Davis Relativity Model (Debate/discussion edition)

I won't try to answer all of this, but will tackle just bits - otherwise, it just gets too long.

Quote from: rabinoz on October 13, 2017, 05:11:00 AM

The nett result of this is that there is no simple relativity of angular motion and there are certainly situations in which it leads to ridiculous results - more on this later.

Alright, think of it this way. If observer A is in a rotating room with respect to observer B, which is observing the rotating room. Could the analysis of observer A considering a rotating shell of the universe dragging for inertial forces match observer B's analysis of the independent acceleration of that rotating room? If so, then we can postulate that inertia may not be intrinsically absolute.

Einstein would have preferred a theory that fitted with Mach's Principle (Einstein's term for it), but found that he was unable to adhere to the relativity of angular motion.
And it was a thought experiment similar to yours that forced Einstein to depart from adherence to Mach's Principle with regard to angular motion.
He reasoned this way: << in the following I have tried to format it as Norton did, maybe unsuccessfully? >>

Relativity of Inertia ("Mach's Principle")

What also attracted Einstein in this analysis was that it promised to remedy a defect he perceived in both Newton's physics and in special relativity. In both, you will recall, it is just a brute fact that certain motions are distinguished as inertial. This, in Einstein's view, was worrisome. It was no better than the original idea that there is an ether state of absolute rest. There seemed to Einstein no good reason for why one state should be the absolute rest state rather than another. Correspondingly, Einstein saw no good reason for why some motions should be singled out as inertial and others as accelerating.

In 1916, Einstein formulated this worry in a thought experiment. He imagined two fluid bodies in a distant part of space. These bodies, the reader quickly infers, are like stars or planets, which form roughly spherical shapes under their own gravity. Einstein further imagined that there is relative rotation between the two bodies about the axis that joins them. This relative rotation is verifiable by observers on each body, who can trace out the motion of the other body. Each would judge the other to be rotating.

It can happen in ordinary Newtonian physics that one of these bodies is not rotating with respect to an inertial frame and the other one is. In that case, the second rotating body will bulge but not the first. This effect arises on the earth. It rotates about the axis of its north and south poles. It bulges slightly at the equator as a result of centrifugal forces that seek to fling the matter of the earth away from this axis.

It would be entirely unacceptable, Einstein now asserted, were this to happen to two spheres in an otherwise empty space. For there is no difference in the observable relations between the two spheres. Each rotates with respect to the other. So why should just one bulge? The supposition of Newton's absolute space or of inertial systems, Einstein protested, was an inadequate explanation. Einstein demanded something observable to make the difference.

Einstein was an avid reader of the physicist-philosopher Ernst Mach. In Mach's writings, Einstein had found what seemed to be a solution to the problem. Mach seemed to be proposing, Einstein thought, that the privileging of certain states of motion is due to the distribution of matter in the universe. Why is our frame of reference inertial? It is because the stars are at rest in our frame.

Why is my wording so careful here? it is not clear that what Einstein reported Mach as saying is what Mach actually said. For more, see John D. Norton, "Mach's Principle before Einstein." in J. Barbour and H. Pfister, eds., Mach's Principle: From Newton's Bucket to Quantum Gravity: Einstein Studies, Vol. 6. Boston: Birkhäuser, 1995, pp.9-57. Download.

When we try to accelerate, we feel inertial forces. These are the forces that make us dizzy when we spin in a fun fair; or they are the forces that throw our coffee in the air when our airplane hits an air pocket.

These forces, Einstein understood Mach to assert, arise from an interaction between the mass of our body (and our coffee) and all the other masses of the universe, distributed in the stars. Einstein first called this idea the "relativity of inertia" and later, in 1918, "Mach's Principle."

In the case of Einstein's two fluid spheres, the bulge of one of them would now be explained by the fact that this bulging sphere was rotating with respect to all the other masses of the universe, whereas the other sphere was not. That would be the observable difference between the two fluid bodies. This analysis was clearly inspired by Mach's famous account of Newton's bucket experiment. Newton had noted that water in a spinning bucket adopts a concave surface. The concavity is a result, Newton urged, of its rotation with respect to absolute space. No, Mach had responded several hundred years later, all one has in the case of Newton's bucket in rotation with respect to the stars. We cannot know anything more than what our direct observations tell us. All they tell us is that these inertial forces arise when we accelerate relative to the stars.

The weakness of this analysis is that there is no account of how rotation with respect to distant masses could produce these inertial forces. In 1907, Einstein hoped that his emerging theory of gravity would provide  the mechanism. It could then satisfy Mach's Principle and, through it, generalize the principle of relativity to acceleration. For in a theory that satisfies Mach's Principle, no state of motion is intrinsically inertial or accelerating. When we see something accelerating, it is not accelerating absolutely in such a theory; it is merely accelerating with respect to the stars. Preferred inertial motions need not enter into the account anymore. All motion, accelerated or inertial, would be relative.

To deliver this sort of account of inertial forces, Einstein's theory would need to break down the strict division between inertial and accelerated motion of his special theory of relativity. The principle of equivalence promised to weaken this division. According to it, whether the physicist in the box was to be judged accelerating or not depended on your point of view. An inertial observer would judge the physicist to be accelerating uniformly in a gravitation free space. The physicist would judge him or herself to be unaccelerated in a gravitational field. It was a first step towards generalizing the principle of relativity to acceleration, Einstein believed.

Learning About Gravitation

By his own later judgment, Einstein did not, in the end, find a theory that fully satisfied Mach's Principle. The immediate benefit of his new principle of equivalence, however, was that it let Einstein learn a lot about gravitation. For the principle delivered to Einstein one special case of a gravitational field that, he believed, conformed with relativity theory and in which all bodies truly fell alike. Einstein's program of research on gravity in the five years following 1907 was simply to examine the properties of this one special case and to try to generalize them to recover a full theory. His early hope was that the generalization of the principle of relativity would somehow emerge in the course of those investigations.

<< You can read the rest of Learning About Gravitation>>
From: Relativity of Inertia ("Mach's Principle")

So no, By his own later judgment, Einstein did not, in the end, find a theory that fully satisfied Mach's Principle.

And as a result claiming that angular motions are purely relative is definitely in conflict with Einstein's GR. Einstein claimed no such thing.

From this it is rather obvious that there is a vast difference between a Geocentric and a Heliocentric Universe.

That's enough for one post. I'll try to do a bit more later.


<< I think my links might have been bad in the earlier post and I hope they are noe right. >>

I won't try to answer all of this, but will tackle just bits - otherwise, it just gets too long.

Quote from: AltSpace on October 13, 2017, 12:27:28 PM

Quote from: rabinoz on October 13, 2017, 05:11:00 AM

The nett result of this is that there is no simple relativity of angular motion and there are certainly situations in which it leads to ridiculous results - more on this later.

Alright, think of it this way. If observer A is in a rotating room with respect to observer B, which is observing the rotating room. Could the analysis of observer A considering a rotating shell of the universe dragging for inertial forces match observer B's analysis of the independent acceleration of that rotating room? If so, then we can postulate that inertia may not be intrinsically absolute.

Einstein would have preferred a theory that fitted with Mach's Principle (Einstein's term for it), but found that he was unable to adhere to the relativity of angular motion.
And it was a thought experiment similar to yours that forced Einstein to depart from adherence to Mach's Principle with regard to angular motion.
He reasoned this way: << in the following I have tried to format it as Norton did, maybe unsuccessfully? >>

Quote

Relativity of Inertia ("Mach's Principle")

What also attracted Einstein in this analysis was that it promised to remedy a defect he perceived in both Newton's physics and in special relativity. In both, you will recall, it is just a brute fact that certain motions are distinguished as inertial. This, in Einstein's view, was worrisome. It was no better than the original idea that there is an ether state of absolute rest. There seemed to Einstein no good reason for why one state should be the absolute rest state rather than another. Correspondingly, Einstein saw no good reason for why some motions should be singled out as inertial and others as accelerating.

In 1916, Einstein formulated this worry in a thought experiment. He imagined two fluid bodies in a distant part of space. These bodies, the reader quickly infers, are like stars or planets, which form roughly spherical shapes under their own gravity. Einstein further imagined that there is relative rotation between the two bodies about the axis that joins them. This relative rotation is verifiable by observers on each body, who can trace out the motion of the other body. Each would judge the other to be rotating.

It can happen in ordinary Newtonian physics that one of these bodies is not rotating with respect to an inertial frame and the other one is. In that case, the second rotating body will bulge but not the first. This effect arises on the earth. It rotates about the axis of its north and south poles. It bulges slightly at the equator as a result of centrifugal forces that seek to fling the matter of the earth away from this axis.

It would be entirely unacceptable, Einstein now asserted, were this to happen to two spheres in an otherwise empty space. For there is no difference in the observable relations between the two spheres. Each rotates with respect to the other. So why should just one bulge? The supposition of Newton's absolute space or of inertial systems, Einstein protested, was an inadequate explanation. Einstein demanded something observable to make the difference.

Einstein was an avid reader of the physicist-philosopher Ernst Mach. In Mach's writings, Einstein had found what seemed to be a solution to the problem. Mach seemed to be proposing, Einstein thought, that the privileging of certain states of motion is due to the distribution of matter in the universe. Why is our frame of reference inertial? It is because the stars are at rest in our frame.

Why is my wording so careful here? it is not clear that what Einstein reported Mach as saying is what Mach actually said. For more, see John D. Norton, "Mach's Principle before Einstein." in J. Barbour and H. Pfister, eds., Mach's Principle: From Newton's Bucket to Quantum Gravity: Einstein Studies, Vol. 6. Boston: Birkhäuser, 1995, pp.9-57. Download.

When we try to accelerate, we feel inertial forces. These are the forces that make us dizzy when we spin in a fun fair; or they are the forces that throw our coffee in the air when our airplane hits an air pocket.

These forces, Einstein understood Mach to assert, arise from an interaction between the mass of our body (and our coffee) and all the other masses of the universe, distributed in the stars. Einstein first called this idea the "relativity of inertia" and later, in 1918, "Mach's Principle."

In the case of Einstein's two fluid spheres, the bulge of one of them would now be explained by the fact that this bulging sphere was rotating with respect to all the other masses of the universe, whereas the other sphere was not. That would be the observable difference between the two fluid bodies. This analysis was clearly inspired by Mach's famous account of Newton's bucket experiment. Newton had noted that water in a spinning bucket adopts a concave surface. The concavity is a result, Newton urged, of its rotation with respect to absolute space. No, Mach had responded several hundred years later, all one has in the case of Newton's bucket in rotation with respect to the stars. We cannot know anything more than what our direct observations tell us. All they tell us is that these inertial forces arise when we accelerate relative to the stars.

The weakness of this analysis is that there is no account of how rotation with respect to distant masses could produce these inertial forces. In 1907, Einstein hoped that his emerging theory of gravity would provide  the mechanism. It could then satisfy Mach's Principle and, through it, generalize the principle of relativity to acceleration. For in a theory that satisfies Mach's Principle, no state of motion is intrinsically inertial or accelerating. When we see something accelerating, it is not accelerating absolutely in such a theory; it is merely accelerating with respect to the stars. Preferred inertial motions need not enter into the account anymore. All motion, accelerated or inertial, would be relative.

To deliver this sort of account of inertial forces, Einstein's theory would need to break down the strict division between inertial and accelerated motion of his special theory of relativity. The principle of equivalence promised to weaken this division. According to it, whether the physicist in the box was to be judged accelerating or not depended on your point of view. An inertial observer would judge the physicist to be accelerating uniformly in a gravitation free space. The physicist would judge him or herself to be unaccelerated in a gravitational field. It was a first step towards generalizing the principle of relativity to acceleration, Einstein believed.

Learning About Gravitation

By his own later judgment, Einstein did not, in the end, find a theory that fully satisfied Mach's Principle. The immediate benefit of his new principle of equivalence, however, was that it let Einstein learn a lot about gravitation. For the principle delivered to Einstein one special case of a gravitational field that, he believed, conformed with relativity theory and in which all bodies truly fell alike. Einstein's program of research on gravity in the five years following 1907 was simply to examine the properties of this one special case and to try to generalize them to recover a full theory. His early hope was that the generalization of the principle of relativity would somehow emerge in the course of those investigations.

<< You can read the rest of Learning About Gravitation>>
From: Relativity of Inertia ("Mach's Principle")

So no, By his own later judgment, Einstein did not, in the end, find a theory that fully satisfied Mach's Principle.

And as a result claiming that angular motions are purely relative is definitely in conflict with Einstein's GR. Einstein claimed no such thing.

From this it is rather obvious that there is a vast difference between a Geocentric and a Heliocentric Universe.

That's enough for one post. I'll try to do a bit more later.


<< I think my links might have been bad in the earlier post and I hope they are noe right. >>

I'm pretty sure it's still an open problem rather than a thing removed by GR, but I'll look more into it.

Frame-Dragging provided a means which the configuration of matter determines inertia, or the rotation of the universe equates to the rotation of an object itself.

Though I presume what this means here is that the object would experience the centrifugal 'bulge' while the universe around it doesn't, but surely such a phenomena could only be determined locally in your reference frame which is experiencing centrifugal force.

Quote from: rabinoz on October 13, 2017, 05:11:00 AM

There is absolutely nothing in Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space".

Curved space-time does, it's flat in non-euclidean space-time.

Not in my book!

The effect on the spacelike component of curved spacetime at the earth's surface is to do not more than change the earth's size by an immeasurable amount.

You might like to read Calculations on space-time curvature within the Earth and Sun, Wm. Robert Johnston
Which concludes with:

9  Conclusion
This exercise produced several expressions of relativistic curvature for solar system objects.  The true diameters of the Sun and Earth are 4.1 km and 4.4 mm greater, respectively, than one would expect from applying Euclidean geometry (C = πd) to the observed surface of these bodies.  These results are significantly affected by the non-uniform internal density variation of these bodies; they are 4 and 1.5 times greater, respectively, than for a equal mass/equal circumference object of uniform density.  In the case of the Sun, this internal space-time curvature affords it a volume 6 parts per million greater than the Sun’s surface
would enclose in Euclidean space.  An embedding diagram was graphed for the case of the Sun.  This demonstrates the contrast between curvature in the Sun and in a uniform density model of the Sun.  Quick calculations for a neutron star, assuming uniform density, showed the relativistic radius and volume to be 10% and 18% greater than the corresponding Euclidean values.

For the earth, curved spacetime changes the diameter by  a whole 4.4 mm in 12,742 km or a part in 2.9 x 10⁹ 

If you can calculate the effect of Einstein's curved spacetime better than Wm. Robert Johnston, please present your calculations forthwith.

So, no there is absolutely nothing in

Quote

By the way, the fact that GR reduces exactly to Newton's Laws of Motion and Gravitation is no accident.

No it doesn't. Newton's gravitation failed to predict the 4D continuum that GR champions, it certainly matches it in many ways, but they are distinct representations.

Please include ALL of what I said next time!

By the way, the fact that GR reduces exactly to Newton's Laws of Motion and Gravitation is no accident.
Some of Einstein's earlier attempts did not reduce to Newton's Laws in the low speed, low mass situation, and were discarded.

And Einstein's General Relativity certainly does "reduce to Newton's Laws in the low speed, low mass situation".

On this topic you might read: SCHWARZSCHILD SOLUTION IN GENERAL RELATIVITY, Marko Vojinovi ́c which contains:

3.3 Newton’s law of gravity
Consider the nonrelativistic motion of a test particle in Schwarzschild geometry. Nonrelativistic means that the particle’s velocity can be considered much less than the speed of light, as measured relative to the origin of Schwarzschild coordinates. It also means that
the source of the gravitational field is weak
<< lots of "hard sums" >>
Finally, if we multiply this with the mass m of the test particle, we can rewrite the result as

which is the familiar Newton’s second law of dynamics, while the force term

is the Newton’s law of gravitation for a body with spherical symmetry and mass M.

So please stop pretending that Einstein's Theory of General Relativity provides and respectible basis for your Ferrari Effect and Davis Relativity Model.
It most certainly does not.

I'm pretty sure it's still an open problem rather than a thing removed by GR, but I'll look more into it.

Sure, Mach's Principle is still being discussed, but it can hardly be called an open problem.
Applied to the universe as a whole, Mach's Principle leads to many totally riduculous situations.

Frame-Dragging provided a means which the configuration of matter determines inertia, or the rotation of the universe equates to the rotation of an object itself.

There is no conflict between frame-dragging and GR. Einstein predicted it!

Einstein's Warped View of Space Confirmed
Earth's spin warps space around the planet, according to a new study that confirms a key prediction of Einstein's general theory of relativity.

After 11 years of watching the movements of two Earth-orbiting satellites, researchers found each is dragged by about 6 feet (2 meters) every year because the very fabric of space is twisted by our whirling world.

The results, announced today, are much more precise than preliminary findings published by the same group in the late 1990s.
Frame dragging
The effect is called frame dragging. It is a modification to the simpler aspects of gravity set out by Newton. Working from Einstein's relativity theory, Austrian physicists Joseph Lense and Hans Thirring predicted frame dragging in 1918. (It is also known as the Lense-Thirring effect.)

Here's how it works:
Any object with mass warps the space-time around it, in much the same way as a heavy object deforms a stretched elastic sheet, explained study leader Ignazio Ciufolini of the Universit? di Lecce in Italy.

The rest in: Space.com, Science & Astronomy, Einstein's Warped View of Space Confirmed

Might be a bit low level.

Though I presume what this means here is that the object would experience the centrifugal 'bulge' while the universe around it doesn't, but surely such a phenomena could only be determined locally in your reference frame which is experiencing centrifugal force.

But, at least in a thought experiment, you could communicate with another observer in a different reference frame.
That was exactly Einstein's thought experimant, and one reason why he felt obliged to abandon that part of Mach's Principle.
By the way, you might note that is always Mach's Principle and has never been looked on as a theory.

Not in my book!

The effect on the spacelike component of curved spacetime at the earth's surface is to do not more than change the earth's size by an immeasurable amount.

You might like to read Calculations on space-time curvature within the Earth and Sun, Wm. Robert Johnston
Which concludes with:

Quote from: Wm. Robert Johnston

9  Conclusion
This exercise produced several expressions of relativistic curvature for solar system objects.  The true diameters of the Sun and Earth are 4.1 km and 4.4 mm greater, respectively, than one would expect from applying Euclidean geometry (C = πd) to the observed surface of these bodies.  These results are significantly affected by the non-uniform internal density variation of these bodies; they are 4 and 1.5 times greater, respectively, than for a equal mass/equal circumference object of uniform density.  In the case of the Sun, this internal space-time curvature affords it a volume 6 parts per million greater than the Sun’s surface
would enclose in Euclidean space.  An embedding diagram was graphed for the case of the Sun.  This demonstrates the contrast between curvature in the Sun and in a uniform density model of the Sun.  Quick calculations for a neutron star, assuming uniform density, showed the relativistic radius and volume to be 10% and 18% greater than the corresponding Euclidean values.

For the earth, curved spacetime changes the diameter by  a whole 4.4 mm in 12,742 km or a part in 2.9 x 10⁹ 

If you can calculate the effect of Einstein's curved spacetime better than Wm. Robert Johnston, please present your calculations forthwith.

So, no there is absolutely nothing in

Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space".

Nowhere in that article does it claim that GR boils down to a sphere in Euclidean space-time. It is comparing the volumes of surfaces under curved space-time and a similar mass in Euclidean space-time.

Because of the curvature of space-time, the volume contained within the surface
of the Sun and the Earth is greater than the volume enclosed by a similar surface
in Euclidean space.

Please include ALL of what I said next time!

I believe I did get what you said.

And Einstein's General Relativity certainly does "reduce to Newton's Laws in the low speed, low mass situation".

On this topic you might read: SCHWARZSCHILD SOLUTION IN GENERAL RELATIVITY, Marko Vojinovi ́c which contains:

Quote from: Marko Vojinovi ́c

3.3 Newton’s law of gravity
Consider the nonrelativistic motion of a test particle in Schwarzschild geometry. Nonrelativistic means that the particle’s velocity can be considered much less than the speed of light, as measured relative to the origin of Schwarzschild coordinates. It also means that
the source of the gravitational field is weak
<< lots of "hard sums" >>
Finally, if we multiply this with the mass m of the test particle, we can rewrite the result as

which is the familiar Newton’s second law of dynamics, while the force term

is the Newton’s law of gravitation for a body with spherical symmetry and mass M.

So please stop pretending that Einstein's Theory of General Relativity provides and respectible basis for your Ferrari Effect and Davis Relativity Model.
It most certainly does not.

It doesn't boil down to the same explanation regardless, with warped space-time geometry and geodesics giving the effect of 'Gravity', not a force since such paths in this field are straight.

Sure, Mach's Principle is still being discussed, but it can hardly be called an open problem.
Applied to the universe as a whole, Mach's Principle leads to many totally riduculous situations.

GR doesn't even have the solutions for all this, it still lacks general two-body solutions.

There is no conflict between frame-dragging and GR. Einstein predicted it!

I never implied there was a 'conflict'.

But, at least in a thought experiment, you could communicate with another observer in a different reference frame.
That was exactly Einstein's thought experimant, and one reason why he felt obliged to abandon that part of Mach's Principle.
By the way, you might note that is always Mach's Principle and has never been looked on as a theory.

Part of being a frame of reference would imply that these happen relative to your coordinates. So, a spinning room could transfer a rotating universe under frame dragging for centrifugal force.
This made sense to me with Mach's principle, though I understand that it hasn't been put into a theory per say, but it isn't contradictory to GR from what I've seen. I've even read that it is a 'philosophical' problem.

But that is not dealing with my point. The Geodesics are straight above the surface, equating the surface to a straight line, as flat.

How is it not dealing with the point?
My example shows a straight line through non-Euclidean geometry, which is straight "above" another line, remaining the same distance "above" the other line and thus by your reasoning that line should be straight.

You didn't show it was anything near wrong, you ignored it and instead use lines along varying longitudes and declination's rather than the surface of the Earth itself as directly across from a geodesic, matching it.

Again, the burden is on you to show it works, not on me to show it doesn't.
Again, I did show what was wrong with your method. Stop pretending your method would only apply to this one specific circumstance.
It is far more general than that.

Also, while I used Earth as an example, that is not the core of my example.
The core of my example is NON_EUCLIDEAN SPACE!!!
Do you understand that?
It isn't just for Earth, it is for a spherical space.

I am showing how your method does not work in non-Euclidean spaces, which would include non-Euclidean space-time.

So no, I wasn't ignoring anything. I was demonstrating that the method you are using does not work in non-Euclidean spaces. You are yet to deal with this FACT! Instead you just repeatedly dismiss it.

Quote

Yes, 4D space-time is a continuum. That doesn't mean it makes sense to describe the surface of an object by appealing to space-time.
Just like you wouldn't try discussing the cross sectional area of an object using 3D space, and instead rely upon the cross section which is intrinsically 2D.

You would need to in this case, since the surface of the Earth is in a unification as a mass by 4D space-time itself, you can't remove time in the description.

No you wouldn't.
Just like you can remove the z axis in a description of a cross sectional area, you can remove the t axis in describing a 3D shape in space.
You can quite easily remove the time axis.
Earth can be described as a 3D shape without appealing to time.
As soon as you start dealing with time you are no longer discussing the shape of objects. Instead, you are discussing trajectories.

The object is 'moving' through time, even if it were to be at 0 velocity through 3D spatial coordinates, so it defines an object as well.

And that doesn't make a bit of difference. Just like if you were to take an infinitely long prism, its cross section remains the same, even though that cross section is "moving through" the 3rd dimension.

The only time time comes into it is when you discussing a changing shape. But in each case, you aren't comparing the trajectory of an test-mass through space-time to the object. You are looking at the shape of the object in space at each instant.

Quote

Exactly which point was wrong? There were several steps. Here they are again, numbered for you:
1 - You have a reference line that is straight.
2 - You have another line which you would like to determine if it is straight.
3 - You start at some point along the straight line, and measure the distance to the other line.
4 - You then move along the green line, measuring the distance distance to the other line.
5 - You note the distance remains the same.
6 - Thus you conclude that the line is straight.

What you did wrong

Are you capable of understanding simple English and making a rational argument?

I provided a method.
What step was wrong?
Can you point that out or can you only make BS statements?

Unless you can tell me exactly what step was wrong and what the correct version is you have no basis to claim that this is not an accurate representation of your method, nor that I haven't disproven your method.

Alternatively, provide your own method like this, making it general, without appealing to any specific thing like a circular orbit, space-time, Earth, etc. Basically you can only appeal to the space itself (the non-Euclidean space, which space-time would be), and things which exist in general for spaces.

taken a geodesic and compared to something not part of the surface of the Earth it crosses through.

What do you mean by this?
Are you once again trying to turn my 2D problem into 3D?
I am using a non-Euclidean surface, a 2D spherical space. All the lines are on the same surface.

I think it's because you know it's straight and choose to ignore it. I really find it hard to believe that you are simply 'missing the point'.

I am not missing the point. I understand the point quite well. I also realise your method is pure BS which relies upon space-time being Euclidean (otherwise the method would not work).
I find it hard to believe that you are simply missing the point and aren't just intentionally spouting whatever BS you can to try and pretend you are correct.

Quote

No. This straight line geodesic is a trajectory through space time, and thus does not describe the surface of Earth (as a shape).
Also, as it passes through a different region of space-time, with different curvature, it does not match the surface of Earth.

This trajectory is following a geodesic that matches the Earth's straight surface.

No it doesn't.
It is no where near Earth's surface. As such, it does not match it at all. It is some orbit up in space.

But guess what? These aren't directly through the center like my example has it.

Going directly through the centre makes no sense.
Are you once again trying to make my 2D problem 3D?
Stop trying to appeal to extra dimensions that do not exist in this problem.
There is no centre of this space. (or alternatively, any point can be considered the centre).
If you like, you can consider the "north pole" of this spherical space as the centre.
In this Euclidean representation of this non-Euclidean space, a straight going towards or away from the north pole will remain visually as a straight line.
Any other straight line will curve, with the extent of curvature being somewhat complex, dependent upon its direction and distance from the "north pole". A straight line around the "equator" would appear as the green circle. Other circles centred on the "north pole" are not straight lines. Instead, all straight lines must cross the "equator".

Note that the terms "equator" and "north pole" are not indicating any higher dimensions, but are merely used as tools to understand this non-Euclidean space as we are far more used to dealing with Euclidean spaces. This space is entirely 2D.

Also note that other than the non-circular straight line, this exact same diagram can be used for your problem, although the plane is now far more complex as instead of starting in 2D, you start in 4D and try to reduce it to 2D.

They are not since your example misses the lines being so that straight connections cross through the same area through the Earth.

Again, stop making it 3D.
The straight connections all cross through the same area, the north pole. But more importantly, they all have the same vector (due north).
So no, they are basically the same.

Again: Stop pretending it does and deal with it or at least admit you can't falsify it.

Why would I lie like that?
It does apply, you are yet to establish that it does not. All you have been able to do is just dismiss it out of hand, or misrepresent it to pretend it doesn't apply or otherwise lie about it.
I have falsified your method.

Earth's surface is in space-time itself as well, just like any other object.

When you are considering it as a surface, rather than as a trajectory, it is in space, not space-time.

You simply can't ignore the 4D space-time continuum with Earth.

Again, when describing the shape of Earth in space, rather than as a trajectory, you MUST ignore the time component, unless you want to discuss how it changes through time.

This trajectory follows the same geodesic that matches the Earth's surface as flat.

Again, are you trying to describe the spatial shape of Earth, or the trajectory of it through time?

I know you are trying so hard to deny it

No. I am trying quite easily to refute it, and suceeding.

No, that would be in the same 4D space-time continuum we all are in, just that one isn't affected by curved space-time.

So an entirely Euclidean space-time which is nothing like the one we live in.
In this, there would be no circular orbits.

The Earth's surface does not remain at the same point in time.

That's right, but if you move on to a new point in time, you are now seeing the surface at that point in time.

Unfortunantly for you, time exists, and defines the Geometry of the Earth as well, so your hilarious roundie excuses won't work here for GR.

No, it defines how the surface of Earth changes over time, and according to GR, it is constantly "expanding".

And the spatial axes too surely, as they are both linked, so an object at 0 velocity will start accelerating with translating curved 4D temporal coordinates to 3D spatial coordinates.

Notice how you had to invoke the time coordinate and have them move through time to have it change?
Now try it with an object which starts at an infinitely velocity and thus can traverse the entirety of space in an instant in time.
Now you don't have the "apparent" acceleration due to gravity and thus you don't get any curvature.

Quote

And in intersection of space time with a purely spatial space to produce space is not space-time.
What you are doing now is akin to saying a line and a plane are the same and that there is no distinction between them.

Straight lines are all parallel to the surface across Earth, it seems the surface would be represented that of a flat plane.

This did not address my point at all.
You are also yet to establish that baseless garbage of yours.

Quote

Again, we are discussing the shape of Earth, not the trajectory of a hypothetical mass through space-time.

These geodesics are quite independent of the masses that follow them.

Again, this does not address the point.

So is the Earth's surface, it's a geodesic itself in space-time.

Partly right, partly wrong.
Earth's surface is going too slowly. That means it isn't a geodesic in spacetime. Instead it is curving "upwards" continually expanding.

No, the surface is flat in Non-euclidean space-time, the round earth fails to match it.

Again, this is discussing it in space, not space-time.
The best, full space-time description of Earth (ignoring the influences of other things) is a continually expanding oblate spheroid.
No point on the surface is flat. It is all going far too slowly.

However, the cross-section through the Earth has straight lines all the way through, which includes the Earth's surface. This clearly shows it's flat. You have yet to address that.

I have addressed that by pointing out it is pure bullshit.
You are yet to show that any point on Earth's surface is a straight line through space time.

The simplest solution here would be to ask Altspace to publish his work and have it peer-reviewed.  I'd be interested what it would take to have this printed in the International Journal of Theoretical Physics.

Quote from: rabinoz on October 13, 2017, 07:53:12 PM

Not in my book!

The effect on the spacelike component of curved spacetime at the earth's surface is to do not more than change the earth's size by an immeasurable amount.

You might like to read Calculations on space-time curvature within the Earth and Sun, Wm. Robert Johnston
Which concludes with:

Quote from: Wm. Robert Johnston

9  Conclusion
This exercise produced several expressions of relativistic curvature for solar system objects.  The true diameters of the Sun and Earth are 4.1 km and 4.4 mm greater, respectively, than one would expect from applying Euclidean geometry (C = πd) to the observed surface of these bodies.  These results are significantly affected by the non-uniform internal density variation of these bodies; they are 4 and 1.5 times greater, respectively, than for a equal mass/equal circumference object of uniform density.  In the case of the Sun, this internal space-time curvature affords it a volume 6 parts per million greater than the Sun’s surface
would enclose in Euclidean space.  An embedding diagram was graphed for the case of the Sun.  This demonstrates the contrast between curvature in the Sun and in a uniform density model of the Sun.  Quick calculations for a neutron star, assuming uniform density, showed the relativistic radius and volume to be 10% and 18% greater than the corresponding Euclidean values.

For the earth, curved spacetime changes the diameter by  a whole 4.4 mm in 12,742 km or a part in 2.9 x 10⁹ 

If you can calculate the effect of Einstein's curved spacetime better than Wm. Robert Johnston, please present your calculations forthwith.

So, no there is absolutely nothing in

Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space".

Nowhere in that article does it claim that GR boils down to a sphere in Euclidean space-time. It is comparing the volumes of surfaces under curved space-time and a similar mass in Euclidean space-time.

Quote from: Wm. Robert Johnston

Because of the curvature of space-time, the volume contained within the surface
of the Sun and the Earth is greater than the volume enclosed by a similar surface
in Euclidean space.

Of course, "Nowhere in that article does it claim that GR boils down to a sphere in Euclidean space-time", it doesn't attempt to.
GR cannot prove the shape of the earth. GR is, among other things, a relativistic theory of gravitation.
So it could no more prove anything about the shape of planets than could Newton's Universal Law of Gravitation.^[1]

But that article shows that the effect of spacetime curvature on the space-like component in the vicinity of the vicinity of the earth is immeasurably small and does not change the shape, just the size.

So, whatever the shape of the earth the curvature of spacetime according to Einstein's GR does not make it look like something else.

This makes your claims of Einstein's GR supporting your Davis Relativity Model totally untenable.

[1] Though, Newtonian Gravitstion or Einstein's GR can show that asteroids (or planets) with radii larger than a few hundred kilometres cannot be flat.
     See "Potato Radius of Asteroids".

Okay, I finally got my math sorted out. Evil degrees radians screwing things up, then figuring out some nice spacing. (Note: not dealing with time dilation, which wont affect it significantly anway.)
Anyway, I have some pictures of "surfaces" in 3D space-time (i.e. 2D of space and 1D of time). Here is an album https://imgur.com/a/DV2kl
The x and y axes have 1 unit=1km, and the z axis is the time axis where 4 units=1s.
So first of all, a simple one to explain it, which isn't actually a surface:


This shows a few different geodesics in spacetime, as well as Earth.
Earth is the grey cylinder.
The cyan (or whatever you want to call it) line is a line purely through space. It is the limit of a hyperbolic trajectory as the velocity tends to infinity (beyond the speed of light).
The blue and green lines are parabolic orbits.
The magenta and yellow lines are circular orbits.

Now this is part of one way to make a surface, where you take all lines passing through a point which is normal to another line.
All of these lines are normal to the line going straight "up" along the x axis (directly out of the screen in the "front" view, and pass through the point which I had centred here but which corresponds to a height of 400 km above Earth's surface.
Continuing this, i.e. putting in more lines, you get a surface like this:


Now the colour corresponds to which of the above orbits it is in between, with red being a missing path which starts off with no motion through space.
Also note that this is for a hypothetical hollow Earth with all its mass as an infinitesimally small point at the centre, or for the region inside the grey cylinder to not exist.

Here is another way to make a surface  (technically 2). In this case, the line has the same angle to the straight line:

So this is the product of translating the cyan line along the magenta or yellow lines.
And you get completely different surfaces.

But here is the real killer of your argument:


First, we have a path through space time for a geostationary satellite. That is one which remains above the same point on Earth. It remains the same distance above this point on Earth, so just like your argument is saying.
This is shown in red.
In purple, there are lines connecting it to the centre of Earth (technically going slightly past it, but who cares).
In blue, we have a circular orbit that remains at the radius of Earth.
Notice how this does not remain below the geostationary orbit?
This is the geodesic that would be on Earth's surface (but not be Earth's surface). It intersects the straight line (or plane) of the geostationary orbit, just like my example with the great circle going to 10 degrees. That is one straight line.

But the real killer is the cyan line.
This is a line that shows a geodesic starting at a point on Earth's surface, as Earth's surface. It starts off at the same velocity as Earth's surface. Notice how it doesn't stay on Earth's surface?
It is a sub-orbital velocity. This would cause it to fall to the centre of Earth.
That is what a straight line through space-time would be for the surface of Earth.

Notice that it doesn't follow the surface of Earth?
It falls away.
So no, Earth's surface, when considered as a surface through space-time, is NOT FLAT!!!
Instead, it is constantly curving upwards/expanding. The reason it doesn't "appear" to be and it instead remains the same size is that the time axis is curving in towards the centre of Earth.

This all goes back to the equivalence principle.
You cannot tell the difference between being in a gravity well (or curved space-time) and an accelerating elevator. The 2 are equivalent.
That means in GR, Earth's surface is constantly accelerating upwards, being forced up by the pressure inside. It is not flat.

So if anything GR means Earth's surface CAN'T BE FLAT!!!

Going to admit you were wrong now?

<< I might get around to answering the rest of this one day. >>

But quite some time back, I posted this in a reply to John Davis (bless his devious soul) on this topic, and I believe exactly the same applies to you:

Quote from: John Davis on July 28, 2017, 12:07:56 PM

The base thought experiment is here.
http://www.theflatearthsociety.org/home/index.php/blog/einsteins-relativity-proves-earth-flat
Ignore the title as it is admittedly inaccurate.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Your non-Euclidean earth is nothing more than a "thought experiment".
There is no experimental basis to it nor the slightest basis for even associating  Einstein's name to it, other than to try to give it an air of false respectability.

Sure, in Einstein's GR spacetime "bends", but nowhere, except in the vicinity of extreme masses such as the extreme concentrations of black holes or whole galaxies, is the "bending" of the spacelike components (our 3-D space) at all significant.

In the vicinity of earth, this "bending" amounts to nothing more than a change of abouy 0.4 centimetre in a diameter of over 12,700 km and even near the sun, only about a 1.4 metre in diameter of almost 1.4 million km.

And here you come on the basis of a mere thought experiment proposing a bending of space that rolls some nebulous flat earth up into the ball that we observe. Total Piffle!

Come up with some experimental evidence or theoretical basis for this massive "rolling-up" of space and it might be a different matter.

Sure Einstein used "thought experiments", but also the experimental work of many others to refine his theories.

But what I cannot follow is that you appear, to my simple mind at least, as though you are now trying to claim that the earth really is flat, but due to the Ferrari Effect it appears is all aspects to look and behave like a Globe.

This would seems completely at odds with your claims of there being massive evidence

for the earth being flat, looking flat and behaving as if it is flat, as in Evidences Of The Flat Earth (On Going @142) « on: September 04, 2016, 03:09:00 PM ».

I might be barking up the wrong tree somewhere but it seems to me that:
All you are trying to do is to obfuscate the whole issue so that any ordinary person will see the whole thing as too deep and complicated to question.
You seem to be trying to emulate Bunthorne from Iolanthe

Quote from: W.S. Gilbert

And ev'ry one will say,
As you walk your mystic way,
"If this young man expresses himself in terms too deep for me,
Why, what a very singularly deep young man
this deep young man must be!"

Seems to fit

Quote from: John Davis on December 17, 2016, 11:53:35 AM

. . . . . . I am not a failed man, but the leading Zetetic scientist of our time. I have advanced our knowledge of the universe more so than any one other person since Rowbotham himself. When the veil is lifted from the eyes of the world, they will sing songs to laud the sacrifices that have led to what we know about the flat earth.
. . . . . . . .

I'm a bit more practical. If the earth

looks like a Globe, measures like a Globe, supports satellites like a Globe and in all respects behaves like a Globe

then the earth really is a Globe.

<< Updated a bit >>

Nothing you have posted does anything more than attempt to increase this obfuscation the whole issue.

See JackBlack's post, Davis Relativity Model (Debate/discussion edition) « Reply #98 on: October 14, 2017, 09:17:49 PM »
for a picture of what these straight lines (Geodesics in spacetime) might look like.

Now lets look at the definition I used:

Quote

it is something based on space and how it relates to the Earth. A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates.
...

I agree that Earth fits this definition, because given 2 points on earth you can take a hypothetical circular orbit on the surface of Earth that connects both of them (hypothetical because it may collide with some land in the way).
What I don't agree on is that this definition generalizes flatness as we know it in Euclidian spaces.
Let's take the parabola y = x² in 2D Euclidian space for example:

It clearly isn't straight/flat.
Now move it in a parabolic route y = -t² in the Euclidean spacetime, producing the hyperbolic paraboloid y = x² - t²:

It clearly isn't flat either.
But let's see if it fits your definition, so let's say I'm in the spatial coordinates (x, x²) on the original parabola for some x, and at time t, so I'm currently in (x, x² - t², t), and I want to get to the spatial coordinates (x + d, (x + d)²) on the original parabola. In this case I can take the line (x + Δ, x² - t² + 2(x - t)Δ, t + Δ). It's clearly a straight line and you can check it's on the paraboloid, when Δ = 0 it is in (x, x² - t², t) and when Δ = d it's on the spatial coordinates of (x + d, (x + d)²) of the original parabola. Also if I want to get to (x - d, (x - d)²) on the original parabola I can choose the line (x - Δ, x² - t² - 2(x + t)Δ, t + Δ) and it works as well. So the paraboloid trajectory of the parabola fits your definition, but it isn't flat, which means your definition is bad.

Quote from: AltSpace on September 26, 2017, 11:20:39 PM

Now lets look at the definition I used:

Quote

it is something based on space and how it relates to the Earth. A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates.

But let's see if it fits your definition, so let's say I'm in the spatial coordinates (x, x²) on the original parabola for some x, and at time t, so I'm currently in (x, x² - t², t), and I want to get to the spatial coordinates (x + d, (x + d)²) on the original parabola. In this case I can take the line (x + Δ, x² - t² + 2(x - t)Δ, t + Δ). It's clearly a straight line and you can check it's on the paraboloid, when Δ = 0 it is in (x, x² - t², t) and when Δ = d it's on the spatial coordinates of (x + d, (x + d)²) of the original parabola. Also if I want to get to (x - d, (x - d)²) on the original parabola I can choose the line (x - Δ, x² - t² - 2(x + t)Δ, t + Δ) and it works as well. So the paraboloid trajectory of the parabola fits your definition, but it isn't flat, which means your definition is bad.

You cannot traverse the parabola in a straight line. You can take a path from (0,0) to (2,4). If you take a straight line you would go through the point (1,2). This point is not on the parabola.
Unless you wanted to model it in a parabolic space, where the parabola would be the straight line, and a path from (0,0), through (1,2) then to (2,4) would no longer be straight.

The issue with his method is the application of an axiom which holds for Euclidean spaces, into non-Euclidean spaces where it does not hold.

So, I got a bit bored while waiting and decided to make another surface.
Note: This surface is not flat.
Instead, it is the surface that has the path that the points on Earth's equator would follow:

Quote from: AFanOfTruth on October 15, 2017, 12:41:48 PM

Quote from: AltSpace on September 26, 2017, 11:20:39 PM

Now lets look at the definition I used:

Quote

it is something based on space and how it relates to the Earth. A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates.

But let's see if it fits your definition, so let's say I'm in the spatial coordinates (x, x²) on the original parabola for some x, and at time t, so I'm currently in (x, x² - t², t), and I want to get to the spatial coordinates (x + d, (x + d)²) on the original parabola. In this case I can take the line (x + Δ, x² - t² + 2(x - t)Δ, t + Δ). It's clearly a straight line and you can check it's on the paraboloid, when Δ = 0 it is in (x, x² - t², t) and when Δ = d it's on the spatial coordinates of (x + d, (x + d)²) of the original parabola. Also if I want to get to (x - d, (x - d)²) on the original parabola I can choose the line (x - Δ, x² - t² - 2(x + t)Δ, t + Δ) and it works as well. So the paraboloid trajectory of the parabola fits your definition, but it isn't flat, which means your definition is bad.

You cannot traverse the parabola in a straight line. You can take a path from (0,0) to (2,4). If you take a straight line you would go through the point (1,2). This point is not on the parabola.
Unless you wanted to model it in a parabolic space, where the parabola would be the straight line, and a path from (0,0), through (1,2) then to (2,4) would no longer be straight.

The issue with his method is the application of an axiom which holds for Euclidean spaces, into non-Euclidean spaces where it does not hold.

Indeed I can't traverse a parabola in a straight line, but I can go on one on a hyperbolic paraboloid, which is the trajectory of a parabola moving in a parabolic path, exactly as there is an orbit on the surface of Earth, but the straight line that connects two points on Earth's surface usually passes through its interior.

I think we don't agree on the definition that was used.
Can AltSpace or Davis clarify what is the correct definition?

Altspace, doesn't it bother you that the paths followed by neutrinos don't conform to this model in any way whatsoever? In that they can be shown to pass directly through the earth and out from somewhere else on its surface?

Indeed I can't traverse a parabola in a straight line, but I can go on one on a hyperbolic paraboloid, which is the trajectory of a parabola moving in a parabolic path, exactly as there is an orbit on the surface of Earth, but the straight line that connects two points on Earth's surface usually passes through its interior.

Yes, there are several straight lines on a hyperbolic paraboloid. But not all of them.

The key part of his definition which he has left to be implied is that you can pick any 2 points.

Consider a generic plane in Euclidean space, with the form z=x+y, which can be rewritten as x+y-z=0.
Now consider any 2 points on the plane. For simplicity, these will be defined by their x and y coordinates as that determines their z coordinates.

So you have the points p1=(x1,y1,x1+y1) and p2=(x2,y2,x2+y2).

Now we want to find the line connecting them. This will be the vector:
<x2-y1,y2-y1,x2+y2-x1-y1>, which I will call V.
So starting at p1, and moving k*V, where k is some arbitrary value (which to strictly match his definition would need to be between 0 and 1, but I wont bother with that for now), we get the position:
p=p1+k*V=(x1,y1,x1+y1)+k*<x2-y1,y2-y1,x2+y2-x1-y1>
=(x1,y1,x1+y1)+(k*x2-k*y1,k*y2-k*y1,k*x2+k*y2-k*x1-k*y1)
=(x1+k*x2-k*y1,y1+k*y2-k*y1,x1+y1+k*x2+k*y2-k*x1-k*y1)

This can be simplified but I will just stick this straight in the equation. For it to lie on the line we need x+y-z=0.
So, x+y+z=x1+k*x2-k*y1+y1+k*y2-k*y1-(x1+y1+k*x2+k*y2-k*x1-k*y1)
=x1+k*x2-k*y1+y1+k*y2-k*y1-x1-y1-k*x2-k*y2+k*x1+k*y1
=x1-x1+k*x2-k*x2-k*y1+k*x1+y1-y1+k*y2-k*y2-k*y1+k*y1
=0

Thus any point between the 2 lies on the line.

Now lets take your hyperbolic parabaloid, this has z=x^2-y^2, thus x^2-y^2-z=0.
We will use the same x and y as before, so we have:
p1=(x1,y1,x1^2-y1^2) and p2=(x2,y2,x2^2-y2^2).
Now V=<x2-x1,y2-y1,x2^2-y2^2-x1^2+y1^2>
Now the generic point
p=p1+kV=(x1,y1,x1^2-y1^2)+k*<x2-x1,y2-y1,x2^2-y2^2-x1^2+y1^2>
=(x1,y1,x1^2-y1^2)+(k*x2-k*x1,k*y2-k*y1,k*x2^2-k*y2^2-k*x1^2+k*y1^2)
=(x1+k*x2-k*x1,y1+k*y2-k*y1,x1^2-y1^2+k*x2^2-k*y2^2-k*x1^2+k*y1^2)
=(x1+k*x2-k*x1,y1+k*y2-k*y1,x1^2-y1^2+k*x2^2-k*y2^2-k*x1^2+k*y1^2)
Now, this needs to be of the form x^2-y^2-z=0.
x^2-y^2-z=(x1+k*x2-k*x1)^2-(y1+k*y2-k*y1)^2-(x1^2-y1^2+k*x2^2-k*y2^2-k*x1^2+k*y1^2)
=(x1^2+2*x1*k*x2-2*x1*k*x1+k^2*x2^2-2*k*x2*k*x1+k^2*x1^2)
-(y1^2+2*y1*k*y2-2*y1*k*y1+k^2*y2^2-2*k*y2*k*y1+k^2*y1^2)
-(x1^2-y1^2+k*x2^2-k*y2^2-k*x1^2+k*y1^2)
=x1^2+2*x1*k*x2-2*x1*k*x1+k^2*x2^2-2*k*x2*k*x1+k^2*x1^2
-y1^2-2*y1*k*y2+2*y1*k*y1-k^2*y2^2+2*k*y2*k*y1-k^2*y1^2
-x1^2+y1^2-k*x2^2+k*y2^2+k*x1^2-k*y1^2
=x1^2-x1^2+2*x1*k*x2-2*x1*k*x1+k*x1^2+k^2*x2^2-2*k*x2*k*x1+k^2*x1^2-k*x2^2
-y1^2+y1^2-2*y1*k*y2+2*y1*k*y1-k*y1^2-k^2*y2^2+2*k*y2*k*y1-k^2*y1^2+k*y2^2
=2*k*x1*x2-k*x1^2+k^2*x2^2-2*k^2*x1*x2+k^2*x1^2-k*x2^2
-2*k*y1*y2+k*y1^2-k^2*y2^2+2*k^2*y1*y2-k^2*y1^2+k*y2^2
=-k*x1^2+k^2*x1^2+2*k*x1*x2-2*k^2*x1*x2+k^2*x2^2-k*x2^2
+k*y1^2-k^2*y1^2-2*k*y1*y2+2*k^2*y1*y2+k*y2^2-k^2*y2^2
=k*((k-1)*x1^2+2*(1-k)*x1*x2+(k-1)*x2^2
+(1-k)*y1^2-2(1-k)*y1*y2+(1-k)y2^2)
=k*(k-1)*(x1^2-2*x1*x2+x2^2-y1^2+2*y1*y2-y2^2)
=k*(k-1)*(x1^2-2*x1*x2+x2^2-(y1^2-2*y1*y2+y2^2))
=k*(k-1)*((x1-x2)-(y1-y2)

And I don't think it can be simplified further.
That means in order for it to be straight we need:
k*(k-1)*((x1-x2)-(y1-y2)=0.
This only happens if k=0, (k-1)=0 (so k=1), or(x1-x2)-(y1-y2)=0, so x1-x2=y1-y2 or y2=y1+x2-x1
This would mean that you can only traverse between some points in a straight line staying on the surface.
For other points, you would either leave the surface or not be able to traverse between the points in a straight line.
You can't go from (0,0) to (1,5) in a straight line on the surface.

But for a given height above Earth's surface, you can go between any 2 points with a circular orbit, which is by definition a geodesic (straight line) through space time.
The issue is this is not a surface in 3D space, it is in 4D space-time, and there are complications with how a surface is defined in non-Euclidean space.

Quote from: AFanOfTruth on October 16, 2017, 09:28:45 AM

Indeed I can't traverse a parabola in a straight line, but I can go on one on a hyperbolic paraboloid, which is the trajectory of a parabola moving in a parabolic path, exactly as there is an orbit on the surface of Earth, but the straight line that connects two points on Earth's surface usually passes through its interior.

Yes, there are several straight lines on a hyperbolic paraboloid. But not all of them.
The key part of his definition which he has left to be implied is that you can pick any 2 points.

Yes, every two spatial points.
My paraboloid is in 3D space-time (2 spatial coordinates, 1 temporal).
I think you missed that point.

But for a given height above Earth's surface, you can go between any 2 points with a circular orbit, which is by definition a geodesic (straight line) through space time.
The issue is this is not a surface in 3D space, it is in 4D space-time, and there are complications with how a surface is defined in non-Euclidean space.

Exactly the same in my parabola-paraboloid example.
For a given parabola moving in a parabolic path in the opppsite orientation to it, you can go between any 2 points on the parabola in a straight line through 3D space-time.
The paraboloid is neither in 2D nor in 3D space, but in 3D space-time, and it matches the definition that was used to "prove" Earth is flat.

Yes, every two spatial points.
My paraboloid is in 3D space-time (2 spatial coordinates, 1 temporal).
I think you missed that point.

No, I didn't.
All I may have done is missed which one you were saying was the temporal one. But you were saying it was Euclidean, unless I misunderstood that and you meant generalising from Euclidean to non-Euclidean.
In that case it lacks a key part of the definition in how a surface is defined in non-Euclidean spaces.

All I may have done is missed which one you were saying was the temporal one.

t, the 3rd one.

But you were saying it was Euclidean, unless I misunderstood that and you meant generalising from Euclidean to non-Euclidean.

My space-time was indeed Euclidean. A definition for sonething on non-Euclidean space-times should generalize the definition for Euclidean ones.

In that case it lacks a key part of the definition in how a surface is defined in non-Euclidean spaces.

What key part?

My space-time was indeed Euclidean. A definition for sonething on non-Euclidean space-times should generalize the definition for Euclidean ones.

No it wasn't.
If your space-time was Euclidean, objects travelling along the time axis would travel in a straight line. It would not travel in a parabola.
In Euclidean space time, it isn't just spatial coordinates. it is any coordinates. You need to be able to go from any point (x,y,z) or (x,y,t) on the "plane" to any other point on the "plane", while travelling in a "straight" line, without leaving the "plane".

If your space-time is Euclidean, time is not special.

What key part?

In Euclidean space, a plane is defined by a point and a normal.
There are 2 main ways (that I can think of) to construct a plane, one is by taking this point, and drawing every line through this point which is normal to the normal vector.
Another way is to take 2 vectors which are normal to the normal, and translate one of these vectors along the direction of the other.
With Euclidean spaces (which would include Euclidean space-time, these produce the same planes, and you can go between any 2 points on this plane while travelling in a straight line, remaining on the plane.

For non-Euclidean spaces, it isn't that simple, you can get a multitude of different planes by following the above methods.
For example, by taking all the vectors through a point which are normal to another vector, you get something like this:

If however you take a line and translate it along another line you can get something like theses 2 surfaces:


So the different methods no longer produce the same lines.
More importantly, you are no longer able to go from one point on the plane to another point on the plane while travelling in a straight line which remains on the plane (note: these need to be straight lines in non-Euclidean space-time).
If you try and make a 2D plane from 4D space-time, you can get a sphere, and I think you can do that in a way where both methods of making the plane agree, but I need to think about it some more to make sure.

Quote from: AFanOfTruth on October 17, 2017, 06:13:11 AM

My space-time was indeed Euclidean. A definition for sonething on non-Euclidean space-times should generalize the definition for Euclidean ones.

No it wasn't.
If your space-time was Euclidean, objects travelling along the time axis would travel in a straight line. It would not travel in a parabola.

But I moved my parabola an a parabola for the sake of my claim.

In Euclidean space time, it isn't just spatial coordinates. it is any coordinates. You need to be able to go from any point (x,y,z) or (x,y,t) on the "plane" to any other point on the "plane", while travelling in a "straight" line, without leaving the "plane".

If your space-time is Euclidean, time is not special.

So it's a minkowski space where straight lines are still straight. It doesn't really matter.

Quote from: AFanOfTruth on October 17, 2017, 06:13:11 AM

What key part?

In Euclidean space, a plane is defined by a point and a normal.
There are 2 main ways (that I can think of) to construct a plane, one is by taking this point, and drawing every line through this point which is normal to the normal vector.
Another way is to take 2 vectors which are normal to the normal, and translate one of these vectors along the direction of the other.
With Euclidean spaces (which would include Euclidean space-time, these produce the same planes, and you can go between any 2 points on this plane while travelling in a straight line, remaining on the plane.

For non-Euclidean spaces, it isn't that simple, you can get a multitude of different planes by following the above methods.
For example, by taking all the vectors through a point which are normal to another vector, you get something like this:

Are you sure that there are two (space-time) points on this surface that the line that connects them isn't on the surface? (maybe most of them are and it's sure that there will be, but I'm not familiar enough with non-Euclidean spaces)

If however you take a line and translate it along another line you can get something like theses 2 surfaces:

Are you sure that you rotated the line correctly? Because you can get the hyperbolic paraboloid by translatg a line along another line and rotate it in a specific way. (for example, the paraboloid z = xy can be produced by taking the x axis, and translate it along the y axis while keeping its slope in the xz plane your y value)
Of course it's a very stupid example, but in non-Euclidean spaces it's not always so sure to which direction you should orient the line.

So the different methods no longer produce the same lines.
More importantly, you are no longer able to go from one point on the plane to another point on the plane while travelling in a straight line which remains on the plane (note: these need to be straight lines in non-Euclidean space-time).

But sometimes it's possible, for example, if your space-time is spherical, the great spheres are planes and every two points on them can be connected by a straight line.

If you try and make a 2D plane from 4D space-time, you can get a sphere, and I think you can do that in a way where both methods of making the plane agree, but I need to think about it some more to make sure.

OK, think about it.
BTW, do you think that Earth's trajectory is flat in space-time?

But I moved my parabola an a parabola for the sake of my claim.

And with your space being euclidean, your surface is not flat, as time is no longer special and would no longer be treated separately to the spatial dimensions and thus spatial coordinates would include time.

Are you sure that there are two (space-time) points on this surface that the line that connects them isn't on the surface? (maybe most of them are and it's sure that there will be, but I'm not familiar enough with non-Euclidean spaces)

Yes.
A straight line through this non-Euclidean space-time will either remain at the same temporal component, or only go forward or only go backwards in time. There are none which start going forward and then go backwards.
The only part which introduces curvature is the temporal component, i.e. if you have a line with a constant temporal component, then not only will it be straight for this non-Euclidean space-time, it will be straight for this representation of this space-time in Euclidean space.
Now then, with that start at the point where all these lines cross.
Pick 2 lines going in opposite directions, such as the 2 parabolic trajectories (pretty much any work, they don't even need to be matched, it will just limit the exact time's you can use).
Now proceed along these until you get to some time value. If you chose a circular orbit, make sure this isn't a half integer multiple (which includes full integer multiples) of the orbital period.
You now have 2 different points in space, at the same time.
As they are at the same time, the path connecting them can have no temporal difference. That is the path connecting them must be a vector purely through space, not through time. This means it can be found by a simple straight line between the 2 points.
But a straight line between these 2 points goes off the surface.
For example:

The cyan (or whatever) line is a straight line through space time, the only one connecting those 2 points (4 actually) on the surface, yet does not remain on the surface.

Are you sure that you rotated the line correctly?

Yes. I'm sure.
The angle between the straight line it is translated along (the helix path of the orbit), and the line which is being translated, remains the same.
If you translated it differently you would get similar issues.

But sometimes it's possible

Yes, sometimes it is possible, but not always.

OK, think about it.

My main thought is to try and force it to be a sphere, so circular orbits around Earth, finding 2 mutually perpendicular vectors perpendicular to this.
So start at the north pole, position (0,0,1,0), then all the vectors will be of the form <vx,vy,0,1/v>, where vx^2+vy^2=v^2, where v is orbital velocity, with "natural" units such that the units are 1, or without that, with units shown you would have <vx*s, vy*s, 0, m/v> where v is orbital velocity in m/s.
As all these vectors make circles, then rotating it or translating it produces the same result, a sphere.
The issue then becomes finding those mutually perpendicular vectors.
One vector could be <0,0,1,0>.
That would mean the other needs to be of the form<a,b,0,c>.
The simplist one to think of would be a vector perpindicular to <1,0,0,1>
But the only one which fits that is <0,1,0,0> which isn't perpindicular to <0,1,0,1>. (which isn't actually all that surprising given <vx,vy,1/v> defines a semicircle offset from the xy plane by 1/v. That means we need 2 vectors with a z component.

In a more general form, as the vector will be <vx,vy,0,1/v>
We need 2 vectors:
<a1,b1,c1,d1>
<a2,b2,c2,d2>
Which are all mutually perpindicular, thus:
a1*a2+b1*b2+c1*c2+d1*d2=0
xv*a1+yv*b1+d1/v=0
xv*a2+yv*b2+d2/v=0
I think giving them an x or y component will be problematic, so I'll try to set a1=a2=b1=b2=0.
Thus it simplifies to:
c1*c2+d1*d2=0
d1/v=0
d2/v=0
Well that doesn't work. (in hindsight, it was stupid, with no x or y coordinate, that would mean that a line purely in x would be perpindicular)

Well, as they both need to have some z component, we can set the z component for both as 1 (which simply scales the vectors):
a1*a2+b1*b2+d1*d2=-1
xv*a1+yv*b1+d1/v=0
xv*a2+yv*b2+d2/v=0
d1/v=-(xv*a1+yv*b1)
d1=-v*(xv*a1+yv*b1)
d2=-v*(xv*a2+yv*b2)

a1*a2+b1*b2+v*(xv*a1+yv*b1)*v*(xv*a2+yv*b2)=-1
a1*a2+b1*b2+v^2*(xv*a1+yv*b1)*(xv*a2+yv*b2)=-1
a1*a2+b1*b2+v^2*(xv^2*a1*a2+xv*yv*a1*b2+xv*yv*b1*a2+yv^2*b1*b2)=-1
a1*a2+b1*b2+v^2*xv^2*a1*a2+v^2*xv*yv*a1*b2+v^2*xv*yv*b1*a2+v^2*yv^2*b1*b2)=-1
a1*a2(1+v^2*xv^2)+b1*b2*(1+v^2*yv^2)+v^2*xv*yv*(a1*b2+b1*a2)=-1

I might try some more later.

BTW, do you think that Earth's trajectory is flat in space-time?

When considered as a point (i.e. all of Earth moving together), yes.
When considered as a collection of points/a surface, no. If it did it would follow a path like this:

and we would constantly be in free fall.

Trying a few examples:
From before:
xv*a1+yv*b1+d1/v=0
Set, xv=v, yv=0:
v*a1+0+d1/v=0
v^2*a1+d1=0
d1=-v^2*a1
Similarly, set xv=0, yv=v:
0+v*b1+d1/v=0
v^2*b1=d1
d1=-b1*v^2

Thus v^2*a1=v^2*b1
Thus a1=b1

Just to check if it is possible, set vx=vy=sqrt(1/2)*v, thus v=sqrt(2*(sqrt(1/2)*v)^2)=sqrt(2*(1/2)*v^2)=sqrt(v^2)=v
sqrt(1/2)*v*a1+sqrt(1/2)*v*a1+d1/v=0
a1*2*sqrt(1/2)*v^2+d1=0
d1=-a1*sqrt(2)*v^2
But from before d1=-a1*v^2
Thus a1*v^2=a1*sqrt(2)*v^2
Thus 1=sqrt(2).
This is an impossibility.

So it looks like I was wrong, and my spherical orbit idea doesn't work after all.

Oh well, back to the drawing board.

Quote from: AFanOfTruth on October 17, 2017, 10:40:56 PM

But I moved my parabola an a parabola for the sake of my claim.

And with your space being euclidean, your surface is not flat, as time is no longer special and would no longer be treated separately to the spatial dimensions and thus spatial coordinates would include time.

If it really matters that much for you, just add a little distortion somewhere not on the paraboloid (like a little hill on a flat surface).

Quote from: AFanOfTruth on October 17, 2017, 10:40:56 PM

Are you sure that there are two (space-time) points on this surface that the line that connects them isn't on the surface? (maybe most of them are and it's sure that there will be, but I'm not familiar enough with non-Euclidean spaces)

Yes.
A straight line through this non-Euclidean space-time will either remain at the same temporal component, or only go forward or only go backwards in time. There are none which start going forward and then go backwards.
The only part which introduces curvature is the temporal component, i.e. if you have a line with a constant temporal component, then not only will it be straight for this non-Euclidean space-time, it will be straight for this representation of this space-time in Euclidean space.
Now then, with that start at the point where all these lines cross.
Pick 2 lines going in opposite directions, such as the 2 parabolic trajectories (pretty much any work, they don't even need to be matched, it will just limit the exact time's you can use).
Now proceed along these until you get to some time value. If you chose a circular orbit, make sure this isn't a half integer multiple (which includes full integer multiples) of the orbital period.
You now have 2 different points in space, at the same time.
As they are at the same time, the path connecting them can have no temporal difference. That is the path connecting them must be a vector purely through space, not through time. This means it can be found by a simple straight line between the 2 points.
But a straight line between these 2 points goes off the surface.
For example:

The cyan (or whatever) line is a straight line through space time, the only one connecting those 2 points (4 actually) on the surface, yet does not remain on the surface.

OK.

Quote from: AFanOfTruth on October 17, 2017, 10:40:56 PM

Are you sure that you rotated the line correctly?

Yes. I'm sure.
The angle between the straight line it is translated along (the helix path of the orbit), and the line which is being translated, remains the same.
If you translated it differently you would get similar issues.

But in 3D it's not only the angle that matters, but also the orientation. If you check my example, you'll see that the translated line always remain perpendicular to the y axis, that it's being translated along.

Quote from: AFanOfTruth on October 17, 2017, 10:40:56 PM

BTW, do you think that Earth's trajectory is flat in space-time?

When considered as a point (i.e. all of Earth moving together), yes.
When considered as a collection of points/a surface, no. If it did it would follow a path like this:

and we would constantly be in free fall.

But I think it matches the 2nd way of making a flat surface – get a circular orbit as the 1st line and a circular orbit in the opposite direction as the 2nd one. In your 3D sketch you're done, but if you are in 4D space-time you'll get the equator and will need a 3rd line, such as a orbits along longitudes.

If it really matters that much for you, just add a little distortion somewhere not on the paraboloid (like a little hill on a flat surface).

And is that distortion making the space non-Euclidean?
If not, you have the same issue of having 2 points you can't go between while staying on the line.
Is so, then it comes back to being a non-Euclidean space making the definition much harder.

But in 3D it's not only the angle that matters, but also the orientation. If you check my example, you'll see that the translated line always remain perpendicular to the y axis, that it's being translated along.

The orientation would be covered by the angle.
In non-Euclidean spaces there is a question of how you translate the lines.
I kept it such that for a Euclidean approximation for the point the lines would be restricted to a plane, where the normal points towards the centre of Earth/the centre of the orbit.

But like I said, I can translate it in other ways as well and get similar issues.
Completely ignoring the nature of the orbit/paths, and instead keeping the orientation of the line the same, translating it along the parabolic trajectory and the circular orbit, you get this:

Quote from: JackBlack on October 18, 2017, 04:17:59 AM

When considered as a point (i.e. all of Earth moving together), yes.
When considered as a collection of points/a surface, no. If it did it would follow a path like this:

and we would constantly be in free fall.

But I think it matches the 2nd way of making a flat surface – get a circular orbit as the 1st line and a circular orbit in the opposite direction as the 2nd one. In your 3D sketch you're done, but if you are in 4D space-time you'll get the equator and will need a 3rd line, such as a orbits along longitudes.

But Earth isn't spinning fast enough to follow that path in space-time.
And, that only works on the equator. Away from the equator it would fall towards the equator.
So at the very least, even if Earth was spinning fast enough, the "flat" trajectory would have the surface collapse to a disk and pop back out.

Quote from: AFanOfTruth on October 18, 2017, 08:01:55 AM

If it really matters that much for you, just add a little distortion somewhere not on the paraboloid (like a little hill on a flat surface).

And is that distortion making the space non-Euclidean?

Yes, like a flat surface with a hill.

Is so, then it comes back to being a non-Euclidean space making the definition much harder.

So is the paraboloid flat then?

Quote from: AFanOfTruth on October 18, 2017, 08:01:55 AM

But in 3D it's not only the angle that matters, but also the orientation. If you check my example, you'll see that the translated line always remain perpendicular to the y axis, that it's being translated along.

The orientation would be covered by the angle.

What?

In non-Euclidean spaces there is a question of how you translate the lines.

Yes, that was what I meant.

I kept it such that for a Euclidean approximation for the point the lines would be restricted to a plane, where the normal points towards the centre of Earth/the centre of the orbit.

OK, sounds good.

Quote from: AFanOfTruth on October 18, 2017, 08:01:55 AM

Quote from: JackBlack on October 18, 2017, 04:17:59 AM

When considered as a point (i.e. all of Earth moving together), yes.
When considered as a collection of points/a surface, no. If it did it would follow a path like this:

and we would constantly be in free fall.

But I think it matches the 2nd way of making a flat surface – get a circular orbit as the 1st line and a circular orbit in the opposite direction as the 2nd one. In your 3D sketch you're done, but if you are in 4D space-time you'll get the equator and will need a 3rd line, such as a orbits along longitudes.

But Earth isn't spinning fast enough to follow that path in space-time.
And, that only works on the equator. Away from the equator it would fall towards the equator.
So at the very least, even if Earth was spinning fast enough, the "flat" trajectory would have the surface collapse to a disk and pop back out.

But I showed you a way to create Earth's surface, as a collection of events, as a flat plane defined by lines. Why does it matter if we don't follow these lines?

Quote from: Copper Knickers on October 11, 2017, 02:45:50 AM

Your link (or at least the replies to the question therein) are clear that they relate to curved space-time. I was writing about curved space.

There is no distinction, they are both one and the same continuum.

Quote

Nevertheless, JackBlack has pointed out that I may be not wholly accurate in my claim. However, within the scale of the earth, light won't deviate from curved space geodesics by very much, and not remotely by enough to detract from my point. Which is that the earth is not flat in curved space.

*space-time

The photon sphere concept points out how wrong this is:
Where photons (light) can be in orbit: r = 3GM/c²
r is the Schwarzschild radius (radius of the event horizon), G is the gravitational constant, M is the mass (black hole mass in this case), and c is the speed of light in a vacuum.
With Earth:
(3*6.674×10^-11*5.9726x10^24)/(299792458*299792458) =1.330545 cm.
The Earth would need to have a less than 1.33 cm radius for light orbit.
A mass like the ISS or a human could follow geodesics, but light goes beyond escape velocity on Earth. So, you can hardly use that reasoning as 'close enough'.

Quote

So what does a 'surface' in space-time mean?

It means the upper most layer of the Earth, with land and water mainly in this case (not exactly the atmospheric strata). If you want a specific defining case, then imagine the Earth being covered completely in water, water serves as an excellent level defining the surface, it follows the path across, and that would represent the surface.

You seem reluctant to acknowledge that space-time is a geometry of events rather than objects and therefore fundamentally different from space. I don't know whether this is wilful ignorance to support your 'model' or that you don't quite get it.

Anyway, this isn't a bad link if you're genuinely interested: http://www.ws5.com/spacetime/

So is the paraboloid flat then?

Again, the definition is harder. I'm not sure. I would need to know more about the space, and how to define a "flat" surface in this space.

Quote from: JackBlack on October 18, 2017, 02:01:00 PM

The orientation would be covered by the angle.

What?

The angle between the lines can be considered as just a single angle, or you can consider how you change the path between the 2 lines. For example, you can have the circular orbit, and keep the line at an angle of 45 degrees left, instead of 45 degrees up.

Quote from: JackBlack on October 18, 2017, 02:01:00 PM

But Earth isn't spinning fast enough to follow that path in space-time.
And, that only works on the equator. Away from the equator it would fall towards the equator.
So at the very least, even if Earth was spinning fast enough, the "flat" trajectory would have the surface collapse to a disk and pop back out.

But I showed you a way to create Earth's surface, as a collection of events, as a flat plane defined by lines. Why does it matter if we don't follow these lines?

Because this surface is a trajectory through space time. As such, to be on this surface you need to be following that trajectory.

Quote from: AFanOfTruth on October 18, 2017, 09:31:16 PM

So is the paraboloid flat then?

Again, the definition is harder. I'm not sure. I would need to know more about the space, and how to define a "flat" surface in this space.

IDK how to define "flat" in this space-time or in non-Euclidean space-times in general. My whole point was that the definition used by AltSpace/Davis was problematic because things such as this paraboloid fit it while they clearly aren't flat

Quote from: AFanOfTruth on October 18, 2017, 09:31:16 PM

Quote from: JackBlack on October 18, 2017, 02:01:00 PM

The orientation would be covered by the angle.

What?

The angle between the lines can be considered as just a single angle, or you can consider how you change the path between the 2 lines. For example, you can have the circular orbit, and keep the line at an angle of 45 degrees left, instead of 45 degrees up.

But what does left/up mean in your space-time and why are they the same direction from the line (like lines with the same vectors are the same direction from a line in Euclidean space)?

Quote from: AFanOfTruth on October 18, 2017, 09:31:16 PM

Quote from: JackBlack on October 18, 2017, 02:01:00 PM

But Earth isn't spinning fast enough to follow that path in space-time.
And, that only works on the equator. Away from the equator it would fall towards the equator.
So at the very least, even if Earth was spinning fast enough, the "flat" trajectory would have the surface collapse to a disk and pop back out.

But I showed you a way to create Earth's surface, as a collection of events, as a flat plane defined by lines. Why does it matter if we don't follow these lines?

Because this surface is a trajectory through space time. As such, to be on this surface you need to be following that trajectory.

I thought we are talking about space-time which consists of events, not space-time-speed which consists of some kind of states or something.