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*To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Subject*: Re: Summarry of the momentum wars and idea.*From*: Steve VanDevender <stevev@efn.org>*Date*: Mon, 4 Dec 1995 01:40:36 -0800*Cc*: Steve VanDevender <stevev@efn.org>, KellySt@aol.com, T.L.G.vanderLinden@student.utwente.nl, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, David@interworld.com*In-Reply-To*: <Pine.3.89.9512022315.A12245-0100000@maroon.tc.umn.edu>*References*: <199512030256.SAA02132@tzadkiel.efn.org><Pine.3.89.9512022315.A12245-0100000@maroon.tc.umn.edu>

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 :-).

**References**:**Re: Summarry of the momentum wars and idea.***From:*Steve VanDevender <stevev@efn.org>

**Re: Summarry of the momentum wars and idea.***From:*Kevin C Houston <hous0042@maroon.tc.umn.edu>

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