Re: Eidetic Anti Matter

Dan Clemmensen (
Mon, 21 Jul 1997 19:08:54 -0400

Nigel Jacob wrote:
> On Sun, 20 Jul 1997, Dan Clemmensen wrote:
> > Unless you find some new laws of physics, your will use the energy
> > to power a reaction drive, sending photons in one direction and
> > your ship in the other. You will not exceed the speed of light.
> > Therefore, you cannot reach Pluto in less than 39 hours (as I recall.)
> >
> > If you intend to use the energy to accelerate to .5c and then
> > decelerate,
> > your matter-antimatter drive will need to to use twice the mass of
> > the ship. That's a lot of antimatter. Production of that amount
> > of antimatter will take several orders of magnitude more energy than
> > the human race has generated in all of history.
> >
> > Conclusion: find some new laws of physics.
> >
> In fact, an anit-matter drive employing equal amounts of reactable
> matter and antimatter is not as efficient as some related alternatives.
> The magnitude of the radiated (non-usable) EM flux would mean lots of
> shielding, thus incurring payload/fuel ratio name just just
> one basic inefficiency of such a drive system. A far more efficient and
> more technically feasible drive would employ a small amount of anitmatter
> with a larger amount of thrust material, such as water. The generated
> thrust of such a drive is very high and the associated payload/fuel ratio
> is also correspondingly high.
> Cheers,
> Nigel.

Yes, I was assuming perfect conversion efficiency, so I should have
stated that twice the mass is a lower bound. Since you won't allow
me perfect efficiency, I'll need even more mass, and so will you.
In theory I can reach near-perfact efficiencies without new
laws of physics. I chose .5c because the relativistic effects are
fairly small, so we can solve the following two equations:
ke=1/2*m*v*v (kenetic energy is half mass times velocity squared)
e=m*c*c (the matter-antimatter conversion of mass to energy)

remembering that we have to accelerate and then decelarate, and that
half of the energy goes onto the reaction mass, and that you must
carry your reaction mass with you.

In fact, I didn't really do the calculation. I just took the result
on faith from Krause, "The Physics of Star Trek".

The point is, even with perfect mass-to-energy conversion and perfect
conversion into kenetic energy, you need a LOT of mass to go