wrote in news:b8df66b1-af2c-4a12-97f7-
:
.....
- Show quoted text -

Centrifugal motion slows time.

You seem to have a superfluous word there.

It is not the fact that the motion is centrifugal that is important. It is
the fact that there is motion.

The particle could oscillate back and forth along a straight line.

As long as the average velocity as the same, the 'slowing of time' would be
the same.

The particle could move in a straight line and the 'slowing of time' would
be the same, with the additional complication of comparing times at distant
locations.
That problem is avoided in the case of circular motion and in the case of
linear oscillatory motion.


--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"N:dlzc D:aol T:com (dlzc)" escreveu na mensagem
...
Dear El Enrrabadore-mor:

"El Enrrabadore-mor" wrote in message
...
...
With constant acceleration you can double, triple, or
10^101 the distance, that nothing changes about the
acceleration caused by the effect.

With Hubble expansion, you double the distance, you
double the effect. If you want to apply it in some other
fashion, you need another model. You should know
this. You Have No Model.

If I double the distance, then I double the effect?
This one is very funny.
It is not Hubble constant a CONSTANT?
If it's a constant effect, how can it double?

http://www.astro.ucla.edu/~wright/cosmo_02.htm
... especially
v = H_o * D_now

... and ...
1+z = exp(v/c)
... with (1+z) = wavelength_measured / original_wavelength

Double distance, more than double wavelength observed.

I cannot believe you don't even remember the formula Ho came from.


Obviously wrong.
Who are you trying to fool?

Read the math in your link carefully:
1) v = H_o * D_now
2) D_now = (c/H_o) ln(1+z)
3) (1+z) = exp(v/c)

Now, replace 3) into 2):
D_now = (c/H_o) ln(exp(v/c))
Hence:
D_now = (c/H_o)(v/c)
D_now = v / H_o
So:
v = H_o * D_now

The term:
ln(exp(v/c)) = v/c
is true, unless my computer program is wrong about
computing the logarithm of the exponential function.
Computer says: ln(exp(x)) = x (a straight line).

Who are you trying to fool with the (1+z) argument?
Hubble Law says:
v = H_o * D_now
It is a LINEAR function.

What you've done was to come up with an exponential
function and didn't mentioned the logarithmic function that
cancels it out.

You can square the function and later take the square root.
But if you do so, you cannot claim it is a square function just
because you've squared it.


Yet the "blue shift" (not "red shift") never varies as the distance
increases.

Obviously not.
It's a linear function by DEFINITION of OBSERVATION.
Don't try to be a magician, playing with math with me.


Since you are now in the "duck and hide" stage, I will leave you to your
personal misery. Good luck with whatever it is you think you accomplish
with showing off your ignorance this way.

Yeah. Thanks for the advise.
I like my ignorance compared to your knowledge. Both happy.


How can someone work with the tools of others, and have no conception of
what they were created for? "Convenience" seems too easy an answer...

Yep. You're right about that.
Humans and primate are the only species that recognise
themselves in a mirror. I wonder if Humans looking at
the Universe...

....never mind.


David A. Smith

"Darwin123" escreveu na mensagem
...

In any case, learn some Newtonian physics before you start
praising him. The best praise for Newton is learning his theory.
Calling him infallible without knowing the caveats to his theory is
false praise.
To put it another way, don't hide behind Newton or Galileo. Name
dropping won't make your arguments any stronger.


I guess we agree after all.

I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).

All bodies in the Universe are seen to spin.
Nobody ever pointed one that isn't.
All them are top/gyroscopes. Nevertheless...

The truth is that I've been loosing faith on Physics.
....and the process is exponential.
Ignorance is a bless.
I'd like to be ignorant again, back to time I loved
to read about relativity.

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
Do you have an equation, dimensionally consistent, where one
could see what you are talking about?

Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame,
and t is the time coordinate of the frame); v(t) can be an arbitrary
function of time. Its elapsed proper time between t=T1 and t=T2 is:

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

(This is easily obtained by integrating the metric along the path of the
object.)

One trivially obtains \tau=T2-T1 for an object at rest in this frame.
And one clearly obtains \tauT2-T1 for nonzero v(t). Note that |v(t)| is
constrained to be less than c, so there is never a numerical problem
with the sqrt(.).


Tom Roberts

"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
Do you have an equation, dimensionally consistent, where one
could see what you are talking about?

Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame, and
t is the time coordinate of the frame); v(t) can be an arbitrary function
of time. Its elapsed proper time between t=T1 and t=T2 is:

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Meanwhile:
integral_sqrt(1-v(t)^2/c^2) dt =
= 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)

Something of the form:
1/2*x*sqrt(1-x^2) + 1/2*arcsin(x^2)
being: x = v(t)^2/c^2
v(t) = r omega = r w
w - angular velocity

Let me see where this leads, thanks.

On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...


I guess we agree after all.
It doesn't sound like that to me. I was suggesting that you don't
understand Newtonian physics, and should study it more carefully. Your
comments seem to reinforce my hypothesis.

I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

All bodies in the Universe are seen to spin.
Nobody ever pointed one that isn't.
All them are top/gyroscopes. Nevertheless...
...the knowledge of how these spinning objects is understood on
a practical level. Technicians, engineers, and scientist launch probes
from a spinning object (the earth), and predict the trajectories with
pin point accuracy. Gyroscopes are used on everything from submarines
to wheel chairs. They seldom break down, because engineers can model
the forces on these gyroscopes. Nevertheless, you somehow claim that
no one understands the top-gyroscope.

The truth is that I've been loosing faith on Physics.
...and the process is exponential.
Ignorance is a bless.
Than you have plenty of bless (i.e., bliss).
I'd like to be ignorant again, back to time I loved
to read about relativity.
However, without any mathematics, right? And without studying
forces, right?
You were not really reading about relativity.

"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...


\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

"Darwin123" escreveu na mensagem
...
On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:

I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

Do you think you know top/gyroscopes?
I've already faced tens of sharks around here with that
challenge and I'm fishing right now (why you think I've
changed my name - I'm fishing).

Do you want to try some basic general questions yourself?

Wait, you know something about precession.
(And only precession of an horizontal gyroscope for sure).
What about a tilted gyroscope, 360 degrees around?
What about nutation?
What about nutation under precession?
What about precession under nutation?
What about energy conservation?
What about gyroscope harmonic motion (clearly seen
combined with external mass inertias - or the usual
top harmonic oscillations)?
What about gyroscope impedance?
Just to mention a few...

I've already spent more then 4,000 hours in the past 3 Years,
looking for gyroscope equations that could predict
basic experimental results.
And I'm not doing this as a hobby. I really need then.
I got:
- Newton method done (T=dL/dt)
- Lagrangian method done
- Vectorial method done
All them full solutions done by myself, zero simplification.
(I did that because couldn't find a book, link, whatever).
What I need is a genius that could explain me
how to get energy conservation out of it.

Euler's equations of motion don't qualify.
Just try them for I1 = I2 = I3 (sphere).

"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...


\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):
\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
Do you have an equation, dimensionally consistent, where one
could see what you are talking about?
Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame, and
t is the time coordinate of the frame); v(t) can be an arbitrary function
of time. Its elapsed proper time between t=T1 and t=T2 is:

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

T1 is the lower limit and T2 is the upper limit of the \integral.

Around here, "_" introduces a subscript and "^" introduces
a superscript (which can be a power).


Meanwhile:
integral_sqrt(1-v(t)^2/c^2) dt =
= 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)

That's hopeless and wrong -- you cannot begin to do the integral until
v(t) is specified. For circular motion, v(t) is constant, the integral
is trivial:

\tau = sqrt(1-|v|^2/c^2) (T2 - T1)

Note that there is no term related to acceleration, and no term related
to "centrifugal force"; all that matters is SPEED (|v|) relative to the
inertial frame.


Tom Roberts