So, it is time to stop this mess...Let me do it. I promise, if MacM will not avoid reply or somebody will reconstruct his arguments, you will see evident mathematical and simplest physical mistakes this guy does…
1. Whole this threat is based on the wrong calculation. In so called "MacM's 3Clock case" the right numbers are the following:
Per A:
B = Tb = Ta * [1-Vb^2]^1/2 = 9.949874______at Tb = 10__and Vb=0.1_
C = Tc = Ta * [1-Vc^2]^1/2 = 4.358899______at Tc = 10__and Vc=0.9_
Per B:
1)
A = Ta = 9.949874* [1-Vb^2]^1/2 = 9.899999___at Ta = 10_ and Vb=0.1_
C = Ta = 9.949874* [1-Vcb^2]^1/2 = 3.959004___at Ta = 10_ and Vc=0.9_
and
2)
A = Ta = 10 * [1-Vb^2]^1/2 = 9.949874____at Tb = 10__ and Vb=0.1_
C = Ta = 10 * [1-Vcb^2]^1/2 = 3.978949___at Tb = 10___ and Vc=0.9_
Per C:
1)
A = Ta = 4.358899 * [1-Vc^2]^1/2 = 1.900000___at Ta = 10_ and Vb=0.1_
C = Ta = 4.358899 * [1-Vcb^2]^1/2 = 1.734384___at Ta = 10_ and Vc=0.9_
and
2)
A = Ta = 10 * [1-Vc^2]^1/2 = 4.358899_____at Tc = 10____ and Vb=0.1_
B = Ta = 10 * [1-Vcb^2]^1/2 = 3.978949____at Tc = 10____ and Vc=0.9_
_
because B and C see velocity
Vcb = (0.1+0.9)/(1+01*09) = 1/1.09 = 0.917431
for each other.
For "Per B" and "Per C" answers are given in two cases:
1) If all observers (by some unknown reason) read all clocks at same moment when their clocks are showing the time, which observer A calculates they should have when he has 10 hrs past the moment of clocks synchronization.
2) If each observer read clocks when his own clock shows 10 hrs past moment of synchronization.
Now, what wrong comes out those numbers, which can be used as proof SRT is wrong or describes reality in wrong way?
1. Whole this threat is based on the wrong calculation. In so called "MacM's 3Clock case" the right numbers are the following:
Per A:
B = Tb = Ta * [1-Vb^2]^1/2 = 9.949874______at Tb = 10__and Vb=0.1_
C = Tc = Ta * [1-Vc^2]^1/2 = 4.358899______at Tc = 10__and Vc=0.9_
Per B:
1)
A = Ta = 9.949874* [1-Vb^2]^1/2 = 9.899999___at Ta = 10_ and Vb=0.1_
C = Ta = 9.949874* [1-Vcb^2]^1/2 = 3.959004___at Ta = 10_ and Vc=0.9_
and
2)
A = Ta = 10 * [1-Vb^2]^1/2 = 9.949874____at Tb = 10__ and Vb=0.1_
C = Ta = 10 * [1-Vcb^2]^1/2 = 3.978949___at Tb = 10___ and Vc=0.9_
Per C:
1)
A = Ta = 4.358899 * [1-Vc^2]^1/2 = 1.900000___at Ta = 10_ and Vb=0.1_
C = Ta = 4.358899 * [1-Vcb^2]^1/2 = 1.734384___at Ta = 10_ and Vc=0.9_
and
2)
A = Ta = 10 * [1-Vc^2]^1/2 = 4.358899_____at Tc = 10____ and Vb=0.1_
B = Ta = 10 * [1-Vcb^2]^1/2 = 3.978949____at Tc = 10____ and Vc=0.9_
_
because B and C see velocity
Vcb = (0.1+0.9)/(1+01*09) = 1/1.09 = 0.917431
for each other.
For "Per B" and "Per C" answers are given in two cases:
1) If all observers (by some unknown reason) read all clocks at same moment when their clocks are showing the time, which observer A calculates they should have when he has 10 hrs past the moment of clocks synchronization.
2) If each observer read clocks when his own clock shows 10 hrs past moment of synchronization.
Now, what wrong comes out those numbers, which can be used as proof SRT is wrong or describes reality in wrong way?