On 10/13/2020 3:12 PM, Lawrence Crowell wrote:
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
And it's further complicated by the Earth being non-spherical. The
calculations find the lat/long of a WGS84 ellipsoid. But of course the
real Earth isn't exactly an WGS84 ellipsoid either and there have to be
local corrections in look-up tables. Off the coast of California where
I used to be involved in developing sea-skimming targets the WGS84 "sea
level" is about 120ft under water.
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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