"Paul B. Andersen" wrote in message
...
| Sorcerer wrote:
| "Paul B. Andersen" wrote in message
| ...
| | 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.
|
| Quite.
| Your point of view is irrelevant and of
| no interest to anybody.
| So shut up.

100/173.2 = 0.577, moron!
Your schoolmarm arithmetic is incompetent and of no interest to anybody.
Explain 0.5c or shut the **** up, embarrassed imbecile!
Why don't you just **** off back to tusseland, stooopid idiot.
Androcles

| ================================================== ===================
|
| 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.

I dont agree. One has to be very carefull on how and to what to apply
that relation.
In our situation, x is NOT 100LY in B's FoR. The 100LY is in A's FoR as
posed by the problem.

B goes away at a speed of 0.5c wrt A (as measured by A, in A's FoR).
B goes away a distance of 100LY wrt A (as measured by A, in A's FoR).
Therefore, B has traveled a time of 100LY/0.5c = 200 Y wrt A (as
measured by A, in A's FoR)

Agree with that?


But dont feel bad, I dont agree with Andersen's proof either.

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.)

Irrelevant. The I believe that the intent of the origianl poster was to
solve the problem posed without accelerations. The twin 'paradox' can
be posed without accelerations by using a third traveller...


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.

Agreed.


B accelerates at 1 (light year per year) per year for one year (c/year),
SNIP

Irrelevant as explained above.

SNIP

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.

That needs to be justified. The 173.2 Y is a result obtianed in
scenario A.
You can not, a priori, use results from scenario A to prove scenario
A.
Can you justify it?

So, can you redo the twin paradox explanation without invoking
accelerations?

L8r

"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.

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/di...insEvents.html
to see how to do it properly.

Go ahead and see how far you get...

Dirk Vdm

"Sorcerer" wrote in message k...

"Paul Cardinale" wrote in message
ups.com...
|
| Sorcerer wrote:
|
| [snip]
|
|
| "An observer accelerating at a will" sounds like a **** up to me.
|
| Yes, we know that's what it sounds like to you. But everybody else
| understands it.
|
Then the Tusselad should explain it.

Then there is the other small matter of his claim
100 ly in 173.2 years = 0.5c

You ****wit parrots can't do simple arithmetic, can you?

I'm sure he can't do *this* kind of simple arithmetic:
http://users.telenet.be/vdmoortel/di...s/LogsHuh.html

Dirk Vdm

"Dirk Van de moortel" 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.
|
| 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.


Did''t they teach you to write E'glish in Belgia' k'itti'g classes?
How old are you?

| 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.

HAHAHAHAHAHAHA!

It's an excellent idea, you STOOPID moron!
You are as ****ing crazy as that schoolmarm Tusselad
who reckons travelling 100 years in 173.2 years is 0.5c!
Can't even handle simple arithmetic, can you?
Bad idea, ******!

And he called it DORK3!

ROFLMAO!
There has to be a fortune in laughs here, I'll save this one!
Androcles.







This distance is x

"Dirk Van de moortel" wrote
in message ...
|
| "Sorcerer" wrote in message
k...
|
| "Paul Cardinale" wrote in message
| ups.com...
| |
| | Sorcerer wrote:
| |
| | [snip]
| |
| |
| | "An observer accelerating at a will" sounds like a **** up to me.
| |
| | Yes, we know that's what it sounds like to you. But everybody else
| | understands it.
| |
| Then the Tusselad should explain it.
|
| Then there is the other small matter of his claim
| 100 ly in 173.2 years = 0.5c
|
| You ****wit parrots can't do simple arithmetic, can you?
|
| I'm sure he can't do *this* kind of simple arithmetic:
| http://users.telenet.be/vdmoortel/di...s/LogsHuh.html
|

Nope, he sure can't!

1/2 = 0.577 -- Tusselad the schoolmarm, endorsed by Dork Van de merde
the local village dog tord!
(t,x) is a bad idea!

HAHAHAHAHAHA!

dstupid/dm = V
dstupid = Vdm + Pentcho Vulva the Einstein Wussy, aka "Ron Samon".

ROFLMAO!

Androcles

wrote in message
ups.com...
|
| | ================================================== ===================
| |
| | 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.
|
| I dont agree.

ROFLMAO!


| One has to be very carefull on how and to what to apply
| that relation.

Two has to be carefull how to spell 'careful'.
Three has to learn that v = x/t, so t = x/v.


| In our situation, x is NOT 100LY in B's FoR.


Oh, it doesn't take light 100 years to travel 100 LY?

The 100LY is in A's FoR as
| posed by the problem.

Ok, how long DOES it take light to travel 100 LY then?


| B goes away at a speed of 0.5c wrt A (as measured by A, in A's FoR).

Yes, so A goes away from B at a speed of 0.5c wrt B (as measured by B, in
B's FoR).

| B goes away a distance of 100LY wrt A (as measured by A, in A's FoR).

Yes, so A goes away a distance of 100LY wrt B (as measured by B, in B's
FoR).



| Therefore, B has traveled a time of 100LY/0.5c = 200 Y wrt A (as
| measured by A, in A's FoR)
| Agree with that?


Therefore, A has traveled a time of 100LY/0.5c = 200 Y wrt B (as
measured by B, in B's FoR)
Agree with that?

And that, son, is why it is called the twin paradox.

The reason behind it is this:
http://www.androcles01.pwp.blueyonde...minoEffect.GIF
http://www.androcles01.pwp.blueyonde...ket/Rocket.htm
http://www.androcles01.pwp.blueyonde...mart/Smart.htm

And this is the bull****:
http://www.androcles01.pwp.blueyonde...winParadox.htm


| But dont feel bad, I dont agree with Andersen's proof either.

Why should I feel bad? If you don't agree t = x/v it is you that
needs an aspirin for your headache, not I.

****wit schoolmarm Andersen thinks 1/2 = 0.577, so my grandson
will NOT be attending Agder College, Kristiansand, Norway.
You couldn't PAY me to send him there, but Dork Van de merde
paid to go. Now that moron is trying to teach **** to Nic Vroom.

Androcles

wrote in message
oups.com...
|
| 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.)
|
| Irrelevant. The I believe that the intent of the origianl poster was to
| solve the problem posed without accelerations. The twin 'paradox' can
| be posed without accelerations by using a third traveller...
|
|
| 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.
|
| Agreed.

SUCKER!
v = 100LY in 173.2 years = 0.577c.

B goes away at a speed v = 0.867c
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) = 99.66 years.
B got there before the light from A.
Faster Than Light (FTL) is 0.867c
ROFLMAO!



|
|
| B accelerates at 1 (light year per year) per year for one year
(c/year),
| SNIP
|
| Irrelevant as explained above.
|
| SNIP
|
| 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.
|
| That needs to be justified. The 173.2 Y is a result obtianed in
| scenario A.
| You can not, a priori, use results from scenario A to prove scenario
| A.
| Can you justify it?
|
| So, can you redo the twin paradox explanation without invoking
| accelerations?
|
| L8r

Tusselad's a ****wit schoolmarm, not an engineer. He doesn't actually
think!

This is what a Norwegian troll looks like
http://www.androcles01.pwp.blueyonde...M/Tusselad.jpg


Androcles

Three has to learn that v = x/t, so t = x/v.

I have not disagred with that.

| In our situation, x is NOT 100LY in B's FoR.


Oh, it doesn't take light 100 years to travel 100 LY?

Yes, but the problem does not involve the voyage of light. It involves
a distance as measured by A.
It can be considered a rod "on the ground" in A's frame and it has a
length of D meters, where D is 100LY. Lets remove LY and just say the
twin B goes away at a distance of 299792458*3600*24*365*100 meters.


The 100LY is in A's FoR as
| posed by the problem.

Ok, how long DOES it take light to travel 100 LY then?

A 100 years, but the travel of light is not at issue here.


| B goes away at a speed of 0.5c wrt A (as measured by A, in A's FoR).

Yes, so A goes away from B at a speed of 0.5c wrt B (as measured by B, in
B's FoR).

Yes. Speeds are symmetric in SR.


| B goes away a distance of 100LY wrt A (as measured by A, in A's FoR).

Yes, so A goes away a distance of 100LY wrt B (as measured by B, in B's
FoR).

That, I disagree. Measurements of distances are not symmetric in SR.

Therefore, A has traveled a time of 100LY/0.5c = 200 Y wrt B (as
measured by B, in B's FoR)
Agree with that?

No, due to my above comment. A has not traveld a distance of 100LY as
measured (wrt) by B.
That conclusion is from SR's principles (and definitions).

Why should I feel bad? If you don't agree t = x/v it is you that
needs an aspirin for your headache, not I.

I do agree that t=x/v.

****wit schoolmarm Andersen thinks 1/2 = 0.577,

Then let him think it...

L8r