Table 5.2 teaches the people who can read physics (certainly, not you) that the standard deviation is of the order of $$10^{-5}mgal=10^{-8}gal$$. This is in perfect agreement with the AmJPhys paper. You and arfa brane should write to the people at NIST, they are doing everything wrong (according to you two).
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Gravitational collapse
- Thread starter arfa brane
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Table 5.2 teaches the people who can read physics (certainly, not you) that the standard deviation is of the order of $$10^{-5}mgal=10^{-8}gal$$. This is in perfect agreement with the AmJPhys paper. You and arfa brane should write to the people at NIST, they are doing everything wrong (according to you two).
Because the paper teaches the people who want to learn (this excludes you) that the surface of the geoid is equipotential, i.e. the same in every point, hence the resulting gravitational acceleration is the same in every point.
Except the geoid you keep referring to is not a reflection of the Earth's surface. It! in one case is defined as the mean sea level extended even under land masses (which provides a reference for elevations) and in another as an ellipsoid that closely approximates the mean gravitational field. Neither of which is generally applicable to someone standing on the surface of the earth.
However, NIST has probably realised that the physical surface of the earth is not a surface of equipotential.
You should definitely write to them and correct their ignorance. I am quite sure they will be grateful for your deep insights.
Except the geoid you keep referring to is not a reflection of the Earth's surface. It! in one case is defined as the mean sea level extended even under land masses (which provides a reference for elevations) and in another as an ellipsoid that closely approximates the mean gravitational field.
You don't get that the standard deviation is $$10^{-8}gal$$, do you? I didn't think so.
Neither of which is generally applicable to someone standing on the surface of the earth.
Get together with arfa brane and write to the people at NIST, they were doing things incorrectly all along.
And, once more, everyone can see that Tach has no answers.
On the one hand, Wikipedia states categorically that g is smaller at the equator than at the poles. On the other, the resident "expert" tells us this is wrong.
It seems to have something to do with a surface of equipotential, and with the surface of the earth, and whether these are the same thing. Tach doesn't seem to know so will most likely not attempt to sort it out (embarrassment can be a real bitch).
But you can describe a surface of equipotential anywhere around the earth (around the moon too). It isn't restricted to being close, or as close as possible, to the actual surface (like the oceans are).
On the one hand, Wikipedia states categorically that g is smaller at the equator than at the poles. On the other, the resident "expert" tells us this is wrong.
It seems to have something to do with a surface of equipotential, and with the surface of the earth, and whether these are the same thing. Tach doesn't seem to know so will most likely not attempt to sort it out (embarrassment can be a real bitch).
But you can describe a surface of equipotential anywhere around the earth (around the moon too). It isn't restricted to being close, or as close as possible, to the actual surface (like the oceans are).
Correction: Tach has been incorrectly interpreting what NIST is doing (all along).Tach said:Get together with arfa brane and write to the people at NIST, they were doing things incorrectly all along.
You don't get that the standard deviation is $$10^{-8}gal$$, do you? I didn't think so.
Get together with arfa brane and write to the people at NIST, they were doing things incorrectly all along.
Here I think is your first reference to the geoid bit.
Actually, the net gravitational acceleration (when you take into consideration that the Earth is a geoid, not a spheroid and that it rotates) is the same at the poles and at the Equator. See here. This also explains why all clocks in the NIST tick at the same rate.
Read the paper, the math is self-explanatory.
The problem is the earth is not in any practical sense a geoid. And yes the hypothetical geoid whether you define it as mean sea level or gravitational potential is a usefull tool, for a number of purposes. But it does not tell you what the gravitational potential at any specific location on the Earth's surface is.., wait for it.., because the earth does not have a uniform density. Even at sea level the gravitational potential varies from one location to another... Though it likely most consistent over deep ocean...
Only when you are modeling the Earth's gravitational field as an astronomical object does treating the earth as a uniformly gravitating body workout. Even then only to varying degrees.
And Tach, NIST demonstrated that all clocks do not tick at the same rate.., unless their clock rate has been tuned to their location and elevation.
It seems to have something to do with a surface of equipotential,
You are starting to get it, keep at it and one day you'll manage to learn.
Here I think is your first reference to the geoid bit.
Yes, I only posted it for you and arfa brain 4 times.
The problem is the earth is not in any practical sense a geoid.
The problem is that you two are in (deep) denial of mainstream science. Try to find the exact statement in the paper:
"Despite the Earth’s complicated shape with mountains and valleys, its gravitational field can be modeled to a fractional accuracy of $$10^{-14}$$ $$by^7$$".
And Tach, NIST demonstrated that all clocks do not tick at the same rate..,
Sorry to dispel your crank ideas, all clocks on the geoid tick at the same rate. No "adjustment" necessary since the geoid is the surface of equi-gravitational potential. The paper explains that very well.
unless their clock rate has been tuned to their location and elevation.
You two (you and arfa brane) should write to them explaining that they (NIST) have been doing things wrong all along.
This does not say the model corresponds to the actual surface, why do you think it does?Tach said:The problem is that you two are in (deep) denial of mainstream science. Try to find the exact statement in the paper:
"Despite the Earth’s complicated shape with mountains and valleys, its gravitational field can be modeled to a fractional accuracy . . . ".
Of course all clocks at the same gravitational potential tick at the same rate, that does not mean all clocks on the surface of the earth tick at the same rate. This is because the physical surface of the earth is not an equipotential surface.Sorry to dispel your crank ideas, all clocks on the geoid tick at the same rate. No "adjustment" necessary since the geoid is the surface of equi-gravitational potential. The paper explains that very well.
Get the idea yet?
Ignoring Tach and his digression, what does GR have to say about the density variations?
My pick is: nothing, in GR a gravitating body is just a mass-energy density. Ricci curvature is about changes in volume due to gravity, Einstein's equation says nothing about the shape or distribution; there is just mass in a given volume.
My pick is: nothing, in GR a gravitating body is just a mass-energy density. Ricci curvature is about changes in volume due to gravity, Einstein's equation says nothing about the shape or distribution; there is just mass in a given volume.
This does not say the model corresponds to the actual surface, why do you think it does?
Very simple , because the NIMA paper I linked earlier says it DOES, within a standard deviation of $$10^{-8}$$, contrary your persistent crank protestations.
Of course all clocks at the same gravitational potential tick at the same rate, that does not mean all clocks on the surface of the earth tick at the same rate.
Look arfa,
It is very simple: the paper demonstrates that the geoid gravitational potential at the pole is equal to the geoid gravitational potential at the equator. So, contrary to your crankish protestations, the clocks on the geoid at the pole and at the equator tick at the same rate. The main reason is that there are two effects , not one, at work: the gravitational attraction and the centrifugal force. The surface of the geoid is "arranged" in such a way that the resulting effective potential , $$\Phi_{eff}$$, is the same everywhere on its surface.
Ignoring Tach and his digression, what does GR have to say about the density variations?
My pick is: nothing,
I do not understand why you keep making up stuff, when you clearly do not know GR. In GR, contrary to your claim, there are specific equations that deal with anisotropic cases.
Einstein's equation says nothing about the shape or distribution; there is just mass in a given volume.
This is, like most everything you post on the subject, false.
That is because the geoidal surface is a surface of equipotential. That does not mean the physical surface is a surface of equipotential.Tach said:It is very simple: the paper demonstrates that the geoid gravitational potential at the pole is equal to the geoid gravitational potential at the equator.
Please explain how the previous sentence is wrong.
How does GR deal with the shape of a gravitating body? Does it 'predict' spherical symmetry?In GR, contrary to your claim, there are specific equations that deal with anisotropic cases.
?.. the exact statement in the paper:
"Despite the Earth’s complicated shape with mountains and valleys, its gravitational field can be modeled to a fractional accuracy of $$10^{-14}$$ $$by^7$$".
Sorry to dispel your crank ideas, all clocks on the geoid tick at the same rate. No "adjustment" necessary since the geoid is the surface of equi-gravitational potential.
Tach you are now not just being argumentative, you are beginning to get stupid!
Stop and think before you post!
The geoid, is a hypothetical surface that extends beneath land masses and represents mean sea level. Read your own references and then cite the specific reference experimentally proving your above statement. You can't because there are very few locations on the earth where the surface of the earth coincides with the geoid.., and even then variations in the density and distribution of the Earth's mass results in variations in the gravitational force or potential at even those locations. Those who work and contribute to the literature from the NIST know all of this.
Get real Tach. You aren't stupid!.. Just argumentative!
That is because the geoidal surface is a surface of equipotential. That does not mean the physical surface is a surface of equipotential.
Incorrect on multiple levels. Let's start with the simplest level, maybe you'll get it: the sea level is part of the geoid. Since oceans form 2/3 of the Earth surface....Now both the equator and the pole seal level are part of the geoid, so you have been wrong all along. If you insist on "learning" from wiki, be prepared to deal with the fact that they are wrong in many instances.
Please explain how the previous sentence is wrong.
I did, you made up your mind that GR says nothing about density variation". I pointed out that the EFEs work just the same for anisotropic environments. Since you do not understand GR, this is difficult to explain to you, suffice to say that density does not need to be constant. The EFEs are simpler in this case (uniform density) but this is not the only case.
How does GR deal with the shape of a gravitating body? Does it 'predict' spherical symmetry?
This is a different, unrelated question. Do you intend to learn GR by posting questions? (and then arguing that the answers do not conform to your prejudices?). This is not the way to learn physics, especially a difficult subject like GR. Enroll in a class.
Tach you are now not just being argumentative, you are beginning to get stupid!
I understand that you have difficulty in following the mathematical formalism. Now, it becomes clear that you cannot even follow the English narrative. The sentence is pretty clear but you continue to argue in favor of your fringe ideas.
The geoid, is a hypothetical surface that extends beneath land masses and represents mean sea level.
"Mean sea level" is about 2/3 of Earth surface. It includes the equator and the poles.
You can't because there are very few locations on the earth where the surface of the earth coincides with the geoid..,
Err, how about 2/3 of Earth surface? (actually, more like 70%).
Get real Tach. You aren't stupid!
I have to agree with you on this one. But someone else is.
Tach said:This is, like most everything you post on the subject, false.me said:Einstein's equation says nothing about the shape or distribution; there is just mass in a given volume.
Tach said:This is a different, unrelated question.me said:How does GR deal with the shape of a gravitating body? Does it 'predict' spherical symmetry?
How is it different or unrelated? GR uses the Ricci tensor, right? Ricci curvature is about changes in volume, not about the shape of a volume.
According to Baez the equation: $$
\frac {\ddot V} { V} {\mid _{t = 0}} = -{1\over 2} (\rho + P_x + P_y + P_z) $$ is Einstein's equation. Where is the information that describes differences in density, or shape?
As for your argument about the geoid: suppose you define a surface of equipotential several thousand k above the earth's surface, and locate several identical clocks on this surface. Then all the clocks will run at the same rate. How does using the surface of the ocean change this? Answer: it doesn't except that clocks closer to the centre of the earth will run slower.
Now Newton's law of universal gravitation is: $$ F\; =\; \frac {-Gm_1 m_2} {r^2} $$. This says the force between two gravitating masses (m[sub]1[/sub], m[sub]2[/sub]) depends on the square of the distance between them. So if the equator is further from the centre than the poles, the force will be less at the equator than at the poles (unless Newton made some kind of mistake).
How is it different or unrelated?
Simple, you demonstrated that you are unable to tell the difference between density and shape.
No point in progressing to advanced subject like EFEs and tensors until after you get the basics.
As for your argument about the geoid: suppose you define a surface of equipotential several thousand k above the earth's surface, and locate several identical clocks on this surface. Then all the clocks will run at the same rate.
Finally you are starting to realize that all clocks tick at the same rate on an equipotential surface. So why are you playing back to me what I've been telling you for the last two pages? Are trying to demonstrate that you are finally getting it?
How does using the surface of the ocean change this? Answer: it doesn't except that clocks closer to the centre of the earth will run slower.
You still don't get it: the pole is closer to the center of the Earth than any point at the Equator. Nevertheless, the clocks on the Equator tick at the same rate as the clock at the pole.
Now Newton's law of universal gravitation is: $$ F\; =\; \frac {-Gm_1 m_2} {r^2} $$.
One last time: the explanation has nothing to do with "Newton's law of universal gravitation". It has to do with the effective potential. Can't you follow a simple paper?
This says the force between two gravitating masses (m[sub]1[/sub], m[sub]2[/sub]) depends on the square of the distance between them. So if the equator is further from the centre than the poles, the force will be less at the equator than at the poles (unless Newton made some kind of mistake).
Err, a few persistent errors:
1. The Earth is rotating, so there is centrifugal force (that you keep missing).
2. The ticking of the clocks is not tied to force but to potential
3. Though the pole is closer to the Earth center than the Equator (this is the only thing that you get right), the effective potential at the pole is equal to the effective potential at the Equator, hence the clocks tick at the same rate.
Oh bullshit. I've demonstrated no such thing. Try again.Tach said:Simple, you demonstrated that you are unable to tell the difference between density and shape.
Where is the INFORMATION in Einstein's equation that describes the shape of a body with mass?
Is that because I've only mentioned that the poles are closer to the centre of the earth than the equator, less than 10 times?You still don't get it: the pole is closer to the center of the Earth than any point at the Equator.
1. I know the earth is rotating.1. The Earth is rotating, so there is centrifugal force (that you keep missing).
2. The ticking of the clocks is not tied to force but to potential
3. Though the pole is closer to the Earth center than the Equator (this is the only thing that you get right), the effective potential at the pole is equal to the effective potential at the Equator, hence the clocks tick at the same rate.
2. The swinging of a pendulum (clock) is tied to g--if g is smaller at the equator then a pendulum will accelerate less than at the poles.
3. Clocks tick at the same rate at the equator and the poles because relativistic effects cancel gravitational effects (but is this true for every oblate spheroid?).
I've seen an article that states the surface of the earth is an equipotential surface which is why clocks tick at the same rate everywhere. But this can't be true; it doesn't appear to allow for height above the surface, and that the surface is not smooth (have you noticed?). Further, there are other articles which say something else: the geoid is a hypothetical surface that coincides with sea level but ignores tides (and some other things).
I've seen an article that states the surface of the earth is an equipotential surface which is why clocks tick at the same rate everywhere.
I know you've seen it, I have pointed it out to you about 4 times, unfortunately you are incapable to understand it. You have demonstrated it repeatedly.
But this can't be true; it doesn't appear to allow for height above the surface, and that the surface is not smooth (have you noticed?).
You are adopting the typical crank position: "it can't be true because...". Take it with the referees and the editors of AmJPhys.