B What is the origin of the Gibbons-Hawking temperature?

1. May 3, 2017

Tim Miller

I recently submitted the following question to Dr. Michio Kaku on his Facebook page regarding an unfamiliar concept in his book "Parallel Universes":

Dr. Kaku,

In your book "Parallel Universes", pg. 301, you mention something called the Gibbons-Hawking temperature, a theoretical low temperature limit of 10 negative log 29 degrees (is that Celsius of Kelvin?). You cited a work by Lawrence Krauss & Glenn Starkman. Can you give me the full citation for this work? Is this it: http://iopscience.iop.org/articl...? How does this relate to the absolute zero of classical physics (−273.15 °C; −459.67 °F), a theoretical point at which nearly all atomic/molecular motion ceases?

2. May 3, 2017

Staff: Mentor

I believe this temperature (in Kelvin) is the temperature associated with the cosmological horizon due to the nonzero cosmological constant in our universe. Nothing in our universe (more precisely, in the causal patch within our cosmological horizon) can ever be at a lower temperature than this.

3. May 4, 2017

Tim Miller

Thanks for your reply! I'm still wondering how this number was arrived at and why scientists don't now call this number absolute zero instead of −273.15 °C / O°K.

4. May 4, 2017

Staff: Mentor

Any horizon has a corresponding temperature associated with it. For the cosmological horizon in de Sitter spacetime, the temperature is proportional to the cosmological constant. I can't dig up an exact formula on line at the moment, but AFAIK the temperature $10^{-29}$ Kelvin is what you get when you plug the best current value for the cosmological constant in our universe into the appropriate formula.

Because it isn't. The temperature associated with a horizon is not zero; it's positive. (Note that any time you see a temperature in physics or cosmology, it will be an absolute temperature unless it's explicitly stated otherwise, since the absolute temperature scale is the one that makes all the formulas involving temperature look simple.) The idea that absolute zero is "the lowest possible temperature" is only a heuristic; the actual physical definition is different.

5. May 4, 2017

kimbyd

Is that last sentence strictly true? My thought was that this temperature was similar to the CMB temperature in its impact on temperature of other bodies, and there are objects out there with temperatures lower than the CMB (e.g. the Boomerang Nebula). Nothing can stay at a lower temperature than the CMB for too long, because eventually it'll reach equilibrium. But they can definitely go below that temperature temporarily.

6. May 4, 2017

Staff: Mentor

I'm actually not sure whether that is true or not. If it is true, then yes, I think it would be possible to temporarily cool something below this temperature. But there might be other effects involved with the temperature associated with a horizon that are not present with the CMB.

7. May 4, 2017

kimbyd

The way I might imagine it could be true would be if this temperature were a fundamental property of the space-time, rather than just photons created at the horizon that fill the universe.

8. May 4, 2017

Staff: Mentor

I'm actually not sure that either of these is a good description of the temperature associated with a horizon. We don't have a good microphysical model of how such a temperature arises, so we can't be sure.

9. May 4, 2017

kimbyd

My suspicion is that ultimately, it doesn't actually matter. I'm pretty sure the universe where this temperature is relevant is an empty universe. That's a de Sitter universe, which is a stationary state. If such a universe is in a stationary state, then the finite temperature would just be an apparent temperature that has nothing to do with the underlying space-time. E.g. there would be no temperature fluctuations. This is the universe described here: https://arxiv.org/abs/1405.0298

10. May 4, 2017

Staff: Mentor

Just to be clear, the paper you linked to is a hypothesis about de Sitter spacetime; it is not a known fact about de Sitter spacetime. I don't know that the conclusions of the paper are generally accepted among physicists.

11. May 5, 2017

Tim Miller

12. May 5, 2017

Tim Miller

If the temperature is positive, the why is it expressed with a negative exponent?

13. May 5, 2017

Staff: Mentor

$10^{-29}$ is greater than zero, so it's a positive number.

14. May 5, 2017

kimbyd

Because it's less than one. E.g. $10^{-1}=0.1$, $10^{-4}=0.0001$.

15. May 5, 2017

Tim Miller

Thanks for the refresher. My memory is not what it used to be.