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Energy:momentum ratio

To Steve,

>Timothy van der Linden writes:
> > Yes, I believe this pressure thing was something I imagined, and later it
> > seemed that the Bussard engine did not work that way.
> > The main reason for that pressure thing was to even out the velocities, this
> > way we would not get very fast and very slow particles all together.
> > Particles with high velocity have a worse momentum:energy ratio (a new
> > term?). This means that if you have some energy and want to make the most
> > velocity (momentum) from it, you get the most of it if you use low exhaust
> > velocities. Unfortunately this also means that more mass is needed, which is
> > not preferrable.
>This statement bugs me because it is completely contradictory to
>something I worked out a while ago and that you seemed to agree with,
>which is that higher exhaust velocities are best, and the ideal case is
>turning all your fuel into zero-mass photons moving at the speed of

Yes, that was a long time ago, that's why some time ago (2 months) I asked
you if you could find that letter. Since you couldn't find it I started
digging... deep, finally I found it, but now I lost it.... So dug it up
again, Sept 1, 1995 that's the date of your letter.

There you say:

  Choice 3:  The fabled Photon Rocket, exhaust velocity r = 1,  f = 0.268

  Assuming you could make a machine that eats matter and produces a
  stream of pure photons (probably not physically impossible, but
  still very difficult) then you get the best fuel efficiency of
  all, about 3 parts fuel for each part payload.

  (note: f = fraction of mass that can be payload)

Your definition of fuel-efficiency is completely different from mine
(momentum:energy-ratio), your f is the same as the ship:fuel-ratio.

>You do want to get all your exhaust products moving backwards, which
>implies confining the fuel reaction and reflecting any forward-moving
>products backwards.  You're absolutely going to lose energy to heat if
>you try to equalize the exhaust velocity.  In any case a gas at a
>particular temperature doesn't have a uniform set of particle
>velocities; it's still a statistical distribution about a mean that is
>characteristic of the gas temperature.

Yes, I'm aware of this heat-loss problem, it was a theoretical model, of
which I in the beginning had not thought enough of the practical problems.


P.S. This is something I derived some time ago:

Momentum/Energy (p/E) ratio:

for photons:    1/c

for mass   :    1/c {v/(c-Sqrt[c^2-v^2])} if you know the velocity
             or 1/c Sqrt[U(U+2 m c^2)] if you know mass and kinetic energy

where {v/(c-Sqrt[c^2-v^2])} is always bigger than 1
(for v->c it goes to 1, for v->0 it goes to infinity)

So if you have some energy and want to get the most momentum of it you could
best use mass, the efficiency is always better.

Note, while the p/E ratio for mass may decrease for increasing v, this does
not mean that the total momentum decreases.