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Okay, I give up ( well, not exactly ;)
- To: Kevin C Houston <email@example.com>
- Subject: Okay, I give up ( well, not exactly ;)
- From: Steve VanDevender <firstname.lastname@example.org>
- Date: Sun, 26 Nov 1995 16:25:36 -0800
- Cc: Steve VanDevender <email@example.com>, Timothy van der Linden <T.L.G.vanderLinden@student.utwente.nl>, KellySt@aol.com, firstname.lastname@example.org, RUSSESS@cellpro.cellpro.com, email@example.com, firstname.lastname@example.org, David Levine <David@InterWorld.com>
- In-Reply-To: <Pine.3.89.9511261243.A6291email@example.com>
- References: <199511260653.WAA09366@tzadkiel.efn.org><Pine.3.89.9511261243.A6291firstname.lastname@example.org>
Kevin C. Houston writes:
> > The only way to cancel out a quantity of momentum is
> > to accept it from another source; you cannot take a single source
> > of momentum and use one part of it to cancel another part of it.
> Agreed that the way i proposed won't work, why can't a single source be
> decomposed and used to cancel itself? It can be done with light, a
> single beam can be split, half the beam can have it's phase delayed by
> 1/2 wavelength, and then recombined with the other half of the beam, the
> net result is _nothing_ no beam. they destructively interfere. Since
> any particle or energy can be considered a wave, (debroglie wave for
> momentum) why can't this same trick be used?
No matter what, the momentum has to go somewhere. You can't
split a momentum vector in half and turn one of the halves around
without creating another momentum vector of magnitude and
direction equal to the original to preserve conservation.
If you use a mirror to reflect half of the light back on itself,
then you get all the momentum of the original beam -- the mirror
(and whatever it's attached to) acquires 2 * (1/2 * p) = p
momentum for reflecting half of the beam backwards.
If you take half the beam and delay it by half a wavelength
before retransmitting it forward to combine with the other half
of the beam, you don't get the momentum, but you also don't get
any of the energy. Unfortunately I'm at a bit of a loss to say
where the momentum goes in this case.
I also don't think that out-of-phase photons annihilate when they
meet. In the region that the out-of-phase photons overlap, they
cancel, but they can't just disappear forever.
> > "I canna change the laws of physics, Captain."
> > -- Chief Engineer Scott
> "any science, sufficiently advanced, is indistinguishable from magic"
> -- Arthur C Clarke
Any science, sufficiently advanced, still cannot violate its own
> so here's what I need:
> a simple easy to use, relativisticlly correct formula that
> tells me how much energy I need to accelerate a given exhaust
> mass to a given speed.
As far as I know, that can be done with the standard relativistic
kinetic energy formula:
KE = m * c^2 * ((1 / sqrt(1 - v^2 / c^2)) - 1)
> I'll be using Me=G*(Ms+Mf) * gamma/(Ve *C)
> Ve is exhaust Velocity expressed as a fraction of C.
> Me is Exhaust Mass.
> Ms is ship's Mass.
> Mf is Reaction Mass.
> G is ship's acceleration.
> gamma is SQRT(1 - Ve^2) I know Tim likes gamma= 1/sqrt(1-Ve^2), but then
> he divides by it instead of multipling.
> all mass is invarient or rest mass.
> Ve is with respect to the ship, and I'm thinking it'll be in the .80-.9?
> perecent of C range.
Whoa. Where did this come from? How is this related to your
You should know that _every_ text I've read says that gamma is
1 / sqrt(1 - v^2), so your use is unconventional and may confuse
Could you at least give some information about how you derived
this? I can say right off that it doesn't look right because it
doesn't have consistent units; the right-hand side evaluates to
units of kg/s.
> every time I try to calculate this, I get flamed for using the
> wrong formula. Timothy gave me one, but then said that you
> (Steve) had problems with it. so I thought I'd head off any
> potential problems by asking for the right formula ahead of
I don't remember when I might have said this or what formula it
was about any more.
I'll get back to you on the formula question. I can adapt some
of the work I did before, but I think it would be more useful if
I cast it in terms of reaction mass and payload mass rather than