p) = \int^{t_p}_{t_s} \frac{a(t_p)\ c\ dt}{a(t)}$.

A sufficient condition for the target particle to escape from the light signal is given by

$\Large \frac{d D(t_p)}{d t_p} \ge \frac{d P(t_p)}{d t_p}$ for all $t_p$.

Substituting into the above condition we find

$\Large \frac{\dot{a}(t_p) d_s}{a(t_s)} \ge c$

Let us assume that $\dot{a}(t)$ is a monotonically increasing function i.e. that $\ddot{a} \ge 0$ so that the expansion of the Universe is accelerating.

Then a sufficent condition for the particles not to annihilate is given by

$\Large \frac{\dot{a}(t_s)d_s}{a(t_s)} \ge c$

As the Hubble parameter at time $t_s$ is given by

$\Large H(t_s) = \frac{\dot{a}(t_s)}{a(t_s)}$

and the Hubble radius at time $t_s$ is given by

$\Large R(t_s) = \frac{c}{H(t_s)}$

we find, finally, that in an accelerating Universe a sufficient condition for the particles to avoid annihilation is that their initial separation at time $t_s$, $d_s$, should obey the relationship

$\Large d_s \ge R_s$

where $R_s$ is the Hubble radius at time $t_s$.

John

Last edited: Mar 27, 2012
2. Mar 27, 2012

### Naty1

Interesting first line premise. [That's as far as I read to avoid the math intracacies....

Do you know how your assumptions apply to an Unruh or Black Hole horizon?? [I do not.]

On one hand I don't buy it because the Hubble radius, if it's the distance beyond which objects recede from the observer at a rate greater than the speed of light, because supposedly it's not a 'horizon'.....but I also have doubts about that perspective....
The Hubble radius seems a lot like the Unruh effect 'horizon' to a novice like me so I'll be interested if experts weigh in here and offer some insights. Your argument also seems to work 'intutively' at the horizon of a black hole regarding Hawking radiation, so I'll try to do a little homework on this.

Maybe I should go read just what a horizon 'is'...
edit: oh,yeah...here is a source I have seen before...

http://en.wikipedia.org/wiki/Event_horizon#Particle_horizon_of_the_observable_universe

which I did not find terribly useful.

Tt would be wonderful if someone could compare the Hubble radius with, say a Schwarszchild radius and Unruh horizon distance.....as similarities and distinctions, if they are known.

Last edited: Mar 27, 2012
3. Mar 27, 2012

### Naty1

If you haven't seen this discussion:

What is a particle:

check out my first originating post....It briefly summarizes a Rovelli paper 'What is a particle? from a QFT perspective....and seems to provide some complementary insights regarding your question here....

4. Mar 28, 2012

### Chalnoth

Sounds like you've got sort of the right idea, but not exactly. The issue is that whether or not you get Hawking Radiation depends upon whether or not you have a cosmological horizon. And whether or not you have a cosmological horizon depends upon the contents of the universe: it isn't a function of the instantaneous expansion rate at any given time, but on the entire expansion history.

So what this comes down to in the end is, is there a cosmological horizon, and how big is it? The temperature of the Hawking Radiation that results is, just as with black holes, inversely proportional to the square root of the area of the horizon.

Last edited: Mar 28, 2012
5. Mar 28, 2012

### johne1618

The Wikipedia article:

http://en.wikipedia.org/wiki/Event_horizon

defines an event horizon as:
I am venturing a definition of a cosmological horizon as the boundary of a sphere centred on us whose proper radius cannot ever be traversed by a light signal sent from us.

Provided the Universe always has a non-negative acceleration (this is the condition on the expansion history) the radius of this horizon is equal to the Hubble radius.

I think my definition gives a concrete meaning to the Hubble radius.

Last edited: Mar 28, 2012
6. Mar 28, 2012

### johne1618

The discussion about global particle states only being well-defined in a flat spacetime is very interesting.

I strongly favour the "coasting cosmology" in which the Universal scale factor, $a(t)$, is strictly linear.

This allows us to have an expanding Universe with a flat spacetime with no Big Bang singularity at all.

I presume global particle states would therefore be well defined in such a cosmology.

In such a cosmology one would have a cosmological horizon whose radius is constant in co-moving cordinates.

The Universe could then be continuously filled with particles created by the Unruh-like mechanism that I have described above.

Last edited: Mar 28, 2012
7. Mar 28, 2012

### Chalnoth

No, this isn't true. It reduces to this only in a universe that has nothing but cosmological constant. We also have matter. Since our universe is currently about 70% cosmological constant (assuming it is a cosmological constant...), this isn't too bad of an approximation at the moment. But it would have been a horrible approximation billions of years ago.

8. Mar 28, 2012

### Chalnoth

A coasting cosmology is ruled out by observation. Also by the simple fact that our universe has matter in it.

9. Mar 28, 2012

### johne1618

A coasting cosmology can have matter in it as long as the resulting gravitational attraction is balanced by the gravitational repulsion of a scalar field with an equation of state given by

$\Large p = -\frac{\rho c^2}{3}$

It is true that a linear cosmology is disfavoured by the current data but I understand that an open cosmology with a power law scale factor with $\alpha \approx 1.3$ fits the current data well.

See:

http://arxiv.org/abs/0804.3491

10. Mar 28, 2012

### johne1618

Here is a maximum entropy argument in favour of a linear (or near linear) cosmology.

Let us assume the scale factor is given by a power law

$\large a(t) \propto t^\alpha$,

where $\alpha \ge 1$ for a non-decelerating Universe.

The Hubble radius is given by

$\large R_H = \frac{c}{H(t)}$.

For a power law scale factor we have

$\large H(t) = \frac{\dot{a}}{a} = \frac{\alpha}{t}$

so that the Hubble radius is given by

$\large R_H = \frac{ct}{\alpha}$.

If our Universe is accelerating (or strictly non-decelerating) then I have proposed that it is bounded by a spherical cosmological horizon at radius $R_H$.

According to the ideas of Bekenstein et al. the maximum entropy, $S_{max}$, inside a volume is proportional to its surface area.

Therefore

$\large S_{max} \propto R_H^2$

and thus

$\large S_{max} \propto \frac{t^2}{\alpha^2}$.

If we want to assume that this maximum entropy bound is itself maximised then we should assume that $\alpha = 1$ so that we are led to a linear scale law:

$\large a(t) \propto t$

according to which the radius of the cosmological horizon, $R_H$, being linear itself, is constant in co-moving co-ordinates and thus can act as a "true" boundary to our Universe containing all its constituents forever.

Last edited: Mar 28, 2012
11. Mar 28, 2012

### Naty1

johne...great subject..keep going!!

As I noted already, the relationship between your post, Unruh and Hawking radiation has stumped me for quite a while.....

Chalnoth posted:
Sounds good, seems correct [according to Hawking and Beckenstein, good enough for me!] and seems to correlate to this idea:

http://en.wikipedia.org/wiki/Pair_creation

From these I conclude that not much radiation would be expected from a huge 'Hubble horizon' area# ....What gravity is there seems feeble to 'wrench' apart much......and from the principle that there are no preferred observers it would seem to follow that an accelerating Unruh observer would have a similar if not identical 'cosmological horizon' as proposed by johne.....?? Why would these be different??

#Penrose says black hole Hawking radiation is 'ridiculously' small so I shall call Hubble sphere radiation 'ridiculously, ridiculously' small!]

Is it a coordinate based difference:
Apparently the typical Unruh effect involves Rindler coordinates while a [no spinning no charge] black hole is typically Scwarszchild coordinates.....while Hubble stuff relates to the Lamda CDM model (right? ] ..and aparently the FLRW metric ...which is not even specifically tied to the equations of general relativity.... Coupling this with the idea that globally particles may not be so well defined in curved spacetime gets my brain cells spinning....

What constitutes 'inside' a horizon, where real particle energy is negative, and outside where real particle energy is positive?? How do I know which side I am on... With Hubble spheres, your outside might be my inside...do we care???

johne:
There was a discussion here , this year I think, about that issue at a black hole horizon, it's interpretation, but I can't find it....Can anybody link to the location?

Last edited: Mar 28, 2012
12. Mar 28, 2012

### Chalnoth

No, this isn't true. A universe with a scalar field of that sort would tend towards a coasting cosmology in the future. In the previous matter and radiation-dominated regimes, it would be decelerating just as our universe was in the past. Such a universe cannot describe the observational evidence we see, which constrains $w$ to be very close to -1.

Trivial curve fitting like this is just nonsensical when we have real, physical models for how the universe behaves. I would also like to point out that they make use of the weakest set of cosmological data we have available to make this claim. I can basically guarantee you that if you combined this with the much stronger constraints from the CMB and baryon acoustic oscillations, you would get a very different result (i.e. that this cosmology just flat-out does not work).

13. Mar 28, 2012

### johne1618

I am assuming a model of the Universe in which space is always flat, the cosmological constant is zero, and the scale factor is given by

$a(t) \propto t$.

The Friedmann equations then imply that the density of mass/energy is given by

$\rho \propto \frac{1}{a^2}$.

This is not the standard Big Bang model in which a fixed amount of matter and radiation is created at the beginning of the Universe and is then diluted by the expanding 3-d space so that $\rho_m \propto 1/a^3$ and $\rho_{rad} \propto 1/a^4$ respectively.

Instead, in the linear model, matter/radiation is created continuously throughout the Universe's lifetime. In fact one can show that the total mass of the Universe in this model, $M$, is proportional to the Hubble radius which in turn is proportional to the scale factor so that we have

$M \propto a$

so that the density of mass/energy is given by

$\rho = \frac{M}{V} \propto \frac{a}{a^3} \propto \frac{1}{a^2}$.

Last edited: Mar 28, 2012
14. Mar 29, 2012

### Chalnoth

Okay. This is not the universe we live in.

And the universe we observe has matter and radiation in it. We can measure their amounts.

15. Mar 29, 2012

### johne1618

Yes, but I think that the standard model is assuming that the total amount of matter/radiation was fixed at the time of the Big Bang.

The linear model that I am describing implies that the total amount of matter/radiation increases linearly with time - a continuous creation rather than an instantaneous creation.

16. Mar 29, 2012

### Chalnoth

And that model doesn't fit reality.

17. Mar 29, 2012

### bapowell

Gravitational particle production by accelerated expansion is what gives rise to the primordial perturbations (except in modern parlance we no longer speak of particles, but fluctuations.) The idea is usually pitched in terms of particle/anti-particle pairs that pop out of the vacuum, only to get pulled apart by the rapidly inflating background. Eventually they are pulled outside the horizon, never to meet again. Of course, this description has the same conceptual shortcomings as the description of the Hawking effect in terms of particle/anti-particle pairs -- it isn't quite right.

The easiest way to "see" how inflationary spacetimes lead to particle production (more precisely, have a de Sitter temperature), is to observe that physical length scales, $\lambda\sim a(t)$, grow at a faster rate than the Hubble scale, $d_H \sim H^{-1}$ when the universe is accelerating ($w < -1/3$)
$$\frac{d}{dt}\left(\frac{d_H}{\lambda}\right) = \frac{d}{dt}\left(\frac{1}{aH}\right) \propto 1+3w$$
In the case of pure de Sitter expansion, H = const, so that $d_H$ is shrinking in comoving coordinates and the above expression is < 0. Physically, the $\lambda$ here is the wavelength of a Fourier mode comprising a quantum fluctuation: once it gets stretched so that $\lambda>d_H$, it becomes frozen-in as a classical perturbation.

That's the quick story. It's a nice exercise though to work through the evolution of a scalar field fluctuation during inflation, from its birth in the vacuum out to superhorizon scales if you haven't done it. What you find once you've done this is that you end up with a spectrum of perturbations across a range of length scales. For a massless scalar, you find that the amplitude of the fluctuation at horizon crossing is $H/2\pi$. This should look familiar.

18. Mar 29, 2012

### Naty1

right, I have seen a statement from Hawking that said the explanation, which he used, was an 'intuitive argument' not one that flows directly from his own mathematics.

Does your online website have any descriptive background? Would you post that link again, I have it somewhere in my notes...
found it: http://www.lightandmatter.com/html_books/0sn/ch07/ch07.html

and can you suggest any reading that is not purely mathematical on horizons and particle production.....
the concept that gravity, or equivalently, acceleration [which seems even crazier] can 'create matter' [particles] is interesting enough that I'd splurge on another used Amazon book!!!!

I've never seen any material that compares the three horizon's production of particles..only the 'similarity' between Hawking black hole and Unruh particles 'popping out'....I've read that a black hole horizon is 'global' but I am unsure what that means mathematically, and have no idea if that's true for other horizons we discuss...Unruh and cosmological...

edit:

"... the evolution of a scalar field fluctuation during inflation, from its birth in the vacuum out to superhorizon scales ..."

Is this 'cool physics jibber jabber' for 'particle'???

"For a massless scalar, you find that the amplitude of the fluctuation at horizon crossing is H/2π . This should look familiar..."

If so, I am in BIG trouble!!!!

Last edited by a moderator: May 5, 2017
19. Mar 29, 2012

### bapowell

I think you might have me confused with bcrowell (understandable, of course.) I'm Brian. There are several good books out there on the subject of gravitational particle production, but I think they're all rather mathematical (although not "purely"). One that I've not personally read but have heard is good is "Introduction to Quantum Effects in Gravity" by Mukhanov and Winitzki. Then there is the classic "Quantum Fields in Curved Space" by Birrell and Davies -- they have several examples of particle creation from expanding spacetimes. Almost all recent cosmology texts that cover inflationary perturbations will likely discuss this result, but I'm not sure of any off-hand that actually go through the calculation. "Cosmological Inflation..." by Liddle and Lyth talk about it, but I'm not the biggest fan of that book.

20. Mar 29, 2012

### Naty1

definitely!!.....thanks for the suggestions....and I have almost gotten you two mixed up previously....