3
$\begingroup$

This is how I understand the Unruh effect: if a person (say, Bob) is moving with acceleration $a$ in his spaceship, then he will see an apparent event horizon around him--that is, the vacuum will appear to have some temperature $T$ from his point of view, where $T$ is given by

$$T=\frac{\hbar a}{2\pi ck_B}$$

Now, let's imagine that Bob is stationary in his spaceship, and Alice is coming towards him at acceleration $a$. Would Bob see the vacuum around Alice have a finite temperature, or does the Unruh effect only occur if the observer themselves are accelerating? I would think that Bob might see space-time "warm" around Alice because she's accelerating towards him from the viewpoint of his reference frame, in the same way that Bob's surroundings are accelerating towards him when he's accelerating. Nevertheless, I am confused on this point.

$\endgroup$

1 Answer 1

5
$\begingroup$

This is a good (and notoriously difficult) question. I'm going to follow the explanation given by Crispino, Higuchi, and Matsas in their review 0710.5373, but you should be aware there are different answers out there and also there is no (uncontroversial) experimental test of this effect.

Having said all of that, the basic picture I have (and is given in the review, especially section III.A) is that the inertial observer does not see any thermal radiation. The accelerated observer sees thermal radiation, and so if the accelerated observer has a particle detector with them (in the review this is taken as a two state system, with a ground state and an excited state), the thermal radiation can cause the particle detector to become excited. From the perspective of the inertial observer, these transitions do not occur because of any thermal radiation--the inertial observer doesn't see thermal radiation. Instead, the acceleration causes the detector to interact with the vacuum in a way that can cause a transition in the detector. More or less, the inertial observer sees the particle detector experiencing a time-dependent hamiltonian because of the acceleration. The inertial observer sees the accelerated observer putting work into maintaining their acceleration, and this work is the ultimate source of the energy causing transitions in the detector.

To state things in more direct terms, the Unruh effect is often taken to mean that if we accelerate a pot of water, it can boil. In the accelerated reference frame, this happens because the water experiences the thermal bath. In the inertial frame, this happens because the accelerated pot can interact with the vacuum, with the energy ultimately coming from the work needed to accelerate the pot.

$\endgroup$
3
  • $\begingroup$ Thank you for the article. However, what if we refine the thought experiment I suggested: that is, put an Unruh-DeWitt detector right outside of Alice's spaceship. Wouldn't Bob see the particle in the detector jump out of the ground state as Alice is accelerating, and therefore detect the Unruh radiation? $\endgroup$ Commented Jul 9, 2015 at 1:21
  • 1
    $\begingroup$ Yes, the radiation emitted by Alice's detector in Bob's frame is computed in the article I cited. I wouldn't say that is a detection of the Unruh radiation per se--Bob isn't seeing a thermal bath of radiation--but yes, I agree that Bob will see Alice's detector radiating as it transitions out of its ground state (up to the fact that everything in this subject seems to be controversial). $\endgroup$
    – Andrew
    Commented Jul 9, 2015 at 1:24
  • 1
    $\begingroup$ Oh sorry, you mean you want to put an Unurh-DeWitt detector in Bob's frame but near Alice's ship at some instant in time? Let's call that Bob's detector. I agree Bob's detector can undergo a transition if Alice has a detector that is emitting radiation using the process above. In other words, Bob is capable of detecting the radiation from Alice's detector. $\endgroup$
    – Andrew
    Commented Jul 9, 2015 at 1:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.