SR/GR use absolute time to synchronize the GPS clocks with the ground clock.
in [Physics]
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From: kenseto on 16 May 2010 20:20 On May 14, 3:17 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On May 13, 11:41 am, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> kenseto <kens...(a)erinet.com> writes: > >> >Sorry A predicts B's time interval for an inertval of delta(t_A) in > >> >his frame as follows: > >> >delta(t_B)= delta(t_A)/gamma_a > >> >Sorry B predicts A's time interval for an inertval of delta(t_B) in > >> >his frame as follows: > >> >delta(t_A)= delta(t_B)/gamma_b > > >> Neither is what SR states. SR states: > >> delta(t') = gamma*delta(t). delta(t) is a local event (A observing a > >> tick of the second hand of A's watch or whatever) > >Yes A observed the elapse of 1 second on his clock. > > >> delta(t') is the *same* event as observed by another observer (B) moving > >> at relative velocity v, and gamma = 1/sqrt(1-v^2/c^2). > >No....A predicts that his 1 second is worth 1/gamma second on the B > >clock. From B's point of view he predicts that 1 of his clock second > >is worth gamma second on the A clock. > > A predicts she'll see B's clock tick at delta(t') = gamma*delta(t), Yes.....one of t' second is worth gamma second on the t clock. That means that the t' clock is slow compared to the t clock. > where delta(t) is a time interval local to B (B's watch's second hand) > computes gamma = 1/sqrt(1-v^2/c^2) and observes that she'll see B's clock > as slow as well. What are you saying here? are you saying that the t' clock predicts that the t clock is slow?? if that's what you are saying then it is wrong. The t' clock must predict that the t clock is running faster than the t' clock. >"Isn't that interesting. Even though B sees my clock > as running slow, I see B's clock as running slow." no....not intersting at all....You made a bogus assertion that B(t') predicts that A(t) is runing slow. > > > > >> Since there is the constraint that v<c, gamma *must* be a real number > >> greater than or equal to 1, so delta(t') is *always* greater than > >> delta(t). =A0In other words, *every* observer moving relative to A > >> will see A's clock as slowed. > >Yes gamma is always greater than 1. But no it is wrong to assert that > >every observer moving wrt A will see A's clock as slowed...an observer > >on A can only conclude that it is running slow by a factor of 1/gamma > >or it is running fast by a factor of gamma. > > This is contradictory. If gamma is always greater than or equal to 1, > and, according to SR, all moving observers observe A's clock as moving at > the rate of gamma*their clock rate, then all moving observers *must* see > A's clock as running slow. No....no observer knows whether his clock is running slow or fast compared to a moving clock. All he can say is that a moving clock can run fast by a factor of gamma or runs slow by a factor of 1/gamma. > > Your bit about "running slow by a factor of 1/gamma" is your own creation, > and not part of SR, but something that exists only in your own mind. > For someone who claims to know more about SR than anyone else here, > you sure don't understand SR. SR is wrong and incomplete....the reason is that every SR observer assumes that he is in a state of rest and a clock moving wrt him are doing the moving....this led him to conclude that all clocks moving wrt him are rinning slow. > > >> There is simply no (non-imaginary) velocity v that can result in a value > >> for gamma < 1. Therefore it is impossible for an observer to see > >> another's clock as running fast due to SR effects. > >I didn't say that there was. > > So the only possible conclusion is that any observer will see the clock > in another relatively moving frame as running slow. No this is true only for an observer who is in a state of absolute rest.....no object in the universe is in a state of absolute rest. Ken Seto - Hide quoted text - > > - Show quoted text -
From: Inertial on 16 May 2010 20:58 "kenseto" <kenseto(a)erinet.com> wrote in message news:8ee8b9b3-7a84-4ec5-ad89-1dd807e3bd4c(a)z33g2000vbb.googlegroups.com... > On May 14, 3:17 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > wrote: >> kenseto <kens...(a)erinet.com> writes: >> >On May 13, 11:41 am, moro...(a)world.std.spaamtrap.com (Michael Moroney) >> >wrote: >> >> kenseto <kens...(a)erinet.com> writes: >> >> >Sorry A predicts B's time interval for an inertval of delta(t_A) in >> >> >his frame as follows: >> >> >delta(t_B)= delta(t_A)/gamma_a >> >> >Sorry B predicts A's time interval for an inertval of delta(t_B) in >> >> >his frame as follows: >> >> >delta(t_A)= delta(t_B)/gamma_b >> >> >> Neither is what SR states. SR states: >> >> delta(t') = gamma*delta(t). delta(t) is a local event (A observing a >> >> tick of the second hand of A's watch or whatever) >> >Yes A observed the elapse of 1 second on his clock. >> >> >> delta(t') is the *same* event as observed by another observer (B) >> >> moving >> >> at relative velocity v, and gamma = 1/sqrt(1-v^2/c^2). >> >No....A predicts that his 1 second is worth 1/gamma second on the B >> >clock. From B's point of view he predicts that 1 of his clock second >> >is worth gamma second on the A clock. >> >> A predicts she'll see B's clock tick at delta(t') = gamma*delta(t), > > Yes.....one of t' second is worth gamma second on the t clock. That > means that the t' clock is slow compared to the t clock. > >> where delta(t) is a time interval local to B (B's watch's second hand) >> computes gamma = 1/sqrt(1-v^2/c^2) and observes that she'll see B's clock >> as slow as well. > > What are you saying here? are you saying that the t' clock predicts > that the t clock is slow?? if that's what you are saying then it is > wrong. The t' clock must predict that the t clock is running faster > than the t' clock. > >>"Isn't that interesting. Even though B sees my clock >> as running slow, I see B's clock as running slow." > > no....not intersting at all....You made a bogus assertion that B(t') > predicts that A(t) is runing slow. >> >> >> >> >> Since there is the constraint that v<c, gamma *must* be a real number >> >> greater than or equal to 1, so delta(t') is *always* greater than >> >> delta(t). =A0In other words, *every* observer moving relative to A >> >> will see A's clock as slowed. >> >Yes gamma is always greater than 1. But no it is wrong to assert that >> >every observer moving wrt A will see A's clock as slowed...an observer >> >on A can only conclude that it is running slow by a factor of 1/gamma >> >or it is running fast by a factor of gamma. >> >> This is contradictory. If gamma is always greater than or equal to 1, >> and, according to SR, all moving observers observe A's clock as moving at >> the rate of gamma*their clock rate, then all moving observers *must* see >> A's clock as running slow. > > No....no observer knows whether his clock is running slow or fast > compared to a moving clock. All he can say is that a moving clock can > run fast by a factor of gamma or runs slow by a factor of 1/gamma. > >> >> Your bit about "running slow by a factor of 1/gamma" is your own >> creation, >> and not part of SR, but something that exists only in your own mind. >> For someone who claims to know more about SR than anyone else here, >> you sure don't understand SR. > > SR is wrong and incomplete....the reason is that every SR observer > assumes that he is in a state of rest and a clock moving wrt him are > doing the moving....this led him to conclude that all clocks moving > wrt him are rinning slow. > >> >> >> There is simply no (non-imaginary) velocity v that can result in a >> >> value >> >> for gamma < 1. Therefore it is impossible for an observer to see >> >> another's clock as running fast due to SR effects. >> >I didn't say that there was. >> >> So the only possible conclusion is that any observer will see the clock >> in another relatively moving frame as running slow. > > No this is true only for an observer who is in a state of absolute > rest.....no object in the universe is in a state of absolute rest. > > Ken Seto After all these years, Ken still simply fails to understand physics, or the consequences of his own 'theory'. Sad.
From: Sam Wormley on 16 May 2010 22:56 On 5/16/10 7:20 PM, kenseto wrote: > What are you saying here? are you saying that the t' clock predicts > that the t clock is slow?? if that's what you are saying then it is > wrong. The t' clock must predict that the t clock is running faster > than the t' clock. > You know, Seto, we've been over this many times over the years and you still can seem to understand the fact relativity is relative. How could your parent raise you to be this way? Please look at the following carefully. Assume that A and B have identical atomic clocks. That means they tick at the same rate when together. Now let us suppose that A and B have relative motion, such that their velocity (closing or opening) with respect to each other is, v > 0, and that dv/dt = 0 . 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. Who's clock measures slow is observer dependent. Seto do you know what observer dependent means?
From: kenseto on 17 May 2010 09:02 On May 16, 8:58 pm, "Inertial" <relativ...(a)rest.com> wrote: > "kenseto" <kens...(a)erinet.com> wrote in message > > news:8ee8b9b3-7a84-4ec5-ad89-1dd807e3bd4c(a)z33g2000vbb.googlegroups.com... > > > > > > > On May 14, 3:17 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > > wrote: > >> kenseto <kens...(a)erinet.com> writes: > >> >On May 13, 11:41 am, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >> >wrote: > >> >> kenseto <kens...(a)erinet.com> writes: > >> >> >Sorry A predicts B's time interval for an inertval of delta(t_A) in > >> >> >his frame as follows: > >> >> >delta(t_B)= delta(t_A)/gamma_a > >> >> >Sorry B predicts A's time interval for an inertval of delta(t_B) in > >> >> >his frame as follows: > >> >> >delta(t_A)= delta(t_B)/gamma_b > > >> >> Neither is what SR states. SR states: > >> >> delta(t') = gamma*delta(t). delta(t) is a local event (A observing a > >> >> tick of the second hand of A's watch or whatever) > >> >Yes A observed the elapse of 1 second on his clock. > > >> >> delta(t') is the *same* event as observed by another observer (B) > >> >> moving > >> >> at relative velocity v, and gamma = 1/sqrt(1-v^2/c^2). > >> >No....A predicts that his 1 second is worth 1/gamma second on the B > >> >clock. From B's point of view he predicts that 1 of his clock second > >> >is worth gamma second on the A clock. > > >> A predicts she'll see B's clock tick at delta(t') = gamma*delta(t), > > > Yes.....one of t' second is worth gamma second on the t clock. That > > means that the t' clock is slow compared to the t clock. > > >> where delta(t) is a time interval local to B (B's watch's second hand) > >> computes gamma = 1/sqrt(1-v^2/c^2) and observes that she'll see B's clock > >> as slow as well. > > > What are you saying here? are you saying that the t' clock predicts > > that the t clock is slow?? if that's what you are saying then it is > > wrong. The t' clock must predict that the t clock is running faster > > than the t' clock. > > >>"Isn't that interesting. Even though B sees my clock > >> as running slow, I see B's clock as running slow." > > > no....not intersting at all....You made a bogus assertion that B(t') > > predicts that A(t) is runing slow. > > >> >> Since there is the constraint that v<c, gamma *must* be a real number > >> >> greater than or equal to 1, so delta(t') is *always* greater than > >> >> delta(t). =A0In other words, *every* observer moving relative to A > >> >> will see A's clock as slowed. > >> >Yes gamma is always greater than 1. But no it is wrong to assert that > >> >every observer moving wrt A will see A's clock as slowed...an observer > >> >on A can only conclude that it is running slow by a factor of 1/gamma > >> >or it is running fast by a factor of gamma. > > >> This is contradictory. If gamma is always greater than or equal to 1, > >> and, according to SR, all moving observers observe A's clock as moving at > >> the rate of gamma*their clock rate, then all moving observers *must* see > >> A's clock as running slow. > > > No....no observer knows whether his clock is running slow or fast > > compared to a moving clock. All he can say is that a moving clock can > > run fast by a factor of gamma or runs slow by a factor of 1/gamma. > > >> Your bit about "running slow by a factor of 1/gamma" is your own > >> creation, > >> and not part of SR, but something that exists only in your own mind. > >> For someone who claims to know more about SR than anyone else here, > >> you sure don't understand SR. > > > SR is wrong and incomplete....the reason is that every SR observer > > assumes that he is in a state of rest and a clock moving wrt him are > > doing the moving....this led him to conclude that all clocks moving > > wrt him are rinning slow. > > >> >> There is simply no (non-imaginary) velocity v that can result in a > >> >> value > >> >> for gamma < 1. Therefore it is impossible for an observer to see > >> >> another's clock as running fast due to SR effects. > >> >I didn't say that there was. > > >> So the only possible conclusion is that any observer will see the clock > >> in another relatively moving frame as running slow. > > > No this is true only for an observer who is in a state of absolute > > rest.....no object in the universe is in a state of absolute rest. > > > Ken Seto > > After all these years, Ken still simply fails to understand physics, or the > consequences of his own 'theory'. Sad. No after all these years you still got your head stuck in your arsehole - Hide quoted text - > > - Show quoted text -
From: kenseto on 17 May 2010 09:04
On May 16, 10:56 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/16/10 7:20 PM, kenseto wrote: > > > What are you saying here? are you saying that the t' clock predicts > > that the t clock is slow?? if that's what you are saying then it  is > > wrong. The t' clock must predict that the t clock is running faster > > than the t' clock. > >   You know, Seto, we've been over this many times over the years and >   you still can seem to understand the fact relativity is relative. >   How could your parent raise you to be this way? Please look at the >   following carefully. > >   Assume that A and B have identical atomic clocks. That means they >   tick at the same rate when together. Now let us suppose that >   A and B have relative motion, such that their velocity (closing or >   opening) with respect to each other is, v > 0, and that dv/dt = 0 . > >   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. Who's >   clock measures slow is observer dependent. Seto do you know what >   observer dependent means? Wormy you are an idiot. All you do is copy and paste. |