Cosmological horizon

1. Jun 4, 2010

Naty1

Here I had been thinking the Unruh horizons and specifically black hole horizons were the only (thermal) radiators.

Now I notice Leonard Susskind says in THE BLACK HOLE WAR that the cosmological horizon also radiates...but "inward" towards us. If a thermometer on a string were extended close to the cosmological horizon, he says ".....we would discoverer the temperature increases eventually approaching the infinite temperature at the horizon of a black hole..."

What do you think??

If that's accepted or at least a rational theoretical idea, would a suspended (stationary) observer inside a black hole horizon also observe such radiation....thermal type???? Seems like one would but I've never seen that stated anywhere.

2. Jun 5, 2010

czes

Susskind is an advocate of the holographic principle in the String Theory. His idea is different than this of Hawking. Each volume is surrounded by its surface and this surface contains all information contained in that volume. Therefore the inside observer is just a part of the hologram created by the surface's screen. If the observer sees a thermal radiation it means due to Davies-Unruh he is in an accelerating motion relative to that screen or eqivalently his apparent horizon increases.
I am not sure if it is correct. May be a professional physicist can write it properly.

3. Jun 5, 2010

Naty1

czes: yes, in general that's correct....but I am not so sure about inside a black hole or cosmological horizon...butI think the same ideas hold....Hawking is a relativisit, Susskind a quantum/string theory oriented physicst.

Last edited: Jun 6, 2010
4. Jun 10, 2010

Naty1

Seems like with Unruh and Black hole horizon heat, the observer must accelerate; with the cosmological horizon, seems like it's accelerating away from any observer...it's doing all the work.....so seems even more likely it would appear "hot".

And makes me wonder about my inside a black hole question....direction of acceleration relative to a horizon matters in such observations.

5. Jun 14, 2010

Naty1

By dumb luck I stumbled across this Verlinde video...

http://streamer.perimeterinstitute.ca/mediasite/viewer/NoPopupRedirector.aspx?peid=37ff156a-33d5-40d8-9986-d5ec82d96d91&shouldResize=False# [Broken]

At minute 4:30 it shows the temperature of the cosmological horizon [in Desitter space] and Verlinde mentions as I suspected that it's the " cosmological constant" ....acceleration.... ....that underlies the temperature....

So it seems the cosmological horizon does indeed radiate.....it HAS a temperature.

Last edited by a moderator: May 4, 2017
6. Jun 15, 2010

George Jones

Staff Emeritus
7. Jun 15, 2010

friend

So mere expansion is creating real particles, particles responsible for the temperature? I thought the cosmological constant was the zero point energy of virtual particle fields.

It sounds like we have spacetime popping into and out of existence right along with virtual particles. It sounds like one is made of the other. Are there virtual spacetime geometries popping into and out of existence, a portion of which become real as the universe expands?

Last edited by a moderator: May 4, 2017
8. Jun 15, 2010

czes

The virtual particle-antiparticle lives to short to be observed. Therefore there is a vacuum made of virtual particles-antiparticles.
What if the virtual particle-antiparticle pair interact with an another pair in that very short time ? It could be in very strong gravitational field (Black Hole) or at gamma ray or at relativistic collision in accelerator. Is it possible that such a permanent relation creates a real particle and antiparticle ?

9. Jun 19, 2010

bapowell

The reason that a de Sitter universe has a temperature is because of the accelerated expansion: comoving distances grow at a faster rate than the Hubble radius. The wavelength of quantum vacuum modes stretch along with the background, eventually surpassing the causal horizon. This is the mechanism by which inflation generates large scale fluctuations -- from gravitational particle production. If one simply starts with a free scalar field and drops it in de Sitter space, one finds that there is indeed a nonzero autocorrelation function $$\propto H^2/2\pi$$.