LAURENCE@SU-CSLI.ARPA (05/21/85)
From: Laurence R Brothers <LAURENCE@SU-CSLI.ARPA> I'm curious: does anyone out there claimm to understand the modern physics concept of "renormalization". As far as I can tell, this involves an equation (extremely simplified) of the form a + I = J where I and J are transfinite. a, however is a relatively normal number which relates to a mathematical feature of mass. Obviously in any normal terms, a is not a quantity which can be solved for, but it IS in renormalization. Admittedly this might not have too much to do with SF, except as it might relate to the basis of a hard SF story, but relativity WAS being discussed.... -Laurence -------
ran@ho95b.UUCP (RANeinast) (05/23/85)
>From: Laurence R Brothers <LAURENCE@SU-CSLI.ARPA> > >I'm curious: does anyone out there claim to understand the modern physics >concept of "renormalization". As far as I can tell, this involves >an equation (extremely simplified) of the form a + I = J where I and >J are transfinite. a, however is a relatively normal number which relates >to a mathematical feature of mass. Obviously in any normal terms, >a is not a quantity which can be solved for, but it IS in renormalization. A simple example of renormalization (even without "modern physics"): Consider a point mass; the gravitational potential is V= -a/r. ["a" is some constant]. If you use this potential to find the potential of an infinite line, you find yourself evaluating the integral I = int from 0 to inf { 1/sqrt(r^2+x^2)} dx, which is infinite. However, you realize that the potential has no real physical significance, just differences between potentials at different places, so you can subtract off a constant without affecting anything. So you replace I with int from 0 to L { 1/sqrt(r^2+x^2)} dx - ln(L), and let L go to infinity. You've now subtracted two infinite numbers to get a finite result, which is V= -b*ln(r), which is the answer you'd have gotten if you'd just used Gauss' law on the line. How does this apply to modern physics? An example is the mass of the electron. The "bare mass" is infinite, but the uncertainty principle allows electron-positron pairs to pop in and out of vacuum (this is also an infinite effect). This effectively shields the original electron, the two infinities cancel, and the measured mass of the electron is perfectly finite. Most physicists believe that this effect is an artifact of the way the calculation is done (as above); if we knew the "correct" way [the equivalent of Gauss' law, above] (not perturbation theory--an approximation), we could do the whole calculation without infinities. -- ". . . and shun the frumious Bandersnatch." Robert Neinast (ihnp4!ho95b!ran) AT&T-Bell Labs
brooks@lll-crg.ARPA (Eugene D. Brooks III) (05/24/85)
> the two infinities cancel, and the measured mass of the electron > is perfectly finite. Most physicists believe that this effect > is an artifact of the way the calculation is done (as above); > if we knew the "correct" way [the equivalent of Gauss' law, above] > (not perturbation theory--an approximation), we could do the whole > calculation without infinities. There are several examples of this in solvable quantum field models. One of them is the case of a very massive Fermion coupled to a Bose field via a Yukawa coupling. The "Bare" interacting particles have infinities when pertubation theory is used. The model can be analytically solved to obtain the "Dressed" physical particles and a finite theory. The result is a free Bose field and Fermions that interact through a static potential. This model is a very good example of renormalization via a solvable theory. A complete chapter on the model can be found in S. Schweber, "Relativistic Quantum Field Theory," around p. 339-351