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.physics Subject: Compton effect & lightsail Message-ID: <145@ihlpa.UUCP> Date: Sat, 9-Mar-85 21:31:39 EST Article-I.D.: ihlpa.145 Posted: Sat Mar 9 21:31:39 1985 Date-Received: Sun, 10-Mar-85 08:10:51 EST Distribution: net Organization: AT&T Bell Laboratories Lines: 29 Mike Augeri posted a derivation of the momentum transfer to a lightsail which I claim is all wet. He based it on the Compton effect, but I don't see why that should apply here. Specular (mirror like) reflection is all that is necessary - as long as you are in the mirror (or lightsail) frame of reference. Also, please note that Mike's derivation took no account of the speed of the lightsail. I think it bothers some people that the photon doesn't lose energy in the lightsail's rest frame. How can the lightsail gain if the photon doesn't lose? Let's resort to a classical analogy to explain this. Suppose we are accelerating an aircraft carrier (in space!) by shooting BBs at its deck. In the ship's rest frame we can assume no loss of energy by the BBs. This is because the energy gained by the ship is: M/2 * dv * dv where M is the ship's mass and dv is its incremental speed. In the "stationary" or shooter's frame, the ship gains energy: M * v * dv + M/2 * dv * dv so when we integrate we can ignore the second order term. Hence when calculating in the ship's frame we can ignore the transfer of energy. This seems paradoxical, but I believe I'm on solid (certainly familiar!) ground here. Lew Mammel, Jr. ihnp4!ihlpa!lew