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*To*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Subject*: Re: Energy:momentum ratio*From*: Steve VanDevender <stevev@efn.org>*Date*: Mon, 10 Jun 1996 08:58:49 -0700*Cc*: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com, neill@foda.math.usu.edu, 101765.2200@compuserve.com, MLEN3097@Mercury.GC.PeachNet.EDU*In-Reply-To*: <199606101036.AA12688@student.utwente.nl>*References*: <199606101036.AA12688@student.utwente.nl>

Timothy van der Linden writes: > The derivation is as follows: > > The momentum of a particle with velocity v : p=gamma m v > The energy of a particle with velocity v : E=m c^2 (gamma-1) That's kinetic energy, not total energy. For many purposes kinetic energy is not particularly useful. Most importantly, when I say p/E is v, I use E to mean total energy (m*c^2*gamma in conventional units). > p/E = v gamma / (c^2 (gamma-1)) > > Simplifying the above gives: > > p/E = 1/c {v/(c-Sqrt[c^2-v^2])} > > All units are according to S.I. > > (When mentioning energy, I only mean kinetic energy) If you want to use kinetic energy as a quantity in your relativistic dynamics problems, be prepared for your analysis to get _a lot_ more complicated. I think I went into some detail in my original derivation about how I analyzed the relativistic dynamics of a ship reacting and ejecting a quantity of fuel. You might want to take a look at it again. The most unlikely result of your derivation is your claim that lower exhaust velocities are better. Does that mean that zero exhaust velocity is best, or is there instead an optimal velocity that is non-zero? > Timothy

**References**:**Re: Energy:momentum ratio***From:*T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)

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