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
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