On Sat, Mar 8, 2014 at 2:03 PM, Edgar L. Owen <edgaro...@att.net> wrote:
> Jesse, > > OK, Assume c=1 and start with your sqrt((t2 - t1)^2 - (x2 - x1)^2) to > calculate what you say is the proper time on a time-like interval. Using > your method, which I assume is correct I do see that A's proper time will > be greater than B's. The reason is basically that A has to travel further > in space to get from t1 to t2 and consequently must also travel less far in > time. Correct? > It's true in C's rest frame that A travels a greater distance than B, but this needn't be true if you used a different frame. For example, if you analyze the problem using an inertial frame D in which A is at rest during the first blue leg of his trip, then A and B should travel the same distance between departing and reuniting, because neither ever turns around and travels in the "wrong" direction in this frame, they are both always at rest in this frame or traveling in the -x direction towards the position in this frame where they will reunite. So if you imagine that A and B are cars that are driving along a piece of flat ground at rest in frame D, and both start out with odometers reading 0 when A first departs from B, then A and B's odometer will show the same reading when they reunite, since they have both traveled in a consistent direction (but varying speed) along the same straight road between the position in frame D where they departed and the position where they reunited. > > > To confirm, consider a simplified twin example with only straight lines so > we can ignore accelerations. A remains at rest with a straight vertical > line from t1 to t3. B travels away from t1 in a straight oblique line, > reverses direction midpoint (call this t2) and travels in a straight > oblique line back to t3. > > The two halves of B's trip are symmetric (have the same velocities away > from and back towards A) therefore B's proper time, calculated by A, will > be = 2 x sqrt((t2 - t1)^2 - (x2 - x1)^2). In other words we have to > multiply by 2 to get the proper time of B for the entire trip. Correct? > Yes, that's correct. > > > OK, now consider another case with A and B just moving with constant > relative motion and their world lines crossing at t1 and then diverging. > There is NO acceleration. > > In this case using the Lorentz transform both A and B will observe each > other's time running slow relative to their own. And using your formula > above both A and B will also observe each other's proper times SLOWED > RELATIVE TO THEIR OWN. > > But doesn't this mean that since A and B get different results about each > other's proper times that this method of calculating proper times is NOT > INVARIANT, and thus is not actually calculating proper times which you say > are invariant? > The method gives an invariant answer for the proper time between any two specific events on an inertial worldline, events which are known to have coordinates (x1,t1) and (x2,t2) in whatever frame you're using. But in your example, they haven't agreed on a specific pair of points on each worldline to calculate the proper time between. Suppose their worldlines cross at the moment of their births, when they are both 0 years old, and subsequently they move apart with arelative velocity of 0.6c so the time dilation factor is 0.8c. If A wants to use his own rest frame to predict how old B will be "at the same moment" that A turns 20, he is picking the event b1 on B's worldline that is SIMULTANEOUS IN A's REST FRAME with A turning 20, and calculate the proper time between the event of B's birth and b1, which is 16. On the other hand, if B wants to use his own rest frame to predict how old he'll be at the "same moment" that A turns 20, he must pick the event b2 on B's worldline that is SIMULTANEOUS IN B'S REST FRAME with A turning 20, and calculate the proper time between B's birth and b2, which is 25. Both frames agree that the proper time between B's birth and b1 is 16, and that the proper time between B's birth and b2 is 25, they just disagree about whether b1 or b2 is simultaneous with A turning 20. So that's why they disagree about whether A is older or younger than B at any specified point on A's worldline, like A turning 20 (and of course the logic works the same if you specify a point on B's worldline and ask about A's age "at the same moment"). Of course, this sort of ambiguity about what events to choose doesn't arise in a twin-paradox type scenario where the twins depart from each other at one specific point on their worldlines, and reunite at some other specific point on their worldlines. If you want further evidence that the method gives an invariant answer, you can use the Lorentz transformation to check that this is so. Pick two events on the worldline of an inertial clock of arbitrary velocity in the frame you're using, and assume that the spacetime origin is chosen so that the first event is labeled with coordinates x1=0, t1=0. Then the second event can be anything (so long as the absolute value of t2 is larger than the absolute value of x2, meaning the worldline is that of a slower-than-light clock)--for example if the clock is moving at 0.8c we might choose x2=8, t2=10. Then use the equation to calculate the proper time in this frame, in this case sqrt(10^2 - 8^2) = 6 years. Now use the Lorentz transformation to transform the events into another primed frame with velocity v relative to the first--if the second frame is moving at v along the x-axis of the first, and the their origins coincide (so x=0,t=0 in the unprimed frame coincides with x'=0,t'=0 in the primed frame), then the coordinates of any event in the second frame will be related to those in the first by these formulas (again assuming units where c=1): x' = (x - v*t)/sqrt(1 - v^2) t' = (t - v*x)/sqrt(1 - v^2) Since we chose the first event to be x1=0,t1=0 the corresponding coordinates of the first event in the primed frame will be x1'=0,t1'=0, but you will have to use these equations to calculate x2' and t2'. But no matter what v you choose for the velocity of the primed frame relative to the unprimed frame, and no matter what coordinates x2 and t2 you picked for the second event, if you calculate x2' and t2' using the Lorentz transformation formulas and then calculate the proper time using the primed frames' coordinates, i.e. sqrt((t2'-t1')^2 - (x2'-x1')^2), you will get exactly the same answer for the proper time as when you calculated sqrt((t2-t1)^2 - (x2-x1)^2) in the unprimed frame's coordinates. And you could get a general proof that this would work for arbitrary choices of event pairs and frames if you didn't pick specific numbers, but just left the notation abstract but substituted in (x2 - v*t2)/sqrt(1 - v^2) in for x2', and (t1 - v*x1)/sqrt(1 - v^2) in for x1', etc., then used algebra to simplify the resulting expression after the substitution. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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