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?

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? 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? I agree that this method correctly calculates how A and B observe each other's clock times, but not sure that's the same as the other's actual proper times. Edgar On Saturday, March 8, 2014 9:31:24 AM UTC-5, jessem wrote: > > > > On Fri, Mar 7, 2014 at 8:37 PM, Edgar L. Owen <edga...@att.net<javascript:> > > 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 li > ... -- 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.