Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84 exptools; site ihlpa.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!ihnp4!ihlpa!lew From: lew@ihlpa.UUCP (Lew Mammel, Jr.) Newsgroups: net.space,net.physics Subject: reflection of photons from a lightsail Message-ID: <141@ihlpa.UUCP> Date: Wed, 6-Mar-85 00:50:24 EST Article-I.D.: ihlpa.141 Posted: Wed Mar 6 00:50:24 1985 Date-Received: Thu, 7-Mar-85 03:49:59 EST Distribution: net Organization: AT&T Bell Laboratories Lines: 41 Xref: watmath net.space:3670 net.physics:2205 I just spent an inordinate amount of time digging through my "archive" ( a box ) of netnews hardcopy and finally found my article of Oct 1983 on the relativistic lightsail problem. Quoting myself: We can calculate the momentum of the reflected photons in the following way. First, calculate the momentum of the incident photons in the sail frame. Second, reverse the sign of the momentum (reflection from sail which is stationary in this frame.) Third, calculate the momentum of the reflected photons in the rest frame. If p is the initial momentum, these steps yield: 1) gamma*(1-beta)*p /* redshift */ 2) -gamma*(1-beta)*p /* reflect */ 3) -gamma^2*(1-beta)^2*p /* red shift again */ ... this gives delta(v) = 2*p/(1+beta) [end of quote] I went on to evaluate the equation of motion. I found that the time scale of the problem was given by T = (m*c^2) / (2*I * p*c) m = mass of ship p = momentum of photon I = photons per second striking sail ... that is, the rest energy of the ship divided by twice the impinging power. A beam of 1 megawatt/meter2 and a sail of 1 gram/meter2 gives T = 1e8 sec, or about 3 years. My solution gave this table of times required to reach the given speeds: v/c t/T .5 1.065 .9 15.316 .95 43.048 .99 474.26 .999 14917.6 Lew Mammel, Jr. ihnp4!ihlpa!lew