# Inflation and temperature

1. Feb 16, 2010

### matteo210

hi everyone,

i am not a physicist, so please forgive me and help me to clarify some basic concepts.
in the context of the inflation of the universe, as i understand it, it was the energy density of the void that expanded space-time, so that the energy density remaining (almost) constant coupled with an increase of volume, created a negative pressure and this in turn led to an exponential increase of the volume with the energy density of the void staying (almost) constant until falling into a lower state of energy at the end of inflation.
Is this the reason why during inflation we don't see an exponential decrease of the temperature of the universe? if that is the case, how is the energy density of the void related to the temperature of the universe?
if that is not the case, why didn't the universe cool down in an exponential way during inflation, as all bodies do when they become bigger in volume?

matteo

Last edited: Feb 16, 2010
2. Feb 17, 2010

### Chalnoth

We do. Before inflation ended, the temperature of our universe was essentially zero. This is why the end of inflation is called "reheating": when the field that drove inflation decayed, its large energy density became a temperature, causing the universe to become extremely hot.

3. Feb 17, 2010

### Dmitry67

No. Because of HUP, the minumum temperature was limited by the life of Universe (delta E par particle calculated based of T using HUP), or, by the Delta X and the maximum radius of observable universe at that time.

4. Feb 17, 2010

### Chalnoth

We've been over this. It doesn't make any sense.

The temperature of the universe scales as 1/a. During inflation, the scale factor 'a' increased by a factor of at least around 10^30, meaning that basically whatever temperature our universe had when inflation began, it was effectively zero by the time inflation ended.

5. Feb 17, 2010

### Dmitry67

Yes, it does not make any sense.
In early universe with Rh, say, 0.0001mm you can not, even theoretically, define a particle with wavelength more then 1mm. Hence, for example, in such universe, you can not imagine 'cold' photons of any radio-wave length. Any photon in such universe has no choice but to be very hot!

6. Feb 17, 2010

### Chalnoth

Well, this has two problems. First, at the start, you're presuming that it must start at the Planck temperature. The uncertainty principle only talks about the minimal time an energy transition can take (and even then it's not entirely clear how this would translate to quantum gravity, which would definitely be a significant factor at those times). Thus you would need a starting point, and it's just not clear that starting point would be the Planck temperature.

Then there's the issue that even if you have a Hubble horizon of 0.0001mm, you can most definitely have photons with much larger wavelengths. The way that this happens is that if you have a vacuum energy, or something that acts similarly, then the horizon size stays nearly constant. But if you start off with a photon that is smaller than the horizon, it quickly grows to be larger than the horizon.

In any case, even if you started out with Planck-scale temperatures, after the universe expanded by a factor of 10^30, the temperature would still be negligible. So after inflation, the universe is effectively at absolute zero, and must be reheated by the decay of the field that drove inflation.

7. Feb 17, 2010

### Dmitry67

Let me prove it to you differently.

In expanding universe the horizon is

$$r_H= \sqrt{3/ \Lambda}$$

and the temperature of hawking radiation from these horizons is

$$T= \frac{1}{2 \pi r_H}$$

For example, when Rh was 0.0016mm the temperature was 225K
This is a minimum temperature available in the Universe.

Interestingly enough, if you try to derive temperature from HUP, as I did before, you get:

$$T= \frac{1}{2 t_H}$$

Which is the same except $$\pi$$
Isn't that a wonderful 'coincidence'?

8. Feb 17, 2010

### Chalnoth

Huh? Why is the Hubble time in there?

9. Feb 17, 2010

### Dmitry67

By $$t_H$$ is mean the age of the Universe.
Of course it would be better to omit _H, it is confusing.

10. Feb 17, 2010

### Chalnoth

In that case, those two quantities are fantastically different in an inflating universe.

11. Feb 17, 2010

### Dmitry67

Correct.
In the era of catastrophic, exponential expansion, $$r_H << t$$ (in planks units) - tidal forces were enormous, ripping everything apart and creating consomological horizons micrometers away.

Classically, space was expanding so rapidly, that even if there was some energy density, it would almost instantly disspear because of the rate of expansion. The faster expansion - the colder universe. This is your idea.

And that idea is correct if we look at it classically. If we take hawking radiation into account, then on the contrary, the faster expansion, the closer horizons, the hotter radiation!

12. Feb 17, 2010

### Dmitry67

But in any case, if we go back in time, $$r_H$$ could not be > age of the Universe, and this guarantees that the very beginning was very hot.

13. Feb 17, 2010

### bapowell

This is true only for an inflating spacetime -- you need a horizon for there to be a de Sitter temperature. So, this has little to do with your point about a maximum wavelength which you seem to suggest holds in general. Of course, the horizon in the early universe doesn't set any limit on the wavelength of photons that exist in the universe -- it merely defines our causally connected region of the universe. As Chalnoth says, particles can (and do) have wavelengths larger than the Hubble radius during inflation.

During inflation, the universe does plunge to near absolute zero before reheating. The temperature of the vacuum (the de Sitter temperature), while setting a minimum temperature for the spacetime, is associated with the horizon.

14. Feb 17, 2010

### Chalnoth

Er, that's not true. The rate of expansion doesn't relate directly to temperature. The amount of expansion does. Thus it's the total amount of expansion during inflation that is important for this argument, not the rate of expansion during that time.

Also, I think you're incorrect about the Hawking Radiation from the horizon of de Sitter space. I'm pretty sure that is only there for accelerated observers, though I'd have to look it up.

15. Feb 17, 2010

### bapowell

No, that's true for everyone -- as you probably know, during inflation the causal horizon is an event horizon. Just like that of a black hole, it is a property of the structure of the spacetime, not a property of any observer's world line. The horizon seen by an accelerating observer (by virtue of his motion) is of a different sort.

16. Feb 17, 2010

### Chalnoth

Well, no, different observers most definitely see different cosmological horizons (as seen by the trivial example of considering an observer 10 billion light years away). I was talking about the Hawking Radiation from the horizon, however. I'd have to think about it a bit more, but I'm not convinced just now that cosmological horizons necessarily have temperatures.

But in any case, if there is some temperature due to the cosmological event horizon during inflation, that temperature is still minuscule compared to the energy density in the field that drives inflation. Consider, for a moment, that inflation happens at the TeV scale. This is extremely low, but we can throw it out there just for a thought experiment. If we have some material with an energy density of 1 (TeV)^4/(hbar^3 c^5), then the Hubble expansion rate would be H = 3.6 * 10^11 s^-1. This would give a temperature from Hawking radiation, by the previously-provided formula, of:
T = hbar/k * H/2pi = 0.4K

For an energy scale of 1TeV, a temperature of 0.4 Kelvin is so absurdly small as to be completely negligible. Furthermore, if you double the energy density, the temperature only increases by a factor of sqrt(2), so for more realistic inflaton energies, the temperature is even smaller compared to the energy in the inflaton field. This is assuming I did all my unit conversions correctly.

17. Feb 17, 2010

### bapowell

Cosmological event horizons most certainly do have temperatures. It's a famous result in quantum fields in curved spacetime. The result is:

$$T = \frac{H}{2\pi}$$.

This was a major part of my dissertation. In fact, the growth of cosmological perturbations generated during inflation is a result of this fact (eg $$H/2\pi$$ is also the amplitude of scalar perturbations generated during inflation).

I agree that this temperature is indeed small during inflation, and so it has no direct bearing on matteo210's original question. However, while it plays no significant role in the evolution of the universe (ie it has negligible contribution to the stress tensor relative to the inflaton), it sets up the primordial perturbations that eventually grow to form galaxies. Pretty far out...