"Daryl McCullough" wrote in message
...
| Androcles says...
|
| "That is, we can reverse the directions of the frames
| which is the same as interchanging the frames,
| which - as I have told you a LOT of times,
| OBVIOUSLY will lead to the transform:
| t = (tau - xi*v/c^2)/sqrt(1-v^2/c^2)
| x = (xi - v*tau)/sqrt(1-v^2/c^2)
| or:
| tau = (t + xv/c^2)/sqrt(1-v^2/c^2)
| xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
|
| " -4.0 = 0.866 * 4.619 " -Paul B. Andersen
|
| First of all, when you write something inside quotes,
| it must be *literally* what was said. Paul Andersen
| did not say that -4.0 = 0.866 * 4.619, did he?

He literally said:
http://groups.google.co.uk/groups?q=...r.co.uk&rnum=2

"Correct calculation:"
x v t tau=(t+xv)/sqrt(...) ---- copy of
my line
" -4.00 0.86 4.65 2.37" ---- Andersen's
modification

I had forgotten it was to two places of decimals instead of three.
I was using the Andersen Transform and he didn't like the result, so he
lied.


|
| Second, why don't we go through the correct calculation
| to determine the elapsed time for the travelling twin.
| Rather than writing xi and tau, let me just use primes:
|
| (x,t) = coordinate system for stay-at-home twin
Ok.

| (x',t') = coordinate system for travelling twin on outward journey
Ok.

| (x'',t'') = coordinate system for travelling twin on return journey

Why is this different?
By the PoR, it is the Earth that moves away and back again.
We only need two coordinate systems, not three.
..
|
| Let t1 = time travelling twin departs earth (according to stay-at-home
twin)

Ok. Time here on Earth. Here we are and we are all familiar with it.


| Let t2 = time travelling twin turns around (according to stay-at-home
twin)
Ok.

| Let t3 = time travelling twin returns to earth (according to stay-at-home
twin)
Ok.

| Then the elapsed time according to the travelling twin is given by:
|
| T_total = T_outward + T_return
| = (t2' - t1') + (t3'' - t2'')
Sure.

|
| The numbers for our particular example a
|
| t1 = x1 = 0
Ok.

| t2 = 4.619, x2 = 4

so x2 = 4 = 0.866 * 4.619, ok

| t3 = 9.238, x3 = 0

Ok. Welcome home. x3 = x1. So we don't need x3.

| gamma = 2
Ok.
| v = .866

Err...No.
v = 0.866 outbound and inbound it is -0.866.
Other coordinates remain correct.
|
| The transformation equations are
|
| t' = gamma (t - vx/c^2)
| t'' = gamma (t + vx/c^2)
Nope. You are going the wrong way.
For small v, you have to approximate the Galilean Transform.
you said : (x,t) = coordinate system for stay-at-home twin

The sign of v changes, and the direction as well.
Now we travel outbound a distance (x2-x1)
at velocity +v, and since time = distance/velocity,
t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma


For the return trip, the interval of time t'' is given by

t'' = [(t2-t1) + (x3-x2)/(t3-t2)*(x3-x2)/c^2] * gamma

Note that (x3-x2) is the same as (x1-x2) and is negative.
(Because this is a coordinate system.)
so
t'' = [(t2-t1) - v *(x3-x2)/c^2] * gamma
|
| So
| t1' = t1'' = 0
Oh?
If the departure is at 12:01 pm on Wednesday, 26 May 2004,
then t1' isn't zero. What you mean is t1-t1' = 0, the clocks are
synchronized.
However, the simplification helps. We'll set both clocks to zero.
I don't see how t'' = 0 though. How did you get that?

| t2' = 2 (4.619 - .866 * 4) = 2.31
Nope, sorry. You are going the wrong way, v isn't negative for the outbound
trip.
Try setting it up using a spreadsheet. Using a computer takes out any
blunders
you make with signs. That's where Andersen went wrong. He tried to tell me
my
computer is wrong.
The time of arrival is t2 = 4.619, found by (t2-t1) and t1 = 0, according to
Earth.
The time of arrival is given by
t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma
t2' = (0 + (4.0 / 4.619) * 4.0 /1) * 2
= (0 + 0.866 * 4.0) * 2
= 6.928

| T_out = 2.31 - 0 = 2.31
Good heavens no. You went the wrong way, v isn't negative.
That's the time for the return trip.

| t2'' = 2 (4.619 + .866 * 4) = 16.166

Ah... now you've found the time for the outbound trip.
We add the two times together for the complete trip, and
t3' = 2.31+16.166 = 18.476.
I don't know what you mean by t2''.
Check it properly with spreadsheet.




| t3'' = 2 (9.238 + .866 * 0) = 18.476
Yeah, that's right, whatever t3'' is.

| T_return = 18.476 - 16.166 = 2.31

Good grief! Time runs backwards!

|
| T_total = T_out + T_return = 4.62
|
| That's the *correct* calculation, Androcles.

Oh... Thanks for telling me.
I'll add this to my web page.
(along with a spreadsheet).

| --
| Daryl McCullough
| Ithaca, NY
Androcles.

Androcles says...


"Daryl McCullough" wrote

Paul Andersen
did not say that -4.0 = 0.866 * 4.619, did he?

He literally said:
http://groups.google.co.uk/groups?q=...r.co.uk&rnum=2

"Correct calculation:"
x v t tau=(t+xv)/sqrt(...) ---- copy of
my line
" -4.00 0.86 4.65 2.37" ---- Andersen's
modification

I had forgotten it was to two places of decimals instead of three.
I was using the Andersen Transform and he didn't like the result, so he
lied.

As I said, he did not literally say "-4.0 = 0.866 * 4.619",
so you were incorrect to put it in quotes.

| (x,t) = coordinate system for stay-at-home twin
Ok.

| (x',t') = coordinate system for travelling twin on outward journey
Ok.

| (x'',t'') = coordinate system for travelling twin on return journey

Why is this different?
By the PoR, it is the Earth that moves away and back again.
We only need two coordinate systems, not three.

The principle of relativity says no such thing. The principle
of (special) relativity says that all *inertial* observers are
equivalent. An inertial reference observer is one that maintains
constant velocity, and never accelerates. The travelling twin
is not inertial throughout the trip---he accelerates halfway.


.
|
| Let t1 = time travelling twin departs earth (according to stay-at-home
twin)

Ok. Time here on Earth. Here we are and we are all familiar with it.


| Let t2 = time travelling twin turns around (according to stay-at-home
twin)
Ok.

| Let t3 = time travelling twin returns to earth (according to stay-at-home
twin)
Ok.

| Then the elapsed time according to the travelling twin is given by:
|
| T_total = T_outward + T_return
| = (t2' - t1') + (t3'' - t2'')
Sure.

|
| The numbers for our particular example a
|
| t1 = x1 = 0
Ok.

| t2 = 4.619, x2 = 4

so x2 = 4 = 0.866 * 4.619, ok

| t3 = 9.238, x3 = 0

Ok. Welcome home. x3 = x1. So we don't need x3.

| gamma = 2
Ok.
| v = .866

Err...No.
v = 0.866 outbound and inbound it is -0.866.

Okay, fine. Let's let v_out be the outward velocity
(+0.866c) and v_return the return velocity (-0.866c).
Then the Lorentz transformations are

t' = gamma (t - v_out x/c^2) = 2 (t - 0.866 * x)
t'' = gamma (t - v_return x/c^2) = 2 (t + 0.866 * x)

Other coordinates remain correct.
|
| The transformation equations are
|
| t' = gamma (t - vx/c^2)
| t'' = gamma (t + vx/c^2)
Nope. You are going the wrong way.
For small v, you have to approximate the Galilean Transform.
you said : (x,t) = coordinate system for stay-at-home twin

The sign of v changes, and the direction as well.

The sign of v *is* the direction of v. v 0 means
travel in the +x direction. v 0 means travel in the
-x direction.

Now we travel outbound a distance (x2-x1)
at velocity +v, and since time = distance/velocity,
t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma

No, the formula is (on the outward trip)

t' = (t - v_out * x/c^2) * gamma

where x = the distance from the stay-at-home twin, *not*
the distance travelled by the travelling twin. As I said,
(t1 = 0, x1 = 0, t2 = 4.619, x2 = 4, t3 = 9.238, x3 = 0,
v_out = +8.66, v_return = -8.66, gamma = 2)

t1' = (t1 - v_out * x1/c^2) * gamma
= 0
t2' = (t2 - v_out * x2/c^2) * gamma
= (4.619 - .866 * 4) * 2
= 2.31
T_out' = t2' - t1' = 2.31

For the return trip, the interval of time t'' is given by

t'' = [(t2-t1) + (x3-x2)/(t3-t2)*(x3-x2)/c^2] * gamma

No, it's not. The formula for the return trip is

t2'' = (t2 - v_return * x2/c^2) * gamma
= (4.619 - (-0.866) * 4) * 2
= 16.166

t3'' = (t3 - v_return * x3/c^2) * gamma
= (9.238 - 0) * 2
= 18.476

T_return'' = t3'' - t2'' = (18.476 - 16.166) = 2.31

However, the simplification helps. We'll set both clocks to zero.
I don't see how t'' = 0 though. How did you get that?

t'' = gamma (t - v_return * x/c^2)

when t=0 and x=0, this gives t'' = 0.

| t2' = 2 (4.619 - .866 * 4) = 2.31
Nope, sorry. You are going the wrong way, v isn't negative for the outbound
trip.

The Lorentz transformation is

t' = gamma (t - vx/c^2)

So the term -vx/c^2 is negative when v is positive, and
is positive when v is negative.

The time of arrival is t2 = 4.619, found by (t2-t1) and t1 = 0, according to
Earth.
The time of arrival is given by
t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma

No. The time of arrival is given by

t2' = (t2 - v_out x2/c^2) * gamma

t2' = (0 + (4.0 / 4.619) * 4.0 /1) * 2
= (0 + 0.866 * 4.0) * 2
= 6.928

No, that's incorrect.

| T_out = 2.31 - 0 = 2.31
Good heavens no. You went the wrong way, v isn't negative.

You have a sign wrong in the Lorentz transformations. The
correct transformation is this:

t' = gamma (x - vt)


That's the time for the return trip.

| t2'' = 2 (4.619 + .866 * 4) = 16.166

Ah... now you've found the time for the outbound trip.

No. The time for the outbound trip, as I've said is

t2' - t1' = 2.31 years

We add the two times together for the complete trip, and
t3' = 2.31+16.166 = 18.476.
I don't know what you mean by t2''.

I thought I explained it: There are three inertial
coordinate systems involved in the twin paradox:
C1. The coordinate system of the stay-at-home twin,
C2. The coordinate system of the travelling twin on his way out.
C3. The coordinate system of the travelling twin on his way back.

t2 = time of turnaround, as measured in C1.
t2' = time of turnaround, as measured in C2.
t2' = time of turnaround, as measured in C3.

To accurately compute elapsed time, you *must* stick to just one
coordinate system.

Here's a rough analogy: Suppose I leave England at 12:00 local time,
and travel east to Sweden. When I get to Sweden, the local time is
6:00. Does that mean that I've been travelling for 5 hours? No, because
England and Sweden are not in the same time zone. What I have to do
to figure out how long I've been travelling is to divide my trip
into segments, such that each segment takes place in a single time
zone. So, let t1' = time I start my trip, in English time zone,
t2' = time I leave the English time zone (as measured in English time),
t2'' = time I enter Swedish time zone (as measured in Swedish time),
t3'' = time my trip ends (as measured in Swedish time). Then the
total time for my trip is

T_total = (t2' - t1') + (t3'' - t2'')

It is completely incorrect to subtract times in different time zones. So
I can't say

T_total = (t2'' - t1') + (t3'' - t2'')

which is what you are doing.

--
Daryl McCullough
Ithaca, NY

"Paul B. Andersen" wrote in message
om...
| (Daryl McCullough) wrote in message
...
| Androcles says...
|
| "That is, we can reverse the directions of the frames
| which is the same as interchanging the frames,
| which - as I have told you a LOT of times,
| OBVIOUSLY will lead to the transform:
| t = (tau - xi*v/c^2)/sqrt(1-v^2/c^2)
| x = (xi - v*tau)/sqrt(1-v^2/c^2)
| or:
| tau = (t + xv/c^2)/sqrt(1-v^2/c^2)
| xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
|
| " -4.0 = 0.866 * 4.619 " -Paul B. Andersen
|
| First of all, when you write something inside quotes,
| it must be *literally* what was said. Paul Andersen
| did not say that -4.0 = 0.866 * 4.619, did he?
|
| Of course he didn't. :-)

He's a liar too.
He literally said:
http://groups.google.co.uk/groups?q=...r.co.uk&rnum=2

"Correct calculation:"
x v t tau=(t+xv)/sqrt(...) ---- copy of
my line
" -4.00 0.86 4.65 2.37" ---- Andersen's
modification

Note :- Positive velocity, positive time, negative distance.

He also says
dtau/dt = 0 1.


| I said that if the moving twin starts at the coordinate
| x = 0 at the time t = 0, and is moving in the negative
| x-direction at the speed v = 0.86c,

'v' is a velocity, not a speed in my calculation.
That's why the thread is titled "How to use a minus sign."
Andersen has just demonstrated yet again that he doesn't know how.
If the speed is in the negative direction, then the velocity is -0.866.

| he will reach
| the coordinate x = -4LY at the time t = 4.619 years.
|
| But Androcles insists that when the twin is moving in
| the negative x-direction,

the velocity is negative, so that the time is positive.
As always, Andersen is trying to squirm his way out of his blunder.
There is nothing wrong with
-4.0 = -0.866 * 4.619
But Andersen said
-4.0 = 0.866 * 4.619 so that he could change the 2.37 time from
the calculate value, according to the Andersen Tranforms he said
I could use,
tau = (t+vx/c^2)/sqrt(1-v^2/c^2)
so that

tau = (4.619+0.866*4.0)*2 = 16.166
(or tau = (4.619+ (-0.866) *(-4.0) *2 = 16.166

and the faster you go, the later you arrive.

But Andersen insists that when the twin is moving in
the negative x-direction, time runs backwards because
t = -x/+v = -4.619 years. So he will arrive at the destination
years before he started.

| My failure to understand that has made Androcles
| roll farting on the floor. He thinks that by doing so,
| he is making the "SRians" look foolish.
|
| Paul
Androcles
ROFLMAO!

"Daryl McCullough" wrote in message
...
| Androcles says...
|
|
| "Daryl McCullough" wrote
|
| Paul Andersen
| did not say that -4.0 = 0.866 * 4.619, did he?
|
| He literally said:
|
http://groups.google.co.uk/groups?q=...physics.relati
vity&hl=en&lr=&ie=UTF-8&selm=Xg22c.2958%24h_.1743%40news-binary.blueyonder.c
o.uk&rnum=2
|
| "Correct calculation:"
| x v t tau=(t+xv)/sqrt(...) ---- copy
of
| my line
| " -4.00 0.86 4.65 2.37" ---- Andersen's
| modification
|
| I had forgotten it was to two places of decimals instead of three.
| I was using the Andersen Transform and he didn't like the result, so he
| lied.
|
| As I said, he did not literally say "-4.0 = 0.866 * 4.619",
| so you were incorrect to put it in quotes.

Picking nits isn't going to change Andersen's Gigantic Blunder.
You asked, and I corrected my error. Andersen will squirm over his.

|
| | (x,t) = coordinate system for stay-at-home twin
| Ok.
|
| | (x',t') = coordinate system for travelling twin on outward journey
| Ok.
|
| | (x'',t'') = coordinate system for travelling twin on return journey
|
| Why is this different?
| By the PoR, it is the Earth that moves away and back again.
| We only need two coordinate systems, not three.
|
| The principle of relativity says no such thing. The principle
| of (special) relativity says that all *inertial* observers are
| equivalent.

My turn to pick nits.
Absolutely NOWHERE in
ON THE ELECTRODYNAMICS
OF MOVING BODIES
By A. Einstein
June 30, 1905
is the term "inertial" used, let alone *inertial*
so you can quit trying to tell me about inertial observers.
In fact is says the very opposite. Quote: Thence we conclude that a
balance-clock at the equator must go more slowly, by a very small amount,
than a precisely similar clock situated at one of the poles under otherwise
identical conditions.
So why don't you learn (special) relativity instead of making up your own?



An inertial reference observer is one that maintains
| constant velocity, and never accelerates. The travelling twin
| is not inertial throughout the trip---he accelerates halfway.
See the Relativity FAQs.

|
| .
| |
| | Let t1 = time travelling twin departs earth (according to stay-at-home
| twin)
|
| Ok. Time here on Earth. Here we are and we are all familiar with it.
|
|
| | Let t2 = time travelling twin turns around (according to stay-at-home
| twin)
| Ok.
|
| | Let t3 = time travelling twin returns to earth (according to
stay-at-home
| twin)
| Ok.
|
| | Then the elapsed time according to the travelling twin is given by:
| |
| | T_total = T_outward + T_return
| | = (t2' - t1') + (t3'' - t2'')
| Sure.
|
| |
| | The numbers for our particular example a
| |
| | t1 = x1 = 0
| Ok.
|
| | t2 = 4.619, x2 = 4
|
| so x2 = 4 = 0.866 * 4.619, ok
|
| | t3 = 9.238, x3 = 0
|
| Ok. Welcome home. x3 = x1. So we don't need x3.
|
| | gamma = 2
| Ok.
| | v = .866
|
| Err...No.
| v = 0.866 outbound and inbound it is -0.866.
|
| Okay, fine. Let's let v_out be the outward velocity
| (+0.866c) and v_return the return velocity (-0.866c).
| Then the Lorentz transformations are
|
| t' = gamma (t - v_out x/c^2) = 2 (t - 0.866 * x)
| t'' = gamma (t - v_return x/c^2) = 2 (t + 0.866 * x)

You still don't get it, do you?


|
| Other coordinates remain correct.
| |
| | The transformation equations are
| |
| | t' = gamma (t - vx/c^2)
| | t'' = gamma (t + vx/c^2)
| Nope. You are going the wrong way.
| For small v, you have to approximate the Galilean Transform.
| you said : (x,t) = coordinate system for stay-at-home twin
|
| The sign of v changes, and the direction as well.
|
| The sign of v *is* the direction of v. v 0 means
| travel in the +x direction. v 0 means travel in the
| -x direction.

Almost correct. However, x itself is a coordinate on the X- axis.
You should have said:
| The sign of v *is* the direction of v. v 0 means
| travel in the +X direction. v 0 means travel in the
| -X direction.


|
| Now we travel outbound a distance (x2-x1)
| at velocity +v, and since time = distance/velocity,
| t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma
|
| No, the formula is (on the outward trip)
|
| t' = (t - v_out * x/c^2) * gamma

You still don't get it, do you?
Einstein said:

"If we place x'=x-vt, ..." (Note the 'IF' at the start.)then lots of
calculation,
ending up with
xi = x-vt (divided by sqrt(1-v^2/c^2)
In other words, the Lorentz Transform approximates the Galilean Transform,
x' = x-vt, for small v.
What we have is x' = x+vt for the outward trip.
so we now have
x' = (x+vt)*gamma (because you don't like to use xi, being whimsical).
so
t' = (t +v_out * x/c^2) *gamma.
That is the Andersen Transform I've given you.
If you do not use it, you are going the wrong way.
|
| where x = the distance from the stay-at-home twin,

No. x is a coordinate. If you set x1 = 0, then x2 IS the same as the
distance.


*not*
| the distance travelled by the travelling twin.
Of course not. The distance travelled by the travelling twin is given by
x', and is always zero.
It is the Earth that moves away, the travelled twin goes nowhere
in his own frame. I get in my car, London comes to me in my
"inertial frame of reference". Same thing if I get on a plane. I see the
ground
moving backwards beneath me. The pilot remains the same distance from
my seat at take-off and landing.



As I said,
| (t1 = 0, x1 = 0, t2 = 4.619, x2 = 4, t3 = 9.238, x3 = 0,
| v_out = +8.66, v_return = -8.66, gamma = 2)

But you are wrong.


|
| t1' = (t1 - v_out * x1/c^2) * gamma
| = 0
Ok, clocks are synch at the origin.

| t2' = (t2 - v_out * x2/c^2) * gamma
| = (4.619 - .866 * 4) * 2
| = 2.31
You are going the wrong way. You can't get to +x2 by going backwards.
t2' = (t2 + v_out*x2/c^2) * gamma
= (4.619 + .866 * 4) * 2
= 16.166
Its only math, after all. That's why I suggested you use a spreadsheet, and
why
I titled the thread "How to use a minus sign".



| T_out' = t2' - t1' = 2.31

Nope: T_out' = t2' - t1' = 16.166
|
| For the return trip, the interval of time t'' is given by
|
| t'' = [(t2-t1) + (x3-x2)/(t3-t2)*(x3-x2)/c^2] * gamma
|
| No, it's not. The formula for the return trip is
|
| t2'' = (t2 - v_return * x2/c^2) * gamma
| = (4.619 - (-0.866) * 4) * 2
| = 16.166
Wrong. THIS is where you should use the Lorentz Transform.
Remember, Einstein said:
"If we place x'=x-vt, ..." (Note the 'IF' at the start.)then lots of
calculation,
ending up with
xi = x-vt (divided by sqrt(1-v^2/c^2)
In other words, the Lorentz Transform approximates the Galilean Transform,
x' = x-vt, for small v.
This time you are going away from x2', which is still there, of course, but
returning to
x1. And 16.166 years have already elapsed, so
t3'' = 16.166+ (t2 - v_return * (x1-x2)/c^2) * gamma
= 16.166 + 2.31
= 18.476
(t2'' is also a coordinate).




| t3'' = (t3 - v_return * x3/c^2) * gamma
| = (9.238 - 0) * 2
| = 18.476
Ok, that's the right answer, even if you got the outgoing and returning
calculations reversed.
|
| T_return'' = t3'' - t2'' = (18.476 - 16.166) = 2.31
Ok, well, the time for the return trip 2.31, the time for the outbound trip
is 16.166, so the total is 18.476.

|
| However, the simplification helps. We'll set both clocks to zero.
| I don't see how t'' = 0 though. How did you get that?
|
| t'' = gamma (t - v_return * x/c^2)
|
| when t=0 and x=0, this gives t'' = 0.
|
| | t2' = 2 (4.619 - .866 * 4) = 2.31
| Nope, sorry. You are going the wrong way, v isn't negative for the
outbound
| trip.
|
| The Lorentz transformation is
|
| t' = gamma (t - vx/c^2)
|
| So the term -vx/c^2 is negative when v is positive, and
| is positive when v is negative.
No no. Refer to
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Einstein clearly states:
"IF x' = x-vt..."

IF x' = x+vt, as it does for the outbound journey,
then
xi = (x+vt)*gamma



|
| The time of arrival is t2 = 4.619, found by (t2-t1) and t1 = 0, according
to
| Earth.
| The time of arrival is given by
| t' = [t1 + (x2-x1)/(t2-t1)*(x2-x1) /c^2 ] * gamma
|
| No. The time of arrival is given by
|
| t2' = (t2 - v_out x2/c^2) * gamma

You keep insisting this. Assertion carries no weight, sorry. You'll have to
prove it.

|
| t2' = (0 + (4.0 / 4.619) * 4.0 /1) * 2
| = (0 + 0.866 * 4.0) * 2
| = 6.928
|
| No, that's incorrect.
Not at all, you are incorrect.
Why don't you read what Einstein said, instead of quoting some other source?

|
| | T_out = 2.31 - 0 = 2.31
| Good heavens no. You went the wrong way, v isn't negative.
|
| You have a sign wrong in the Lorentz transformations. The
| correct transformation is this:
|
| t' = gamma (x - vt)

Huh?
Sorry, that is hopeless. I'll assume it's a typo.
x' = gamma (x-vt) if we start with x' = x-vt.


|
| That's the time for the return trip.
|
| | t2'' = 2 (4.619 + .866 * 4) = 16.166
|
| Ah... now you've found the time for the outbound trip.
|
| No. The time for the outbound trip, as I've said is
|
| t2' - t1' = 2.31 years
And I've told you that you are wrong, and told you why you are wrong.
Its up to you to prove it. Just saying it isn't good enough.


| We add the two times together for the complete trip, and
| t3' = 2.31+16.166 = 18.476.
| I don't know what you mean by t2''.
|
| I thought I explained it:
Never mind.



There are three inertial
| coordinate systems involved in the twin paradox:
| C1. The coordinate system of the stay-at-home twin,
| C2. The coordinate system of the travelling twin on his way out.
| C3. The coordinate system of the travelling twin on his way back.
|
| t2 = time of turnaround, as measured in C1.
| t2' = time of turnaround, as measured in C2.
| t2' = time of turnaround, as measured in C3.
|
| To accurately compute elapsed time, you *must* stick to just one
| coordinate system.
|
| Here's a rough analogy: Suppose I leave England at 12:00 local time,
| and travel east to Sweden. When I get to Sweden, the local time is
| 6:00. Does that mean that I've been travelling for 5 hours? No, because
| England and Sweden are not in the same time zone.
Oh Pu'lease!
[remainder snipped]
Androcles

"Paul B. Andersen" wrote in message
om...
| "Androcles" wrote in message
...
|
| "Paul B. Andersen" wrote in message
| om...
| | "Androcles" wrote in message
| | . ..
| | Four years ago I derived the Andersen Transforms:
| |
| | No, you did not.
| |
| | Four years ago, you derived the Adroclean transform in the posting:
| |
|
http://groups.google.com/groups?q=g:....pa.h ome.com
| | From where I quote:
| | |
| | | Watch the signs, carefully.
| | |
| | | 1/2[tau(0,0,0,t) + tau(0,0,0,t + x'/(c+v) + x'/(c-v))] =
| | | tau(x',0,0,t+x'(c+v))
| | | (The mirror moves the opposite way.)
| | | Hence, if x' be chosen infinitesimally small,
| | | 1/2(1/(c+v) + 1/(c-v) ) @tau/@t = @tau/@x' + 1/(c+v) @tau/@t'
| | | or
| | | @tau/@x' - v/(c^2 -v ^2)@tau/@t = 0
| | |
| | | and (ta daaaaa)
| | | beta = 1/sqrt(1+v^2/c^2)
| | |
| | | You understand Einstein's derivation? What a joke!
| | | You probably dont even
| | | know that sqrt(1) has two answers, 1 and -1.
| | | The correct answer to the
| | | question I gave, and you guessed at, is
| | | tau = (t + vx/c^2)/sqrt(1 (PLUS) v^2/c^2)
| | | xi = (x + vt/c^2)/sqrt(1 + v^2/c^2)
| | | eta = y
| | | zeta = z
| |
| | Ta daaaa - that's how to use a minus sign.
| |
| | Well done, Androcles.
| |
| | Paul
|
| I did, didn't I? That was how I got you to produce the Andersen
Transforms.
|
| Quite.
| The posting you wrote July 7, 2001 made me produce
| the following more than a year before.
| Time is known to run backwards in Androcles' world.
| That's how to use a minus sign.
|
| Paul B. Andersen wrote Mars 3, 2000:
| | The _form_ of the Lorentz transform
| | depend on how the frames of references are chosen.
| | There are a number of different possibilities:
| |
| | Choice #1:
| | =============
| | All axes parallel, origo of K' moving in positive direction
| | along the x-axis of K
| |
| | ----|---------------- x' - v
| | ----|---------------- x
| |
| | With this choice , the LT is:
| | forward, K-K' inverse, K'-K
| | x' = g*(x - v*t) x = g*(x' + v*t')
| | t' = g*(t - x*v/c^2) t = g*(t' + x'*v/c^2)
| |
| | Choice #2:
| | =============
| | All axes parallel, origo of K moving in positive direction
| | along the x'-axis of K'
| |
| | ----|---------------- x - v
| | ----|---------------- x'
| |
| | With this choice, the LT is:
| | forward, K'-K inverse, K-K'
| | x = g*(x' - v*t') x' = g*(x + v*t)
| | t = g*(t' - x'*v/c^2) t' = g*(t + x*v/c^2)
| |
| | Choice #3:
| | [ etc.]
| | ....
| |
| | However, Einstein used "choice #1" in his paper.
| | When you interchange the frames of references,
| | you get "choice #2" above.
| | Finding it self contradictory that that the transform equations
| | change as shown above is - not very smart.
|
| Well done me.
|
| Indeed.
| The credit is all yours, obviously.
|
| Paul
Smarter than someone that claims
dtau/dt = 0 1, eh?
Androcles.

"Androcles" wrote in message ...

"Daryl McCullough" wrote in message
...
| Androcles says...

[snip Androcles' stomach content]

A piece of advice for Daryl (and for everyone who might be interested):

When you are dealing with a stupid person of the malicious kind, then
*always* make sure that, in every transaction you have with them, you
EITHER
- explain no more than 1 simple concept.
OR
- ask no more than 1 simple question.

Of course, even then, do *not* expect an intelligent reply, but you will
be sure that you will be able to properly handle it.

This concludes lesson #7 of my free course
on Applied Village Idiot Psychology.

Dirk Vdm

Androcles a écrit :
My turn to pick nits.
Absolutely NOWHERE in
ON THE ELECTRODYNAMICS
OF MOVING BODIES
By A. Einstein
June 30, 1905
is the term "inertial" used, let alone *inertial*
so you can quit trying to tell me about inertial observers.
In fact is says the very opposite. Quote: Thence we conclude that a
balance-clock at the equator must go more slowly, by a very small amount,
than a precisely similar clock situated at one of the poles under otherwise
identical conditions.
So why don't you learn (special) relativity instead of making up your own?

You know little child that it is not the words (neither their sound nor
their spelling) which matter. It is their meaning. "Inertial" or
"Galilean" means exactly what A.E. said in two sentences at the very
beginning of his paper :

"Let us take a system of co-ordinates in which the equations of
Newtonian mechanics hold good [...]"

"[...] two systems of co-ordinates in uniform translatory motion [...]"

Got it ? Not likely : it is far more complicated than coping with
the xor logical operetor.

"Dirk Van de moortel" wrote in message
...

"Androcles" wrote in message ...

"Daryl McCullough" wrote in message
...
| Androcles says...

[snip Androcles' stomach content]

A piece of advice for Daryl (and for everyone who might be interested):

When you are dealing with a stupid person of the malicious kind, then
*always* make sure that, in every transaction you have with them, you
EITHER
- explain no more than 1 simple concept.
OR
- ask no more than 1 simple question.

Of course, even then, do *not* expect an intelligent reply, but you will
be sure that you will be able to properly handle it.

This concludes lesson #7 of my free course
on Applied Village Idiot Psychology.

Dirk Vdm

By the way, Androcles, there is no need for *you* to reply to the
previous post, since it obviously contains much more than 1 simple
concept. Since *this* one contains no more than 1 simple concept,
feel free to have a go at it.
(I will not ask you to pay a visit to the toilet first, because in your
condition that is a very difficult concept)

Dirk Vdm

"YBM" wrote in message ...
Androcles a écrit :
My turn to pick nits.
Absolutely NOWHERE in
ON THE ELECTRODYNAMICS
OF MOVING BODIES
By A. Einstein
June 30, 1905
is the term "inertial" used, let alone *inertial*
so you can quit trying to tell me about inertial observers.
In fact is says the very opposite. Quote: Thence we conclude that a
balance-clock at the equator must go more slowly, by a very small amount,
than a precisely similar clock situated at one of the poles under otherwise
identical conditions.
So why don't you learn (special) relativity instead of making up your own?

You know little child that it is not the words (neither their sound nor
their spelling) which matter. It is their meaning. "Inertial" or
"Galilean" means exactly what A.E. said in two sentences at the very
beginning of his paper :

"Let us take a system of co-ordinates in which the equations of
Newtonian mechanics hold good [...]"

"[...] two systems of co-ordinates in uniform translatory motion [...]"

Got it ? Not likely : it is far more complicated than coping with
the xor logical operetor.

Ouch.... you explained two difficult concepts and asked a tough question.
This is not going to work. Mark my words...

Dirk Vdm

Androcles wrote:
Smarter than someone that claims
dtau/dt = 0 1, eh?
Androcles.

Too old to learn.
Too stupid to learn.
Too proud to learn.

Sad.