SR/GR use absolute time to synchronize the GPS clocks with the ground clock.
in [Relativity]
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From: kenseto on 10 May 2010 09:24 On May 9, 11:31 am, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/9/10 9:32 AM, kenseto wrote: > > > ROTFLOL....you are the one who need to learn SR....what you said is an > > erroneous assumption of the PoR that every SR observer is in a state > > of absolute rest. This is wrong. The rate of two relative clock A and > > B is as follows: > > 1. A is running fast compared to B. > > 2. B is running slow compared to A. > >   Wrong again Seto! > >   Let's be a bit more precise here: > >   Assume that A and B have identical atomic clocks. That means they >   tick at the same rate. Now let us suppose that A and B have relative >   motion, such that their velocity with respect to each other, v > 0, >   and that dv/dt = 0 . "Closing speed" or "Separation speed", if you >   like. > >   Correcting for any Doppler shift, A measures B's time interval as >    ât_B' = γ ât_B > >   and B measures A's time interval as >    ât_A' = γ ât_A > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   Therefore, A measures B's time interval to be longer than her own. >   And B measures A's time interval to be longer than his own. No wormy....there is no measurement at all. A predicts that B's clock is running at a rate of 1/gamma or gamma....and B predicts that A is running at a rate of 1/gamma or gamma. Only one of A's prediction is correct and only one of B's prediction is correct. So if A is truly running at a faster rate of gamma than B then B is truly running at a slower rate of 1/gamma than A. What you said is based on the erroneous assumption that each onserver is in a state of rest (abslute rest) and thus each observer sees (predicts) the observed clock runs slow. You really need to learn what SR is really saying, wormy. Ken Seto
From: kenseto on 10 May 2010 09:33 On May 9, 3:38 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On May 7, 1:56 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> Sam Wormley <sworml...(a)gmail.com> writes: > >> >On 5/7/10 10:32 AM, Michael Moroney wrote: > >> >> My posts are made on the assumption that all GR effects on theGPS > >> >> satellites can be simplified to the gravitational well effect of > >> >> ~45uS/day and the motion effects of ~7uS/day. > >> > GTR covers everything! SR may be more convenient for some calculations > >> > but GTR encompasses SR. Both theories need not be applied separately. > > >> I know that. But the satellite in orbit is constantly accelerating, > >So why did you treat the GPS clock as inertial and then claim that the > >GPS sees the ground clock as 7 us/day slow and from the ground clock > >point of view it also sees the GPS as 7 us/day slow?????? > > If you had read what I wrote after that instead of just ignoring it, you'd > see that, for the case of this discussion, I'm simplifying things to the > gravitational effects (~38 uS) and SR effects (~7 uS) and treating them > as if they were separate effects that can be added. In an earlier post I said that if you accelerate away from me then your clock is running slower than my and thus there is no mutual time dilation because you experienced acceleration. Now with the GPS clock you claimed that there is mutual time dilation for the SR effect between the GPS clock and the ground clock even though the GPS was accelerated away from the ground clock. This proved that you just made thing up as you go along. Krn Seto > > > Do you just > >make up stuff as you go along???? > > No, that's your work.
From: kenseto on 10 May 2010 10:13 On May 9, 3:55 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On May 7, 11:51=A0am, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> >Hey idiot theGPSwas accelerated from thegroundclock.... > > >> So are you now saying the satelliteclock"remembers" that it was once > >> accelerated into space, and will therefore run at a different rate > >> from then on? > >Yes...after acceleration the SR effect on the satellite clock is > >permanently 7 us/day slow. > > >> What if, I took some cesium, accelerated it to near c in a particle > >> accelerator, then stopped it and mixed it with cesium that was never > >> accelerated, and used it in a cesiumclock? =A0Will the two "kinds" of > >> cesium atom try to make theclocktick at two different rates? > >Hey idiot when you make the Cs atoms come to rest with your Cs atoms > >then they will operate the same as your Cs atoms. > > This is contradictory. First you claim the satellite clock is permanently > changed by the acceleration somehow, but Cs atoms accelerated much more > strongly aren't. Esp. since the basis of timekeeping are the Cs atoms in > the clock. Hey idiot....a period of a moving CS 133 radiation will take a different time to complete than you Cs 133 radiation. Whenn the moving Cs 133 atom returns to you a period of its radiation will take the same time to complete as your Cs 133 atom. > > >> >so why did > >> >you claim that from theGPSpoint of view the SR effect on theground > >> >clockis also 7 us/day slow? > > >> Please learn some SR. An observer observes aclockmoving relative > >> to him as running slow. > >ROTFLOL....you are the one who need to learn SR....what you said is an > >erroneous assumption of the PoR that every SR observer is in a state > >of absolute rest. This is wrong. The rate of two relative clock A and > >B is as follows: > >1. A is running fast compared to B. > >2. B is running slow compared to A. > > Again, learn some SR. You learn some SR. SR is incomplete....the reason is that every SR onbserver assumes that he is in a state of rest (absolute rest) and thus he assumes that all clocks moving wrt him are running slow. In real life when comparing two clocks the following possibilities exist: 1. A runs fast compared to B. 2. B runs slow compare to A. >If clock B is moving with velocity v with respect > to clock A, A observes B's clock as running slow, by the ratio: > delta t' = delta t /sqrt(1-v^2/c^2). If you swap A and B, you could argue > the sign of v reverses, but because of the squaring, the results are > unchanged. B observes A's clock as running slow, by the same ratio. No this is wrong and incomplete. A predicts that B runs slow by a factor of 1/gamma_a OR A predicts that B runs fast by a factor of gamma_a. Similarly B predicts that A runs alow by a factor of 1/ gamma_b OR B predicts that A runs fast by a factor of gamma_b. Once it is dertermined that A is truly running fast compare to B then it follows that B is truly running slow conpare to A. What this mean is that there is no such thing as mutual time dilation. > > >> >> >The redefinedGPSsecond > > >> >> THERE IS NO SUCH THING!!!! > >> >Answer me this question: When theGPSwant to signal the passage of a > >> >second to the gorundclockdoes it sends the signal after > >> >9,192,631,770 periods of Cs 133 radiation or does it sends the signal > >> >after the passage of (9,192,631,770 + 4.15) periods of Cs 133 > >> >radiation? > > >> The onboardclockwill tick at 9,192,631,770 periods of Cs 133 radiation > >> per second, exactly (by definition). > >But this definition is not used by the GPS. > > If it needs an exact 1 second timer on board, it does. No that definition is no longer operative. > > > It uses the redefined > >second of (7,192,631,770 +4.15) periods of Cs 133 radiation. > > You still have never made a case for using the word "redefined". Since > the second is _*defined*_ as 7,192,631,770 ticks of a Cs atom, it doesn't > make sense to call a second as something else Sure I did. The GPS uses the redefined second to communicate with the ground clock. Ken Seto .. > > >>AGPSsatelliteclockwill also > >> generate a signal that ticks at one pulse per 9,192,631,774.15 periods of > >> Cs 133, which will be received on thegroundas a highly accurate 1 pulse > >> per secondclock. > >So it uses the redefined second to send signals....that means that the > >GPS second is redefined. > > Now you're stating it's redefined because it's redefined. Foolishness.- Hide quoted text - > > - Show quoted text -
From: Sam Wormley on 10 May 2010 13:10 On 5/10/10 8:24 AM, kenseto wrote: > On May 9, 11:31 am, Sam Wormley<sworml...(a)gmail.com> wrote: >> On 5/9/10 9:32 AM, kenseto wrote: >> >>> ROTFLOL....you are the one who need to learn SR....what you said is an >>> erroneous assumption of the PoR that every SR observer is in a state >>> of absolute rest. This is wrong. The rate of two relative clock A and >>> B is as follows: >>> 1. A is running fast compared to B. >>> 2. B is running slow compared to A. >> >> Wrong again Seto! >> >> Let's be a bit more precise here: >> >> Assume that A and B have identical atomic clocks. That means they >> tick at the same rate. Now let us suppose that A and B have relative >> motion, such that their velocity with respect to each other, v> 0, >> and that dv/dt = 0 . "Closing speed" or "Separation speed", if you >> like. >> >> Correcting for any Doppler shift, A measures B's time interval as >> ∆t_B' = γ ∆t_B >> >> and B measures A's time interval as >> ∆t_A' = γ ∆t_A >> >> where ∆t represent a time interval, v is the relative velocity >> between A and B, and γ = 1/√(1-v^2/c^2) . >> >> Therefore, A measures B's time interval to be longer than her own. >> And B measures A's time interval to be longer than his own. > > No wormy....there is no measurement at all. A predicts that B's clock > is running at a rate of 1/gamma or gamma....and B predicts that A is > running at a rate of 1/gamma or gamma. Only one of A's prediction is > correct and only one of B's prediction is correct. So if A is truly > running at a faster rate of gamma than B then B is truly running at a > slower rate of 1/gamma than A. Oh no, Seto! Major misunderstanding on your part! Correcting for any Doppler shift, A measures B's time interval as ∆t_B' = γ ∆t_B and B measures A's time interval as ∆t_A' = γ ∆t_A where ∆t represent a time interval, v is the relative velocity between A and B, and γ = 1/√(1-v^2/c^2) . You REALLY DON'T UNDERSTAND relativity at all! Do you even know what the word "relative" means?
From: kenseto on 11 May 2010 08:29
On May 10, 1:10 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/10/10 8:24 AM, kenseto wrote: > > > > > > > On May 9, 11:31 am, Sam Wormley<sworml...(a)gmail.com>  wrote: > >> On 5/9/10 9:32 AM, kenseto wrote: > > >>> ROTFLOL....you are the one who need to learn SR....what you said is an > >>> erroneous assumption of the PoR that every SR observer is in a state > >>> of absolute rest. This is wrong. The rate of two relative clock A and > >>> B is as follows: > >>> 1. A is running fast compared to B. > >>> 2. B is running slow compared to A. > > >>   Wrong again Seto! > > >>   Let's be a bit more precise here: > > >>   Assume that A and B have identical atomic clocks. That means they > >>   tick at the same rate. Now let us suppose that A and B have relative > >>   motion, such that their velocity with respect to each other, v>  0, > >>   and that dv/dt = 0 . "Closing speed" or "Separation speed", if you > >>   like. > > >>   Correcting for any Doppler shift, A measures B's time interval as > >>    ât_B' = γ ât_B > > >>   and B measures A's time interval as > >>    ât_A' = γ ât_A > > >>   where ât represent a time interval, v is the relative velocity > >>   between A and B, and γ = 1/â(1-v^2/c^2) . > > >>   Therefore, A measures B's time interval to be longer than her own. > >>   And B measures A's time interval to be longer than his own. > > > No wormy....there is no measurement at all. A predicts that B's clock > > is running at a rate of 1/gamma or gamma....and B predicts that A is > > running at a rate of 1/gamma or gamma. Only one of A's prediction is > > correct and only one of B's prediction is correct. So if A is truly > > running at a faster rate of gamma than B then B is truly running at a > > slower rate of 1/gamma than A. > >   Oh no, Seto! Major misunderstanding on your part! > >   Correcting for any Doppler shift, A measures B's time interval as >    ât_B' = γ ât_B No idiot A predicts D's time interval....ât_B = γ ât_A > >   and B measures A's time interval as >    ât_A' = γ ât_A No idiot B predicts A's time interval.....ât_A = γ ât_B > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   You REALLY DON'T UNDERSTAND relativity at all! Do you even know >   what the word "relative" means? Looks like its you who don't understand relativity. Ken Seto - Hide quoted text - > > - Show quoted text - |