RE: starship-design: Relativity

[Steve VanDevender <stevev@efn.org>]

L. Parker writes:
 > I was in Chapter 5, specifically, Figure 5-8. It seems illogical that the
 > worldline indicated could approach zero, but that does seem to be what is
 > represented.
 > 
 > Lee

It's not so much that the interval along the curved worldline approaches
zero; it's that the curved worldline has _less_ interval than the
straight one, or that a clock carried along the curved worldline
experiences less time than a clock carried along the straight one, even
though both start and end with the same events.  Curved worldlines
correspond to paths of objects undergoing acceleration.

In general a more curved worldline has less interval than a less curved
one.  So yes, the higher the acceleration and the closer to c the the
object gets during its round trip, the less time it experiences.

Mathematically, this counterintuitive result is the consequence of the
non-Euclidean geometry of spacetime; drawn on a sheet of paper with
(locally) Euclidean geometry, it's initially hard to get used to the
idea that the longer line is the shorter interval.  Note the equation
that they show with figure 5-8; it's the basis of doing a non-Euclidean
line integral (if you're familiar with calculus -- otherwise I'm
probably about to lose you).

Let's say you have a graph with the axes x and t, and a parameterized
function f(t') = (x(t'), t(t')) (that is, a function that returns an
(x, t) coordinate given the parameter t').  In Euclidean geometry, the
length of the curve swept out by f(t') is

integral (sqrt((d x(t')/dt')^2 + (d t(t')/dt')^2)) dt'

(pardon the ASCII math).  Note that the function being integrated really
represents the length of an infinitesimal segment of the curve, i.e.
sqrt((dx)^2 + (dt)^2).

But spacetime geometry is different.  There the spacetime interval along
an infinitesimal segment of a curve is sqrt((dt)^2 - (dx)^2), and so the
"interval integral" for the path along a worldline is

integral (sqrt((d t(t')/dt)^2 - (d x(t')/dt')^2)) dt'

The instaneous speed along the worldline at a particular point is
(d x(t'))/(d t(t')), so the higher the speed at that point, the less
the contribution to the integral.

That's why I started with the somewhat unconventional notation of
labeling the axes x and t instead of x and y, to match the notation used
in spacetime diagrams.  t' represents the proper time along the
worldline, which is a relativistic invariant, and x(t') and t(t')
represent the spacetime coordinates of a point in a frame at which a
clock carried along the worldline reads t'.  You can actually represent
any worldline as a parameterized function of t', and the "interval
integral" produces the same result in all frames for the same worldline,
even though the functions x(t') and t(t') depend on the frame the
observations are being taken in.