Re: starship-design: Re: FTL travel

[Steve VanDevender <stevev@efn.org>]

STAR1SHIP@aol.com writes:
 > In a message dated 3/26/00 9:02:14 PM Pacific Standard Time, stevev@efn.org 
 > writes:
 > 
 > > STAR1SHIP@aol.com writes:
 > >   > I previously pointed out why and how the relativistic rocket equation 
 > MV 
 > >   > times gamma of payload=MV times gamma of exhaust converts to the real 
 > >   > equation MV of exhaust= MV of payload by canceling the gamma factor out.
 > >  
 > >  You can't cancel out the gamma factors because the payload and the fuel
 > >  are travelling at different velocities, hence have different gamma
 > >  factors.
 > 
 > Nonsense- The velocity used is that of the payload wrt the exhaust and the 
 > second gamma velocity is that of the exhaust wrt the payload. The same value. 
 > Do the math the both gamma variables calculate to the same numeric value. 
 > Quit making unsubstaniated claims.

Let me quote back to you what you said earlier:

 > Taking for example the general equation MVof exhaust=MVof payload.
 > to accelerate a given payload mass of 5 tons twice light speed requires I 
 > accelerate 100 tons of exhaust to 1/10 light speed (at 1 g acceleration).
 > Using the equation E(kinetic)=1/2 MV we get the energy required.
 > using the equation E=Mc^2 we get the mass that needs to be converted 
 > providing the energy.

Here you say that the exhaust travels at 0.1 c and the payload travels
at 2 c.  Setting aside for the moment that no normal interpretation of
special relativity allows for something to travel at 2 c, what you
originally said did have the exhaust traveling at a different velocity
than the payload.

More fundamentally, any relativistic analysis requires that you look at
a situation in a single reference frame.  You can pick the reference
frame -- one in which the payload is at rest and the exhaust moving, one
in which the exhaust is at rest and the payload moving, or even one in
which both are moving, but from the viewpoint of an observer who
considers himself to be at rest.  But whatever frame you choose, you
need to do the analysis consistently in that one frame.

Your original statement implies a frame in which the rocket and its fuel
are initially at rest before the rocket is ignited.  In that frame,
after the rocket burns out (which, by your choice of formula, must be
essentially instantaneously, since no consideration is made for a
gradually decreasing fuel mass over an extended burn period), the
payload velocity and the exhaust velocity are by your stipulation not
equal, and therefore you cannot cancel the gamma factors in:

Mp / sqrt(1 - Vp^2) = Me / sqrt(1 - Ve^2)

 > Laws of math - you can multiply both sides of any equation by any variable or 
 > constant and the equation remains equal. -used to solve problems and clear 
 > denominators of fractions and decimals. If a=b and b= c then a=c and a/c=1 
 > making the factor cancelable.

Sure, but you're not canceling equal variables, since Vp and Ve are not
equal.

 > Law of physics-
 > Gamma is a complex variable with mass, velocity, time and spacial dimensions.

Gamma is actually dimensionless.  v has units of distance/time; c has
units of distance/time, hence (v / c) is unitless, and so
1/sqrt(1 - v^2/c^2) is also unitless.

 > Momentum (p)is conserved.

At least we can agree on this.

 > always P=MV

Not in the relativistic case; p = m * v is only the Newtonian
approximation.

 > Thus momentum of the payoad=momentum of exhaust and gammas cancel.

The payload of the momentum does equal the momentum of the exhaust in
the center-of-momentum frame, but in that case the gammas don't cancel.

 > Problem solved

If I had pulled the kind of wild handwaving and sloppy thinking you are
trying to pull here, I would have flunked all my math and physics
classes.