Re: special relativity: accelerated frames (long)

[Michael Bowen <fizzbowen@MINDSPRING.COM>]

"John S. Denker" wrote:
> The canonical homework exercise that goes with this bit of physics is as
> follows:
>
>     Suppose an interstellar spaceship starts from rest, and
>     accelerates such that the passengers feel one Gee (980 gal)
>     for one year.  How fast are they going at the end of the year?
>
> Huge hint:  V=tanh(rho) where rho is called the "rapidity".  Find rho.
>
> The answer is cute and worth remembering.

OK, John, please check or comment on my model here.

My conceptual understanding of rapidity (vaguely remembered from
Taylor and Wheeler) is that it "adds" the way that velocity does in
Newtonian physics. In other words, if you shoot a torpedo forward at
0.8c relative to a starship traveling at 0.4c past a planet, the
velocities do NOT add to give a torpedo velocity of 1.2c relative to
the planet, but the _rapidities_ corresponding to 0.4c and 0.8c do add
to give the _rapidity_ of the torpedo relative to the planet. This can
then be back-solved to give the torpedo's velocity (0.909c). (This is
an alternative approach to the addition-of-velocity formula; its
validity may be thought of as arising from the structural similarity
of the a-of-v formula with that for the tanh of a sum.)

As an approximation, then, I'll let N be the number of seconds in a
year (about 3.2E7) and assume there are N + 1 instantaneously comoving
observers (numbered consecutively from 0 to N), with observer number k
being the comoving observer at the instant for which the spaceship has
been accelerating for k seconds (as measured by a clock traveling with
the ship, which I shall assume is essentially synchronous at time k
with the clock of observer k, due to its modest acceleration). At time
(k + 1) seconds, the spaceship is comoving with observer (k + 1), and
moving at a non-relativistic speed of essentially v = 980 cm/s
relative to observer k, who in turn is moving at v = 980 cm/s relative
to observer (k - 1), and so on. The velocity V (in caps) we are
looking for is that of observer N with respect to observer 0.

If this were a Newtonian problem we would simply add v = 980 cm/s to
itself N times to get this relative velocity, or Nv. A quick
calculation shows, however, that the final result is moderately
relativistic, so this approach does not work. However, the rapidity of
observer (k + 1) relative to observer k is the same as that of
observer k relative to observer (k - 1), and so on, and the rapidities
do add (right?). So one way to get the answer would be to add the
relative rapidity of nearest-neighbor observers to itself N times to
get John's (rho), and then use John's formula to get V. The
nearest-neighbor relative rapidity is close to zero since v = 980 cm/s
is so small compared to c. So we can use the hyperbolic version of the
"small-angle formula" (tanh x ~ x for x ~ 0) and set the relative
rapidity of nearest-neighbor observers roughly equal to v/c. The
rapidity of observer N compared to observer 0 is then Nv/c, which
works out numerically to be very close to unity (within about 3
percent; is this the cute part?). The corresponding velocity V is
about 0.76c.

I cheated a bit here by taking advantage of an ambiguity in John's
question: by whose clocks should the one-year period be measured? To
make it easy on myself I assumed that the year of elapsed time was
spaceship time. Does someone else want to try to figure out if, how,
and by how much the answer would change if we insisted on one year of
earth time?

Note also that I used the term "nearest neighbor observers" in a
semantically peculiar way to mean observers with consecutive values of
k. In this context, there is no implication that "nearest neighbor
observers" are (or were) located physically close to each other, only
that their relative velocity is v.

--MB