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*To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Subject*: Re: Engineering Newsletter*From*: Steve VanDevender <stevev@efn.org>*Date*: Sat, 25 Nov 1995 22:53:57 -0800*Cc*: Steve VanDevender <stevev@efn.org>, Timothy van der Linden <T.L.G.vanderLinden@student.utwente.nl>, KellySt@aol.com, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl*In-Reply-To*: <Pine.3.89.9511252041.A4280-0100000@maroon.tc.umn.edu>*References*: <199511260108.RAA08463@tzadkiel.efn.org><Pine.3.89.9511252041.A4280-0100000@maroon.tc.umn.edu>

Kevin C. Houston writes: > > We'll see. You must either absorb the photon and thereby absorb > > its momentum, or reflect the photon and get momentum to balance > > the change in direction of the photon. The only way to not get > > any momentum change is to either not interact with the photon or > > reflect it so as to leave it travelling in the same direction it > > came with the same energy. > > no, first the photon is reflected at an angle, this dumps some of the > momentum into the perpendicular directionthe photon leaves the first sail > element (reflection) and is absorbed in the second, inner element. as > far as that element is concerned, the photon came from the _front_ of the > ship. Look, I'm not doing this to dump on your parade, but it's a very simple check that I'm doing on your proposed solution to verify that it satisfies the laws of physics. We are looking at a system consisting of some quantity of photons with total momenergy [ p p 0 0 ] and a spacecraft with momenergy [ m 0 0 0 ], as seen in the frame where the spacecraft is initially at rest. No matter what, whether or not these two components interact, then the final total momenergy of all the resulting components must sum to [ m+p p 0 0 ]; in other words, the system has net forward momentum p always and forever. This is fundamental and inescapable. If you start out with a big box containing the photons and the spacecraft, then that box always has the same forward momentum for as long as all the original components remain in the box and nothing new is put in the box. Even worse for your argument, you are specifying that the photons are entirely absorbed by the spacecraft, meaning that the only component that is left after the interaction is the spacecraft, which must then have all the forward momentum. I know that my argument barely even uses math and must seem too simple to refute all the work you've put into this, but your solution fails to conserve momentum and is therefore physically invalid. > > If the system consisting of the photon beam and the spaceship > > plus sail assembly doesn't have the same momentum after the > > interaction as before, then you'll have to go back to the drawing > > board. Perhaps after a brief refresher course in physics. > > it does have the same momentum, just that some of that momentum has been > made to cancel out. equal but opposite momentums cancel out I think what your analysis is failing to account for, at some level, is that momentum is a vector quantity. Your think you have found a way to turn around part of the momentum and pit it against itself. Unfortunately that can't satisfy conservation of momentum. Not just the magnitude of the total system momentum, but the direction, must be conserved. You can't "dump momentum in the perpendicular direction.". Any forward momentum that you take away from the beam _must_ be transferred to the ship, whether the beam is reflected or absorbed, no matter the angle of the reflecting or absorbing surface; it cannot be transferred to momentum in any other direction. The structural loading caused by the beam does not absorb momentum continuously, because momentum implies motion -- unless you unhook the reflector or cut it into pieces, the reflector cannot accept momentum separately from the ship because it cannot move relative to the ship; a tiny amount of momentum goes into stretching the ship's structure when the beam is first turned on, and accounts for no more absorption after that. The only way to cancel out a quantity of momentum is to accept it from another source; you cannot take a single source of momentum and use one part of it to cancel another part of it. Again, let's start with what you've got -- photons with momenergy [ p p 0 0 ], and a spacecraft with momenergy [ m 0 0 0 ], for a total system momenergy [ m+p p 0 0 ]. When the photons bounce off the conical reflector tilted at an angle a, it changes the total photon momenergy from [ p p 0 0 ] to [ p*cos(2*a) -p*cos(2*a) 0 0 ]. The spacecraft then changes momenergy to [ m+p*(1-cos(2*a)) p*(1+cos(2*a)) 0 0 ]. The total system momenergy is still [ m+p p 0 0 ]. You have pitted the photon beam against itself, because the beam has been decollimated and individual photons now have sideways momentum that sums to zero for the entire beam, but that doesn't mean that the spacecraft doesn't gain momentum and energy to conserve the system momenergy [ m+p p 0 0 ]. Now the photons with momenergy [ p*cos(2*a) -p*cos(2*a) 0 0 ] travel back and hit the absorber (which need not be tilted or conical itself; it could just be a long column down the axis of the conical reflector). They transfer their momenergy into the collector, so the spacecraft momenergy changes to [ m+p p 0 0 ]. The spacecraft has net forward momentum. Momenergy has been conserved at all times throughout. There is no way to absorb the photons without absorbing their momentum. "I canna change the laws of physics, Captain." -- Chief Engineer Scott

**Follow-Ups**:**Okay, I give up ( well, not exactly ;)***From:*Kevin C Houston <hous0042@maroon.tc.umn.edu>

**References**:**Re: Engineering Newsletter***From:*Steve VanDevender <stevev@efn.org>

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

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