On Fri, Mar 7, 2014 at 8:37 PM, Edgar L. Owen <edgaro...@att.net> wrote:
> Jesse, > > I guess I'm supposed to take that as a yes? You do agree that A's world > line is actually shorter than C's (even though it is depicted as longer) > because A's proper time along it is less than C's from parting to meeting? > Correct? Strange how resistant you are to ever saying you agree when we > actually do agree. Remember we are not counting points here, at least I'm > not, we are trying to find the truth.... > I'm not "resistant" in general, I have said "I agree" to a number of agree/disagree questions you asked in the past. But in this one case I was expressing irritation because from your question it seemed pretty obvious you either hadn't read, or hadn't paid any attention to, my discussion of "lengths" in the post you were responding to. If you really, really can't deduce my opinion on this from statements like this: "in terms of proper times C > B > A which is the opposite of how it works with spatial lengths" or: "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" ...then just tell me why you think these statements are ambiguous and I will then tell you whether I "agree that A's world line is actually shorter than C's (even though it is depicted as longer) because A's proper time along it is less than C's from parting to meeting". But if reading those statements does answer your question, then I would suggest that part of "trying to find the truth" is actually reading through the responses you get, not skimming/skipping over parts of it. > > First, note you don't actually have to calculate anything. A and C just > compare clocks when they meet and that gives the actual world line lengths. > Sure, but I'm talking about the theoretical analysis. > > But, if you want to calculate to predict what that comparison will be, > then you have to be careful to do it correctly. > > C can't just use the Pythagorean theorem on A's world line from his > perspective on the x and y distances, he has to use it on the time > dimension as well squareroot((y2-y1)^2 + (x2-x1)^2 - c(t2-t1)^2). > You have the right idea, although that formula (with c^2 rather than c) actually calculates proper length on a spacelike interval, if you want proper time on a timelike interval the equivalent formula would be squareroot((t2-t1)^2 - (1/c)^2*(x2-x1)^2 - (1/c)^2*(y2-y1)^2). And since we were talking about a 2D spacetime diagram where all motion was along a single spatial axis, I dropped the y-coordinate, and as I mentioned I was also assuming units where c=1, like years for time and light-years for distance. That's why I just wrote the formula as sqrt((t2 - t1)^2 - (x2 - x1)^2). > It is the subtraction of this time term that will reduce the length of the > slanting blue lines of A and B to THEIR PROJECTIONS ON C'S OWN WORLDLINE. > That statement appears wrong, although you'd have to give me a definition of what you mean by "projections on C's own worldline" for me to be sure. It seems to me that by the normal definition of "projection", projecting one of the slanted blue line segments onto the C's vertical worldline would give a new segment parallel to the vertical axis, whose length is just equal to the vertical separation between the ends of the original slanted blue segment. If so, the length of that sort of "projection" is NOT equal to the proper time of the original slanted blue segment, instead it's equal to the coordinate time between its endpoints, in C's rest frame. For example, looking at the diagram at http://www.jessemazer.com/images/tripletparadox.jpg , let's say the bottom blue segment on A's worldline begins in 2001 in C's rest frame, and ends in 2008, and it has a velocity of 0.6c. In that case, by the normal meaning of "projection", projecting this segment onto C's vertical worldline would just create a vertical segment that goes from 2001 to 2008, and thus has a coordinate time of 7 years (and for any vertical segment of a worldline parallel to the time coordinate axis, proper time is supposed to be equal to coordinate time). But because of time dilation, the proper time along the original blue segment (before it was "projected" to make it vertical) is less than the coordinate time by a factor of sqrt(1 - (0.8c/c)^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6, so relativity says the correct proper time along that original slanted blue segment is 7*0.6 = 4.2 years. Do you agree or disagree with these numbers? > > I think that is what you are saying as well, but my point is that that > NULLIFIES any effect on the length of the world lines by the SLANTING of > the blue lines NO MATTER WHAT THEIR LENGTHS, and LEAVES ONLY the effects of > the red curves. > No, you're simply wrong about this. Let's actually do a numerical example. Suppose that both A and B go through the same sequence of 3 accelerations: ACCELERATION 1: starting from rest in C's frame, they accelerate for 2 years of coordinate time in C's frame with constant proper acceleration 0.375 light years/year^2 in the +x direction. After the 2 years is over, they stop accelerating, with a new velocity of 0.6c in the +x direction in C's frame. ACCELERATION 2: Starting from 0.6c in the +x direction in C's frame, they accelerate for 4 years of coordinate time in C's frame with constant proper acceleration 0.375 light years/years^2 in the -x direction. After the 4 years is over, they stop accelerating, ending with a new velocity of 0.6c in the -x direction in C's frame. ACCELERATION 3: Starting from 0.6c in the -x direction in C's frame, they accelerate for 2 years of coordinate time in C's frame with constant proper acceleration 0.375 light years/year^2 in the +x direction. After the 2 years is over, they stop accelerating, with a new velocity of 0 in C's frame. Using the relativistic rocket equations at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html we can verify that an acceleration a of 0.375 light/years^2 for 2 years of coordinate time t does lead to a change in v of 0.6c. We can also use the equation T=(c/a)*sh-1(at/c) to calculate the proper time T elapsed in these accelerations (sh-1 is the inverse hyperbolic sine function, which can be calculated with the sinh-1 button on the calculator at http://keisan.casio.com/has10/Free.cgi ), it works out to about 1.84839 years of proper time for ACCELERATION 1 and ACCELERATION 3, and twice that, or 3.69678 years of proper time, for ACCELERATION 2. However, since both A and B undergo the same sequence of 3 accelerations, their acceleration phases add the same amount of proper time to their total, so we don't really need to worry about the exact value. So, referring again to the diagram at http://www.jessemazer.com/images/tripletparadox.jpg , let's say that A's worldline consists of the following five segments (between the point at the bottom of the diagram where A departs from B&C, and the point at the top where A&B reunite with C), with the coordinate times of the beginning and end of each segment given in terms of C's rest frame: Segment 1 (red): ACCELERATION 1 from t=1999 to t=2001 Segment 2 (blue): Moving inertially at 0.6c in the +x direction, from t=2001 to t=2008 Segment 3 (red): ACCELERATION 2 from t=2008 to t=2012 Segment 4 (blue): Moving inertially at 0.6c in the -x direction, from t=2012 to t=2019 Segment 5 (red): ACCELERATION 3 from t=2019 to t=2021 And B's worldline consists of the following five segments: Segment 1 (blue): Remaining at rest in C's frame, from t=1999 to t=2009 Segment 2 (red): ACCELERATION 1 from t=2009 to t=2011 Segment 3 (blue): Moving inertially at 0.6c in the +x direction, from t=2011 to t=2013 Segment 4 (red): ACCELERATION 2 from t=2013 to t=2017 Segment 5 (blue): Moving inertially at 0.6c in the -x direction, from t=2017 to t=2019 Segment 6 (red): ACCELERATION 3 from t=2019 to t=2019 Assuming you agree that the ACCELERATION segments contribute equal amounts of proper time to A and B's worldlines, if we want to figure out whether they experience the same TOTAL proper time we just have to calculate the proper time along each of the blue constant-velocity segments of their worldlines. Can you do this, adding up both the total proper time along blue segments for A and the total proper time along the blue segments for B? In general if a segment of a worldline begins at t1 and ends at t2 and has a constant velocity v throughout the segment, the proper time T along that segment is given by: T = (t2-t1)*sqrt(1 - (v/c)^2) I promise that if you do the math correctly, you'll find that the total proper time along B's blue segments is LARGER than the total proper time along A's blue segments. Try it and see! > > This must be the case because NON-accelerated relative motion DOES NOT > affect proper time rates. This is because it is exactly the same from the > perspective of A and C moving relative to each other, thus it cannot affect > the lengths of their world lines. > > I'm trying to parse your last paragraph. Your diagram shows ONLY how A's > and B's world lines appear in C's comoving frame. It does NOT show the > proper LENGTHS of A's and B's world lines. I think we agree the lengths > depicted are NOT the actual world line lengths. > > I claim the blue slanting lines of A and B, one set longer than the other, > have NO EFFECT on the actual lengths of A's and B's world lines. Because > when we calculate just their proper lengths subtracting the time term as I > do above, their proper lengths reduce to their VERTICAL PROJECTIONS on C's > vertical world line. In other words there is no difference in proper time > rates of A, B or C during the intervals of the slanting blue lines. > But B also has a blue segment between the bottom and top of the diagram that is NOT slanted, the one I labeled "segment 1" for B above. During this segment, B remains at rest in C's frame while A accelerates and then moves away inertially (segments 1 and 2 for A). Not sure how you define "proper time rates", but do you think the proper time rates for A and B are the same when B is on segment 1 (at rest in C's frame) and A is on segment 2 (moving inertially in C's frame)? > > Thus, in my view, we are left with ONLY the effects of the curving red > accelerations, and these are exactly the same for A and B. And when the > lengths of those red acceleration segments are calculated we find that A's > and B's world lines will both be SHORTER than C's world line AND by the > SAME AMOUNT and that A's and B's world line lengths will be EQUAL due only > to their equal accelerations. > > > Perhaps to make this clearer consider just two blue lines of A and B > slanted with respect to each other and crossing at P. From A's perspective > B's line will be slanted, but from B's perspective A's line will be slanted > in the other direction by an equal amount AND since this is NON-accelerated > inertial motion only, both views are EQUALLY VALID. When we do the > Pythagorean world line length calculation we get EXACTLY THE SAME RESULTS > from both frame views. So both world line lengths are exactly equal. > > Thus slanted blue lines of ANY LENGTH have NO EFFECT AT ALL on world line > lengths, and only curved red line accelerations do. > > If you disagree I can give you another example. > > I definitely disagree, but before you give me another example try actually CALCULATING the proper time along the blue segments of A and B's worldlines using the numbers I suggested above. You will see that in fact the total proper times along blue segments are NOT equal for A and B. 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 firstname.lastname@example.org. 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