Re: EUREKA!!!!!!!!!! (ha ha ha ha ha ha) >:-)

[Steve VanDevender <stevev@efn.org>]

Kevin C. Houston writes:

 > On Wed, 15 Nov 1995 KellySt@aol.com wrote:
 > 
 > > I still have questions about momentum transfer vs power absorbsion. This
 > > neat conversion of absorbed power and momentum bothers me.  I would think
 > > that converting the power would use up the energy that would generate the
 > > thrust.  Maybe I'm confused?
 > > 
 > 
 > Not at all, if you reflect, you get twice the momentum that you'd get if 
 > you absorbed.  so there is no violation of energy conservation.

This is not strictly true.  In fact, I've been quite dubious of
Kevin's claim that his configuration of solar sails would really
produce deceleration.  In fact, I believe that you can't use a
static sail (that is, one permanently attached to the spacecraft)
to decelerate, as to produce deceleration the reflected photon
must increase in momentum to compensate for the spacecraft
decreasing in momentum; the static sail is tied to the spacecraft
so it cannot change its momentum separately from the spacecraft.
The previously-mentioned "Dragonfly" sail (from Robert L.
Forward's _Flight of the Dragonfly_) accomplishes deceleration by
splitting the sail into a retro-reflecting portion that focuses
energy back to the remaining portion of the sail attached to the
spacecraft; in this case the retro-reflecting sail increases in
momentum to compensate for the spacecraft's decrease in momentum,
and thereby ends up carrying all of the momentum from the beamed
power by the time the spacecraft has come to "rest".

Consider a photon with momenergy [ p, p, 0 ] (that is, energy p,
momentum p in the x direction, no momentum in the y direction),
incident on a mirror with initial momenergy [ m, 0, 0 ] tilted at
an angle theta, such that when theta = 0 the mirror reflects the
photon straight back along its original path, and positive theta
means counterclockwise rotation of the mirror.  After reflection,
the photon has momenergy

[ p, -p * cos(2 * theta), -p * sin(2 * theta) ],

and to conserve momenergy the mirror must then have momenergy

[ m, p * (1 + cos(2 * theta)), p * sin(2 * theta) ].  

(the total system momenergy remains [ m + p, p, 0 ]).

You'll note that the expression (1 + cos(2 * theta)) is always
greater than or equal to 0.  In other words, the x-component of
the mirror's momentum after reflection is always forward, or at
best nil.  So you can steer by tilting the sail, but you can't
slow down.

If you use multiple reflections, then you can analyze the
behavior simply by considering the final photon momentum after
the reflections.

In the case of a double reflection that leaves the photons
travelling in the same direction they originally were (maybe
parallel to their original course) then there is _no_ net change
in spacecraft momentum.  If you change the direction of the
photons, then you must _always_ leave them with a smaller
x-component of momentum than they originally had, and this
results in increased x-momentum for the spacecraft.  You may be
able to decrease the efficiency of the sails as much as you want,
but you will never produce deceleration with multiple-reflection
schemes as long as the sails are attached to the spacecraft and
therefore tied to the spacecraft's momenergy.

A really thorough treatment of multiple reflections would have to
consider both the finite speed of light and the finite speed of
propagation of changes in velocity through the spacecraft
structure; for very high rates of acceleration these would be
significant effects.