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*To*: "L. Parker" <lparker@cacaphony.net>*Subject*: RE: starship-design: Relativity*From*: Steve VanDevender <stevev@efn.org>*Date*: Tue, 11 Jan 2000 20:17:58 -0800 (PST)*Cc*: <starship-design@lists.uoregon.edu>*In-Reply-To*: <000801bf5cae$4b396220$0401a8c0@broadsword>*References*: <14459.54198.620583.617471@darkwing.uoregon.edu><000801bf5cae$4b396220$0401a8c0@broadsword>*Reply-To*: Steve VanDevender <stevev@efn.org>*Sender*: owner-starship-design@lists.uoregon.edu

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.

**Follow-Ups**:**RE: starship-design: Relativity***From:*"L. Parker" <lparker@cacaphony.net>

**References**:**RE: starship-design: Relativity***From:*Steve VanDevender <stevev@darkwing.uoregon.edu>

**RE: starship-design: Relativity***From:*"L. Parker" <lparker@cacaphony.net>

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