From: George Matsas Date: Wed, 3 Jun 2020 15:13:08 -0300 Hi Bill, Thank you for your message. I do not disagree. I guess that, at least partially, what you say is reflected in the fact that the particle emission rate from a detector with arbitrarily small energy gap (19) differs from the corresponding one for a structureless scalar source (20) (by a factor 2). (Still, one can obtain the right result for a structureless scalar source by making \Delta E=0 (structureless condition) before the integral (18) is performed.) Recovering my cautionary remark 6, it may have been a bad idea to try to rephrase the original papers in terms of the more familiar detector language. After all, in Ref. [1] the same conclusion of item 1 is reached using structureless electric charges -- although it is technically more elaborated. Best wishes, George George E. A. Matsas Professor of Physics Institute for Theoretical Physics/ S=C3=A3o Paulo State University S=C3=A3o Paulo State Academy of Science https://urldefense.com/v3/__https://professores.ift.unesp.br/george.matsas/= Em 02/06/2020 23:12, Stephen A. Fulling escreveu: > 1) From: Bill Unruh I guess the reason I have problems with this line of argument is that the essence of the detector is that there must be a change of state of the internal degrees of freedom of the detector for it to be a detector, and the coupling of the field is solely to this change in the internal degree of freedom. The ultimate longer term behaviour of the detector is a scatterer of the em radiation comined with these changes in internal state of the detector. For a charged particle there are no internal states, and thus no change in those internal states. One could for example imagine the internal states are a harmonic oscillator. As the freq goes to zero, there is no spontaneous emission, so the system has no desire to prefer one state (the ground state) over any other, and it will randomly wander through that infinite internal set of states as it interacts with the field. but even in that limit, the emission of a photon is always associated with a change in state. The emission of two photons (entangled) is associated with the detector in the same state it started from. For a charged object, there is no scattering. It acts as a direct source of radiation. There are no internal states. There is no change in those internal states associated with the the interaction. > William G. Unruh __| Canadian Institute for|____ Tel: +1(604)822-3273 > Physics&Astronomy _|___ Advanced Research _|____ Fax: +1(604)822-5324 > UBC, Vancouver,BC _|_ Program in Cosmology |____ unruh@physics.ubc.ca > Canada V6T 1Z1 ____|____ and Gravity ______|_ > =C2=A0https://urldefense.com/v3/__http://www.theory.physics.ubc.ca =========================================================== From: Don Page Date: Wed, 3 Jun 2020 15:23:24 -0600 Dear George and Steve (to circulate to all), Thanks for your point about the acceleration being very long, say given by the dimensionless quantity a T >> 1. The switching rate alpha gives two other dimensionless ratios, alpha/Delta E and alpha/a. Presumably you need a T >> f(alpha/Delta E, alpha/a) for some function f of these latter two dimensionless ratios. Do you know what f is? In the limit that not only a T >> 1 but also a >> Delta E, the late time von Neumann entropy of the 2-state detector should be very near ln 2, with the field then also having that same von Neumann entropy, which is very small for the huge number of field degrees of freedom. Could this partly help solve some of the issues Bill raised about the difference between an Unruh-DeWitt detector and a point charged particle with no internal degrees of freedom, since in comparison with the field degrees of freedom, the UD detector has so very few degrees of freedom? Another issue I did not mention yesterday is that in Point 3 you wrote, "Because zero-energy Rindler photons concentrate on the horizon, (physical) Rindler observers do not have any access to them. This harmonizes the fact that uniformly accelerated charges radiate for inertial observers but do not for coacelerating (Rindler) ones, who only see a distorted Coulomb field." But then how do the Rindler Unruh-DeWitt detectors absorb or emit the zero-energy Rindler photons, if Rindler observers moving along similar constant-acceleration worldlines do not have any access to them? Best wishes, Don >From: George Matsas Date: Wed, 3 Jun 2020 14:24:55 -0300 Thank you for your message. Concerning item 21, indeed, this a nontrivial point. I am using the results of paper [2] where an explicit calculation is done using a particular switching on/off function. My physical intuition is that although alpha is larger than the acceleration, a, the second dominates the detector transition *rate* because the time interval, T, along which the detector stays accelerated is much larger than any other time scale of the problem. (Loosely speaking, while the switching on/off contributes much to the transition probability but only during the switching on/off time-lapse, the acceleration a contributes little but along a time interval which is larger than any other time scale in the problem.) Concerning your second point, this is an interesting one. Indeed, the calculations are at the lowest order tree level -- which leads to the standard transition rate (18) proportional to the squared coupling constant, which is the same as the usual Larmor radiation (20). Andre (Landulfo) will discuss it in more detail and show explicitly how the classical (Larmor) radiation result (20) from a uniformly accelerated structureless source can be entirely built from the zero-energy Rindler modes -- see Ref. [6]. Briefly speaking, in the presence of a uniformly accelerated source (and preparing the field in the Minkowski vacuum in the asymptotic past), a nonperturbative calculation shows that the field will appear as a coherent multiparticle state in the asymptotic future [6]. We are still looking on questions as these ones. (Gabriel Cozzella, Andre, and I are analyzing Lin and Hu's nonperturbative calculations (https://journals.aps.org/prd/abstract/10.1103/PhysRevD.73.124018), where the source is a harmonic oscillator.) Best wishes, George George E. A. Matsas Professor of Physiscs Institute for Theoretical Physics/ S=C3=A3o Paulo State University S=C3=A3o Paulo State Academy of Science https://urldefense.com/v3/__https://professores.ift.unesp.br/george.matsas/= __;!!KwNVnqRv!RGeXwQasAvv2fqBDAFJhpy__L8nC3QCzR_j9nV4Il8v5NhCE7rNj025gLuszq= boKQ2w$=20 Em 02/06/2020 13:38, Don Page escreveu: > Dear Stephen (to circulate to all) and George, > > Thanks for this paper.=C2=A0 I have a couple of questions: > > In Item 21 and Eq. (16), since the switch on/off rate alpha is large=20 > compared with the acceleration a, why don't the probabilities depend=20 > more strongly upon alpha than upon a? > > On another point, the formulas for the excitation and de-excitation=20 > probabilities given seem to be lowest nontrivial order (quadratic) in=20 > the amplitudes given by Eq. (5).=C2=A0 So how can this apply to the case = in=20 > which the Unruh detector accelerates for a very long time so that it=20 > reaches thermal equilibrium and emits and absorbs a huge number of=20 > Rindler particles?=C2=A0 I would have thought that one would need to go t= o=20 > a very high order in the amplitudes. > > Best wishes, > > Don > ============================================================ Date: Wed, 3 Jun 2020 11:59:43 -0500 (CDT) From: Benito Juarez-Aubry Thanks for the paper. I have found this very interesting, especially given the comments on experimental verification of the Unruh effect (or it's shadow a la Plato's cave, cf. Paragraph 37). Perhaps touching on Bill Unruh's comment, in fact in the original paper linking Bremsstrahlung to absorption or emmission of Rindler particles, [1] A. Higuchi, G. E. A. Matsas and D. Sudarsky, Bremsstrahlung and Fulling-Davies-Unruh thermal bath, Phys. Rev. D 45, R3308 (1992) a central part of the discussion is regularising the point charge as a slowly oscillating dipole with frequency E, and the radiation acceleration and transition rates of photons coincide in the limit E -> 0 and as the dipole "collapses" to a single charge. I found this to be analogous to the E->0 limit in the UDW detector discussion shared by George Matsas. In fact, in my view the criticism of Unruh could be extended to [1] as in "we shouldn't think of a point charge as the limit of a dipole", but I think that the result is so interesting that I wouldn't want to dismiss it immediately. I actually think that the UDW detector is interesting in that it sort of makes everything fully quantum, as opposed to the classical accelerated charge vs quantum photons discussion in [1], and perhaps it's a stepping stone towards a full QED description of the problem of acceleration radiation. Some comments: 1. Is alpha being small (fast switching) actually relevant in condition (16)? I would expect that provided that T is very, very (very!) large the switching is totally irrelevant. Or is it because the computation leading to eq. (17) is done for simplicity in the sharp-switching limit, so that c(\tau) is a Heaviside step function (times c_0)? 2. Do you have any interpretation of why one should perform "limit + integration" and not "integration + limit" in order to arrive to eq. (20) from eq. (18)? This is somewhat surprising. Usually when we look at asymptotic regimes (E -> 0 being an example) we have to do so at the level of the "final expression" (this is why we use convergence theorems when we can't integrate, so that "limit + integration" can be shown to be equivalent to the desired "integration + limit"), so for me integrate-then-take-the-limit is more natural, but that seems to give the wrong answer. 3. All this discussion on the, let's say, "Larmor shadow of the Unruh effect" got me thinking about whether one can think of "shadows" of analogue Hawking radiation. Have you thought about this? Best, Benito