Re: The Twin Paradox explained from the moving twin ?


"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message news:9BfPg.81516$RS2.1038472@xxxxxxxxxxxxxxxxxxxxxxxx

"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> schreef in bericht
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"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) )

I made a typo here (- this is the same result as mine)
That should be
[B2]: ( t, x ) = ( v/(c-v), c v/(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.

And that is correct.
Starting from A1 it takes a time v/(c-v) to reach return point B


Yes, it takes the signal a time c/(c-v) to reach point [B].
If you "start at [A1]" then
[A1]: ( t, x ) = ( 0, 0 )
[A0]: ( t, x ) = ( -1, 0 )
and
[B2]: ( t, x ) = ( v/(c-v), c v/(c-v) )
so the time for the *signal* to reach event [B2] is
t[B2] - t[A1] = v/(c-v)
and the time for *B* to reach event [B2] is
t[B2] - t[A0] = v/(c-v) - (-1) = c/(c-v)

Starting from A0 it takes a time c/(c-v) to reach return point B.

Yes, it takes the signal a time c/(c-v) to reach point [B].
If you "start at [A0]" then
[A0]: ( t, x ) = ( 0, 0 )
[A1]: ( t, x ) = ( 1, 0 )
and
[B2]: ( t, x ) = ( c/(c-v), c v/(c-v) )
so the time for the *signal* to reach event [B2] is
t[B2] - t[A1] = c/(c-v) - 1 = v/(c-v) {as before}
and the time for *B* to reach event [B2] is
t[B2] - t[A0] = c/(c-v) - 0 = c/(c-v) {as before}


If you start from A1 you have to add 1 in order to calculate the total time
v/(c-v) + 1 = v/(c-v) + (c-v)/(c-v) = (v + c - v)/(c-v) = c/(c-v)
And that is your answer,
which is correct

And so is my answer.

Yes, but be aware of the fact that your statement
| 1. When the clock of A reaches 1 time unit (t=1), A issues a light
| sinal to B.
explicitly means that you start with [A0], not with [A1].


Your method is more general

I wouldn't say more general.
It's just standard basic 2D analytic geometry.
And so is your method. We just take different origins.
I use the common method to give the origin both zero-coordinates:
[A0]: ( t, x ) = ( 0, 0 )

Dirk Vdm




| 2. When B receives this signal, this defines the turnpoint P, B returns back home

Dirk Vdm


It is interesting to compare this with the example by Paul B Andersen
(excluding acceleration of 1 year)

What you have to do is to multiply my example with 100
and then you see that both are the same.

Nicolaas Vroom
http://users.pandora.be/nicvroom




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