# B Distribution and direction of Unruh radiation

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1. Aug 27, 2016

### GeorgeDishman

If an observer accelerates through a simple vacuum, it is often said that they see Unruh radiation with acceleration of $2.5*10^{20} m/s^2$ equivalent to a temperature of 1K, but I haven't seen the polar distribution described, one might assume it was from 'ahead' of the direction of acceleration, is that correct?

If so, by the equivalence principle, if I am at a constant radius from a mass, I am accelerating 'upwards' relative to a free-fall observer. Does that mean that looking up when standing on the Moon, I should see a thermal contribution to the sky at a temperature a little less than $10^{-20} K$?

One source on this (Scholarpedia) states:
• "From the point of view of Rindler quantization (2) the detector is responding to the particles whose presence was calculated in (4). From the point of view of Minkowski quantization (1) the excitation of the detector is correlated with emission, not absorption, of particles (Unruh and Wald, 1984); thus a stationary (or inertial) observer "sees" the detector radiating, .."
This makes me think perhaps an observer on the Moon wouldn't see a thermal sky, rather an observer at infinity would see thermal radiation from the surface, i.e. this is Hawking Radiation?

2. Aug 27, 2016

Staff Emeritus
1. That's not what the equivalence principle says.
3. I do not believe you get Hawking radiation from non-black holes. The standard derivation relies on a horizon, and without a BH you don't have one. If you get radiation, it's from some other means.

3. Aug 27, 2016

### GeorgeDishman

Of course, that's the key, thank you! When accelerating through Minkowski space, the horizon is behind you so that's where the Unruh radiation must appear to originate, not from in front.
Yes, if you are accelerating upwards relative to a free-fall observer, the radiation would appear to come from 'below' which is then consistent with Hawking Radiation, that solves my puzzle.
That you would say that surprises me, in what way have I got it wrong?

4. Aug 27, 2016

### tionis

Hi George. See below:

''The radiation will be isotropic. Ie, it will obey detailed balance-- the radiation received depends both on the radiation there and the cross section of the detector for radiation from that direction. The cross section is the same for emission and reception, so in directions where it is difficult to emit, it is also difficult to receive. But taking into account the anisotropy of the detector, the thermal spectrum is isotropic. Note that the detector is also in a strong gravitational field, so the Tolman relation says that the temperature has a gradient, and if the detector is sensitive to that gradient, it will see "up" as different from sideways.

No, on earth or the moon there is no such radiation. Note that the wavelength for radiation with a temp of 10^-20 is about 10^18 m (ie about 100 light years), so the radiation will be sensitive to changes in space on that kind of scale. Also the moon has no horizon which also prevents it from having such radiation. An observer at infinity also will not see radiation from the moon.''

5. Aug 27, 2016

Staff Emeritus
The equivalence principle does not say you can swap gravity for acceleration. It takes 16 numbers to describe gravity and 3 to do acceleration. It's really more the reverse - it tells how to tell apart gravity from acceleration.

6. Aug 27, 2016

### GeorgeDishman

Ah OK. The descriptions I've seen are that acceleration is equivalent to a uniform gravitational field hence it is valid where the region is small enough that tidal components are negligible and I intended the comment that way. I should have been explicit about that.

7. Aug 27, 2016

### GeorgeDishman

Yes, my silly mistake was getting the horizon on the wrong side of the observer, there's no limit going upwards ;-)

8. May 7, 2018

### craigthone

Would yould like to say more about the meaning of equivalence priciple?

9. May 8, 2018