Date: Tue, 28 Jul 2020 09:37:43 -0500 (CDT) From: "Stephen A. Fulling" To: Acceleration Radiation Community: ; Subject: Four comments by three authors ============================================================== 1. From: Don Page Date: Sun, 26 Jul 2020 16:33:02 -0600 Subject: Comment on one of Bill's comments See below for my response to Bill's comment on my P.S. > 2. From: Don Page > > Date: Thu, 23 Jul 2020 12:45:09 -0600 > > ... > > > > P.S. There was some mention in a comment I received today of a possible > > divergence if the acceleration of the charge worldline changed suddenly. > > However, the 4D form F = (e/r^3) N ^ [(1 - a.N) V + r a] does not show > any > > divergence in F, but just a discontinuity between different finite > values of > > F if a jumps suddenly. > > I was basing my comment on the belief that The vector potential had a > discontinuity in that case which would give a delta function in the EM > field. > For example the field at the charge goes as the time derivatative of a > which > is a delta function if the acceleration is a step function. > But the Lienard-Wiechert 4-vector potential (in units with c = 4 pi epsilon_0 = 1) is A = e V/(V.N) = eV/r, where r = V.N with V the 4-velocity of the charge worldline at the emission event and N the null vector from the emission event to the event where the vector potential is evaluated and the EM field is to be observed. So long as V is continuous along the charge worldline, A is continuous, though if the 4-acceleration a of the charge worldline jumps discontinuously, the retarded EM Faraday 2-form F = (e/r^3) N ^ [(1 - a.N) V + r a] that one gets from the Lienard-Wiechert potential A can also jump discontinuously. However, so long as a remains finite, so does F, and hence so does the stress-energy tensor of the EM field. So I don't see any need to require a to change smoothly; one could have the charge worldline have a = 0 for t < t_0 so that one has a Coulomb electric field for t < t_0 (boosted if the charge 4-velocity V is not purely timelike in the chosen coordinate system), and then suddenly have a change to a nonzero 4-vector (of course remaining orthogonal to the charge 4-velocity V), with the stress-energy tensor having a jump across the light cone of the charge event where its 4-acceleration jumped, but remaining finite. Best wishes, Don =========================================================== 2. From: Steve Fulling (fulling@math.tamu.edu) Date: July 28, 2020 Subject: Comment on Don's comment above I don't dispute anything that Don said, but to avoid confusion on the part of casual readers, I want to emphasize that Don is talking about a charge density with uniformly spatially bounded support. The retarded field of the eternally accelerated charge, as I described in my presentation (following Bondi and Gold and later authors) does have a discontinuous A on the past horizon, hence a distributional F there, and hence a stress tensor that includes terms proportional to the square of a delta function. Of course, that last fact is disturbing. Boulware (pp. 184-185) argues that the delta-squared term is "formally conserved by itself" and the remainder is rigorously conserved when the (delta times theta) terms are properly interpreted. Of course, efforts to understand these formulas by taking limits of less singular problems are all to the good. As I mentioned, Boulware made a good start on that on pp. 178-180. As others have mentioned, one reason for wanting to avoid such regularizations is to avoid contaminating the radiation field by transients from the switching-on-and-off, and as still others have mentioned, one can hope to isolate the transient effects by taking the epoch of uniform acceleration to be very long with its length T as a parameter, and/or by exploiting the fixed speed of light (Huygens' principle). Huygens doesn't hold for, say, a massive scalar field or a Proca field, nor in odd dimensions, or even for a massless field in some curved backgrounds. So I think it is worthwhile to establish the "physical" correctness of the idealized EM solution (even though we can never perform such an experiment) so that similar distributional techniques can be employed with some confidence in these other situations. =============================================================== 3. From: Don Page Date: Mon, 27 Jul 2020 11:59:57 -0600 Subject: Static observers at finite distance can see zero net outward energy flux from an accelerating charge People have discussed starting with a finite acceleration time for the charge and then taking the limit as that time goes to infinity. They have also discussed taking a finite-size charge and then taking the limit as the size goes to zero. Now I wish to consider the case in which the observers measuring the radiation are on an everlasting static finite-radius sphere and then take the limit as the radius R of the sphere is taken to infinity. Assuming sufficient time-reversal symmetry for the worldline of the charge, I seem to be getting zero total outward energy flux through the static sphere of observers for all R, and hence zero energy in the limit that R is taken to infinity. Use Minkowski coordinates, and let the location of the sphere of observers be at x^2 + y^2 + z^2 = R^2 for all -\infty < t < \infty. Assume that the charge worldline has time-reversal symmetry, t(tau) = -t(-tau), x(tau) = x(-tau), y(tau) = y(-tau), z(tau) = z(-tau). Although I don't think this is necessary, for simplicity assume that for |tau| sufficiently large, the charge worldline has its r^2 = x^2 + y^2 + z^2 >> R^2, the square of the radius of the sphere of observers, so at very early and at very late times the charge is far outside the sphere of observers, and there is only a finite range of tau, say |tau| < tau_0, for which r^2 < R^2 so that the charge is inside the sphere. Now consider the EM field of the charge. The retarded + advanced field is time symmetric and hence gives zero total energy flux (integrated over the whole worldtube of the persisting sphere of observers, so it has units of energy, not energy per area). The retarded - advanced field is a homogeneous solution to the Maxwell equations that has a flux passing inward and then outward through the observer sphere worldtube, so it also gives zero total flux. Hence the retarded field by itself also gives zero total flux through the observer sphere worldtube. If one then takes the limit as R is taken to infinity, since for each finite R the total energy flux is zero, it remains zero when one takes the limit R -> \infty. Of course, what differs in this description from the usual assumption that the observer sphere worldtube is the time-asymmetric 2-sphere cross affine parameter on scri-plus is that here one takes a static sphere of observers at finite R, so that as the charge approaches the observer sphere from the outside, it can radiate energy into the sphere (negative net outward energy flux when just integrated over this part of the observer sphere worldtube). (It also carries Coulomb electric field energy into the sphere, but that is canceled by the Coulomb energy carried out when the charge leaves the sphere in this time-symmetric motion in which the charge leaves with the same speed and hence gamma factor relative to the observers at which it entered. For a point charge, the Coulomb energy entering and leaving are both infinite, but these infinities can be taken to cancel, as one can get, if necessary, by starting with a finite-size charge and then after calculating the zero difference taking the limit of a point charge.) So it seems to be important for getting a nonzero total energy flux through a spherical worldtube of observers not to have this worldtube be a static sphere at finite radius R, but to have it be at scri-plus, at infinite R and also moving at the speed of light inward rather than being static. Of course, then for a point charge undergoing constant acceleration forever, since it goes to scri-plus with infinite gamma factor there, it takes an infinite Coulomb energy to scri-plus, even if one makes the charge have a finite size in its rest frame. Since one is not counting the energy flux through an observer worldtube at scri-minus, in this case there is no canceling of the outgoing infinite Coulomb energy by any incoming infinite Coulomb energy. I can imagine that it might be a bit ambiguous how to define precisely what the infinite energy is being carried to scri-plus by the Coulomb energy of the charge that has by then been boosted to infinite gamma factor, so what one gets for the radiation energy after subtracting off this infinite part might also be ambiguous, whether it is still infinite, whether it is finite but nonzero (which seems implausible; what would set the scale for this finite energy?), or whether it is zero. So unless one can unambiguously split up the infinite energy into a Coulomb part and a radiation part, it might be inherently ambiguous whether a charge with constant acceleration forever radiates zero or infinite energy to scri-plus. But for time-symmetric charge motion, the total energy flux through a finite-radius but everlasting static spherical worldtube of observers seems to be zero. Best wishes, Don =============================================================== 4. From: Robert Wald Date: Mon, 27 Jul 2020 20:50:01 -0500 Subject: more comments It has gotten way too complicated for me to try to follow all the back and forth comments, but let me make a few remarks based on the discussion that, in principle, should put everything to rest (although I realize that, in practice, it undoubtedly won't). 1) Electromagnetic energy and radiation: The stress-energy properties of the electromagnetic field are given by its stress-energy tensor T_{\mu \nu}. The definition of "energy" requires the specification of a Killing vector field K^\mu. If one has specified such a Killing field, then one can define an energy current $J_\mu = -T_{\mu \nu} K^\nu$. One can integrate $-J_\mu n^\mu$ over a Cauchy surface to get the total energy or one can integrate it over a timelike or null surface to get an energy flux. If people use the word "radiation" (in such sentences as "a charge does/does not emit radiation") to mean "energy flux" they should say what Killing field they are using and what surface they are integrating over. If people are using "radiation" in some other sense, they should define what they mean. 2) Retarded solution for a charged particle: If, in Minkowski spacetime, the particle worldline $\gamma$ has the property that its future $I^+(\gamma)$ is the entire spacetime, then the retarded solution (= the Lienard-Wiechert solution) is globally well defined (and smooth away from the particle worldline). The retarded solution is uniquely determined by having vanishing initial data on past null infinity, scri^-. The general asymptotically flat solution is obtained by prescribing arbitrary data at scri^-. It is given by the Lienard-Wiechert solution plus a term involving the initial data on the sphere at scri^- corresponding to the intersection of the past light cone of the observation event with scri^-. 3) Idealized cases such as the charge that accelerates uniformly forever: There is no reason, a priori, why there should be a well defined "retarded solution" in this case. There is certainly a well defined retarded solution to the chronological future of the particle worldline and there is certainly a well defined (and vanishing) retarded solution outside the closure of this region, but I see no reason, a priori, why these solutions should (distributionally) extend through the null plane boundary. It's nice that it can be distributionally extended, but I don't see why this idealized solution is important for anything anyway. If the past light cone of the observation event intersects the worldline during a period of uniform acceleration, the solution is given by the uniform acceleration Lienard-Wiechert solution (plus an integral over the sphere at scri^- if one wishes to allow nontrivial incoming radiation). It makes no difference what the charge was doing before this time. ==============================================================