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

"Tom Roberts" escreveu na mensagem
...
bz wrote:
Look at the particle that has just made one trip around the ring.
It is now back at its starting point in space but has moved in time.
The particle that stayed stationary at the center of the ring has NOT
moved in space but it, too has move in time. The integral along both
trajectories must have the same length. Therefore the moving particle has
traveled less distance along 'its time axis' (but the same distance along
the 'Lab' time axis.

This is a fruitful way to look at it, but your statement "The integral
along both trajectories must have the same length." is inadequate --
integral of WHAT? What you mean is that the elapsed time in the lab frame
is the same for both.

Here's a semi-mathematical way of looking at it:

[Notation: \tau is always the proper time of the particle
being discussed. t and x are always lab coordinates, with
x measured tangent to the ring.]

For the particle at rest in the lab, the lab time axis and the particle's
time axis are the same, so dt/d\tau=1. While time T elapses in the lab
frame, the particle experiences T elapsed proper time.

For the particle going around the ring, dx/d\tau is nonzero (this is a
spatial component of its 4-velocity). Because 4-velocity is normalized to
1, this implies that dt/d\tau1 for this particle [#]. So while the time T
elapses in the lab frame, the particle experiences less than T elapsed
proper time.

[#] U^i = dx^i/d\tau. |U|=g_ij U^i U^j.
Remember that {g_ij} = diag(-1,1,1,1).

Yep.
Even better then bz.

Do you have an equation, dimensionally consistent, where one
could see what you are talking about?

The problem is pretty simple, only time and the angle theta
are variables. The radius is constant and no changes on z exist.
Can you please make the equations meaningful?

Does this equation (more or less from Physics FAQ) qualify?
(c*dtau)^2 = (c*dt)^2-dz^2-dr^2-r^2d(theta)^2/(1-r^2omega^2/c^2)
From: "The Rigid Rotating Disk in Relativity"
http://hermes.physics.adelaide.edu.a...igid_disk.html
Those clearly look to be cylindrical coordinates, whe
ds = dz + dr + r d(theta)

I'm to believe that relativity is the art to write
sound bites that no one can understand.

"dlzc" escreveu na mensagem
...
Dear El Enrrabadore-mor:

On May 7, 4:14 am, "El Enrrabadore-mor"
wrote:
"N:dlzcD:aol T:com (dlzc)" escreveu na
...
"El Enrrabadore-mor" wrote in message
...
....
Horse****?
So far this one have passed 3 or 4 times.
You 're the first to insist that Hubble
constant combined with a constant speed
of light doesn't generate an "Hubble
acceleration".

Sure, that same crowd that thinks that
"Planck mass" is meaningful. They failed
"dimensional analysis", and seek to
confuse numerology with science.

Dimensional analysis is fundamental.
Anything that doesn't meet dimensional
analysis is crap. That's why I told you
that the link you provided about
"The Rigid Rotating Disk in Relativity"
snip link now broken by GG
is horse dung.
I shall said BIG SMELLY HORSE MANURE in
Physiscs FAQ. The equation given:
d(tau)^2 = dt^2-dz^2-dr^2-r^2d(theta)^2/(1-r^2omega^2)
not even meets simple dimension requirements.

The analysis was done. t and tau are "time times c". Had you read
the page...

Usually I don't read crap, but yes, I've read.

Even so, to get the dimension right you also need do
divide *r^2omega^2* per *c^2*, otherwise you get
the same horse manure, smaller, but manure anyway.

If I'm right, and it is *c^2* dividing *r^2omega^2*,
the formula is the trivial linear equation, written
in cylindrical coordinates.

And what about the missing 2pi term?
I smell a missing 2pi =6.28... term.
See, 2pi*r is the perimeter that gives w.

So, the length contraction is on "theta".
Let me guess! The circle doesn't measure 2pi
radians anymore?


Why did you post such crap?

You asked about crap.

I didn't say "rotating disk" anywhere, nor the
problem given has any kind of rigidity.

What I've asked in first place sounds impossible but can
be made very real, for smaller velocity and smaller
acceleration figures.
For instance, planets around the Sun have different
velocities and different gravitational fields, not only
from the sun, but also due to its own gravity plus the ...
....centrifugal force (here we go).
Imagine some perfect atomic clocks. Place one by each
of all planets of the solar system.
One day we go get all those clocks and bring then
to Earth again.
If you can predict the time clocks will show in this
real question, you can also answer my original question,
because the original question is just the limite case.


Do you some times require dimensional
analysis and some other times not?

Can you read?

...
snip you harping about what you do not know about dimensional
analysis, because you appear unable to read

I don't need to read, I just smell it.

....
Your "mental gyrations" (snipped) show that you have
no clue about the topic. You have no answer for
doubling the distance but the effect is the same.

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?


You have no answer for Cassini not showing the
effect.

I've already told you it is a matter of distance.
Cassini didn't went that far, nor known data exists.
What else can I tell about that?

Cassini did not show the effect, when Pioneer (and others) did.
Pioneer showed the effect *before* it got as far as Saturn.

Are you sure?
(Of course you're sure. But remember Pioneer took about 18
Years to be known. And I guess no one want to loose the job.)


You have no answer for a wrong sign.

The wrong sign happens to be the right sign
for a non-existent effect computed in first
place.

So the Pioneer anomaly is just a math error?

Relativity and/or Hubble effect was in the computer program
in first place. Why shouldn't be?


You may or may not have an answer for
planets not being affected.

Planets orbit, don't go away.
The argument makes no sense at all.

You have an anomalous acceleration at some "place", starting just
before or just after the orbit of Jupiter. Yet these planets don't
accelerate anomalously, maintaining their orbits over hundred of
years.

I know that you know. Now you are just kidding due to lack
of better arguments.
If planets feel such effect, then I presume the main consequence
will be that you compute wrong masses for planets.

Actually, my point is that Universe expansion is crap.
It should be you to explain why planets don't suffer
from universe expansion, but...never mind.


What sense could your planets argument make?
Care to explain?

Pioneer and Saturn shared the same space. Pioneer linearly
accelerated Sunward, Saturn did not. Same story for all the other
outer planets.

Saturn feels happy to move where the sun gravity pull
equates the centrifugal force due to its angular speed.
If you increase the gravity pull, no problem, saturn
fall to a lower orbit and in the processes increases its
speed accordingly (since it's coming down).


If you are going to invent your own set of
"magic numbers", you don't need to play with
me. I mistakenly thought that you wanted to
discuss physics.

I've just presented a fact, coincidence or not,
the facts are that the anomalous acceleration
experienced by Pioneer spacecrafts is a PERFECT
MATCH when one combines:
- Hubble Constant
- Speed of light constant
That's all.

Which is not physics.

Not physics?
I've converted distance into time, like Physics had
converted the speed of light into distance to define
the meter standard, and you claim is not physics?

And I got dimension analysis right:
m/s /s = m/s^2 = acceleration


It does not present a model. It does not allow
for falsifiability. It discounts what effects this "application"
would have on:
- the probe's size and internal processes
- on the fact that the effect does not vary by distance... required
when you invoke Hubble's constant.

A constant acceleration whose effect vary by distance is
a lovely argument.
Does anyone need a model to calculate the time it takes
for light to travel a distance of 1 Mpc? You are kidding
with me, for sure.


So you play number games, and waste time. This is known as
"numerology".

David A. Smith

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.

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

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.

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...

David A. Smith

On May 5, 11:36 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...



On May 5, 8:10 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na
...

Your problem is that you don't know ANY physics, including
Newtonian. Learn Newtonian physics and get back to me.

OK, Darwin, I promise I will study Newton one day.
I'm only 17 Years old, you know... I'm a liar too.
When I grown up I'm going to be a relativistic
rocket ship constructor, you know?

Thanks for explaining me Physics, specially that
centrifugal force doesn't exist.
I'm going to pick my friend motorcycle (100CV)
and I'm going to make the coffee turn (a 90 degrees
curve) at 250 Km/h. No problem, it doesn't exist.

Make sure you have enough centripetal force, though. Centripetal force
does exist. If you don't have enough centripetal force, you won't make
that coffee turn. You will go in a straight line which will intersect
a wall. Then you will be dead. So make sure about that centripetal
force.

All right Darwin.
To have that centriptal force you're talking about,
required to balance a new born centrifugal force,
you need a fixed point, with infinite rigidity, at the center
of rotation.
When such point exists, the centrifugal force exists.
The centrifugal force only exists in an accelerated reference
frame, as defined by Newton. Newton describes an absolute space and
time, which means a coordinate system where his three laws apply.
Alas, centrifugal force does not exist in this coordinate system.
Because it doesn't exist in the absolute coordinate system, it can't
balance the centripetal force.
Another way to look at it is to ask yourself the following
question. What object in the universe is applying that centrifugal
force? In the case of earth orbiting the sun, I know what object is
applying force to the earth. The object applying centripetal force to
the earth is the sun, by means of its gravitational field. In fact, I
can say that the centripetal force is the gravitational force of the
sun. However, I can't say the same for the centrifugal force. What
object is applying centrifugal force to the earth? There is no such
object.
I'll take the epistemological view that only observers in an
inertial frame experience "real" forces. from that point of view,
there are no real forces. Because if an observer is in an inertial
frame, he can't experience a centrifugal force. If you break your head
in an motorcycle accident, obviously you are not in an inertail frame.
You were in an inertial frame until you hit your head.
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.

"N:dlzc D:aol T:com (dlzc)" wrote in message
...
| 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.
|
| Yet the "blue shift" (not "red shift") never varies as the
| distance increases.
|
| 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.

ABUSE! Report it!
Oh wait... He's a ****ing idiot that doesn't know his own name anyway.
Carry on calling the **** ignorant.




| 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...
|
| David A. Smith
|
|

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.