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

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.

The url shows that before the point of return
both observers will observe the same.
The "trick" is that the moving observer does not stay
at rest in his original frame.

Nicolaas Vroom.

http://users.telenet.be/nicvroom/

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

No The reason you would consider one clock moving and the
other stationary is a clue as to why.


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.

No... there is no mechanism in SR to alter the appearance of
clocks except the length of the light path. When
the twins are together. that length is zero.


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.

Correct. The motion of the clocks wrt the dielectric
media is zero.


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

Yes.... just so the motion is zero. For other values it
doesn't work.


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

Not true If they watch Dr. Who via equal lengths of video cable,
they need pay no attention to each other.


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

ROFL
Putting the handle of a malicious subject line
trasher is not a good way to get someone to
click on a URL.



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.

I will take your word that it is wrong. Possibly
viral too.


The url shows that before the point of return
both observers will observe the same.
The "trick" is that the moving observer does not stay
at rest in his original frame.

Yes, the matter that interacts in a real-world light
path has to be considered to avoid the paradox.

Propagation in a dielectric medium
http://farside.ph.utexas.edu/teachin...es/node98.html
"Retarded potential"
http://farside.ph.utexas.edu/teachin...es/node50.html

... Einstein's resolution using
gravitational time dilation suffers from logical and
physical flaws, and gives incorrect answers in a general
setting. The counter examples imply the need to reconsider
many issues related to the comparison of transported
clocks. The failure of the accepted views and
resolutions is traced to the fact that the special relativity
principle formulated originally for physics in empty
space is not valid in the matter-filled universe.

C. S. Unnikrishnan
Gravitation Group,
Tata Institute of Fundamental Research,
Homi Bhabha Road, Mumbai 400 005, India
http://www.iisc.ernet.in/currsci/dec252005/2009.pdf
-----

Sue...


Nicolaas Vroom.

http://users.telenet.be/nicvroom/

"Nicolaas Vroom" wrote in message
news
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.

All observers have to agree that the *average* clock rate (in ticks per
whatever inertial reference second) was less than the inertial clock rate
while the clocks were separated .

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.

The moving observer makes the same comparison as the one "at rest", and
reaches the same conclusion.

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

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.

The url shows that before the point of return
both observers will observe the same.
The "trick" is that the moving observer does not stay
at rest in his original frame.

Of course - and anyone can choose whatever inertial frame he wants as
reference frame.

Harald

Nicolaas Vroom.

http://users.telenet.be/nicvroom/

Nicolaas Vroom wrote:
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

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.

The url shows that before the point of return
both observers will observe the same.
The "trick" is that the moving observer does not stay
at rest in his original frame.

Nicolaas Vroom.

http://users.telenet.be/nicvroom/


OK.
I will show a scenario where we see it from the travelling
twin's point of view.

But first:
An observer accelerating at a will measure the rate of
an instantly stationary clock at a distance d in the direction
of the acceleration to be: (1 + a*d/c^2)
(A clock higher up in the gravitational field runs fast.)

Now to the promised scenario.

Let us first describe it from the point of view of the "home twin" A.
================================================== ===================

A B- v
|-----------------------------------------|

B goes away at a speed v = 0.5c
When B has travelled a distance 100 LY,
he starts his rocket engine.
At that time A's clock shows 200 years,
and B's clock show 200*sqrt(1 - v^2/c^2) = 173.2 years.

B accelerates at 1 (light year per year) per year for one year (c/year),
thus changing his speed by c and going at 0.5c in the other direction.
(This acceleration is in the order of 1g.)
A's clock is now showing 201 years.
B is again 100 LY away, heading home.
B's clock is showing 174.2 years.

When B is back, A's clock shows 401 years,
B's clock shows 2*173.2+1 = 347.4 years.

From "travelling twin" B's point of view:
=========================================
A goes away at a speed v = 0.5c.
When B's clock shows 173.2 years, he starts his rocket.
At that time B will observe A's clock to show
173.2*sqrt(1 - v^2/c^2) = 150 years.

B accelerates for one year at c/year.
While he is doing so, he will observe A's clock to run at the rate:
(1 + a*d/c^2) = (1 + (c/year)*(100 light year)/c^2) = 101.
Thus will B observe that A's clock advances 101 years during
the year he is accelerating.
So when the acceleration is done, B's clock shows 174.2 years,
and he will observe A's clock to show 150+101 = 251 years.

A is approaching at 0.5 c.
When they meet, B's clock will show 347.4 years.
A's clock will show 251+150 = 401 years.

None of the twins are surprised when they see the other
twin's clock.

Important notice:
Note that it is B's acceleration that causes B to _observe_
(measure) A's clock to run fast.
But nothing ever happened to clock A, it ticked
along at its normal rate the whole time. The acceleration
of B can obviously not affect A in any way.
It can however affect B's measurements of A.

(Yes, to keep the math simple, I have done a couple of
simplifications which the more knowledgeable readers will spot.
Please don't nitpick. :-) )

Paul

Paul B. Andersen wrote:

[...]
Now to the promised scenario.
[...]

Can you show us the one about the bellhop and the
two dollars. I'd rather have more money than than
more time? . ;-)

Sue...

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

IMO this is not possible.

It is possible. There are numerous explanations/proofs out there.

The basic idea is that the situation is not symmetric. If B goes 100Ly
away wrt A, that means that A is less than a 100Ly away from B wrt B.
IOW, A does not go as far so he is quicker to arrive at B hence B's
time is less. Moreso that A's clock is slowed wrt B's clock.

Lookup this NG. Some time ago I explained it that way with calcultaions.

"Paul B. Andersen" wrote in message
...
| Nicolaas Vroom wrote:
| 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
|
| 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.
|
| The url shows that before the point of return
| both observers will observe the same.
| The "trick" is that the moving observer does not stay
| at rest in his original frame.
|
| Nicolaas Vroom.
|
| http://users.telenet.be/nicvroom/
|
|
| OK.
| I will show a scenario where we see it from the travelling
| twin's point of view.


Points of view are not relevant, ****head.



|
| But first:
| An observer accelerating at a will



An observer at a will?
Your frequent blunders are...
Hilarious, yes?



| measure the rate of
| an instantly stationary clock at a distance d in the direction
| of the acceleration to be: (1 + a*d/c^2)
| (A clock higher up in the gravitational field runs fast.)

No way, ****head. Clocks higher up in the gravitation field
run at 1 second per second.


|
| Now to the promised scenario.
|
| Let us first describe it from the point of view of the "home twin" A.

Points of view are not relevant, ****head.



| ================================================== ===================
|
| A B- v
| |-----------------------------------------|
|
| B goes away at a speed v = 0.5c
| When B has travelled a distance 100 LY,
| he starts his rocket engine.

So B's clock read 200 years in B's FoR by definition t = x/v.


| At that time A's clock shows 200 years,
| and B's clock show 200*sqrt(1 - v^2/c^2) = 173.2 years.

HAHAHA!
B's clock read 200 years by definition t = x/v.

Hilarious, yes?


B sees A's clock showing 100 years by definition
of Doppler.

A's clock has slowed down from B's PoV, ****head.
Hilarious, yes?



|
| B accelerates at 1 (light year per year) per year for one year (c/year),
| thus changing his speed by c and going at 0.5c in the other direction.
| (This acceleration is in the order of 1g.)
| A's clock is now showing 201 years.

B sees A's clock showing 101 years. B's clock shows 201 years.
Hilarious, yes?


| B is again 100 LY away, heading home.

You got one right, ****head.
Hilarious, yes?

| B's clock is showing 174.2 years.

Nope. B's clock shows 201 years.
Hilarious, yes?



|
| When B is back, A's clock shows 401 years,
| B's clock shows 2*173.2+1 = 347.4 years.

Nope. B's clock shows 401 years, by definition of t = -x/-v.


|
| From "travelling twin" B's point of view:

We already did that.
Here is the derivation of the cuckoo misformations.

http://www.androcles01.pwp.blueyonde...ket/Rocket.htm
http://www.androcles01.pwp.blueyonde...mart/Smart.htm

Highly embarrassing for a ****head schoolmarm troll like you.
Hilarious, yes?
Androcles

"harry" wrote in message
...
|
| "Nicolaas Vroom" wrote in message
| news | 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.
|
| All observers have to agree that the *average* clock rate (in ticks per
| whatever inertial reference second) was less than the inertial clock rate
| while the clocks were separated .

This observer doesn't agree, ****head.
http://www.androcles01.pwp.blueyonde...ket/Rocket.htm
http://www.androcles01.pwp.blueyonde...mart/Smart.htm


Androcles

"Nicolaas Vroom" :
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.

http://home.scarlet.be/~pin12499/MyS...yTravtwin2.TIF

As you see, the moving twin "acknowledges" home twin's clock ticking (slowly
at that) only between A and D, and between E and C. Home twin's interval DE
is skipped during moving twin's turnback, and so, 'lost' in moving twin's
time count of his homestaying kin.

More at
http://home.scarlet.be/~pin12499/paratwin.htm

guido

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

IMO this is not possible.

It is possible. There are numerous explanations/proofs out there.

So called 'proofs' require you ignore the theory's own admonition
about imaginary operators in Minkowski space and if they
involve gravity or acceleration similar mal-operation is required
in the four space:
http://www.nrao.edu/~smyers/courses/...edoflight.html

It is all every bit as convincing as the bellhop paradox. ;-)

Sue...



The basic idea is that the situation is not symmetric. If B goes 100Ly
away wrt A, that means that A is less than a 100Ly away from B wrt B.
IOW, A does not go as far so he is quicker to arrive at B hence B's
time is less. Moreso that A's clock is slowed wrt B's clock.

Lookup this NG. Some time ago I explained it that way with calcultaions.