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We have:OK.
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
And what's missing:This is nonsense, and demonstrates that you
tA'(e2) = 2.25
The problem with B is that B is in B, that is to say somewhere other than in the conjunction AA'.Is the "conjunction of AA'" the same as event E1?
We have synchronized the watches, A and A', but how do we do it for B?????????!!!!!!!!!!!!!!
Of course there are idiots, like Python, who will say, we just have to synchronize anyhow. But they are crazy. Don't worry, we are waiting for them, we will burn them all.Yes, there are idiots, and you are one of them. Python is not.
There are two main ways to synchronize B and B', namely the synchronize on A and A' by noting tB(e1)=-1 and tB'(e1)=-1;Good grief!
or either in practicing M type synchronization (Einstein synchronization),
noting tB(e1)=0 and tB'(e1)=0.
We are going to use Einstein synchronization to please Jean-Pierre.
For synchronization on AA', and not on M(R) and M'(R'), it will suffice to add Δt=-1.
We then have:
tB(e1)=0 {-1}I think Jean-Pierre is pleased to see that all he
tB'(e1)=0 {-1}
tB(e2)=0.75 {-0.25}
tB'(e2)=0.25 {-0.75)
We are now going to annoy Jean-Pierre,
add the e3 event which is the conjunction BB'
which will happen at a certain time.
We apparently have:
tA(e3)= 0.50
tA'(e3)= 1.50
tB(e3)= -1.50 tB'(3)= -0.50
For these last two data, we notice that they are negative, but that's normal.
For B and for B', the synchronization is done after the conjunction BB'.
Thank you for your attention.
R.H.Above I asked you to read my paper again.
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