Is the Unruh radiation isotropic? The Unruh effect is the prediction that what appears to be a Minkowski vacuum to an inertial observer, would appear to be a thermal bath of particles to an observer accelerating uniformly in the same spacetime. My question is: does the accelerating observer see an isotropic blackbody distribution? As he is accelerating in a certain direction, he would be breaking isotropy. Does this not matter?  
Thanks.
 A: The radiation is isotropic for a point detector, meaning it looks the same in all directions locally, since it is in equilibrium with any small object at the appropriate temperature. If it were not isotropic, it would not be in thermal equilibrium, since you could make a heat-engine simply by reflecting the particles in the direction they came, and get momentum from this.
The raiation is not isotropic in the sense of translation invariance, since it is getting hotter as you approach the horizon. You should think of the radiation as coming from the horizon--- if you place a refrigerated barrier between you and the horizon, you won't see any radiation past the barrier (at least not until it heats up). The reason is that the temperature of the barrier at the end furthest from the horizon forms the boundary condition for the Rindler Hamiltonian (which is the generator of Minkowksi boosts) and if it has a very long period in imaginary time, so does all the spacetime further along the Rindler x coordinate (the transformation to Rindler coordinates and the Unruh Hamitlonian is described under Unruh effect on Wikipedia)
A heuristic reason the radiation is isotropic is that the quanta coming out of the horizon classically free fall back through after a finite time, and you can catch them on either leg of their path. For massless particles this is a little glib, but you can think of the occupied modes as having both momenta (coming out and going in) because the occupation of the field mode is symmetric between both the future and past horizon.
