Dear El Enrrabadore-mor:

"El Enrrabadore-mor" wrote in
message ...

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

On May 6, 8:10 am, "El Enrrabadore-mor"

wrote:

Hubble constant is an acceleration, as everybody
knows.

km/sec / megaparsec = distance / time / distance
= 1/time

You might want to start a study in "dimensional analysis".
The 1700's and 1800's were a productive time.

Maybe I went a little bit to fast claiming "as everybody
knows". Even so, "Hubble acceleration" gives 730 hits
for an exact query.

And is horse**** still.

Hubble constant is an acceleration because, as I've
already explained, it's a velocity taken per a
distance (Megaparcec).

So it is a velocity at a distance. It is also a frequency change
at a distance. But it is no "acceleration".

So that, we know that during the time
it takes for light to travel that distance (1 Mpc) the
velocity of celestial bodies increase 85 m/s - we
have a "dv". So the time it takes for light to travel
the distance of 1 Mpc can be calculated - it will
be our "dt". dv / dt = 85m/s / (time for light to
travel 1Mpc) = Pioneer 11 acceleration.

Wrong. It is pretty close the the recession rate of the Moon
too. But that is not Hubble expansion either.

What I didn't mention was that Hubble constant
only becomes an acceleration when combined
with the velocity of light. Sorry.

It is not an acceleration even then. Sorry.

Seriously, you need to work on a different tack. You have played
this one out, and your chewing gum and string will no longer hold
it together.

David A. Smith

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

"El Enrrabadore-mor" wrote in message
...

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

On May 6, 8:10 am, "El Enrrabadore-mor"
wrote:

Hubble constant is an acceleration, as everybody
knows.

km/sec / megaparsec = distance / time / distance
= 1/time

You might want to start a study in "dimensional analysis".
The 1700's and 1800's were a productive time.

Maybe I went a little bit to fast claiming "as everybody
knows". Even so, "Hubble acceleration" gives 730 hits
for an exact query.

And is horse**** still.

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".


Hubble constant is an acceleration because, as I've
already explained, it's a velocity taken per a
distance (Megaparcec).

So it is a velocity at a distance. It is also a frequency change at a
distance. But it is no "acceleration".

Well, this is only a matter of putting neurons working
together.
You know that per each Mpc distance, the speed of celestial
objects increase by 85 m/s, as seen by earth telescopes and
by means of light. This effect was called Hubble constant.
So, a stopped object at 1 Mpc distance will be seen at
a speed of 85 m/s.
A stopped object at 2 Mpc distance will be seen at
a speed of 170 m/s. And so on.
In fact, it doesn't matter if the object is stopped or moving.
Depending on the distance to Earth telescopes, one needs to
add a velocity increase of 85 m/s per Mpc.

Since the speed of light is constant and the standard
taken to define distance (the meter and all astronomical
distances), working a little bit further with our brain, one
can easily figure that Hubble constant states there is velocity
increase of celestial objects dependent on the time it
takes for light emitted by those celestial objects to
reach the Earth.
H = 85m/s / Mpc
If instead of units of Mpc for distance, you chose to use
such distance expressed in terms of time it takes for
like to travel that distance, what changed?
Nothing changes, since the speed of light is a
Universal constant taken to measure distances,
nothing changes.

Therefore, I can call Hubble constant an acceleration
of about 8.5x10^-10 m/s^2, and no one can ever
prove me wrong. Unless you don't agree that the
velocity of light is constant in vacuum and a good
standard for distances.

The effect is the same, but the value is not.
Call it H' = 8.5x10^-10 m/s^2 = Pioneer anomalous
acceleration.

Dear El Enrrabadore-mor:

"El Enrrabadore-mor" wrote in
message ...

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

"El Enrrabadore-mor" wrote in
message
...

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

On May 6, 8:10 am, "El Enrrabadore-mor"

wrote:

Hubble constant is an acceleration, as everybody
knows.

km/sec / megaparsec = distance / time / distance
= 1/time

You might want to start a study in "dimensional analysis".
The 1700's and 1800's were a productive time.

Maybe I went a little bit to fast claiming "as everybody
knows". Even so, "Hubble acceleration" gives 730 hits
for an exact query.

And is horse**** still.

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.

Hubble constant is an acceleration because, as I've
already explained, it's a velocity taken per a
distance (Megaparcec).

So it is a velocity at a distance. It is also a
frequency change at a distance. But it is no
"acceleration".

Well, this is only a matter of putting neurons
working together.

I have tried to point you at "dimensional analysis" a few times
now. When will you get the point?

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. You have no answer for Cassini not
showing the effect. You have no answer for a wrong sign. You
may or may not have an answer for planets not being affected.

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.

David A. Smith

On May 6, 3:20 pm, Eric Gisse wrote:
On May 6, 1:05 pm, Koobee Wublee wrote:

[snip whatever]

There is no need to snip whatever trollish actions. Here is the post
that you have so much trouble on.

http://groups.google.com/group/sci.p...58e4926ae7ed82

A brief reading of your repeated spew shows that you still do not
understand that the Lorentz transform is only valid for inertial
frames and that the acceleration [v --- -v] by the returning twin
breaks the symmetry that you try to use with the PoR.

Your brief reading is so much fouled. That is why you remain a multi-
year super-senior at the very prestigious University of Alaska
majoring in basket weaving. shrug

"El Enrrabadore-mor" wrote in
:


"bz" escreveu na mensagem
98.139...
"El Enrrabadore-mor" wrote in
:


"Tom Roberts" escreveu na mensagem
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One can analyze their experiment (including comparison to muon decay
at rest) in two different ways:
a) use the overall inertial frame of their storage ring
and apply SR.
b) use the equivalence principle of GR, and treat the LOCAL
acceleration of the stored muons as a gravitational field
and compute the gravitational time dilation in LOCAL
coordinates in which the stored muon is at rest.
These obtain the same answer.

Moreover:
Your a) appeals on velocity as the cause of
time dilatation.
Your b) appeals on acceleration (or gravity
by equivalence principle) to be the cause on
time dilatation.

It is not the velocity or the acceleration (in SR) that explains the
time difference.
It is the different trajectory.
Trajectory through space-time.

And that space-time is a Gaussian coordinate
system made of curved lines like two orthogonal
mirror spirals?

Please explain Bob.

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.






Physics say:
c) Acceleration is the time derivative of velocity.

My c) proves your a) and b) to be incompatible,
since time used on the derivative is ABSOLUTE
TIME.

a = dv/dt says nothing about ABSOLUTE time.

Where did you get the impression that it did?

dt is CHANGE in time.


I wonder how derivatives will be if the said changing
time had been already affected by time dilatation.

Just imagine that velocity changes.
Time dilatation will change too.
You get a derivative where time himself changes.
What a mess I presume.

Use partial derivatives when the function is a function of more than one
thing.




--
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

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...

Why did you post such crap?

You asked about crap.

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

...
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.

*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.

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?

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.

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.

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. 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.

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

David A. Smith

On May 7, 8:30*am, bz wrote:
"El Enrrabadore-mor" wrote :







"bz" escreveu na mensagem
. 198.139...
"El Enrrabadore-mor" wrote in
:

"Tom Roberts" escreveu na mensagem
...

One can analyze their experiment (including comparison to muon decay
at rest) in two different ways:
a) use the overall inertial frame of their storage ring
* *and apply SR.
b) use the equivalence principle of GR, and treat the LOCAL
* *acceleration of the stored muons as a gravitational field
* *and compute the gravitational time dilation in LOCAL
* *coordinates in which the stored muon is at rest.
These obtain the same answer.

Moreover:
Your a) appeals on velocity as the cause of
time dilatation.
Your b) appeals on acceleration (or gravity
by equivalence principle) to be the cause on
time dilatation.

It is not the velocity or the acceleration (in SR) that explains the
time difference.
It is the different trajectory.
Trajectory through space-time.

And that space-time is a Gaussian coordinate
system made of curved lines like two orthogonal
mirror spirals?

Please explain Bob.

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 one of the nicest explanations of the effect. :-)

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).


Tom Roberts

Tom Roberts wrote in
:

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).


Yes, you are correct. I should have been a bit more formal in my
explaination. 'Trajectory' seemed to be throwing him a curve and I hoped
to make it a bit clearer. Thanks for cleaning up the loose ends.




--
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

On May 7, 12:57*pm, bz wrote:
Tom Roberts wrote :





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).

Yes, you are correct. I should have been a bit more formal in my
explaination. 'Trajectory' seemed to be throwing him a curve and I hoped
to make it a bit clearer. Thanks for cleaning up the loose ends.

--
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- Hide quoted text -

- Show quoted text -

Centrifugal motion slows time.