Re: The Twin Paradox explained from the moving twin ?
- From: "Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 17 Sep 2006 15:47:56 GMT
"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message news:TjbPg.81143$XS2.1040521@xxxxxxxxxxxxxxxxxxxxxxxx
"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> schreef in bericht
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"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message news:O89Pg.80940$Uh2.1177712@xxxxxxxxxxxxxxxxxxxxxxxx
"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> schreef in bericht
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"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
It does not.
In http://users.telenet.be/nicvroom/dirk3.htm you say:
| 1. When the clock of A reaches 1 time unit (t=1), A issues a light signal to B.
So your t has value t = 1 at event [A1].
If it started at event [A1] then it would hav value t = 0 at [A1].
So your t does not start from [A1].
But if indeed your t *would* start from [A1] then
[A1]: ( t, x ) = ( 0, 0 )
[A0]: ( t, x ) = ( -1, 0 )
and then you have to solve the system
{ x = c t
{ x = v ( t - (-1) )
which gives
[B2]: ( t, x ) = ( c/(c-v), c^2/(c-v) )
But that is not what your first two assumptions suggest:
| 1. When the clock of A reaches 1 time unit (t=1), A issues a light signal to B.
| 2. When B receives this signal, this defines the turnpoint P, B returns back home
Dirk Vdm
.
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