SR/GR use absolute time to synchronize the GPS clocks with the ground clock.
in [Relativity]
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From: Sam Wormley on 24 May 2010 23:28 On 5/24/10 2:02 PM, kenseto wrote: > On May 24, 9:46 am, Sam Wormley<sworml...(a)gmail.com> wrote: >> >> Boy are you confused, Ken. Let A be the observer and B be a clock >> in relative motion, then from the perspective >> of A. >> >> Now choosing the perspective of B, Let B be the observer and A be a >> clock in relative motion, then ∆t' of A = γ ∆t of A. > > Hey idiot....I corrected you many times. How can B predicts that an > interval of A's clock Delta(t'_A) on the A clock be equal to > gamma*delta(t_A) on the same A clock???? B observes (measures) what SR predicts, and that is ∆t_A' = γ ∆t-A Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html > > > Ken Seto > >> >> You can have ONLY ONE PERSPECTIVE SIMULTANEOUSLY... sorry, my caps key >> stuck on momentarily. You can only have one perspective, either that >> of A where ∆t_B' = γ ∆t_B , >> or B where ∆t_A' = γ ∆t-A . >> >> Seto, you have failed, and I do mean FAILED, to understand this simple >> concept of relativity for many years. There is no contradiction, as >> one can only have ONE perspective.- Hide quoted text - >> >> - Show quoted text - >
From: kenseto on 25 May 2010 08:56 On May 24, 11:28 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/24/10 2:02 PM, kenseto wrote: > > > On May 24, 9:46 am, Sam Wormley<sworml...(a)gmail.com>  wrote: > > >>   Boy are you confused, Ken. Let A be the observer and B be a clock > >>   in relative motion, then from the perspective > >>   of A. > > >>   Now choosing the perspective of B, Let B be the observer and A be a > >>   clock in relative motion, then ât' of A = γ ât of A. > > > Hey idiot....I corrected you many times. How can B predicts that an > > interval of A's clock Delta(t'_A) on the A clock be equal to > > gamma*delta(t_A) on the same A clock???? > >   B observes (measures) what SR predicts, and that is >    ât_A' = γ ât-A Hey idiot...You didn't answer my question. B predicts that an interval of delta(t'_A) on the A clock is worth: Delta(t'_A)= gamma*Delta(t_B) Ken Seto > >   Physics FAQ: What is the experimental basis of special relativity? >    http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html > > > > > > > Ken Seto > > >>   You can have ONLY ONE PERSPECTIVE SIMULTANEOUSLY... sorry, my caps key > >>   stuck on momentarily. You can only have one perspective, either that > >>   of A where ât_B' = γ ât_B , > >>   or B where ât_A' = γ ât-A . > > >>   Seto, you have failed, and I do mean FAILED, to understand this simple > >>   concept of relativity for many years. There is no contradiction, as > >>   one can only have ONE perspective.- Hide quoted text - > > >> - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Sam Wormley on 25 May 2010 09:45 On 5/25/10 7:56 AM, kenseto wrote: > > Hey idiot...You didn't answer my question. B predicts that an interval > of delta(t'_A) on the A clock is worth: > Delta(t'_A)= gamma*Delta(t_B) > Ken Seto Seto, you confuse A and B -- Know wonder you never understand what is going on! You cannot assume that when A and B are separated and in relative motion that ∆t_A is identical to ∆t_B. Nor can you assume that ∆t_A' is identical to ∆t_B'. Seto, you mistakenly think these are all interchangeable. They are not! ____________________ A and B are observers with identical clocks. That is A and B's clocks ticked synchronously when they were together. ∆t represent a time interval between tick of the clocks. Special relativity predicts that observer A will measure that ∆t_B' = γ ∆t_B where ∆t represent a time interval, v is the relative velocity between A and B, and γ = 1/√(1-v^2/c^2) . Furthermore, special relativity predicts that observer B will measure that ∆t_A' = γ ∆t_A Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
From: Sam Wormley on 25 May 2010 10:02 On 5/25/10 8:45 AM, Sam Wormley wrote: > On 5/25/10 7:56 AM, kenseto wrote: > >> >> Hey idiot...You didn't answer my question. B predicts that an interval >> of delta(t'_A) on the A clock is worth: >> Delta(t'_A)= gamma*Delta(t_B) >> Ken Seto > > > Seto, you confuse A and B -- Know wonder you never understand > what is going on! > > You cannot assume that when A and B are separated and in relative > motion that ∆t_A is identical to ∆t_B. Nor can you assume that ∆t_A' > is identical to ∆t_B'. Seto, you mistakenly think these are all > interchangeable. They are not! > > ____________________ > > > A and B are observers with identical clocks. That is A and B's > clocks ticked synchronously when they were together. > > ∆t represent a time interval between tick of the clocks. > > Special relativity predicts that observer A will measure that > ∆t_B' = γ ∆t_B > > where ∆t represent a time interval, v is the relative velocity > between A and B, and γ = 1/√(1-v^2/c^2) . > > Furthermore, special relativity predicts that observer B will > measure that > ∆t_A' = γ ∆t_A > > Physics FAQ: What is the experimental basis of special relativity? > http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html > I would like to further clarify something for you, Seto Special relativity predicts that observer A will measure that ∆t_B' = γ ∆t_B Because ∆t_B' is a measured quantity. A can calculate what ∆t_B would be. ∆t_B = ∆t_B'/γ ^ ^ | | Calculated Measured I think one of your problems, all along, is you were confusing ∆t_B with ∆t_A and/or their primes and mistakenly mixing them up in the equations. This would thoroughly confuse any student. ∆t_A' ≠ γ ∆t_B That makes no sense to try to equate them.
From: kenseto on 25 May 2010 13:19
On May 25, 9:45 am, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/25/10 7:56 AM, kenseto wrote: > > > > > Hey idiot...You didn't answer my question. B predicts that an interval > > of delta(t'_A) on the A clock is worth: > > Delta(t'_A)= gamma*Delta(t_B) > > Ken Seto > >   Seto, you confuse A and B -- Know wonder you never understand >   what is going on! No it is you who is confused. If B is the observer he predicts that an interval of Delta(t'_A) on the B clock is worth gamma*delta(t_B) on the B clock. > >   You cannot assume that when A and B are separated and in relative >   motion that ât_A is identical to ât_B. Hey idiot I did not make such assumption. > Nor can you assume that ât_A' >   is identical to ât_B'. Seto, you mistakenly think these are all >   interchangeable. They are not! Hey idiot I did not make such assumption. > >          ____________________ > >   A and B are observers with identical clocks. That is A and B's >   clocks ticked synchronously when they were together. Yes. > >   ât represent a time interval between tick of the clocks. > >   Special relativity predicts that observer A will measure that >    ât_B' = γ ât_B No... A predicts that ât_B' is worth γ ât_A Ken Seto > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   Furthermore, special relativity predicts that observer B will >   measure that >    ât_A' = γ ât_A > >   Physics FAQ: What is the experimental basis of special relativity? >    http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html |