SR is not wrong but it is incomplete
in [Physics]
|
Prev: Taking a Fresh Look at the Physics of Radiation Pressure.
Next: 9/11 nutbag * Hates US * cannot provide cite to his disproven claim that PNAC "wanted the attack" because that is a lie; inside job proven physically impossible
From: Daryl McCullough on 23 Jul 2010 14:15 PD says... > >On Jul 23, 9:44=A0am, kenseto <kens...(a)erinet.com> wrote: >> Then why does the SR effect on the gPS is calculated using the SR >> equations??? > >1. It's not. The GPS time lag is calculated using GR, not SR. I think that what Ken might be thinking of is this: If you take the Schwarzschild metric for the Earth, and then use it to compute the elapsed time on a clock moving in the Earth's gravitational field, in the approximation in which v << c, and GM/r << c^2, you get: dT/dt = 1 - 1/2 (v/c)^2 - GM/(c^2 r) + terms of order (v/c)^4 where T is the elapsed time on the clock, t is the coordinate time (using Schwarzschild coordinates), v is the speed of the clock, M is the mass of the Earth, and G is Newton's constant. You can think of the terms in the following way: 1. Nonrelativistically, the result would be 1 (coordinate time and clock time are the same, even for moving clocks). 2. SR makes a correction to the non-relativistic prediction equal to - 1/2 (v/c)^2 in lowest-order. 3. GR makes a correction to the SR prediction equal to - GM/(c^2 r) in lowest order. Of course, all three terms are present in the GR prediction. -- Daryl McCullough Ithaca, NY
From: PD on 23 Jul 2010 14:28 On Jul 23, 1:15 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > PD says... > > > > >On Jul 23, 9:44=A0am, kenseto <kens...(a)erinet.com> wrote: > >> Then why does the SR effect on the gPS is calculated using the SR > >> equations??? > > >1. It's not. The GPS time lag is calculated using GR, not SR. > > I think that what Ken might be thinking of is this: > If you take the Schwarzschild metric for the Earth, and then > use it to compute the elapsed time on a clock moving in the > Earth's gravitational field, in the approximation in which > v << c, and GM/r << c^2, you get: > > dT/dt = 1 - 1/2 (v/c)^2 - GM/(c^2 r) + terms of order (v/c)^4 > > where T is the elapsed time on the clock, t is the coordinate > time (using Schwarzschild coordinates), v is the speed of the > clock, M is the mass of the Earth, and G is Newton's constant. > > You can think of the terms in the following way: > > 1. Nonrelativistically, the result would be 1 (coordinate time > and clock time are the same, even for moving clocks). > > 2. SR makes a correction to the non-relativistic prediction equal > to - 1/2 (v/c)^2 in lowest-order. > > 3. GR makes a correction to the SR prediction equal to - GM/(c^2 r) > in lowest order. > > Of course, all three terms are present in the GR prediction. > > -- > Daryl McCullough > Ithaca, NY Understood. Newton's first law is a special case of Newton's second law. Now consider an object accelerating under the influence of a force, for which you would usually apply Newton's second law rather than the first, for which there is the result s = (v_0)t + 1/2(F/m)t^2 the first term being something expected from Newton's first law and the second coming from the second law. Follow my drift?
From: Michael Moroney on 23 Jul 2010 15:28 stevendaryl3016(a)yahoo.com (Daryl McCullough) writes: >I think that what Ken might be thinking of is this: >If you take the Schwarzschild metric for the Earth, and then >use it to compute the elapsed time on a clock moving in the >Earth's gravitational field, in the approximation in which >v << c, and GM/r << c^2, you get: >dT/dt = 1 - 1/2 (v/c)^2 - GM/(c^2 r) + terms of order (v/c)^4 >where T is the elapsed time on the clock, t is the coordinate >time (using Schwarzschild coordinates), v is the speed of the >clock, M is the mass of the Earth, and G is Newton's constant. >You can think of the terms in the following way: >1. Nonrelativistically, the result would be 1 (coordinate time >and clock time are the same, even for moving clocks). >2. SR makes a correction to the non-relativistic prediction equal >to - 1/2 (v/c)^2 in lowest-order. >3. GR makes a correction to the SR prediction equal to - GM/(c^2 r) >in lowest order. >Of course, all three terms are present in the GR prediction. That is essentially what I was doing in my earlier replies. The correction terms were small enough (parts per billion) that the equation could be broken into what I called an SR term (7uS/day) and a GR term (45uS/day) that could be treated independently as the higher terms ((v/c)^4 or higher) could be ignored. Of course it's all GR and I should have called it a motion component and a gravitational well component. Given the caveat that I'm ignoring higher order terms, is there anything wrong with the following logic: An observer on a GPS satellite would see a ground clock as running slow by ~53 uS/day, 7uS/day slow due to relative motion of the earth (as far as the satellite is concerned) and 45uS/day slow due to the fact the ground clock is lower in a gravitational well than the satellite.
From: BURT on 23 Jul 2010 18:37 On Jul 23, 12:28 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > stevendaryl3...(a)yahoo.com (Daryl McCullough) writes: > >I think that what Ken might be thinking of is this: > >If you take the Schwarzschild metric for the Earth, and then > >use it to compute the elapsed time on a clock moving in the > >Earth's gravitational field, in the approximation in which > >v << c, and GM/r << c^2, you get: > >dT/dt = 1 - 1/2 (v/c)^2 - GM/(c^2 r) + terms of order (v/c)^4 > >where T is the elapsed time on the clock, t is the coordinate > >time (using Schwarzschild coordinates), v is the speed of the > >clock, M is the mass of the Earth, and G is Newton's constant. > >You can think of the terms in the following way: > >1. Nonrelativistically, the result would be 1 (coordinate time > >and clock time are the same, even for moving clocks). > >2. SR makes a correction to the non-relativistic prediction equal > >to - 1/2 (v/c)^2 in lowest-order. > >3. GR makes a correction to the SR prediction equal to - GM/(c^2 r) > >in lowest order. > >Of course, all three terms are present in the GR prediction. > > That is essentially what I was doing in my earlier replies. The > correction terms were small enough (parts per billion) that the > equation could be broken into what I called an SR term (7uS/day) > and a GR term (45uS/day) that could be treated independently as > the higher terms ((v/c)^4 or higher) could be ignored. Of course > it's all GR and I should have called it a motion component and a > gravitational well component. > > Given the caveat that I'm ignoring higher order terms, is there > anything wrong with the following logic: An observer on a GPS > satellite would see a ground clock as running slow by ~53 uS/day, > 7uS/day slow due to relative motion of the earth (as far as the > satellite is concerned) and 45uS/day slow due to the fact the ground > clock is lower in a gravitational well than the satellite.- Hide quoted text - > > - Show quoted text - The well goes to even strength. Mitch Raemsch
From: kenseto on 24 Jul 2010 10:23
On Jul 23, 3:28 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > stevendaryl3...(a)yahoo.com (Daryl McCullough) writes: > >I think that what Ken might be thinking of is this: > >If you take the Schwarzschild metric for the Earth, and then > >use it to compute the elapsed time on a clock moving in the > >Earth's gravitational field, in the approximation in which > >v << c, and GM/r << c^2, you get: > >dT/dt = 1 - 1/2 (v/c)^2 - GM/(c^2 r) + terms of order (v/c)^4 > >where T is the elapsed time on the clock, t is the coordinate > >time (using Schwarzschild coordinates), v is the speed of the > >clock, M is the mass of the Earth, and G is Newton's constant. > >You can think of the terms in the following way: > >1. Nonrelativistically, the result would be 1 (coordinate time > >and clock time are the same, even for moving clocks). > >2. SR makes a correction to the non-relativistic prediction equal > >to - 1/2 (v/c)^2 in lowest-order. > >3. GR makes a correction to the SR prediction equal to - GM/(c^2 r) > >in lowest order. > >Of course, all three terms are present in the GR prediction. > > That is essentially what I was doing in my earlier replies. The > correction terms were small enough (parts per billion) that the > equation could be broken into what I called an SR term (7uS/day) > and a GR term (45uS/day) that could be treated independently as > the higher terms ((v/c)^4 or higher) could be ignored. Of course > it's all GR and I should have called it a motion component and a > gravitational well component. > > Given the caveat that I'm ignoring higher order terms, is there > anything wrong with the following logic: An observer on a GPS > satellite would see a ground clock as running slow by ~53 uS/day, > 7uS/day slow due to relative motion of the earth (as far as the > satellite is concerned) and 45uS/day slow due to the fact the ground > clock is lower in a gravitational well than the satellite. You were corrected that from the GPS point of view the ground clock is not ~53us/day slow. Why? Because: 1. it invokes the bogus SR concept of mutual time dilation. Evenn PD diagreed with you....he said that mutual time dilation does not apply to the GPS. 2. The redfined GPS second (N+4.46) periods of Cs 133 radiation would not agree with your assertion that the Ground clock is ~53 us/day slow. Ken Seto - Hide quoted text - > > - Show quoted text - |