sue jahn wrote:
This is science and science costs money.
The taxpayer decides how it is used.
No balanced equations = No money. ;-)

No, in the case of Special Relativity, science was dirt
cheap---Einstein did it in his spare time. It didn't cost taxpayers one
cent.

The point is that there is no inconsistency involved in the twin
paradox.

--
Daryl McCullough
Ithaca, NY

Androcles wrote:
tau = (t-vx/c^2)/sqrt(1-v^2/c^2) one way and
tau = (t+vx/c^2)/sqrt(1-v^2/c^2) coming back.

No, relativity doesn't say that. The Lorentz transformations are
coordinate transformations between two *inertial* reference frames. A
rocket ship that turns around is *not* inertial. The Lorentz
transformations cannot be used to describe how things look aboard the
rocket. That doesn't mean that you can't figure out how things look to
the astronauts aboard the rocket, but you can't use the Lorentz
transformations alone for that purpose.

--
Daryl McCullough
Ithaca, NY

Perspicacious, religious idiot and self-aggrandizing nitwit:
Rage Bilge wrote:
Why can't you make yourself understood?

It's a problem in you and in many others:
"There are none so blind as those who
refuse to see."

Well then, the solution is simple: Listen when we tell you that,
the problem would be solved. Unfortunately, you're simpler than
the solution.

"SDaryl" wrote in message
oups.com...
|
| Androcles wrote:
| tau = (t-vx/c^2)/sqrt(1-v^2/c^2) one way and
| tau = (t+vx/c^2)/sqrt(1-v^2/c^2) coming back.
|
| No, relativity doesn't say that.

I know it doesn't, but you'll have to take that up with Paul Andersen.


| The Lorentz transformations are
| coordinate transformations between two *inertial* reference frames.

[quote]
If at the points A and B of K there are stationary clocks which,
viewed in the stationary system, are synchronous; and if the clock
at A is moved with the velocity v along the line AB to B, then on its
arrival at B the two clocks no longer synchronize, but the clock
moved from A to B lags behind the other which has remained at B
by (1/2)tv^2/c^2 (up to magnitudes of fourth and higher order),
t being the time occupied in the journey from A to B.
It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points
A and B coincide.
If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its arrival
at A will be (1/2)tv^2/c^2 slow.
[end quote]
A circle is not an inertial reference frame. Wrong again, McCullough.


If you want to do it in an inertial frame using coordinates, then
an ant doesn't walk from 0 to 30 cm on a ruler, flip the ruler 180
degrees
and walk from 0 to 30 back again. He walks from 30 to 0 at -v.

tau = (t-vx/c^2)/sqrt(1-v^2/c^2) one way and
tau = (t+vx/c^2)/sqrt(1-v^2/c^2) coming back.

See Paul Andersen, he's a dumb relativist, but not quite so dumb as you.

"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


|A
| rocket ship that turns around is *not* inertial.

So what?

The Lorentz
| transformations cannot be used to describe how things look aboard the
| rocket.

Your cuckoo transforms can't be used for anything, there are no
inertial frames anywhere.

That doesn't mean that you can't figure out how things look to
| the astronauts aboard the rocket, but you can't use the Lorentz
| transformations alone for that purpose.

Your cuckoo transforms can't be used for anything, there are no
inertial frames anywhere.

Androcles

Androcles wrote:
[quote]
If at the points A and B of K there are stationary clocks which,
viewed in the stationary system, are synchronous; and if the clock
at A is moved with the velocity v along the line AB to B, then on its
arrival at B the two clocks no longer synchronize, but the clock
moved from A to B lags behind the other which has remained at B
by (1/2)tv^2/c^2 (up to magnitudes of fourth and higher order),
t being the time occupied in the journey from A to B.
It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points
A and B coincide.
If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its arrival
at A will be (1/2)tv^2/c^2 slow.
[end quote]
A circle is not an inertial reference frame.

You are getting two different concepts confused: (1) coordinate time,
and (2) proper time.

Coordinate time is a function of a coordinate system. The Lorentz
transformations describe how coordinate times of two different inertial
coordinate systems relate to each other. The Lorentz transformations
cannot be used for noninertial coordinate systems.

Proper time is the elapsed time shown on an idealized clock. It doesn't
matter whether the clock is travelling inertially or not.

The formulas for coordinate time and proper time are different:

Coordinate time for a "moving" coordinate system:
t' = gamma (t - vx/c^2)

Proper time for a moving clock:

T = integral of square-root(1-v^2/c^2) dt

--
Daryl McCullough
Ithaca, NY

"SDaryl" wrote in message
oups.com...
|
| Androcles wrote:
| [quote]
| If at the points A and B of K there are stationary clocks which,
| viewed in the stationary system, are synchronous; and if the clock
| at A is moved with the velocity v along the line AB to B, then on
its
| arrival at B the two clocks no longer synchronize, but the clock
| moved from A to B lags behind the other which has remained at B
| by (1/2)tv^2/c^2 (up to magnitudes of fourth and higher order),
| t being the time occupied in the journey from A to B.
| It is at once apparent that this result still holds good if the
clock
| moves from A to B in any polygonal line, and also when the points
| A and B coincide.
| If we assume that the result proved for a polygonal line is also
valid
| for a continuously curved line, we arrive at this result: If one of
two
| synchronous clocks at A is moved in a closed curve with constant
| velocity until it returns to A, the journey lasting t seconds, then
by
| the clock which has remained at rest the travelled clock on its
arrival
| at A will be (1/2)tv^2/c^2 slow.
| [end quote]
| A circle is not an inertial reference frame.
|
| You are getting two different concepts confused: (1) coordinate time,
| and (2) proper time.

BULL****. I've just proved to you that your inertial frame argument
is denied by the chief relativist himself.
You are handwaving. It doesn't work on me. I've got the whole spiel
at my fingertips.


|
| Coordinate time is a function of a coordinate system. The Lorentz
| transformations describe how coordinate times of two different
inertial
| coordinate systems relate to each other. The Lorentz transformations
| cannot be used for noninertial coordinate systems.
|
| Proper time is the elapsed time shown on an idealized clock. It
doesn't
| matter whether the clock is travelling inertially or not.

You are handwaving. Waffle away, McCullough.
[snip]
Two can play snipping.
I'n not interested in your cuckoo transforms until YOU understand
their derivation, so you can quit quoting them to me.
I understand SR better than you ever will. I KNOW it's a joke.
If you are gullible enough to believe it, that's your problem.
Androcles.

Androcles,

I just want to remind you that if something seems like nonsense to you,
there is always two possible explanations: (1) It really *is* nonsense,
and (2) It is not nonsense, but you are misunderstanding it.

Yes, I know it can be really difficult sometimes to tell the difference
between these two cases.

"SDaryl" wrote in message oups.com...

sue jahn wrote:
This is science and science costs money.
The taxpayer decides how it is used.
No balanced equations = No money. ;-)

No, in the case of Special Relativity, science was dirt
cheap---Einstein did it in his spare time. It didn't cost taxpayers one
cent.
Pontificating on the backs of old envelopes is not science.

The point is that there is no inconsistency involved in the twin
paradox.
Thus the name.

Sue...

--
Daryl McCullough
Ithaca, NY

"SDaryl" wrote in message ps.com...
Androcles,

I just want to remind you that if something seems like nonsense to you,
there is always two possible explanations: (1) It really *is* nonsense,
and (2) It is not nonsense, but you are misunderstanding it.

Yes, I know it can be really difficult sometimes to tell the difference
between these two cases.


IOW
What the the great Einstein can't conceive, is fair game for
a negative proof. ;-)

Sue...

"SDaryl" wrote in message oups.com...

Androcles wrote:
tau = (t-vx/c^2)/sqrt(1-v^2/c^2) one way and
tau = (t+vx/c^2)/sqrt(1-v^2/c^2) coming back.

No, relativity doesn't say that. The Lorentz transformations are
coordinate transformations between two *inertial* reference frames.

IDIOT!!!

Frame of reference -- Facts, Info, and Encyclopedia articleSimply put, a set of locally inertial reference frames at varying
distances from the Earth's axis are twisting up kind of like molasses stirred by a central ...
www.absoluteastronomy.com/ encyclopedia/f/fr/frame_of_reference.htm - 27k - Cached - Similar pages

Sue...

A
rocket ship that turns around is *not* inertial. The Lorentz
transformations cannot be used to describe how things look aboard the
rocket. That doesn't mean that you can't figure out how things look to
the astronauts aboard the rocket, but you can't use the Lorentz
transformations alone for that purpose.

--
Daryl McCullough
Ithaca, NY