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

        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-list+unsubscr...@googlegroups.com <mailto:everything-list+unsubscr...@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/e18af997-41f7-4bfe-e572-739355e1cae6%40verizon.net.

Reply via email to