On Tuesday, October 13, 2020 at 4:13:05 PM UTC-5 Brent wrote: > > > On 10/13/2020 1:34 PM, Lawrence Crowell wrote: > > On Tuesday, October 13, 2020 at 3:28:21 PM UTC-5 Lawrence Crowell wrote: > >> On Tuesday, October 13, 2020 at 3:26:14 PM UTC-5 Lawrence Crowell wrote: >> >>> I will try to give a definitive answer. The Schwarzschild metric is >>> >>> ds^2 = c^2(1 – 2m/r)dt^2 – (1 – 2m/r)dr^2 – r^2(dθ^2 – sin^2θdφ^2) >>> >>> for m = GM/c^2. For the motion of a satellite in a circular orbit there >>> is no radial motion so dr = 0. We set this on a plane with θ = π/2 so dθ = >>> 0 and this reduces this to >>> >>> ds^2 =c^2(1 – 2m/r)dt^2 – r^2dφ^2. >>> >>> For circular motion dφ/dt = ω and the velocity v = ωr means this is >>> >>> ds^2 = [c^2(1 – 2m/r) – r^2ω^2]dt^2 >>> >>> and so ds^2 = [c^2(1 – 2m/r) – v^2]dt^2 the term Γ = 1/√[c^2(1 – 2m/r) – >>> v^2] is a general Lorentz gamma factor and in flat space with m = 0 reduces >>> the form we know. ds is an increment in the proper time on the orbiting >>> satellite and t is a coordinate time, say on the ground of the body. >>> >> > Another erratum. The coordinate time t is for a clock very far removed, > not on the ground. On the ground that clock ticks away with a factor Γ = > 1/√[c^2 – v^2] change. So there is a relative time difference. > > > A clock on the ground is also moving with rotation of the Earth, with > different speed at different latitudes. This is taken out of the equations > by comparing the GPS clock to ideal clocks on a fixed (non-rotating Earth) > and then after GPS calculates the location on the non-rotating Earth, it > calculates what point this is on the rotating Earth. > > Brent >
This gets really complicated. I did a lot of post-Newtonian parameter work on this back in the late 80s. A lot of it was numerical, because on the ground there are different values of gravity, and these too can cause drift. Gravitation, thinking of a Newtonian force, is different near a mountain than on the top of it, and the direction can vary some from the radius. It also fluctuates with tides! The surging in and out of a lot of ocean water actually changes the Newtonian gravitation potential and force. LC > > We can do more with this. The ds^2 = [c^2(1 – 2m/r) – r^2dφ^2]dt^2 can be >>> written as >>> >>> 1 = [c^2(1 – 2m/r) – r^2ω^2](dt/ds)^2 >>> >>> Now take a variation on this, where obviously δ1 = 0 and >>> >>> 0 = [c^2δ(1 – 2m/r) – δ(r^2ω^2)](dt/ds)^2 + [c^2(1 – 2m/r) – >>> r^2ω^2]δ(dt/ds)^2. >>> >>> We think primarily of a variation in the radius and so >>> >>> 0 = -[ 2mc^2/r^2 – 2rω^2](dt/ds)^2δr + [c^2(1 – 2m/r) – r^2ω^2] >>> δ(dt/ds)^2, >>> >>> where for the time I will ignore the last term. The first term gives >>> >>> rω^2 = -GM/r, >>> >> I mean rω^2 = -GM/r^2 >> >> >>> and this is just Newton’s second law with acceleration a = rω^2 with >>> gravity. Also this is Kepler's third law of planetary motion. >>> >>> Now I will hand wave a bit here. The term δ(dt/ds)^2 = 1 in the >>> Newtonian limit, but we can feed the general Lorentz gamma factor in that. >>> This will have a correction term to this dynamical equation. This >>> correction is general relativistic. The algebra gets a bit dense, but it is >>> nothing conceptually difficult. >>> >>> LC >>> >>> On Tuesday, October 13, 2020 at 9:17:37 AM UTC-5 agrays...@gmail.com >>> wrote: >>> >>>> >>>> >>>> On Tuesday, October 13, 2020 at 8:06:30 AM UTC-6, Lawrence Crowell >>>> wrote: >>>>> >>>>> I am not sure why you have endless trouble with this. On the Avoid >>>>> list you repeatedly brought up this question, and in spite of dozens of >>>>> explanations you raise this question over and over. You need to read a >>>>> text >>>>> on this. The old Taylor and Wheeler book on SR gives some reasoning on >>>>> this. Geroch's book on GR is not too hard to read. >>>>> >>>>> LC >>>>> >>>> >>>> Actually, I think your memory is faulty, other than to express your >>>> annoyance with my question. In any event, if gravity and acceleration >>>> exist >>>> for a system under consideration, why is SR relevant? Why does Clark claim >>>> that the result of SR must be subtracted for the result of GR to determine >>>> an objective outcome, when the conditions of SR are non-existent? AG >>>> >>>>> >>>>> >>>>> On Tuesday, October 13, 2020 at 12:20:44 AM UTC-5 agrays...@gmail.com >>>>> wrote: >>>>> >>>>>> >>>>>> >>>>>> On Monday, October 12, 2020 at 11:11:33 PM UTC-6, Brent wrote: >>>>>>> >>>>>>> >>>>>>> >>>>>>> On 10/12/2020 9:56 PM, Alan Grayson wrote: >>>>>>> > Why is it that in SR a stationary clock appears to advancing at a >>>>>>> more >>>>>>> > rapid rate than a moving clock, and vice versa -- so the effect is >>>>>>> > relative or symmetric, not absolute -- whereas in GR the effect >>>>>>> seems >>>>>>> > absolute; that is, a ground clock actually advances at a slower >>>>>>> rate >>>>>>> > compared to an orbiting clock? AG >>>>>>> >>>>>>> It's the same as the twin effect. The clock on the ground is >>>>>>> following >>>>>>> a non-geodesic path thru spacetime and so measures less duration, >>>>>>> while >>>>>>> the orbiting clock is following a geodesic path. In relativity the >>>>>>> minus sign in the metric means that the path that looks longer >>>>>>> projected >>>>>>> in space is shorter in spacetime. >>>>>>> >>>>>>> Brent >>>>>>> >>>>>> >>>>>> How does gravity cause the difference between what the theories >>>>>> predict? AG >>>>>> >>>>> -- > > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-li...@googlegroups.com. > > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/1964a545-2eb9-471e-8ce8-76543e2b74e5n%40googlegroups.com > > <https://groups.google.com/d/msgid/everything-list/1964a545-2eb9-471e-8ce8-76543e2b74e5n%40googlegroups.com?utm_medium=email&utm_source=footer> > . > > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/ae6a9158-dcf3-4da8-8065-faf236f04210n%40googlegroups.com.
