On Fri, Mar 7, 2014 at 4:02 PM, Edgar L. Owen <edgaro...@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 rest frame to assign x and t coordinates to the endpoints of any straight blue segment, x1 and t1 for one endpoint and x2 and t2 for the other, and then C can calculate the proper time along that segment as squareroot[(t2 - t1)^2 - (x2 - x1)^2] and get the correct INVARIANT answer (note that I am using units of light-years and years where c=1 so it doesn't appear in the equation, otherwise the second term in the square root would have to be (1/c^2)*(x2 - x1)^2). What the diagram is trying to show is that even though the different paths have identical red acceleration curves, they have different SPATIAL lengths, i.e. the length you'd measure if you printed out the diagram and laid a flexible cloth tape measure along each path to measure the distance ALONG THE PATH between the point at the bottom of the diagram where the paths diverge and the point at the top where they rejoin. 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. If you assign time coordinates to the beginning and end of each acceleration phase, and you specify the proper acceleration involved, then you can calculate the proper time along elapsed on each worldline during both the acceleration phases (using the relativistic rocket equations given at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html ) as well as the proper time during the constant-velocity phases (using the method I mentioned above with squareroot[(t2 - t1)^2 - (x2 - x1)^2] for each segment). If you do this, you do find that in a detailed numerical version of the scenario in the diagram, A ELAPSES LESS TOTAL PROPER TIME THAN B DESPITE HAVING IDENTICAL ACCELERATIONS. I can give the detailed calculations using the relativistic rocket equations if you want, or you can just take my word for it. > > 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. > Yes, that would be another way to calculate proper time along the blue constant-velocity segments: just take the times t2 and t1 of the beginning and end of each segment in C's frame, and multiply by the time dilation factor which depends on the speed v in C's frame during that segment: proper time = (t2 - t1)*sqrt[1 - v^2] This is actually mathematically equivalent to the method I mentioned, since for a constant velocity segment velocity is just distance/time, so we have: v = (x2 - x1)/(t2 - t1) So the above can be rewritten as: proper time = (t2 - t1)*sqrt[1 - (x2 - x1)^2/(t2 - t1)^2] And (t2 - t1) = sqrt[(t2 - t1)^2], so with that substitution the equation is: proper time = sqrt[(t2 - t1)^2] * sqrt[1 - (x2 - x1)^2/(t2 - t1)^2] Then if you multiply together the terms inside each square root you get: proper time = sqrt[(t2 - t1)^2 - (x2 - x1)^2] Which is the same equation I gave before. > > 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. > You can just *calculate* the proper time along each blue segment using C's coordinates with the equations I described, you don't have to actually physically shrink the blue segments in the diagram and rotate them to be parallel to C, I've never seen anyone do that in a relativistic diagram. Anyway, if you do that the diagram will be highly misleading because the diagram will no longer show them all meeting up at the same point in spacetime when B and C finish their final acceleration, even though physically they do all meet at the "same point in spacetime" at that moment, according to the operational definition I gave earlier. > 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. > No, if you do this strange procedure of shrinking and rotating ON THE DIAGRAM, then if you do it correctly they will no longer reunite at the same height ON THE DIAGRAM, even though physically they do reunite at the moment A and B finish their final acceleration. You are correct that both A and B experience the same intervals of proper time during the acceleration phases, so to figure out if they have the same or different TOTAL proper time we just need to find the proper time along the blue segments for each one. This makes the math even simpler, because it means we don't really have to use the relativistic rocket equations for finding proper time during the acceleration phase, we can concentrate on just the blue constant-velocity segments. If you'd like me to show you a numerical example that works in the way of the diagram, and to demonstrate that A and B do NOT have the same proper time along the blue constant-velocity segments of their respective worldlines, this would be simple enough, so just ask. > > 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? > No, since as I said, in this example A and B will have different total proper times despite identical accelerations (and obviously there is no gravity in this example). Jesse 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> 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. > -- 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. 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