Re: The Twin Paradox explained from the moving twin ?


"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> schreef
in bericht news:7QaPg.81097$ZZ2.1210461@xxxxxxxxxxxxxxxxxxxxxxxx

"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message
news:O89Pg.80940$Uh2.1177712@xxxxxxxxxxxxxxxxxxxxxxxx

"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
schreef in bericht news:XbzOg.77748$oT2.1054784@xxxxxxxxxxxxxxxxxxxxxxxx

"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message
news:oJ9Og.75563$t31.1106993@xxxxxxxxxxxxxxxxxxxxxxxx

For more details goto : http://users.telenet.be/nicvroom/dirk3.htm



So the event [B2] where B receives the signal is given by solving the
system
{ x = v t
{ x = c (t-1)
which gives
[B2]: ( t, x ) = ( c/(c-v), v c/(c-v) )

What you do is nicer
For v = c/2 you get:
t = c/(c-v)= 1 / (1-0.5) = 2
x = c(2-1)= c
The same result as I get in the drawing.
B2= (2,1)

But for the time for the signal to go from A to B
I have
t = c/(c-v)
and you have
t = v/(c-v)

So you made a mistake, and it was not a typo.

I did not made any mistake.

Your t and my t are different.

Your t starts from A0
My t starts from A1

In order to get from my t to your t
(It is tea time or are we playing golf)
you have to add 1

v/(c-v) + 1 = c-v + v /(c-v) = c / (c-v)

(simple comme bonjour)

When I original wrote the url I already
realized that you can do the same using your method.



So you see that:

| 3'. The distance from A to point P is v c/(c-v)
| 4'. The time for the signal to go from A to B is c/(c-v)

Do you understand what you did wrong?

Apparently you don't.
We can't continue before you do...

Dirk Vdm




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