This is why time has a minus sign in SR. (I believe the usual way this informally is put is that the space-traveller "trades space for time".)

On 8 March 2014 13:26, Jesse Mazer <laserma...@gmail.com> wrote: > > On Fri, Mar 7, 2014 at 7:20 PM, Edgar L. Owen <edgaro...@att.net> wrote: > >> Jesse, >> >> Do you understand why the world line that is depicted as LONGER in the >> typical world line diagram is ACTUALLY SHORTER? >> >> E.g. in your diagram do you understand why even though A's world line >> looks longer than C's world line, it is ACTUALLY SHORTER? >> >> Edgar >> > > Are you actually reading my posts carefully all the way through, or just > skimming them or something? I spent a whole extended section of my post > discussing just this point, read it again: > > 'It is true that if you just look at the spatial lengths of each path on > the diagram, the ratio between the spatial lengths doesn't actually match > up with the ratio between the proper times that would be calculated using > relativity. If you use any Cartesian spatial coordinate system to draw x-y > axes on the diagram, then you can use this coordinate system to assign x > and y coordinates to the endpoints of any straight blue segment, x1 and y1 > for one endpoint and x2 and y2 for the other, and then calculate the > spatial length of that segment using the Pythagorean theorem: > squareroot[(y2 - y1)^2 + (x2 - x1)^2]. Note that you ADD the squares of the > two terms in parentheses when calculating spatial length, but my earlier > equation showed that you SUBTRACT the square of the two terms in > parentheses when calculating proper time, which explains why this sort of > spatial path length on a spacetime diagram can be misleading. For example, > in spatial terms a straight line is the SHORTEST path between two points, > but in spacetime a straight (constant-velocity) worldline is the one with > the LARGEST proper time between points. > > Nevertheless, the math for calculating the invariant spatial path length > using a Cartesian coordinate system is closely analogous to the math for > calculating the invariant proper time using an inertial frame. The diagrams > show the spatial length of the paths being different despite identical red > acceleration segments, and this remains true if you actually calculate > proper time, even though in terms of proper times C > B > A which is the > opposite of how it works with spatial lengths.' > > > > >> >> >> >> >> >> On Friday, March 7, 2014 5:15:57 PM UTC-5, jessem wrote: >>> >>> >>> >>> >>> On Fri, Mar 7, 2014 at 4:02 PM, Edgar L. Owen <edga...@att.net> wrote: >>> >>> Jesse, >>> >>> Finally hopefully getting a minute to respond to at least some of your >>> posts. >>> >>> I'm looking at the two 2 world line diagram on your website and I would >>> argue that the world lines of A and B are exactly the SAME LENGTH due to >>> the identical accelerations of A and B rather than different lengths as you >>> claim. >>> >>> The length of a world line is the PROPER TIME along that world line. >>> Thus the length of a world line is INVARIANT. It is the length of the world >>> line according to its proper clock and NOT the length according to C's >>> clock which is what this diagram shows. >>> >>> >>> I don't understand what you mean by "the length according to C's >>> clock"--are you just talking about the numbers on the vertical time axis, >>> 2000-2020? That axis represents the coordinate time in C's rest frame, and >>> obviously the coordinate time between "2000" at the bottom of the diagram >>> and "2020" at the top is 20 years regardless of what path you're talking >>> about, so I don't see how it makes sense to call this the "length" of any >>> particular path. But you can also use C's >>> ... >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Everything List" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to everything-list+unsubscr...@googlegroups.com. >> To post to this group, send email to everything-list@googlegroups.com. >> Visit this group at http://groups.google.com/group/everything-list. >> For more options, visit https://groups.google.com/d/optout. >> > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com. > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.