Re: The Twin Paradox explained from the moving twin ?


"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.
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.
Use the letters x and t as *variables*.

Here's a proper diagram:

^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) )

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) )


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. 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.

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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