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From: gwyn@brl-smoke.ARPA (Doug Gwyn )
Newsgroups: sci.physics
Subject: Re: A Question
Message-ID: <5025@brl-smoke.ARPA>
Date: Wed, 29-Oct-86 20:31:40 EST
Article-I.D.: brl-smok.5025
Posted: Wed Oct 29 20:31:40 1986
Date-Received: Thu, 30-Oct-86 07:26:59 EST
References: <230@sri-arpa.ARPA> <572@epimass.UUCP> <2182@ecsvax.UUCP> <8597@sun.uucp> <1388@trwrb.UUCP>
Reply-To: gwyn@brl.arpa (Doug Gwyn (VLD/VMB) <gwyn>)
Organization: Ballistic Research Lab (BRL), APG, MD.
Lines: 69

In article <1388@trwrb.UUCP> galins@trwrb.UUCP (Joseph E. Galins) writes:
>In other words, with a constant acceleration eventually you would be going
>at a speed of 'c' with no more acceleration therefore losing the 'gravity'
>feeling.

No, acceleration at a constant rate "forever" will never exceed the
speed of light (measured with respect to the "stationary" reference frame
from which you started).  (For simplicity let's keep the discussion at
the special-relativistic level and leave cosmology out of it.)

Using primes (') for the accelerating object and no-primes-attached for
the stationary reference frame:

	dv' = a' * dt'		for acceleration a' felt by object

	Beta(t'+dt') = Beta(t') VADD dv'/c

where Beta is the velocity of the traveler measured in the stationary
frame divided by c, and VADD is relativistic velocity-combination

	Beta[A wrt C] = Beta[A wrt B] VADD Beta[B wrt C]
		      =	(Beta[A wrt B] + Beta[B wrt C])
			/ (1 + Beta[A wrt B] * Beta[B wrt C])

One also needs to take into account the relationship between time
measures between the two frames ("time dilation")

	dt' = dt / Gamma

where

	Gamma = sqrt( 1 - Beta^2 )

Combining all this, one gets

	Beta(t'+dt') - Beta(t') = a' * dt'
				  * (1 - Beta(t'+dt') * Beta(t')) / c

In the limit dt' -> 0 this becomes

	dBeta(t') = a' * (1 - Beta(t')^2) * dt'

which (using the initial condition Beta(0)==0) integrates to


	Beta(t') = tanh( a' * t' / c )

[This asymptotically approaches 1, i.e. speed of light, thereby proving
the claim made in my first sentence.]

To finish the analysis:

	dx(t') = v(t') * dt

	       = sinh( a' * t' / c ) * c * dt'

To compute the distance D traveled in ship-board time T', when half
the time is spent decelerating at -a', integrate dx from 0 to D/2
(corresponding to t' from 0 to T'/2; Symmetry! Is The Way Things Have
To Be -- Jane Siberry) to obtain

	D(a',T') = 2 * c^2 * (cosh( a' * T' / (2 * c) ) - 1) / a'

Using light-years, gees, and years for units,

	D(a',T') = 1.938 * (cosh( 0.516 * a' * T' ) - 1) / a'

This is the formula I used to calculate travel distance in my other
posting.