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*To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Subject*: Re: Engineering Newsletter*From*: Steve VanDevender <stevev@efn.org>*Date*: Tue, 21 Nov 1995 17:28:52 -0800*Cc*: Timothy van der Linden <T.L.G.vanderLinden@student.utwente.nl>, KellySt@aol.com, stevev@efn.org, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl*In-Reply-To*: <Pine.3.89.9511211804.A6072-0100000@maroon.tc.umn.edu>*References*: <199511212031.AA00782@student.utwente.nl><Pine.3.89.9511211804.A6072-0100000@maroon.tc.umn.edu>

Kevin C. Houston writes: > To: Timothy > From: Kevin > > >However, I will in the future use: > > >KE=M * C^2 * (gamma - 1) > > >where > > >gamma=1/SQRT(1-Ve^2) > > > > Great, but I think Steve does not like them... :) > > arrgghh! will you two make up your mind! either figure it out, consult > a prof. or a book or something. and when you do get it right, please > tell me and I'll use that method. <sheesh!> I say that "relativistic mass increase" is a misnomer and that you are better off treating mass as invariant. > > OK, let me rewrite this and tell me if you agree: > > > > incoming outgoing > > ray ray > > \ / > > \ / > > \ / > > a (\/ > > ------------------- mirror (sail?) > > || > > || > > \/ > > resulting momentum > > > > The ray reflects at an angle a, this gives the mirror a momentum of > > 2*p*SIN(a) where p the momentum of the incoming and outgoing photon (this > > assumes the mirror does not move). > > The (kinetic) energy gain of the mirror in this proces is 2*p*c*SIN(a) the > > other part of the energy of the photon is still in the outgoing ray. > > Okay, I see, it does give the same result for angles other than 45 degrees. > but it's cos not sin. Timothy uses a different convention for the angle than I did. His math is correct using his convention. I was measuring a relative to normal of the mirror plane rather than relative to the surface. > Not new at all, tbhe width of a EM wave is the same as it's length. Huh? Why is this even relevant? > That's why an absorbed photon gives a momentum normal to the absorbing > surface. BZZZT! Thank you for playing. A reflected photon transfers momentum to a reflector normal to the reflecting surface. An absorbed photon transfers momentum in the direction and magnitude of the the photon's original momentum. This is the only consistent way to preserve conservation of momentum in both cases. Fundamental principle of relativistic mechanics: The momenergy (vector quantity of energy and momentum of a system) is conserved within the system through all interactions of the system components. If an absorbed photon only transferred momentum normal to the surface of the absorber, then the final momentum total for the system (photon and absorber) would _decrease_. This can't be true. You can't even weasel out of it by saying the momentum goes into extra heat or energy or something; momentum and energy are tallied in separate components. The photon energy goes into raising the heat of the absorber. The momentum goes into changing the velocity of the absorber.

**References**:**Re: Engineering Newsletter***From:*T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)

**Re: Engineering Newsletter***From:*Kevin C Houston <hous0042@maroon.tc.umn.edu>

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