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Re: Engineering Newsletter

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

 > > 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