Due to the Unruh effect the vacuum energy for the uniformly accelerated observer looks like as the equilibrium background with the Hawking-Unruh temperature $T=\hbar a/2\pi ck_B$, where $a$ is the acceleration. So we can conclude that the vacuum energy specifies a noninertial frame of reference with respect to which one can define an acceleration of any particle (note that vacuum energy does not specify any inertial frame of reference because it is uniformly distributed in a four-dimensional continuum so that all four directions for it are identical). But as it follow from Freedmann's equations a relative acceleration of two galaxies (observers) which currently are on distance R from each other is equal to $a= (\Omega_{\Lambda}-\Omega_m)H_0^2R/2$. So if in some point of the Universe the Hawking-Unruh temperature is equal to zero for the other points it is not so. Such way the vacuum energy specifies on an existence of a center of the Universe contrary to the cosmological principle which claims that the properties of the Universe are the same for all observers. In this situation it does not matter whether we can observe the Unruh radiation or not. This means that our knowledge about the Universe and the vacuum energy are incompatible.
To solve this conflict one can assume that the Universe is a hypersphere which isotropically expands on the background of 4D Euclidean space. In such case the accelerations (with respect to background) of all points belonging to hypersphere would be equal. As an alternative, we also can assume that vacuum energy does not exist in reality. Unfortunately both of these assumptions lie beyond the standard model of physics. Can someone help me solve this puzzle?
Dear Robert, according to Your logic, the observer which falling on the black hole don't observes radiation predicted by Unruh and Hawking, since such observer travels along geodesic line? (the Unruh's radiation is connected with the Hawking's radiation through equivalence principle).
The answer to the question is No. The reason is that, in curved spacetime, the vacuum of any given observer isn't a meaningful concept, because it's not globally defined. That was what Hawking found for black holes and Unruh generalized to any accelerated observer. Hawking radiation is an observer-independent statement at infinity, i.e. the boundary of the spacetime in question,where it implies a steady flux of particles at a given temperature-which is only meaningful there. It's a major current research issue to understand how to map what a free-falling observer measures to what the observer at infinity measures, beyond the classical case.
Robert, but Unruh claims that if we will replace the force acceleration "g = F/m = GM/R^2" on the kinetic acceleration "a=g", we still will observe thermal bath of particles.
Robert, Hawking claimed that the extremally strong gravity field is able to divide virtual particle-antiparticle pairs near black hole horison. These particles can fly out from black hole and should be detected by any external observer - stationary or free falling.
Unruh, who worked with models of vacuum, concluded that any accelerated with respect to vacuum observer must observe a thermal bath of particles with temperature $T=\hbar a/2\pi ck_B$, where $a$ is an acceleration. If $a$ replace by $g=GM/(Rg)^2$, where Rg - is gravitational radius of black hole, we can obtain Hawking temperature for his radiation.
The generalization is as follows: the stationary observer sees inhomogeneous gravity field and observes Hawking radiation, the free-falling observer don't sees gravity field but moving with acceleration, so he observes Unruh radiation, which is equal to Hawking radiation for stationary observer. So, Hawking radiation and Unruh radiation is the same radiation with respect to the different observers - stationary and free-falling observers correspondingly.
As result one can conclude that any acceleration with respect to vacuum and/or an inhomogeneous gravity field can 'produce' thermal bath of particles.
In these considerations what's being assumed is that the particles thus produced don't affect the spacetime: ``backreaction'' is neglected. This is, surely, not the case for Hawking radiation, since the black hole ``loses'' mass to radiation and its area decreases. What this means is that, since the spacetime is thereby changing, a ``free-falling observer'' may not be a valid notion, during the evaporation process. For ``large'' black holes, that evaporate ``slowly'', this notion can remain valid quite a long time, of course.
Indeed-but there are two definitions: in a static spacetime, a free-falling observer, defined as following a time-like geodesic, also, is in a vacuum of particles-and the two notions are equivalent. In a non-static spacetime, however, is it obvious that they remain equivalent?
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