Re: The Twin Paradox explained from the moving twin ?


"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
Question:
Is it possible to explain the Twin Experiment from the moving point of view ?

IMO this is not possible.

When the two Observers meet they realize that there clocks are not the same.
To be more specific: The clock of the moving observer runs behind.
The clock of the moving observer runs slow.

To explain the Observer at rest will say:
Look on my clock I read 10000 counts.
Based on the speed v (Lorentz Transformation) gamma is 0.5
That means your clock should read 5000 counts.
Which is what we measure. qed.

The question is: Is it possible to explain this also from the
point of view of moving observer ?

IMO this it not possible.
The moving Observer has to accept what the Observer at rest tells him.

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

From that page:

| 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

Up to here everythig is okay.
Now the trouble starts:

| 3. The distance from A to point P is x.
| 4. The time for the signal to go from A to B is t

So you want the event [B2] to have coordinates (t,x). Bad idea.

| 5. x = ct = (1+t)*v
| 6. ct - vt = v
| 7. t = v/(c-v) time units

This is wrong.

This is not wrong.
See below.

It is wrong. I have given you the proof that it is wrong below.


You confuse the coordinates of some event with the variables
so you make a mistake from the very start.
Try to go back your analytic geometry lessons from school.
I wish that would be possible

I have shown you how.


Use the letters x and t as *variables*.
I agree what you do is nicer more general.

Here's a proper diagram:
My diagram is the same.

Yes.



^t
|
|
[A4]
|\
| \
| \
| \
| \
[A3] \
| . \
| . \
| . \
| . \
| .\
[A2]--------[B2]
| ./~
| . / ~
| . / ~
| . / ~
| . / ~
[A1] / ~
| / ~
| / ~
| / ~
| / ~
|/ ~
[A0]---------P------------------------>x

In the coordinates of A, the worldline between [A0} and [B2] B is given by
B: x = v t

The lightline of the signal that is sent out at event [A1] with coordinates
[A1]: ( t, x ) = ( 1, 0 )
is therefore x-0 = c (t-1), giving
line [A1-B2]: x = c (t-1)

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.


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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