# I Unruh effect in full spacetime

1. Jul 9, 2017

### gerald V

In the formula for the Unruh effect, temperature (of a photon gas mostly) is proportional to the acceleration of some classical observer. But how can that be? Temperature is a scalar, while acceleration is a vector.

The cause of the effect obviously preferres one spatial direction, and Rindler coordinates reflect that. I have looked into an article explaining the Unruh effect in some detail (https://arxiv.org/pdf/1304.2833.pdf). This article explicitely states to restrict itself to 1+1 – dimensional spacetime, and so do some other sources I found.

So, for the Unruh experiment taking place in our full spacetime, what is the distribution of the photons in 3-dimensional position space as well as in 3-dimensional momentum space?

In contrast, Hawking temperature appears as quite plausible to me. But I assume that for black holes which are not spherically symmetric, i.e. rotating ones, temperature varies over the latitude. Is that right?

2. Jul 10, 2017

### Staff: Mentor

It's proportional to the magnitude of the acceleration.

The magnitude of a vector is a scalar.

3. Jul 10, 2017

### Demystifier

Here is a simple analogy. Suppose that a body moves with constant velocity through a medium. Due to friction with the medium the body will be heated and, after a certain transient time, the body will have a constant temperature caused by velocity relative to the medium. Yet, velocity has a direction and temperature has not.

4. Jul 11, 2017

### gerald V

Thank you very much for these explanations. However, since I still have problems to comprehend the situation, please let me first try to check whether I understood correctly what the Unruh effect claims. Is it correct that it is claimed that for an observer in hyperbolic motion (having come from sufficiently far back in the past) in flat spacetime, apart from statistical fluctuations
- the situation is stationary
- the "warm gas" homogeneously fills all of 3-space
- at any point in 3-space, the momenta of the "molecules" making up the gas are isotropically distributed in 3-momentum space?

5. Jul 11, 2017

### Demystifier

Yes.

No. The excitations exist only in the accelerated detector. (And I wouldn't call it a "gas".)

Yes, the momentum of excitations is distributed isotropically. (But I wouldn't call it "molecules".)

6. Jul 11, 2017

### LeandroMdO

Careful there. While the mean number of particles registered by the detector is thermally distributed, that does not mean that a real photon gas at finite temperature exists. Other properties of Unruh radiation do not necessarily coincide with those of a real gas. For example, higher moments of the particle number distribution are not necessarily thermal.