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Re: starship-design: Re: FTL travel

In a message dated 4/7/00 12:48:35 AM Pacific Daylight Time, stevev@efn.org 

> STAR1SHIP@aol.com writes:
>   > In a message dated 4/6/00 11:18:40 PM Pacific Daylight Time, stevev@efn.
> org 
>   > writes:
>   > 
>   > >  K = m * ((1 / sqrt(1 - v^2/c^2)) - 1)
>   > >  
>   > >  K = kinetic energy
>   > >  m = object mass
>   > >  v = object velocity
>   > >  c = speed of light
>   > >  
>   > >  K = 1/2 * m * v^2 is only a low-speed approximation. 
>   > 
>   > 1/10 C is low speed so the K calculated is near exact. The relativistic 
>   > effects are insignificant though not zero.
>  You're claiming that a smaller mass with an equal amount of kinetic
>  energy will go faster than c.  That's not what will happen; putting that
>  amount of kinetic energy into the smaller mass makes it go closer to,
>  but not faster than, c.

You are so close but still you get no cigar. I claim there are two different 
velocities (Vreal and Vrel. to be calculated for the payload with one 
velocity (Vrel.) no faster than C as you claim. Focus and concentrate only on 
the problem I gave.

Chose say payload mass of 2 tons and the Vp calculates to 1/2 light speed.
We know that this is so because momentum is conserved as MpVp=Pp
and Pe=MeVe and since Pp=Pe then the calculation is considered consistent 
with universal law as momentum is conserved. 

This conservation occurs with any chosen mass only with the Vreal and not the 
relativistic velocity(Vrel.); the only one you considered and calculated to 
be .89C unless you also transform the values Ve,Me and Mp. If your complete 
transforms are correct then momentum is conserved as Perel.=Pprel. Or Pe'=Pp' 
with prime understood to be relativistic or rel.. Note! subscript consistency 
truly is difficult with a typewriter or keyboard without 1/2 space carriage 
returns. Boolean algebra
even is prone to type errors even by me. Focus and concentrate please only on 
the problem given and make no conclusions like this proves such and such 
until all calculations are made and any errors (yours and mine a given) are 

>   > > Relativistic
>   > >  kinetic energy grows without bound as an object approaches the speed 
> of
>   > >  light.  
>   > 
>   > I agree totally. Real kinetic energy does not.
>  Relativistic kinetic energy _is_ real.

You misunderstand the subscript real. Both velocities real and rel. are valid 
calculations and measurable depending on the frame of reference. Think of it 
like velocity of propellant given as 1/10 C. Of course this is understood to 
be an average velocity as some mass parts are slow and some are fast and so 
can be designated Vavg to calculate.

Vreal=distance rocket traveled/ship time
Vrel=distance rocket traveled/earth time  

>  Newtonian physics is merely a
>  low-speed approximation to relativistic physics; the approximation
>  breaks down when you start dealing with velocities that are significant
>  fractions of c. 

Einsteinian physics does not throw out Newtonian physics but relies on them 
to transform measurements to Einsteinian values and so they can be 
transformed back to Newtonian physics. Transforms are called that because 
they go both ways and both ways are valid depending on the frame of reference 
being observer or observed. 

> If you want calculations that are accurate at any
>  velocity and not just small ones, you need to use the relativistic
>  formula.

A point you need to consider is relativistic formula are correct not because 
of any accuracy. They near Newtonian physics when the part of the gamma 
graphed curve E Vs V can be considered and treated as linear and close to the 
x axis. When the curve becomes close to the c asymptote rapidly rising toward 
infinity (no boundry).

There the effects are more pronounced and the narrow range of velocities are 
commonly called relativistic velocities. Focus and concentrate as this is not 
to be confused with velocities relativistic. 

>   > To be more precise you cannot see the rocket traveling at twice light 
> speed 
>   > from the below viewpoint of earth. Such velocities relativistic are 
> limited 
>   > by the equation you give to below C. The velocity I calculated was 
> velocity 
>   > real as an observer at rest on earth was not part of the problem and I 
> did 
>   > not ask what velocity relative to earth was obtained.
>  No observer sees the rocket traveling faster than c, not even one on the
>  rocket.

The first part of your statement is true. The second is theory 
(unsubstantiated claim) stated as fact. 

Now I will focus and concentrate on the problem given and the proof you 
failed to do even when I told you how to do it. Conserving e-mail paper as 
you keep erasing the stated problem is not a valid excuse for you not solving 
it. Focus as I prove it.

The rocket at twice light speed travels between two stars a distance of 4 
light years.
The time onboard the ship measures two years. Using Einstein's time dilation 
formula for the given masses, two years ship time calculates to be 4.44.. 
years earth time (recalled from memory).

Dividing 4 light years by 4.44.... years and the velocity with respect to 
earth is observed on earth to be as you calculated as .89c (or .9c as I 
recall calculating).

Label .89C as Velocity rel.
Label  2C   as Velocity real.

Therefore an observer on the ship looks at the 4 light years distance between 
the stars he traveled and then looks at his calendar watch and knowing time 
he took to travel the distance (two years) calculates his velocity to be two 

Vrel is valid,
Vreal is valid,
Both are valid,
Your unsubstantiated claim "No observer sees the rocket traveling faster than 
c (true), not even one on the rocket (theory)" is proven invalid.
and (at last) Your unsubstantiated claim (jumping to conclusions)"Energy 
relativistic approaches infinity(unbound) as Vrel approaches C (a true 
statement) then nothing can exceed light speed (theory)" is proven invalid.

Lighting my cigar ;=)'