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.