Re: The Twin Paradox explained from the moving twin ?


"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote
in message news:M5gPg.81564$UT7.1169522@xxxxxxxxxxxxxxxxxxxxxxxx
|
| "Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message
news:9BfPg.81516$RS2.1038472@xxxxxxxxxxxxxxxxxxxxxxxx
| >
| > "Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
schreef in bericht
| > news:M6ePg.81384$AU2.1096819@xxxxxxxxxxxxxxxxxxxxxxxx
| >>
| >> "Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> wrote in message
news:TjbPg.81143$XS2.1040521@xxxxxxxxxxxxxxxxxxxxxxxx
| >>>
| >>> "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
| >>
| >> 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

You lost, Dork!

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You are funnier than Wilson Rabbidge!
You are never going to win, DORK!
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Androcles, inventor of the integration constant.




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