Several parts tonight, starting with a message from myself. I've done my best to repair the corruption introduced by multiple transfers between mail/operating systems, but some of it has me defeated. SAF ================================================================ From S. A. Fulling, June 4, 2020 These comments respond to the messages of Bill Unruh and George Matsas in yesterday's collection, https://www.math.tamu.edu/~fulling/ar/matsas1_comments603.txt . (The texts of those messages are also visible below the message of Eduardo Martin-Martinez, below.) 0) I also have been slightly bothered because the detector (which I'll continue to take as a 2-state one for the moment) still has internal structure in the limit of 0 energy splitting. In addition to that qualitative problem, there are actually three delicate factors of 2 floating around (counting the one in the endpoint delta function in Matsas paragraph 30). 1) At first I thought that [2 states] => [twice as much radiation], but that is wrong. An ordinary electron also has 2 states (spin), but that does not affect the Larmor formula. The difference, as Bill pointed out, is that in the detector case, emission is associated with a change of internal state. That should not change the Larmor formula either. Nevertheless, one can't say that an accelerating charge is "just" a limiting case of a detector, since the latter must have internal multiplicity. This would affect calculations of entropy (cf. Don Page's last message of yesterday), especially for a large number of particles that are secretly Unruh-DeWitt detectors (not to mention those that are Unruh-classic detectors with infinitely many internal harmonic-oscilator states!). 2) George says (par. 18) "we must eventually trace out on the detector internal state." I agree, but doesn't that mean we should AVERAGE over initial states (or take a linear combination of them (as density matrices) with positive coefficients (a,b) that add to 1), rather than adding them? That would mean that the two surviving terms in par. 23 need to be divided by 2 (if the two states are equally populated), or (more generally) the second term in (17) should have been multiplied by a + b = 1 instead of by 2. Then we would get the expected answer (20) already in (19) without need for the endpoint delta function argument in par. 30. 3) That is not the end of the story, however. If we take the energy splitting to 0 at the beginning, then the argument in par. 23 is no longer valid: all 4 terms contribute equally, not just 2. There is no averaging this time. (I remember a mantra from graduate QM, "Average over initial states and sum over final states.") So we're back where we started, with an extra factor of 2. But now the endpoint delta argument in par. 30 seems sounder than before, and we can argue the 2 away (probably -- this is the kind of distributional argument that is not convincing unless you know the right answer beforehand). 4) When Delta-E is literally 0, the unperturbed states are degenerate and the interaction Hamiltonian can be diagonalized (go to a basis where Delta-H is proportional to sigma_z instead of sigma_x). Then the interactimg theory is solved exactly! There are two decoupled states differing slightly in energy, and there are no transitions between them. But in these states the radiation field is correlated with the detector state; in each case the field state is some coherent superposition of the vacuum and a one-photon state. So we have abandoned the "retarded" initial condition by passing to an eigenstate of the full Hamiltonian. ================================================================ From george.matsas@unesp.br Thu Jun 04 10:52:14 2020 Subject: Response to Benito Dear Benito, Thank you for reading the manuscript. *Concerning your main point*: As you say, in Ref. [1] we begin considering an oscillating dipole A-B (ensuring charge conservation). This adds two scales to the problem: 1.L associated with how far A and B are from each other and 2.E associated with how fast the charge oscillates. At the end of the calculations, we make E->0 and L->infty. By making E->0, we freeze the oscillation. By making L->infty, we send B to the infinity and make it inertial. Because inertial charges in the Minkowski vacuum do not radiate, this procedure eliminates any eventual influence from B. In summary, after taking E->0 and L-> infty the results refer to one single uniformly accelerated charge. So, this is not a coincidence that after this procedure our accelerated frame results were in agreement with the inertial frame results, where a uniformly accelerated charge is assumed from scratch. Now, for those who do not find it convincing enough, let me address to Ref. [8]. In Ref. [8] we analyze in the inertial and uniformly accelerated frames a single electric charge uniformly accelerated in the z direction and rotating along the xy plane and find explicitly that the emission of a Minkowski photon corresponds to the absorption or emission of a Rindler photon -- quite expected. Now, the results of Ref. [1] can be straightforwardly recovered by vanishing the charge rotation in the xy plane -- in this sense, Ref. [1] is a particular case of Ref. [8] where no dipole is used. We believed that this should put to rest any doubts on Ref.[1]. *Concerning your observations:* A. Inequality (16) is assumed to allow one to hold on Ref. [2] results. Assuming c(\tau)proportional to a Heaviside function is not an option because switching on/off detectors instantaneously leads to divergent responses. The switching function must be at least C^0 (continuous). B. The physical interpretation is the following. By doing the integration first, I am making sure that I am considering a detector all the way to the end. Thus, by taking the limit \Delta E -> 0 *after that* we get the emission rate for a detector with arbitrarily small energy gap -- but still a detector. On the other hand, by taking \Delta E ->0 before the integral, we are degenerating the internal degrees of freedom making the detector a structureless source. This is why we obtain in this way the right result (20) -- right in the sense that a straightforward calculation assuming a structureless source from scratch gives it [5]. C. Concerning my Plato cave comment, as I see it, any experiment only tests some aspect of a theory. Analog experiments are not an exception. (Of course not all shadows are equally interesting.) Best wishes, George 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 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 ============================================================== From: Don Page Date: Thu, 4 Jun 2020 14:33:03 -0600 Dear George and Steve (to circulate to all), Thanks for your comments. I'll probably have to think more about your first (main) point before I understand it. If for the UD detectors the absorption and induced emission do not vanish, in what sense do other observers static in the Rindler coordinates not have access to these zero-frequency modes? On the second point, I would imagine that if alpha were large enough, the transitions in the detector caused by it would dominate over those caused by a. So presumably for the effects of alpha not to dominate, you must require some inequality where alpha/a is much _less_ than some other dimensionless quantity, such as perhaps some power of a T. However, none of your 4 inequalities in your Eq. (16) give an inequality in that direction (since they only have alpha^{-1} much less than other quantities, or alpha much _more_ than other quantities, so I would think there must be some missing requirement that you have not spelled out. Best wishes, Don On Thu, Jun 4, 2020 at 1:15 PM George Matsas wrote: Dear Don and Steve (to circulate to all), Let me begin with your last question: how do UD 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? Thank you for the question. This is the main point I think. Indeed, the absorption and emission probabilities of a single zero-energy Rindler photon in the *Rindler vacuum* vanish. However, this is compensated by the Unruh thermal bath. This is so because the thermal factor 1/(e^(omega_R/T_U) -1) diverges for omega_R ->0. In other words, spontaneous emission vanishes because of what you said but absorption and induced emission do not. In Ref. [1] we deal with this indeterminacy "0 x Infty" by replacing the uniformly accelerated charge by an oscillating dipole A-B (where everything is well defined) and erasing B at the end of the calculations to end up with a single uniformly accelerated charge. But let me mention that we can recover the same results as in Ref. [1] from Ref. [8] by vanishing the rotation of the charge -- in this sense, Ref. [1] is a particular case of Ref. [8] where no oscillating dipole is introduced. Concerning your first point, we can recast the 4 inequalities (16) in dimensionless forms, e.g., aT >>1>> a/alpha,andaT >> a/Delta E >>a/alpha but I do not see how it could help. I am probably missing the point? Concerning your comments on the von Neumann entropy, I shall think of it. At this point the best answer I have is -- I do not know. Best wishes, George Em 03/06/2020 18:23, Don Page escreveu: > 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.=C2=A0 Presumably you need a T >> f(alpha/Delta E, > alpha/a) for some function f of these latter two dimensionless > ratios.=C2=A0 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.=C2= =A0 > 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: Eduardo Martin-Martinez Date: Thu, 4 Jun 2020 15:29:22 -0400 Jose de Ramon (my PhD student) and I have put together this note responding to George's original note. I am not sure how this works, but I assume that we send it to you and that you distribute it. Is this right? Thank you very much in advance. Look forward to the community's opinion. All the best, Eduardo Editor's note: PLEASE READ EDUARDO'S NOTE AT https://www.math.tamu.edu/~fulling/ar/matsas1_martram.pdf On 2020-06-03 9:35 p.m., Stephen A. Fulling wrote: > 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 >