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Time dilatation in circular motion



 
 
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  #1  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
El Enrrabadore-mor
external usenet poster
 
Posts: 2
Default Time dilatation in circular motion

It is said that a speeding clock shows less elapsed time than
the stay-at-home clock, because (if already speeding) it is
running at a slower rate. Or else, because it had run at a
slower rate when it was speeding, assuming now it is
stopped near the stay-at-home clock.

The funny thing about this is that time and length change
at the same time, but not the ratio between both (velocity).

If we keep length constant, the only possible solution is
uniform circular motion. That is a twin travelling in circles,
of constant radius r, around the first twin assumed to be
stopped at the center of rotation.

Let's say the radius r is a constant value of 100 light-seconds
(r = 100c).
The speeding twin goes on a spaceship at 0.999c.
Therefore, the angular speed 'omega' is v/r = 0.999c/100c
= 0.01 rad/s.
The speeding twin takes 628 seconds to have a complete turn
of 360 degrees.

For small values of t, the speeding twin is almost going in
a straight line, but is fact it has a centripetal force f = m c^2/r
= m c/100, being the centripetal acceleration c/100, towards
the first stopped twin in the center of rotation.

Both twins have powerful antennas that broadcast radio
spherically around the entire space. Both twins are tuned
to each other frequency/radio-station.

Since the distance r = 100c between the emitter and the
receptor is constant, obviously that both twin will hear
each other radio (music) in perfect conditions.

Nevertheless, relativity says that the clock synchronising
the emission of the speeding twin must be running at
a clock rate close to zero. Theoretically, the speeding
twin won't have any trouble receiving the stay-at-home
radio emission, but the stay-at-home twin cannot
receive the speeding twin radio emission, because
the speeding clock is running near zero.
The speeding twin radio emission will take infinite
time to broadcast one single spoken word. The
stay-at-home will be dead by the time the speeding
twin could say a single word.

The trouble seams to be the acceleration:
a = (0.999)^2 c/100 which is about c/100.
(Here the number 100 means 100 seconds).
That's a huge gravity field of 300,000g at
a radius of 100 light-seconds, just imagine
the value it will be at Earth radius based
on the inverse-square Law.)

I presume that such acceleration of 300,000g will
be responsible for a clock speed up rate that
should keep time unchanged after all.

Any comments welcome.







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  #2  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
Androcles[_7_]
external usenet poster
 
Posts: 4,339
Default Time dilatation in circular motion

This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"El Enrrabadore-mor" wrote in message
...
| It is said that a speeding clock shows less elapsed time than
| the stay-at-home clock, because (if already speeding) it is
| running at a slower rate. Or else, because it had run at a
| slower rate when it was speeding, assuming now it is
| stopped near the stay-at-home clock.
|
| The funny thing about this is that time and length change
| at the same time, but not the ratio between both (velocity).
|
| If we keep length constant, the only possible solution is
| uniform circular motion. That is a twin travelling in circles,
| of constant radius r, around the first twin assumed to be
| stopped at the center of rotation.
|
| Let's say the radius r is a constant value of 100 light-seconds
| (r = 100c).
| The speeding twin goes on a spaceship at 0.999c.
| Therefore, the angular speed 'omega' is v/r = 0.999c/100c
| = 0.01 rad/s.
| The speeding twin takes 628 seconds to have a complete turn
| of 360 degrees.
|
| For small values of t, the speeding twin is almost going in
| a straight line, but is fact it has a centripetal force f = m c^2/r
| = m c/100, being the centripetal acceleration c/100, towards
| the first stopped twin in the center of rotation.
|
| Both twins have powerful antennas that broadcast radio
| spherically around the entire space. Both twins are tuned
| to each other frequency/radio-station.
|
| Since the distance r = 100c between the emitter and the
| receptor is constant, obviously that both twin will hear
| each other radio (music) in perfect conditions.
|
| Nevertheless, relativity says that the clock synchronising
| the emission of the speeding twin must be running at
| a clock rate close to zero. Theoretically, the speeding
| twin won't have any trouble receiving the stay-at-home
| radio emission, but the stay-at-home twin cannot
| receive the speeding twin radio emission, because
| the speeding clock is running near zero.
| The speeding twin radio emission will take infinite
| time to broadcast one single spoken word. The
| stay-at-home will be dead by the time the speeding
| twin could say a single word.
|
| The trouble seams to be the acceleration:
| a = (0.999)^2 c/100 which is about c/100.
| (Here the number 100 means 100 seconds).
| That's a huge gravity field of 300,000g at
| a radius of 100 light-seconds, just imagine
| the value it will be at Earth radius based
| on the inverse-square Law.)
|
| I presume that such acceleration of 300,000g will
| be responsible for a clock speed up rate that
| should keep time unchanged after all.
|
| Any comments welcome.



Install an atomic clock at Ross Island, McMurdo Sound.
Antarctica's largest science base, the United States' McMurdo Station, as
well as New Zealand’s Scott Base are located on the island’s south shore.

"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."
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

I haven't heard anyone screaming about how Einstein was proved right and it
would be quite easy to do.



  #3  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
N:dlzc D:aol T:com \(dlzc\)[_1175_]
external usenet poster
 
Posts: 1
Default Time dilatation in circular motion

"El Enrrabadore-mor" wrote in
message ...
....
Any comments welcome.


http://hermes.physics.adelaide.edu.a...eleration.html
http://hermes.physics.adelaide.edu.a...igid_disk.html

David A. Smith


  #4  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
Androcles[_7_]
external usenet poster
 
Posts: 4,339
Default Time dilatation in circular motion



--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"N:dlzc D:aol T:com (dlzc)" wrote in message
...
| "El Enrrabadore-mor" wrote in
| message ...
| ...
| Any comments welcome.
|
|
http://hermes.physics.adelaide.edu.a...eleration.html
|
http://hermes.physics.adelaide.edu.a...igid_disk.html
|
| David A. Smith

The prat Koks writes
"How could a physicist like Born, mathematically sophisticated, have made
such an elementary error?"

What he should have written was

How could a jerk like Einstein, mathematically unsophisticated, have made
such an
obvious error as:
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the time each way is the same?

How could cretins like Koks and Smiffy believe such garbage?










  #5  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
Uncle Al
external usenet poster
 
Posts: 16,279
Default Time dilatation in circular motion

El Enrrabadore-mor wrote:

It is said that a speeding clock shows less elapsed time than
the stay-at-home clock, because (if already speeding) it is
running at a slower rate. Or else, because it had run at a
slower rate when it was speeding, assuming now it is
stopped near the stay-at-home clock.

The funny thing about this is that time and length change
at the same time, but not the ratio between both (velocity).

If we keep length constant, the only possible solution is
uniform circular motion. That is a twin travelling in circles,
of constant radius r, around the first twin assumed to be
stopped at the center of rotation.

[snip]

The experiment was done with Mossbauer spectroscopy in an
ultracentrifuge, rim to hub. READ IT.

http://prola.aps.org/abstract/PR/v129/i6/p2371_1

Fe-57 14.4-keV Mossbauer absorption line traversing an ultracentrifuge
rotor was measured versus angular velocity omega. Fe-57 absorber
placed at 9.3 cm radius. Co-57 source mounted on a hub piezoelectric
transducer. Triangular voltage wave was applied to the transducer to
move source relative to absorber. The entire resonance line was
observed at various omega values. Measured transverse Doppler shift
agreed within 1.1% experimental error with relativity predictions.

Let's say the radius r is a constant value of 100 light-seconds
(r = 100c).

[snip crap]

Precision measurement over an artifical 18.7 million mile radius? Not
clever.

Any comments welcome.


1) Due diligence in a literature search lest you make a public fool
of yourself.
2) Fit performed experiments on a lab bench unless nature is
picking up the tab.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
  #6  
Old May 3rd 08 posted to sci.physics.relativity,sci.physics
mitch.nicolas.raemsch@gmail.com
external usenet poster
 
Posts: 2,849
Default Time dilatation in circular motion

On May 3, 11:52*am, "Androcles" wrote:
This message is brought to you by Androcles
*http://www.androcles01.pwp.blueyonder.co.uk/

"El Enrrabadore-mor" wrote in message

...
| It is said that a speeding clock shows less elapsed time than
| the stay-at-home clock, because (if already speeding) it is
| running at a slower rate. Or else, because it had run at a
| slower rate when it was speeding, assuming now it is
| stopped near the stay-at-home clock.
|
| The funny thing about this is that time and length change
| at the same time, but not the ratio between both (velocity).
|
| If we keep length constant, the only possible solution is
| uniform circular motion. That is a twin travelling in circles,
| of constant radius r, around the first twin assumed to be
| stopped at the center of rotation.
|
| Let's say the radius r is a constant value of 100 light-seconds
| (r = 100c).
| The speeding twin goes on a spaceship at 0.999c.
| Therefore, the angular speed 'omega' is v/r = 0.999c/100c
| = 0.01 rad/s.
| The speeding twin takes 628 seconds to have a complete turn
| of 360 degrees.
|
| For small values of t, the speeding twin is almost going in
| a straight line, but is fact it has a centripetal force f = m c^2/r
| = m c/100, being the centripetal acceleration c/100, towards
| the first stopped twin in the center of rotation.
|
| Both twins have powerful antennas that broadcast radio
| spherically around the entire space. Both twins are tuned
| to each other frequency/radio-station.
|
| Since the distance r = 100c between the emitter and the
| receptor is constant, obviously that both twin will hear
| each other radio (music) in perfect conditions.
|
| Nevertheless, relativity says that the clock synchronising
| the emission of the speeding twin must be running at
| a clock rate close to zero. Theoretically, the speeding
| twin won't have any trouble receiving the stay-at-home
| radio emission, but the stay-at-home twin cannot
| receive the speeding twin radio emission, because
| the speeding clock is running near zero.
| The speeding twin radio emission will take infinite
| time to broadcast one single spoken word. The
| stay-at-home will be dead by the time the speeding
| twin could say a single word.
|
| The trouble seams to be the acceleration:
| a = (0.999)^2 c/100 which is about c/100.
| (Here the number 100 means 100 seconds).
| That's a huge gravity field of 300,000g at
| a radius of 100 light-seconds, just imagine
| the value it will be at Earth radius based
| on the inverse-square Law.)
|
| I presume that such acceleration of 300,000g will
| be responsible for a clock speed up rate that
| should keep time unchanged after all.
|
| Any comments welcome.

Install an atomic clock at Ross Island, McMurdo Sound.
Antarctica's largest science base, the United States' McMurdo Station, as
well as New Zealand’s Scott Base are located on the island’s south shore.

"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."
Ref: *http://www.fourmilab.ch/etexts/einstein/specrel/www/


But centrifigul force cancels gravity weight at the equator.

You weigh less.




I haven't heard anyone screaming about how Einstein was proved right and it
would be quite easy to do.


  #7  
Old May 4th 08 posted to sci.physics.relativity,sci.physics
El Enrrabadore-mor[_2_]
external usenet poster
 
Posts: 173
Default Time dilatation in circular motion


"N:dlzc D:aol T:com (dlzc)" wrote
...
"El Enrrabadore-mor" wrote in message
...
...
Any comments welcome.


http://hermes.physics.adelaide.edu.a...eleration.html
http://hermes.physics.adelaide.edu.a...igid_disk.html


I've read the links. Let me tell you that my horse dung
detector exploded due to overload.
Both are total nonsense.

The first article blames on rapidity.
What is rapidity?

The second, besides all excuses made, says that:
quote:
"Einstein doesn't give an explicit formula for the spacetime
metric of the rigid spinning disk, but here's one obvious
candidate (assuming the angular speed is omega): "
Tell me please, is it d(theta squared), or the square of
d(theta)?
Also:
"The events along a particle's worldline are parametrized
by 'tau'."
Since worldlines are time like curves in spacetime, 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.
Some terms have time dimension and others lenght
dimensions, mixed, and cannot be a valid equation.
Only a fool buys that.



David A. Smith






  #8  
Old May 4th 08 posted to sci.physics.relativity,sci.physics
El Enrrabadore-mor[_2_]
external usenet poster
 
Posts: 173
Default Time dilatation in circular motion


"Uncle Al" wrote
...
El Enrrabadore-mor wrote:

It is said that a speeding clock shows less elapsed time than
the stay-at-home clock, because (if already speeding) it is
running at a slower rate. Or else, because it had run at a
slower rate when it was speeding, assuming now it is
stopped near the stay-at-home clock.

The funny thing about this is that time and length change
at the same time, but not the ratio between both (velocity).

If we keep length constant, the only possible solution is
uniform circular motion. That is a twin travelling in circles,
of constant radius r, around the first twin assumed to be
stopped at the center of rotation.

[snip]

The experiment was done with Mossbauer spectroscopy in an
ultracentrifuge, rim to hub. READ IT.

http://prola.aps.org/abstract/PR/v129/i6/p2371_1

Fe-57 14.4-keV Mossbauer absorption line traversing an ultracentrifuge
rotor was measured versus angular velocity omega. Fe-57 absorber
placed at 9.3 cm radius. Co-57 source mounted on a hub piezoelectric
transducer. Triangular voltage wave was applied to the transducer to
move source relative to absorber. The entire resonance line was
observed at various omega values. Measured transverse Doppler shift
agreed within 1.1% experimental error with relativity predictions.

Let's say the radius r is a constant value of 100 light-seconds
(r = 100c).

[snip crap]

Precision measurement over an artifical 18.7 million mile radius? Not
clever.


If not clever, why did you bring up the Mossbauer argument,
that you have qualify as "not clever" for a 100 second-light radius?

Do you pretend to measure resonance lines too?
Any frequency that is a multiple of the natural frequency will
cause resonance. So the 46 Years old experiment can only
be horse dung.

BTW, what's relativity prediction?



  #9  
Old May 4th 08 posted to sci.physics.relativity,sci.physics
Eric Gisse
external usenet poster
 
Posts: 16,899
Default Time dilatation in circular motion

On May 3, 3:31*pm, "El Enrrabadore-mor"
wrote:
"Uncle Al" ...



El Enrrabadore-mor wrote:


It is said that a speeding clock shows less elapsed time than
the stay-at-home clock, because (if already speeding) it is
running at a slower rate. Or else, because it had run at a
slower rate when it was speeding, assuming now it is
stopped near the stay-at-home clock.


The funny thing about this is that time and length change
at the same time, but not the ratio between both (velocity).


If we keep length constant, the only possible solution is
uniform circular motion. That is a twin travelling in circles,
of constant radius r, around the first twin assumed to be
stopped at the center of rotation.

[snip]


The experiment was done with Mossbauer spectroscopy in an
ultracentrifuge, rim to hub. *READ IT.


http://prola.aps.org/abstract/PR/v129/i6/p2371_1


Fe-57 14.4-keV Mossbauer absorption line traversing an ultracentrifuge
rotor was measured versus angular velocity omega. *Fe-57 absorber
placed at 9.3 cm radius. *Co-57 source mounted on a hub piezoelectric
transducer. *Triangular voltage wave was applied to the transducer to
move source relative to absorber. *The entire resonance line was
observed at various omega values. *Measured transverse Doppler shift
agreed within 1.1% experimental error with relativity predictions.


Let's say the radius r is a constant value of 100 light-seconds
(r = 100c).

[snip crap]


Precision measurement over an artifical 18.7 million mile radius? *Not
clever.


If not clever, why did you bring up the Mossbauer argument,
that you have qualify as "not clever" for a 100 second-light radius?


Mossbauer is clever.


Do you pretend to measure resonance lines too?
Any frequency that is a multiple of the natural frequency will
cause resonance. So the 46 Years old experiment can only
be horse dung.


It doesn't work that way. The energy has to be exactly the energy of
the resonance - not just any "multiple".


BTW, what's relativity prediction?


Well, glad we got that out of the way - you didn't even pretend to
read the paper.
  #10  
Old May 4th 08 posted to sci.physics.relativity,sci.physics
El Enrrabadore-mor[_2_]
external usenet poster
 
Posts: 173
Default Time dilatation in circular motion


"Eric Gisse" escreveu na mensagem
...
On May 3, 3:31 pm, "El Enrrabadore-mor"
wrote:

Any frequency that is a multiple of the natural frequency will
cause resonance. So the 46 Years old experiment can only
be horse dung.


It doesn't work that way. The energy has to be exactly the energy of
the resonance - not just any "multiple".


What the hell is "energy of resonance"?
I've been working with resonant systems for Years
(mechanical and electrical) and never thought about
any "energy of resonance".
A system in resonance is a closed system that exchanges
no energy with surroundings.
Energy is conserved in resonance.
Go back to the classic mechanics class, where you shouldn't
ever get out without knowing the minimum.


 




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