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Energy:momentum ratio

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

There's one problem here.  momentum and energy in conventional units
don't have the same units.  Efficiency is by nature dimensionless.
mass/mass is dimensionless.

 > 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)

I don't know where you got this "v / (c - sqrt(c^2 - v^2))" bit.  How
did you derive this?  Why do you think it's correct?

 > 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.

There's another fundamental flaw in your argument.  You can't convert
energy to momentum.  Momentum is momentum and energy is energy and one
cannot be converted into the other.

When you take something that has momentum, and you get energy out of it
by slowing it down, you are getting the extra energy it has to
compensate for its momentum (m^2 = E^2 - p^2).  If m is constant, then
when p increases E also increases.

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

In the formalism I'm used to, where v is dimensionless, p/E is exactly
the same as v.  If you want to use conventional units then p * c^2 / E
is exactly the same as v.

Frankly, I don't know where you're coming from on this and your units
don't work out in any useful way (momentum / energy as a measure of
efficiency has units seconds per meter?  What the hell is that?).
Efficiency is dimensionless; it is (energy out) / (energy in).