This is showing 2 ships that are passing each other at the moment that T of ship #1 is passing N of ship #2. Each diagram shows one of the ships at rest and the other ship moving.
Special relativity....
Ship #1 at rest at top:
1 light-year
0sec------------0sec
T--------------------N
N-----------T <-------- .8c
0sec 15,147,296sec
.6 light-year
Ship #2 at rest at bottom:
.6 light-year
0sec -15,147,296sec
T---------------N -------> .8c
N--------------------T
0sec------------0sec
1 light-year
T1 is T of ship #1. N1 is N of ship #1. N2 is N of ship #2. T2 is T of ship #2.
T1, N1, T2, and N2 each have their own clock.
At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame. T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2. N2 is displaying 0 seconds and T2 is displaying 15,147,296 seconds when viewed from ship #1.
Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front.
Now to the point of this post. There seems to be a problem here. It seems that T1 is passing N2 twice.
T1 passes N2 when the clock at N1 displays -15,147,296 seconds. T1 passes N2 again when the clock at N1 displays 0 seconds.
Also,
T1 passes N2 when the clock at T2 displays 0 seconds. T1 passes N2 again when the clock at T2 displays 15,147,296 seconds.
What is displayed on the clocks at N1 and T2 when T1 passes N2?
What is the resolution to this problem?
Special relativity....
Ship #1 at rest at top:
1 light-year
0sec------------0sec
T--------------------N
N-----------T <-------- .8c
0sec 15,147,296sec
.6 light-year
Ship #2 at rest at bottom:
.6 light-year
0sec -15,147,296sec
T---------------N -------> .8c
N--------------------T
0sec------------0sec
1 light-year
T1 is T of ship #1. N1 is N of ship #1. N2 is N of ship #2. T2 is T of ship #2.
T1, N1, T2, and N2 each have their own clock.
At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame. T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2. N2 is displaying 0 seconds and T2 is displaying 15,147,296 seconds when viewed from ship #1.
Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front.
Now to the point of this post. There seems to be a problem here. It seems that T1 is passing N2 twice.
T1 passes N2 when the clock at N1 displays -15,147,296 seconds. T1 passes N2 again when the clock at N1 displays 0 seconds.
Also,
T1 passes N2 when the clock at T2 displays 0 seconds. T1 passes N2 again when the clock at T2 displays 15,147,296 seconds.
What is displayed on the clocks at N1 and T2 when T1 passes N2?
What is the resolution to this problem?
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