Re: Summarry of the momentum wars and idea.

[Steve VanDevender <stevev@efn.org>]

Kevin C. Houston writes:
 > Half formed idea follows: (modifications/analysis welcomed)
 > 
 > 
 >                                   __________
 >                                   |RM inlet
 >                     ship's core   /"/
 > Power from Sol     ______________/"/________
 > ~~~~~~~~~~~~~~~~~~~~~~============"="="="="="="="="=""  exhaust and power 
 > ~~~~~~~~~~~~~~~~~~~~~~============="="="="="="="="="="
 > ~~~~~~~~~~~~~~~~~~~~~~============"="="="="="="="="=""
 >                    ______________  _________
 >                                  \"\
 >                                   \"\
 >                                   |RM inlet
 >                                   
 > 
 > ~ Photons going one way
 > = photons going both directions (canceling? hope not)
 > " Reaction Mass
 > 
 > my question is this:
 > does the ship slow down?

As drawn, I don't think so.

The reason a rocket accelerates is that it reacts some material
to produce pressurized gas in a vessel that is open on one end.
The gas escapes from the opening, and the rocket moves in
reaction because the pressure on the vessel (which is attached to
the rocket) is asymmetric -- constant everywhere except at the
opening.  In microscopic terms all of the molecules that get
forward momentum from the reaction bounce off the forward side of
the vessel and transfer momentum to it.

What you've built is sort of like a rocket that's a big open
tube.  Once the reaction mass is dumped into the tube, it's
accelerated by the maser, but there's no way for it to transfer
momentum with the ship.  A chemical rocket built that way would
just spew flame out each end of the pipe and not move.  Here,
whether the reaction mass absorbs or reflects the laser energy,
it does so independently of the ship.

A simple analogy might be to consider the difference between you
throwing a baseball in zero-g and drifting backwards from the
reaction, or placing a baseball in front of you and having
someone shoot it with a rifle. The baseball goes flying away but
you don't move from where you started.

 > I think it should. (but I've been wrong before)  before the interaction, 
 > the ship and the RM were traveling to the right with some speed V.  after 
 > the interaction, the RM is traveling at a greatly increased speed to the 
 > right, and the photon beam is traveling to the left. (complete reflection 
 >  -- don't ask me how, just assume it for the moment.) 

I'll allow the hypothetical complete reflection, and your
analysis of this part looks correct.  The reaction mass
accelerates but the ship doesn't slow down.

 > Now absorb the photon beam and convert it to electricity.  
 > (the ship should act as though the photons came from Tau Ceti, slowing 
 > even more.) -- again, we either need a magical one-way absorber at the 
 > ~/= interface (~ photons enter from the left, = photons are absorbed from 
 > the right), or a complicated series of reflectors.  or we have to abandon 
 > the photons capture and just let all that lovely energy zing back to earth.
 > 
 > the electricity (if we can capture it) can then be used to power a lineac, 
 > acellerating the plasma stream even more. which definitly would slow the ship
 > 
 > I think this is what Kelly was trying to get at with his plasma mirror, 
 > but this puts the mirror inside the ship.  I realize that I'm talking 
 > about some very very complicated twists and turns, but just answer the 
 > following question.  is this system physically possible?  does it 
 > preserve momenergy, and does the ship slow down.  we can worry about the 
 > merely difficult engineering tasks later.

In principle this is really much like the Dragonfly two-piece
lightsail.  It's OK to reflect the photons backwards and use them
to decelerate as long as you give the forward momentum from the
original forward-moving photons to something, whether it's the
Dragonfly forward sail or the reaction mass you're throwing into
the pipe.

 > here's how I see it breaking down,
 > 
 > 1 photons reflecting off RM:
 > 
 > Momenergy (very dangerous of me to toss a word around that I don't fully 
 > understand.) conserved by acceleration of RM.  Ship slows down a little or 
 > not at all  (photons momentum equals RM momentum and ship stays the 
 > same???  but ship now has less mass  <very confusing>)

My thinking is that you are right that the ship won't slow down.

Momenergy is Taylor and Wheeler's invented term.  Their rationale
for creating the concept seemed to be:

1.  Conservation of energy is a major principle of physics.

2.  Conservation of momentum is a major principle of physics.

3.  Momentum and energy are conserved independently of each
other, so energy can be added as a fourth component to a momentum
vector, and normal vector operations will preserve conservation
of the components.

4.  When using total (rather than kinetic) energy for an object,
there is a geometrically compelling interpretation for the
resulting "momenergy" vector -- it points in the same direction
as the velocity vector of the object, and its Lorentz magnitude
is equal to the mass (in Timothy's terms, "rest mass") of the
object.

They then use momenergy vectors as fundamental items in
relativistic kinematics problems.  Conservation is handled
automatically by requiring that the sum of momenergy vectors of
all the components of an interaction remains the same before and
after any interaction.  In simple problems where you use one
spatial dimension and momenergy vectors are two-dimensional, you
can even draw nice diagrams on paper to show relationships, which
is handy for solving problems.

I write full four-dimensional momenergy vectors as:

[ energy x-momentum y-momentum z-momentum ]

The Lorentz magnitude of this vector is:

sqrt(energy^2 - x-momentum^2 - y-momentum^2 - z-momentum^2)

A more leisurely-paced explanation of momenergy vectors is in
Chapter 7 of Taylor and Wheeler's _Spacetime Physics_.  Have you
bought your copy yet?

 > 2 reflected photons absorbed:
 > Momenergy conserved by ship slowing down.

This is OK too, given your previous assumptions.

 > 3 electricity used to further accelerate RM.  ship slows down even more.

This is actually a very interesting idea.  A small problem is
that accelerating the reaction mass forward means that any light
reflected back from it will be lower in energy because of
doppler-shifting, so you get less energy back to accelerate
further reaction mass.

 > I cannot begin to solve the math showing how much the ship slows down, or 
 > how much energy is required, or how much RM is required.  I'm not even 
 > sure the physical model is correct.
 > 
 > help.

Well, I can at least show you how I'd start laying out the
solution using the methods I'm familar with.

First, let's pick a frame for doing the analysis.  I tend to like
using a frame where the spacecraft is initially at rest.

So in this frame, the ship has momenergy [ s 0 ] (since motion in
only one dimension is necessary for this analysis, I'll use 2-d
vectors), and photons with momenergy [ p p ] are being beamed to
the ship.  This gives us the first major constraint on the
solution -- no matter what happens, the total momenergy after any
interaction will be [ s+p p ].

The intended reaction is to eject some quantity of reaction mass
r into the photon beam, diminishing the ship's mass to s-r.  This
reaction mass also reflects the photons backward (in your ideal
case), and the ship absorbs them.  So now we have a system with
these momenergies:

ship: [ se sp ]
reaction mass: [ re rp ]

We also have the relationships:

The invariant mass of the ship after ejecting the reaction mass
is the magnitude of its momenergy vector:
(s - r)^2 = magnitude [ se sp ] = se^2 - sp^2

Similarly, the invariant mass of the ejected reaction mass is the
magnitude of its momenergy vector:
r^2 = magnitude [ re rp ] = re^2 - rp^2

The momenergy vectors of the system components after the reaction
is the same as the sum of the components before the reaction:
[ se sp ] + [ re rp ] = [ s 0 ] + [ p p ] = [ s+p p ]

It's too late for me to want to work on solving this now, but
hopefully with appropriate juggling several of these variables
can be eliminated and the final result can be expressed in terms
of the quantities we consider known:  s, r, p.

 > Kevin
 > 
 > PS, I do appreciate your kind tutuledge Steve,  I know it must be 
 > frustrating trying to pound knowledge into a head as thick as mine, 
 > especially through such a small bandwidth channel like this.

If I keep this up I'd better get tenure in the LIT Physics
department :-).