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 

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.


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 <<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
> ...

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