Re: The Twin Paradox explained from the moving twin ?


"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.

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

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.


^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)

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?
In your examples you continue to work with this error, so I will
stop referring to it here.

But we can continue as follows:
Event [A2] which is simultaneous with event [B2] has coordinates
[A2]: ( t, x ) = ( c/(c-v) , 0 )

Using the coordinates of event B2 and the known light velocity -c,
the light signal this sent at event [B2] toward A, has equation
line [B2-A3]: x - v c/(c-v) = -c ( t - c/(c-v) )

To find the coordinates of event [A3] you need the worldline equation
of A, which is trivially
A: x = 0

So the coordinates of event [A3] can be found by solving the system
{ x = 0
{ x - v c/(c-v) = -c ( t - c/(c-v) )
which gives
[A3]: ( t, x ) = ( (c+v)/(c-v), 0 )


Likewise, the worldline of B between events [B2] and [A4] is
given by
line [B2-A4]: x - v c/(c-v) = -v ( t - c/(c-v) )

The coordinates of event [A4] are found by solving the system
{ x = 0
{ x - v c/(c-v) = -v ( t - c/(c-v) )
which gives
[A4]: ( t, x ) = ( 2 c/(c-v), 0 )

So, taking everything together we have the following events:
[A0]: ( t, x ) = ( 0, 0 )
[A1]: ( t, x ) = ( 1, 0 )
[A2]: ( t, x ) = ( c/(c-v) , 0 )
[A3]: ( t, x ) = ( (c+v)/(c-v), 0 )
[A4]: ( t, x ) = ( 2 c/(c-v), 0 )
[B2]: ( t, x ) = ( c/(c-v), v c/(c-v) )

For A4 you get 4 and I get the same.

At that url there is also a sketch which explains this from the
moving point of view.

But that sketch is wrong so it seems.

Yes.
The next thing you do wrong, is to try to draw the moving twin B
as one line in the diagram on the right side.

I agree that what I do is wrong.
That was the reason for me question.

You can't do that with
a spacetime diagram. Every fixed place in a spacetime diagram
has a straight line. So the moving twin's line must be broken at the
return event.

See my comments today or tomorrow to Paul B Andersen

If you want to describe what happens above "as seen by the
moving twin", you must use two different coordinate systems.
Have a look at
http://users.telenet.be/vdmoortel/dirk/Physics/TwinsEvents.html
to see how to do it properly.

Go ahead and see how far you get...

Dirk Vdm



.

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