On 10/13/2020 4:54 PM, Alan Grayson wrote:
On Tuesday, October 13, 2020 at 5:16:04 PM UTC-6, Brent wrote:
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
*Yes, very complicated to get an exact solution. BUT, what I was
trying to say, before getting a ton of crap from LC and Clark, the
solution depends ONLY on GR since gravity is involved which distorts
the spacetime paths and thus the proper times along these paths. Do
you agree with this statement? TIA, AG*
I wouldn't agree that LC and JKC are providing a ton of crap, but yes
it's just path length thru non-flat spacetime. Looking at in terms of
relative speed (which is a symmetric relation) and higher v. lower
gravitational potential is making approximations; which is OK but not
the way to understand the conceptual basis.
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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