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. So to calculate the length of A's and B's world lines in C's frame (which this diagram represents) we must take the apparent lengths as shown from C's frame view on the diagram, and SHORTEN each section by the apparent slowing of ITS CLOCK relative to C's CLOCK. In other words, the proper time LENGTHS of A's and B's world lines will NOT be as they appear in this diagram which displays their apparent length's relative to C's proper clock. To get the actual length we have to use the readings of A's and B's clock and shorten their apparent lengths by that amount. When we do this all the blue segments of A's and B's world lines become parallel to C's and thus add no length to A's or B's world lines. This is what we would expect since the pure NON-accelerated relative motion of the blue segments doesn't add length to a world line. So when we subtract the apparent length differences of the blue lines all we are left with is the red ones which are equal. Thus the actual LENGTHS of A's and B's world lines are equal. And the only effects which add length to world lines are in fact accelerations as I claimed. The point is that the TRAJECTORIES in spacetime of world lines from some frame like C's in this diagram do NOT properly represent the invariant LENGTHS of those world lines. Because to get the invariant proper time length we must shorten those trajectories by the apparent clock slowing along it to get the actual proper clock interval from start to finish. So when we do this we find that the different LENGTHS of world lines between any two spacetime points are due ONLY TO ACCELERATIONS OR GRAVITATION as I previously stated. Do you agree? Edgar On Thursday, March 6, 2014 12:01:53 PM UTC-5, jessem wrote: > > > > On Thu, Mar 6, 2014 at 11:02 AM, Edgar L. Owen <edga...@att.net<javascript:> > > wrote: > >> Liz, >> >> Sure, but aren't the different lengths of world lines due only to >> acceleration and gravitational effects? So aren't you saying the same thing >> I was? >> >> Isn't that correct my little Trollette? (Note I wouldn't have included >> this except in response to your own Troll obsession.) >> >> Anyway let's please put our Troll references aside and give me an honest >> scientific answer for a change if you can... OK? >> >> It would be nice to get an answer from Brent or Jesse as well if they >> care to chime in...... >> > > > In the case of the traditional twin paradox where one accelerates between > meetings while the other does not, the one that accelerates always has the > greater path length through spacetime, so in this case they are logically > equivalent. But you can have a case in SR (no gravity) where two observers > have identical accelerations (i.e. each acceleration lasts the same > interval of proper time and involves the same proper acceleration > throughout this interval), but because different proper times elapse > *between* these accelerations, they end up with worldlines with different > path lengths between their meetings (and thus different elapsed aging)...in > an online discussion a while ago someone drew a diagram of such a case that > I saved on my website: > > http://www.jessemazer.com/images/tripletparadox.jpg > > In this example A and B have identical red acceleration phases, but A will > have aged less than B when they reunite (you can ignore the worldline of C, > who is inertial and naturally ages more than either of them). > > You can also have cases in SR where twin A accelerates "more" than B > (defined in terms of the amount of proper time spent accelerating, or the > value of the proper acceleration experienced during this time, or both), > but B has aged less than A when they reunite, rather than vice versa. As > always the correct aging is calculated by looking at the overall path > through spacetime in some coordinate system, and calculating its "length" > (proper time) with an equation that's analogous to the one you'd use to > calculate the spatial length of a path on a 2D plane. > > Jesse > -- 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.