Re: Age Correspondences, When Both Travelers Accelerate
- From: Mike Fontenot <mlfasf@xxxxxxxxxxx>
- Date: Tue, 21 Nov 2006 15:22:48 -0700
I've finally finished my study of the "Quasi-Minkowski" procedure for
determining the age correspondences for two travelers who
each accelerate arbitrarily and independently. I suspected that
the procedure might have advantages over the "direct" method
which I have described in previous postings. But I have now
concluded that that hope was ill-founded: it can't work.
The direct method
requires that we determine the intersection of the line of
simultaneity (LOS) of the observer (say, T1) at some instant
in T1's life, with the worldline L2 of the object (say, T2).
My hope was that the quasi-Minkowski (QM) method could eliminate
the need for determining that intersection.
In the direct method, we select an inertial frame (called the
FIRF, for "fixed inertial reference frame"), and plot the
worldlines L1 and L2 with respect to this frame. Call the
axes of the FIRF x0 and t0. We then use x0 for the vertical
axis of the plot, and t0 for the horizontal axis (which is
opposite from the usual convention). So on this diagram,
we plot both L1 and L2.
At any instant t1p in T1's life, there is a "momentarily
stationary inertial reference frame", the MSIRF1(t1p), for
T1. (I.e., at the instant when t1 = t1p, there is an
inertial frame in which T1 is momentarily stationary). Denote
the coordinates for the MSIRF1 as mx1 and mt1.
Likewise, at any instant t2p in T2's life, there is an
MSIRF2(t2p). Denote the coordinates for the MSIRF2 as mx2
and mt2.
For the instant t1p, we can plot the axes mx1 and mt1 on the
(t0, x0) plane. The axis mt1 will make some angle alpha1
with the t0 axis (whose tangent is the speed beta1 of T1 wrt
the FIRF). Positive alpha1 is counter-clockwise for positive
beta1.
The axis mx1 will make the same angle alpha1 with the x0 axis
(positive is clockwise). So the two axes mt1 and mx1, when
plotted on the (t0, x0) plane, will be rotated by equal
amounts toward an imaginary line at +45 degrees on the
(t0, x0) plane (when beta1 > 0). For negative beta1, the
positive mt1 axis is rotated below the positive t0 axis, and
the positive mx1 axis is rotated by the same amount to the left
of the positive x0 axis.
Likewise, we can plot the axes mt2 and mx2 on the same (t0, x0)
plane, and the angles will in this case be equal to the
tangent of beta2.
At t1 = t1p, the line of simultaneity for T1 is a straight line
parallel to the m1x axis, and passing through the point on
the worldline L1 where t1 = t1p. So, in the direct method,
we need to determine the intersection of that line with the
worldline L2.
Now, it is possible to perform a transformation, which rotates
the mt1 axis to horizontal, and the mx1 axis to vertical. On
this new (mt1, mx1) plane, the mx2 and mt2 axes will each be
rotated by some new angle, alpha_new. This is the QM
transformation, which has the desired property that the lines
of simultaneity (t1 = t1p) are now vertical, and it is now
trivial to determine their intersection with any other curve
plotted on this new (mt1, mx1) plane. This is the basis of the
QM method. I had hoped to perform this transformation, at each
small step dt1, and thereby eliminate the need to determine the
LOS intersection that is required in the direct method.
But the fallacy is this: before the transformation can be performed,
both instants t1 = t1p and t2 = t2p in the lives of the travelers
must be specified. Once that is done, then it is indeed true that
the point on the worldline of T2 where t2 = t2p will indeed be
vertically above the point on the horizontal axis where t1 = t1p.
So the determination of the intersection is now trivial. But
the "fly in the ointment" is that we have to already know the
age of t2 which corresponds to the age t1 = t1p, BEFORE we
can do the transformation! I.e., we actually have to KNOW
the intersection in the direct method before we can perform
the transformation. So the transformation is not of any value
in avoiding the determination of the intersection.
This all seems very obvious to me now, but I've been dancing
around this issue, without really understanding it, for
about a month now.
So, the bottom line is that the direct method is (as far as
I know) the only game in town. But that's not really so bad:
the program (called cado2) that I wrote to implement the
direct method works fine, and runs fast enough to be practical
to use. The algorithm I used to determine the intersection
is just "brute-force" and unsophisticated, and not very
efficient, but it apparently is fast enough.
Mike Fontenot
.
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