space@mit-mc (03/02/85)
From: LANTZ@RUTGERS.ARPA If the energy transfered to the sail results in a shift in the wavelength of the photon, is it true that a sail wich is moving faster (producing greater red-shift) receives more energy from each photon? I would guess that this is not the case due to factors having to do with relative velocity of the sail and the observer, but I am not able to put it all together. Brian Lantz (Lantz@Rutgers) -------
space@mit-mc (03/02/85)
From: Rick McGeer (on an aaa-60-s) <mcgeer%ucbkim@Berkeley> Sorry, my message was quite incorrect. As another correspondent has pointed out, it is not the wavelength of the photon that is affected, but rather the vector. Since momentum is a vector quantity, not scalar, the total net momentum transfer to the sail is 2P, where P is the momentum of the entire set of photons impacting the sail. To see how this works, remember that the *vector* momentum is conserved; to a first approximation, we imagine the sail to be travelling in a linear direction x, with 0 momentum in this reference frame. We then imagine a solid photon wavefront with aggregate momentum P travelling in the positive x axis. The photon rebounds in a perfect elastic collision along the x axis... Total Momentum before collision: P (photons) + 0 (sail) Total Momentum following collision: -P (photons) + x (sail) equating and solving, we get x = 2P. (1) Of course, this is idealized. The total momentum transfer to the sail can be obtained by integrating the momenta of the photons before and after the collision. This is related to wavelength by the DeBroglie formula: mv = hf, where h is Planck's constant and f = 1/lambda. The wavelength will have lengthed by Compton's formula: lambda' = lambda + h/mc (1 - cos theta) where m is the mass of the particle that the photon collided with, and theta is the angle of the deflection... In our idealized case, theta = pi, hence (assuming the entire mass of the sail is involved in a collision), we get (note I'm assuming coherent light): lambda' = lambda + 2 h / mc (2) writing lambda as l, the total momentum of the photon wavefront post-collision is: h/(l+2h/mc) hence the total momenta of the photon post-collision is: P[1-2h/(clm+2h)] (3) as a result, there's a net loss of momentum transfer from (1) due to the (trivial) redshift noted in (2). The total net momentum transfer to the sail is then: 2P[1-h/(clm+2h)] Bottom line: yes, as I said yesterday, there *is* a redshifting effect. However, its net effect on the momentum of the light sail is *negative*, not positive, and it is trivial. Sorry about that. Hope this clears everything up. [NB -- the effect here is the *maximum* loss due to the Compton effect...] Rick.