Clock rates; SR v GR

[Alan Grayson]

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

[Brent Meeker]

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

[Alan Grayson]

How does gravity cause the difference between what the theories predict? AG 

[Brent Meeker]

It curves the time axis (mainly).  Don't you have a copy of Epstein's "Relativity Visualized"?

Brent

[Alan Grayson]

I do. And I understand your remark about the Twin Paradox. But I was wondering; do the calculated results, that is the differential in clock rates, differ between SR and GR? AG

[John Clark]

On Tue, Oct 13, 2020 at 1:20 AM Alan Grayson <agrays...@gmail.com> wrote:

> How does gravity cause the difference between what the theories predict? AG 

There is no contradiction about what the theories predict because Special Relativity can make no prediction at all about what a clock will do when it is in a gravitational field or is accelerated in any other way; that's why it's called "special", it's only applicable in certain special circumstances. But General Relativity can handle acceleration and gravity. A GPS Satellite is moving fast compared to a clock on the ground so Special Relativity says the clock on the satellite will lose 7210 nanoseconds a day, but the satellite clock is further from the Earth's center so it's in a weaker gravitational field, and because of that general relatively says the satellite clock will gain 45850 nanoseconds a day relative to the clock on the ground. So the two theories together predict the satellite clock will gain 45850 −7210 = 38,640 nanoseconds a day relative to a clock on the ground. If this were not taken into account the GPS system would be in error by 6 miles every day and the error would add up, soon the entire GPS system would be useless. 

John K Clark 

[Lawrence Crowell]

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

[Alan Grayson]

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

[Alan Grayson]

On Tuesday, October 13, 2020 at 8:17:37 AM UTC-6, Alan Grayson 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

I don't believe I was able, years ago, to clearly express my issue I am now raising. Note that Clark claims SR is irrelevant to the issue of comparing an orbiting clock with a stationary clock on the Earth, but he then factors SR into the result of GR. So Clark seems to contract himself, although maybe you don't agree with his analysis. AG 

[John Clark]

On Tue, Oct 13, 2020 at 10:39 AM Alan Grayson <agrays...@gmail.com> wrote:

> Note that Clark claims SR is irrelevant to the issue of comparing an orbiting clock with a stationary clock on the Earth, 

No, Clark did not make that claim. Clark claims that in some more complex situations, such as situations where gravity becomes important, one number alone, the relative speed of the two clocks, is necessary but not sufficient to be able to calculate what the clocks will do. And Clark's "claim" has been verified many many times experimentally, including the time you used your car's GPS to find the nearest Burger King.

 John K Clark

[Alan Grayson]

Sorry, but I don't see why you need to use both theories to get the actual differential rates of the ground clock vs the orbiting clock. AG 

[John Clark]

On Tue, Oct 13, 2020 at 11:12 AM Alan Grayson <agrays...@gmail.com> wrote:

> Sorry, but I don't see [...]

I've noticed there are a lot of things that you just don't see, and you don't seem to be making any effort to improve your understanding even when things are carefully explained to you.  You just keep asking the exact same questions over and over.

John K Clark

[Alan Grayson]

Are you clerking for Barrick? AG 

[Alan Grayson]

Are you clerking for Barrett? AG 

This would be a non-answer in the style of Amy Coney Barrett, or maybe a sycophant of LC. I never, in the past, was able to pose the question with such precision. Why use a theory, SR, which at the start of your discussion you admit doesn't apply? AG

[Alan Grayson]

And this isn't the first time I've caught you in serious scientific dishonesty. So I find your response inappropriate and disingenuous. I'm outta here. AG 

[Brent Meeker]

On 10/13/2020 6:18 AM, John Clark wrote:

On Tue, Oct 13, 2020 at 1:20 AM Alan Grayson <agrays...@gmail.com> wrote:

> How does gravity cause the difference between what the theories predict? AG 

There is no contradiction about what the theories predict because Special Relativity can make no prediction at all about what a clock will do when it is in a gravitational field or is accelerated in any other way; that's why it's called "special", it's only applicable in certain special circumstances. But General Relativity can handle acceleration and gravity. A GPS Satellite is moving fast compared to a clock on the ground so Special Relativity says the clock on the satellite will lose 7210 nanoseconds a day,

That's a kind of engineering way of looking at it which works well to first order.  But remember "moving fast" is relative and one could naviely conclude the ground clock must run slow compared to the satellite. 

but the satellite clock is further from the Earth's center so it's in a weaker gravitational field, and because of that general relatively says the satellite clock will gain 45850 nanoseconds a day relative to the clock on the ground.

The more fundamental way to look at it is the spacetime is curved, but locally it's Minkowski.  So every clock just measures the proper distance along it's world line.  Acceleration only enter indirectly since it deviates the path from geodesic which is the extremum.  In spacetime the acceleration is just seen as a geometric factor.  In this way of looking at it there are not two theories SR and GR, it's just SR on a curved background.

Brent

So the two theories together predict the satellite clock will gain 45850 −7210 = 38,640 nanoseconds a day relative to a clock on the ground. If this were not taken into account the GPS system would be in error by 6 miles every day and the error would add up, soon the entire GPS system would be useless. 

John K Clark 

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[Lawrence Crowell]

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.

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,

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

[Lawrence Crowell]

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.

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

[Lawrence Crowell]

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.

[Brent Meeker]

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

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[Lawrence Crowell]

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

[Brent Meeker]

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

To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/ae6a9158-dcf3-4da8-8065-faf236f04210n%40googlegroups.com.

[Alan Grayson]

On Tuesday, October 13, 2020 at 2:26:14 PM UTC-6, Lawrence Crowell wrote:

I will try to give a definitive answer.

So regardless of your subsequent corrections, will you now admit, as I was suggesting, that the exact solution can be determined solely by GR, and that Clark's introducing SR is confusing and mistaken. Thank you in advance for your honesty! AG

[Alan Grayson]

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

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[Brent Meeker]

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

To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/abe62f22-954e-418d-a174-0757c74d4838o%40googlegroups.com.

[Alan Grayson]

It's hard to determine who is the greater a'hole; Clark or Lawrence. Clark began his explanation by saying SR makes no prediction, but then used that prediction in saying SR and GR together give the solution (and Lawrence gave his tacit support). Is this what you consider proper English? I say it's a ton of $hit. AG 

To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/abe62f22-954e-418d-a174-0757c74d4838o%40googlegroups.com.

[Alan Grayson]

On Monday, October 12, 2020 at 11:11:33 PM UTC-6, Brent wrote:

The mistake I made in my original question, was comparing a symmetric situation under SR, with an asymmetric situation in GR.  But you're right. The GR situation with an orbital clock being compared with a ground clock, should be compared with the Twin Paradox in SR. In both cases, there is an objective difference in what the clocks measure, and this is attributable to the differential path lengths in spacetime. But then Clark introduced his flawed contradictory explanation, later embellished by his arrogance and failure to use proper English. AG

[Alan Grayson]

Seriously, you fit the profile of an arrogant prick. You start off above by saying SR cannot say anything about clock behavior in a gravity field, but then you use an SR result, with GR, to estimate the differential clock rates. Then, when I ask you to explain this apparent inconsistency, I get a load of $hit in reply. GFY. AG 

[Alan Grayson]

If you had more technical intelligence and less arrogance you'd realize that the relative speed of the orbiting clock can't be used to calculate how its rate is decreased compared to the ground clock BECAUSE the orbiting clock is constantly accelerating. Applying SR is simply invalid. The problem can only be solved via GR exclusively.  AG

[John Clark]

On Tue, Oct 13, 2020 at 3:28 PM Alan Grayson <agrays...@gmail.com> wrote:

> I'm outta here. AG 

You promise?

John K Clark

[Lawrence Crowell]

On Tuesday, October 13, 2020 at 6:47:37 PM UTC-5 agrays...@gmail.com wrote:

On Tuesday, October 13, 2020 at 2:26:14 PM UTC-6, Lawrence Crowell wrote:

I will try to give a definitive answer.

So regardless of your subsequent corrections, will you now admit, as I was suggesting, that the exact solution can be determined solely by GR, and that Clark's introducing SR is confusing and mistaken. Thank you in advance for your honesty! AG

I would not put it that way. Special relativity defines the Lorentz group that is the local set of spacetime symmetries in any small region of spacetime. In teaching relativity the global symmetry condition of flat spacetime and special relativity are taught first.

LC

[Lawrence Crowell]

On Tuesday, October 13, 2020 at 6:16:04 PM UTC-5 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

The world geodetic system might be thought of as a set of Bessel functions and similar polynomials. The Earth is not a perfect ellipsoid and there are other mass distributions going on with the tectonic plates and interior mass distributions. The model is subject to constant refinement.

LC