lew@ihlpa.UUCP (Lew Mammel, Jr.) (03/10/85)
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
gv@hou2e.UUCP (A.VANNUCCI) (03/13/85)
> 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. The fact is that the frame of reference of the lightsail is not an inertial frame *precisely* because the lightsail gains energy. Energy is, obviously, not conserved in a non-inertial frame. In the inertial frame tangent to the lightsail just prior to a photon reflection the reversal in the momentum of the photon is accompanied by a small increase in the momentum of the lightsail, such that both energy and momentum are conserved. G. Vannucci AT&T Bell Labs, Holmdel