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Messages - Rog

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The wiki says that one proof point of FE is that you can't see or feel movement, therefor the earth is stationary.
When we are standing still, we aren’t moving relative to the earth or its atmosphere so why should we feel or see motion?  When we walk, run, drive or whatever, we are in motion relative to the earth and that’s when we feel and see motion.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: March 03, 2022, 06:21:18 PM »
My model has the intrinsic curvature of a globe. (This has been shown at least 3 or 4 different ways)
However what we haven't yet discussed is extrinsic curvature, ie embedding of the surface in a higher dimension.
And in that sense my model is flat.

Gaussian (intrinsic) curvature is quantified as the product of the two principal (extrinsic) curvatures.  If one (as in a cylinder) or both (as in your flat model) extrinsic curvatures are zero, then the intrinsic curvature is zero.

IOW, it's impossible for your model (or anything else) to have intrinsic curvature unless it has extrinsic curvature.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 26, 2022, 05:09:00 PM »
There are loads of different distances. Straight through the earth, angular distance, taxicab. hamming distance if you like a weird one.
It's your responsibility to take one that makes sense: that matches observations. Personally i find it remarkable that physics uses both spherical and euclidean distances within the same model. This alone clearly shows there's no one correct distance per model

There is only one correct shortest distance between two points per model because you have to use different formulas for each.  Haversine will give you the shortest distance between two points on sphere and the PT will give you the shortest distance between points on a plane. 

The taxicab metric assumes right angles are necessary to travel from point A to point B.  So it doesn’t make sense to use it when that’s not necessary. Haversine assumes an angle between the two points created by the curvature of the surface of a sphere. On a plane, so such angle is created because there is no curvature, so it doesn’t make sense to use Haversine on a plane.  But still you insist on doing so.

Those are the distances measured  between Perth and Johannesburg using the correct formulas for each model.

When you use the correct formula for the correct model, you get two different distances, and it’s easy enough to determine which one reflects reality.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 25, 2022, 11:51:46 PM »
You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality

The reason Haversine makes sense and matches reality is because Haversine formula gives you the shortest distance between two points on a sphere. It is specifically designed to measure the distance on a sphere. If you want to find the shortest distance between two points on a plane, you have to use the PT.  That is the formula that will give you the shortest distance between points on a plane.  If you use the same two points in the two different formulas, you will get different distances, because distances on a sphere are measured differently than on a plane.

You seem to lack the fundamental understanding that Euclidean and non-Euclidean geometry are different.  Different rules, different definitions, different ways of measuring. 

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 23, 2022, 02:13:11 AM »
It’s certainly an interesting thought experiment, but agreed, it’s not a model, let alone a ‘fully working’ model.

OP seems to suffer from the same misconception that many have on this site.  That there is something magical about changing coordinates or frames of reference and that it actually effects or changes physical reality. It doesn’t.

The idea that we can never know the true shape of the earth, or that it could be anything, starts with the assumption that there aren’t physical forces at work that determine the shape.  Everything is perception and one perception is as valid as another.  That is such bunk. Perception is not always reality and there are ways to test perception against reality

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 19, 2022, 09:49:32 AM »
except my "disc" is not an euclidean plane. It does not have an orthnormal base.
- You've had me calculate curvature of my space using gauss and it came out indistinguishable from a sphere.
- You then had me do parallel transports, which were indistinguishable from a sphere.
- I've given you the calculation for riemann tensor and ricci scalar, and it was comparable with a sphere.
You never found fault in any of the math, and yet you now call it a euclidean plane which violates all of the above 

It doesn’t matter if your model has an orthonormal basis or not. That isn’t what defines euclidean space.   Where have you posted a graphic of a parallel transport of your model?  How would it differ from the example I posted?  Nor have I seen any calculations of what the Gaussian curvature of your model is or the calculations of the Riemann tensor.  A few days ago you acknowledged that much of the math is over your head.  Now you’re an expert in differential geometry and tensor calculus?

The Riemann tensor is invariant.  If it exists in one coordinate system, it exists in every coordinate system. If it is zero in one coordinate, it is zero in all of them.

This tensor DOES NOT CARE about coordinate systems. Flat space has all components of Ra bcd = 0 irrespective of whether you are working in spherical polar coordinates or cartesian coordinates

In other words, the independent components of the Riemann tensor can be thought of as the n 2 (n 2 − 1)/12 (linear combinations) of second derivatives of the metric tensor that cannot be set to zero by coordinate transformations.

But because the Riemann tensor is a genuine tensor, if it vanishes in one coordinate system then it must vanishes in all of them.

Is it possible to find an inertial coordinate system such that also the second derivatives of the metric tensor vanish at the coordinates origin? The answer is negative: the reason being that the Riemann curvature tensor (see Chapter 5) cannot be made to vanish by use of coordinate transformations.

Proof: if a tensor is zero in any frame, then it is zero in all frames, as a trivial consequence of the transformation law

Do I really need to keep going?

 Mathematically it behaves like a euclidean globe.

There’s no such thing as a euclidean globe. The definition of a globe is that it doesn’t “behave”, or have the all of the same properties of a euclidean space

We imagine trying to draw a map of a region D ⊂ S2. The map is a smooth function  F : D → E 2
     to the Euclidean plane that preserves as much information about the geometry of D as is possible.
I am not working in a euclidean plane. After coords transform my basis is not orthonormal.

One more doesn’t matter if the basis isn’t orthonormal.  If it doesn’t have curvature, it is a euclidean plane.  If it has curvature it is a sphere.  Coordinate transformations don’t make any difference.

The metric is flat if the Riemann curvature tensor is zero. That's true regardless of what coordinates you use. Spherical coordinates can be used in a flat space, just as polar coordinates can be used on a flat plane.

Also i found this definition for isometry:
It's basically any transformation that preserves distance

It doesn’t just preserve distance.  It preserves angles.  You can’t have an isometric transformation between a plane and a sphere because the angles of a plane and a sphere are necessarily different. See below…

My coordinates are in (lat, long). So my distances are calculated with haversine, and will thus give the same result as on your globe.

No they won’t.  Haversine calculates great circle distances.  If you connect three great circles to make a triangle on a sphere, it will have more than 180 degrees.  You can’t have a triangle of more than 180 degrees on a flat surface.  So the area represented by connecting three “great circles” on a flat disc will be different than the area represented by connecting the same three “great circles” on a sphere.

The supposedly same three points won’t meet in the same place on a plane and sphere and the angles connecting the supposedly same three points are different.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 17, 2022, 09:42:41 PM »
If I understand the model, and Troolon, correct me if I'm wrong, I think it's pretty simple. It's a great circle, identical in both the RE and Troolon models. Same route, same distance. Just depicted, portrayed, visually different. Just like how the route would show as curved on a Mercator projection though it appears "straight" on a globe.

A Mercator projection isn’t making the claim that the earth is actually flat. It shows a curved route because that represents the shortest distance on a globe.

If the earth were actually flat the shortest distance between the two points would be a straight line and different route. A route from NY to Moscow, passing the tip of Greenland and coming back down through northern Europe isn’t the shortest distance on a flat earth because it isn’t a straight line.

If there were no distortions in Trooloon’s model, if all the distances and angles were same, the shortest distance would be the same route because all the land masses would be in the same relative position and have the same relative distance between them.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 17, 2022, 03:55:37 AM »
Straight is only defined the way you think it is in an orthonormal basis.

Straight is the shortest distance between two points, regardless of the coordinate system.  Straight is a curve on a sphere.  What would be the shortest distance between NY and Moscow on your model?

I'll try to answer your latest post tonight. But i think in summary my reply will be:
A disc in an orthonormal basis, with orthonormal distances, will have a curvature of 0.
A disc in celestial coords, with spherical distances, will have the curvature of a sphere. (as shown)
I believe you're taking a disc in celestial coords, and expecting it to behave like a disc in an orthonormal basis.
Intrinsic curvature doesn't see the coordinate axis directly, but it does notice distances, angles and tangent planes behaving relative to this axis.
You're basically taking my representation, changing the axis, distances, angles, ... and then exclaiming: look curvature is broken.

No, a disc will always have a curvature of zero and a sphere will always have a non-zero curvature. ]Changing coordinates doesn’t create curvature in an object where none physically exists.  You are conflating curvature of the object and a curvature of the coordinate system.  If you can make the curvature appear or disappear by changing coordinates, the curvature isn’t intrinsic to the space.  It is just an artifact of the coordinate system. And distances, angles and tangent planes won’t be represented the same. 

“Intrinsic” curvature means that it is a physical, immutable, property of the object.  Coordinate systems are just mathematical abstractions.  They don’t influence physical reality anymore than photo shop does. A coordinate transformation is a representation, but it isn’t always an accurate one.  Not all transformations are isometric, which just means a transformation that it doesn’t distort angles or distances. 

Here is the mathematical proof. (starts around slide 20) I can’t help you much working through it, but I suggest that unless you can contradict the math, you really don’t have any basis to say that your model doesn’t have any distortions.

Since the curvature of the sphere does not vanish, it CANNOT BE LOCALLY ISOMETRICALLY MAPPED TO THE EUCLIDEAN PLANE

Luckily, though you don’t need to know or do any complicated math. If a flat circle had intrinsic curvature, you wouldn’t be able to parallel transport a vector around the circumference.

For a funhouse mirror it's quite easy for an outside observer to see the shapes don't match with the owner.
Problem is we can't step outside of the universe to check. We have no absolute references.

Yes, we do have an absolute reference.  See above.

At what point have i tried to "flatten" the sphere?

By trying to project a sphere onto a flat surface without any distortion.

The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.

If it has intrinsic curvature, it doesn't matter how you view it.  It will always have intrinsic curvature and you can always detect it with some simple tests.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 16, 2022, 12:55:30 AM »
But to understand this more intuitively i must go again to to the construction. ...
1. We start with a sphere in cartesian coords:
2. We convert the sphere to geographic coords (lat, lon, dist). The graphic looks the same, angles and distances still measure the same. It still looks like this: 
3. draw latitude on a straight axis instead of a radial one. This is just a change in representation. We do not change coordinates, distances or angles. There's no change in intrinsic curvature as this step is invisible to the maths. I've attached a graphic of what i think the parallel transport looks like but i'm not 100% certain.
Ultimately this graphic is irrelevant as this last step is just a change in representation (like pie charts vs bar charts and not in coordinates, distances or angles -- at least not intrinsically).

You are making this much more difficult than it is.  Intrinsic curvature is coordinate independent.  A sphere has it, regardless of what coordinate system you use..

What is meant is the intrinsic curvature of the space, meaning it is independent of the choice of coordinates. There are clever methods of determining whether and to what extend your space deviates from flat euclidean space, namely Gaussian curvature, and, more importantly, the Riemann tensor.,to%20zero%20in%20flat%20spacetime.

You are correct that transforming from Cartesian to Geo graphic coordinates doesn’t change the intrinsic curvature, but just drawing latitude on a straight axis doesn’t change it either. 

The problem comes when you try and “flatten” the sphere...without any distortion.  It’s been pointed out to you umpteen times.  I’m not sure why you are having difficulty grasping it.  It is a fundamental principle of geometry that has been known for 100s of years that it is impossible to accurately represent a sphere on flat plane.  Trying to do that isn’t just a coordinate change, it is changing the fundamental properties of the geometry of the sphere.  Simple coordinate changes don’t do that.  As you say, a coordinate change is just another way of representing something.  If you try and flatten a sphere you aren’t changing the way something is represented, you are trying to represent something else

A flat disk has no intrinsic curvature.  The Riemann tensor will be zero.  A sphere has a non-zero Riemann tensor and it has intrinsic curvature.  A body can’t be a disk and a sphere at the same time.  It can only be one or the other and it is easy to determine which simply by doing a parallel transport.

On the left, the vectors make the full loop around the triangle, in the same direction, while not deviating from parallel.  On the right, the vector has to change direction in order to stay parallel all the way around the triangle.  If you look at it, its easy to see why. To stay on the surface, the vector has to “bend” around the top of the triangle.  No bending required on a flat triangle.  No complicated math, or measurements involved.

If there is no path on your model where a vector has to change direction in order to make a closed loop and staying parallel, then your model is flat.  It has no intrinsic curvature and is not mathematically, or any other way, equivalent to a sphere. You certainly have the graphic skills, so it shouldn't be hard for you to post evidence of doing a parallel transport.

Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too

You keep contradicting yourself by stating that your model has intrinsic curvature and also stating that we can’t know the shape of the earth.  The very definition of intrinsic curvature is that “the inhabitants” can know.  So which is it?  Does your model have intrinsic curvature or is it impossible to know the shape of the earth?

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 15, 2022, 03:44:05 AM »
1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original.
3. Flattening a sphere into a disk is a coordinate change.
4. Therefor, a disk is mathematically equivalent to a disk.
5. Because a disk is mathematically equivalent to a sphere, the physics is equivalent.
6. Because it is physically equivalent, light must bend however it needs to to change the RE appearance into FE reality.
7. Because the disk and sphere are mathematically equivalent, the equation for distance on a sphere can be used on a disk.
8. If the distance calculation on a disk does not match observed reality, then the process of measurement needs to be flexible to be made to match, measurement is not being done right

1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original. They are only equivalent if the change was isometric.
3. Flattening a sphere into a disk is a coordinate change. Flattening a sphere is more than just a coordinate change because it also changes the metric and the Riemann curvature tensor
4. Therefor, a disk is mathematically equivalent to a disk. When you transform a disc to a sphere the metric changes.  The Riemann tensor changes from 0 to non-zero.  Therefore, a disk and a sphere are not mathematically equivalent.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 15, 2022, 12:00:42 AM »
If you insist on measuring reality with an orthogonal ruler and drawing it on an orthonormal axis, you will indeed get a sphere. But then you are pre-supposing an orthonormal axis.

Orthonormal coordinates are not the only coordinates in Euclidean space.  Just because the coordinate lines are curved, does not mean that the space is curved.  The coordinate system does not define the whether or not a space is curved.

If a space has intrinsic curvature, the Riemann curvature tensor is non-zero.  What is the value in your model?

The existence of non-zero curvature can be easily visualized by parallel transport.

A sphere has it because it is possible to change the direction of vector by forming a loop over the surface.  A flat disk does not have it because you can take any random path and when you return to the origin, the vector will have the same direction. 

It’s an easy test and you don’t have to assume any shape or even take any measurements. Can you show a parallel transport on you model?

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 13, 2022, 07:01:57 AM »
When you take a 2D graph of a line, and transform it to polar coordinates, it will still be a line with the same length and angle as before.
why does this work?
Polar coordinates are non-euclidean: they have curvy axis for one and the coordinates for every point are radically different.

Polar coordinates are Euclidean.(and 2d)  Euclidean coordinates just mean that distances can be measured using the PT.  That is true of Cartesian and Polar coordinates. I don’t know what you mean by the axis is “curvy”,  (I think you might be confusing Polar coordinates with Spherical coordinates...see my comment below) It transforms correctly because lines are Euclidean and 2d and polar coordinates are Euclidean and 2d.

But if you use Polar coordinates on a sphere, this is what you end up with.

The lines are distorted because the PT can’t be used to measure distances on a sphere.

In my transformation i'm doing the exact same thing. (celestial coords are really 3D polar coords)
We draw a line/sphere/plane/... in 3D cartesian space. Then we transform every point to (latitude, longitude, distance), just like polar coords.
The line/sphere/plane is still a line/sphere/plane.  All distances still match with the cartesian drawing.

On a sphere, it will transform correctly because when you transform the coordinates from Cartesian, the metric also transforms and both systems are 3d.

This is just a guess, but I think you might be confusing Polar coordinates with Spherical Polar Coordinates. Polar coordinates are Euclidean, distances can be measured with the PT. Spherical Polar coordinates are not. They use the spherical metric.  If that’s the case, what you are calling “3d polar coordinates” are Spherical coordinates  (sometimes called Spherical Polar Coordinates)...and as you say they are basically the same thing as celestial coordinates.

But whatever you call them, when you transform from Cartesian onto a sphere, it will transform correctly. Both systems are 3d and the metric changes when you transform the coordinates, so distances are measured with the spherical metric.

The problem comes in when you try and project the transformed sphere onto a flat surface. The metric changes back to the PT, and your distances will be off..  You can do it, but there will be distortion.

No map from the sphere to the plane can be both conformal and area-preserving. If it were, then it would be a local isometry and would preserve Gaussian curvature. The sphere and the plane have different Gaussian curvatures, so this is impossible

An isometry means that distances and angles are preserved   You seem to be under the impression that a transformation,  when done correctly, never changes distances or angles. That’s not true.
There are many ways to move two-dimensional figures around a plane, but there are only four types of isometries possible: translation, reflection, rotation, and glide reflection. These transformations are also known as rigid motion.,also%20known%20as%20rigid%20motion.

Unless your transformation is one of those types, it’s not going to preserve angles or distances. You can have isometric  transformations between a plane and a plane and a sphere and sphere (or other 3d object, but the types of transformations are more limited), but you can’t have one between and plane and a sphere.

The reason should be obvious.  Angles and distances on a plane and angles and distances on a sphere are different. That’s why they use different metrics. If you preserve them, you aren’t going from sphere to plane, you are just going from one sphere to another sphere.

That’s all I have time for now. I’ll try and respond to more later.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 11, 2022, 01:18:16 PM »
I've asked you for several posts now where in my coord transform i'm making any mistakes and i've still not received a reply.
So lets try again. Example will be to calculate the width of Australia.
1. start with a globe in cartesian coords. Distance formula is arclength along a greatcircle
2. convert to celestial coords. Distance formula becomes haversine. Width of Australia is still correct
3. Render latitude on a straight rather than a curvy axis. All coordinates and formulas stay the same. The width of Australia is still correct as mathematically nothing changed.

When you transform to celestial coordinates the metric does not become haversine.. It becomes the formula I posted earlier.  Haversine calculates the shortest distance between two points on a sphere.  It is only accurate for defined areas of a sphere because it only takes the radius of the earth into account between the two points in question.  It only transforms the area between those two points.  The spherical metric transforms the whole sphere.

I was thinking about this last night and maybe this will give you some clarity.

A Euclidean coordinate system has two dimensions, length and width.  Non-Euclidean has three, length, width and depth.   There is no way to visually depict depth on a flat surface. So when you transform from 2 coordinates to 3  (or the other way), you are adding or removing a dimension that in reality, doesn’t exist. To account for this extra dimension, shapes or distances are necessarily distorted.

Think about what happens when you break down a cardboard box. In a sense, you are physically transforming from a 3 dimension coordinate system to a 2 dimension one.  When you break it down, the box becomes flat, but it also becomes longer and/or wider. The area that formed the depth has to go somewhere.

You can’t break down a sphere like you can a box, but try this.  Take some clay, form it into a sphere. Draw a # on it so it covers the whole surface of the sphere.  Then use a rolling pin or something to flatten it out so that the # covers the whole surface. The # will distort as the sphere becomes more and more flat. That’s because (at least one reason) a sphere has less surface area per volume than a flat surface.  In order to still cover the whole surface, the # has to “spread out”.

Or you can do it the other way and draw the # on flat clay and then form it into a sphere.  The # becomes curvy.  That’s what happens when you transform from a 2 coordinate system to a 3 coordinate system...the metric changes and straight lines become curvy.  There is less surface area so the lines have to curve to still fit into the same amount space.

You could use the haversine formula to calculate the shortest distance between two points, but the # is still curvy. Just arbitrarily drawing new lines after the fact doesn’t change the fact that the lines had curve in order to still fit into a smaller amount of surface area.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 11, 2022, 04:40:52 AM »
I don't agree pythagorean distances are the only distance metric. In our discussion we're more interested in arclength along a greatcircle distances.
But fine, let's go with straight distances through the earth.
It isn’t the only distance metric, that’s the whole point.  But there are defined metrics for Euclidean and non-Euclidean spaces.  You can’t just make up your own based on how you want it to turn out and expect your model to reflect reality.

That’s the metric for non-Euclidean spaces. It takes arclengths, angles and great circles into account.
I can’t help you with the math, but if you aren’t using that formula for your metric, your model doesn’t reflect reality.
These are not arbitrary formulas. This is the way you're supposed to do coord transforms
Some coordinate systems are inherently Euclidean (they only have two coordinates) and some non-Euclidean (three coordinates) The metric is baked into whatever coordinate system you are using.  If you transform from a spherical coordinate system to another spherical coordinate system, there is no distortion because they both have the same metric. When you transform from a coordinate system that is inherently Euclidean to one that is non-Euclidean, the transformed system becomes non-Euclidean and there is distortion.

If you are measuring the triangle on a sphere you get the measurements on the right, if measuring on flat space, you get the measurements on the left.

Distortions are the result of using a non-Euclidean metric in a Euclidean space. If there is a triangle that in reality, on a sphere earth, looks like the one on the right, it will look like the one on the left on a flat space. It won’t be accurate.  Your model is inherently distorted because you are using celestial coordinates, which have three coordinates,  and projecting them onto a Euclidean space, which is measured in only two coordinates.
The geometric properties of the space depends on the metric chosen, and by using a different metric we can construct interesting non-Euclidean geometries such as those used in the theory of general relativity.
You can apply any random metric to any shape, using any coordinate system and mathematically “change” the underlying geometry but if you are mixing and matching Euclidean and non-Euclidean spaces and metrics, your are always going to have distortion from an actual real, physical structure.
How will you tell?
Both models are indistinguishable. You called them a single model.
There is no measurement or observation of reality that can tell them apart. They both represent is equally well.
Your logic is backwards.  You are starting with the assumption that the earth doesn’t have real, physically defined geometry to begin with and the geometry isn’t defined until some arbitrary, random metric is applied to it. We can't know the correct geometry because we don't know the "correct" metric.
But we do know the correct geometry of the earth. It has intrinsic curvature, which by definition, can be measured by the “inhabitants”.  You keep saying that there is no test or observation we can use, but that is just wrong.  There are lots of them.

10.2  Parallel Transport in a Curved Space, D=2
Consider the scenario shown in figure 12. It starts out exactly the same as the previous scenario. However, in this case we suppose that the arrows exist in a two-dimensional space, namely a sphere, i.e. the surface of the earth.

The yellow arrows along the equator are exactly the same here as in the previous scenario. Even though the arrows are the same, we have to describe them differently. We say they all point north along the earth’s surface. They point toward the earth’s geographic pole, not toward the celestial north pole, because the latter does not exist in the two-dimensional space we are using.
As we move northward along the leg of the triangle that goes through North America, the arrows in figure 11 continue to point north toward the geographic north pole. Relative to the arrows in figure 12, these arrows must pitch down so that they remain within the two-dimensional space. They are confined to be everywhere tangent to the surface of the earth. As we move north, each of the arrows is parallel to the previous arrow, as parallel as it possibly could be.
Let’s be clear: Each new arrow is constructed to be parallel to the previous one, as parallel as it possibly could be. What we mean by “parallel” is discussed in more detail in section 10.3.
After we get to the north pole, we start moving south along a the prime meridian. We move south through Greenwich and keep going until we reach the equator at a point in the Gulf of Guinea. As always, each newly constructed vector is parallel to the previous arrow. All the arrows on this leg point due east.
Finally, we move west along the equator until we reach the starting point. Again each arrow is parallel to the previous one. All the red arrows on this leg point due east.
At this point we see something remarkable: The final arrow is not parallel to the arrow we started with.
From this we learn that in a curved space, there cannot be any global notion of A parallel to B. We must instead settle for a notion of parallel transport along a specified path. That is: the notion of parallelism is path-dependent. It also depends on whether you go around the path clockwise or counter-clockwise.
If you start with a northward-pointing vector in Brazil and parallel-transport it to the Gulf of Guinea, you get a northward-pointing vector. If you start with the same vector and transport it clockwise around two legs of the triangle as shown in figure 12, you get an eastward-pointing vector.

Creatures who live in the curved space can perceive this in a number of ways. Careful surveying is one way. Gyroscopes provide another way. That is, a gyroscope that is carried all the way around a loop will precess relative to a gyroscope that remains at the starting point.

Parallel transport in a flat environment is not path dependent.  It would be a simple matter to test and if it is found that parallel transport is path dependent, then we know we live in a curved environment.  If we want to accurately model it, we know to use a 3 coordinate system, which will give us the correct metric.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 09, 2022, 07:09:08 AM »
You broke physics (actually geometry is probably more accurate) when you decided to just make up your own rules for a metric. I’ve wracked my brain trying to figure out a way to explain to you, in a way that you will understand. It is difficult because you don’t seem to grasp the very basic concept that Euclidean (flat) and non-Euclidean geometry (spherical) do not follow the same rules.  Most importantly, distances are measured differently

In differential geometry, distances are measured by a metric.  Euclidean and non-Euclidean spaces use different metrics. They have to because the definition of  “the shortest distance between two points is different”.  For Euclidean space it is a straight line, for non-Euclidean it is a curve. For Euclidean space the metric is simply the Pythagorean theorem.  In non-Euclidean space it is

Obviously, very different formulas

When you do a coordinate transformation between two Euclidean spaces, the metric remains the same so lengths and angles are preserved.  When you do a coordinate transformation from a Euclidean space to a non-Euclidean space (Cartesian on a flat disc to Celestial on a sphere, for example), you aren’t just transforming the coordinate system, you are also transforming the metric from Euclidean to non-Euclidean. You seem to have just arbitrarily transformed the metric by “updating the formula”.  You can’t do that. The metric transforms when you do the coordinate transformation. You can’t just decide that you don’t like the way it transforms and invent your own metric.

But even when done correctly, because the bases of the two different metrics (the underlying geometry) are different, it will never transform exactly. Non-Euclidean metric defines distances in terms of angles, Euclidean doesn’t.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 06, 2022, 08:35:41 PM »
You continue to miss the point.  If you did the transformation correctly, the distance would be the same in any coordinate system

Mathematically the above picture is a sphere. Just like the 2 graphs below are both representations of the same sine function yet look totally different

Just because the coordinates are the same, doesn't mean that mathematically it is a sphere.  A sphere has the least amount of surface area per volume of any shape.  All you've done is manipulate the amount of surface area to fit the same volume.

Check out this page
Example above: Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same. They have different directions and lengths and hence different cartesian coordinates. The lengths of the magenta vectors are shown at |Vglobe| and |Vfe| respectively [/quote]

The vectors represent a measurement of area.  So even though the coordinates are the same, the coordinates represent a different measurement of area. That’s exactly what is not supposed to happen when you transform coordinates from one system to another.  If you transform the metric tensor correctly, the coordinates in any system should represent the same amount of area in any other system.

Note: Coordinate System Transformations (CS Transf) do not change the length and direction of vectors. They only change the values of the vector components to get the same vector in the corresponding coordinate system.

Translation: A correctly performed transformation doesn't change the amount of surface area.

.Sure, you can manipulate the same volume of space of a sphere and plot it on a disc, but you have to ignore the rules about transforming the metric tensor and therefore aren’t represent reality. You can change the shape of anything to look like anything, but if you ignore the rules of physics doing so, it isn’t much of an accomplishment.

And this brings us back to my original point: physics works regardless of shape.

Not if you follow the rules of physics to determine the shape.

Also, another question.  What mechanism for gravity are you assuming?  If you are sticking with UA, then you should probably be using Rindler coordinates. 

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 06, 2022, 07:51:22 AM »
flat earth can not work with cartesian distances
This is exactly what you are not getting.  There is no such thing as “cartesian distance” or “spherical distance” or “polar distance”.  When you transform the metric tensor correctly, the distances are the same no matter what coordinate system you use
The distance between two points will not change if you change your coordinates. Also the metric tensor itself, as a function, will not change. But the matrix which represents the metric tensor depends on what coordinate system is being imposed on all of the tangent spaces, so the matrix representation of the metric tensor will in fact change with changes of coordinates.
The whole point of coordinate transformations is that you don’t change the physical nature or geometry of the underlying structure.  If your transformation results in different distances between points and you have to use a "bendy ruler" to make the distances work, that is exactly what you have done.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 05, 2022, 05:21:50 AM »
It's like claiming i broke physics because i drew my graph in red pen instead of a black one.
Its a lot more complicated that that
I believe everything works because of the distance metric changing. My guess is that the changes to the distance metric nullify the changes in shape and i believe i can make a compelling case for this.

The question is did you transform the metric correctly? It is the metric, not the coordinate system, that defines curvature. It is the metric that measures how much the object deviates from flat by giving you distances and angles. The metric tensor exists independent of a coordinate system and is an inherent property of the object. Changing the coordinate system doesn’t change the properties of the object. You are just changing the environment it is in. You are correct about that.
But  when you do a coordinate transformation, you have to transform the metric tensor as well.   Actually, you just transform the components, but the tensor as a whole does not change.  But you can’t just do that willynilly and just make up any arbitrary components  or metric that you want.  That seems to be what you have done. There are rules about how to transform the metric.
The most general transformation will transform coordinates, the metric tensor, the torsion tensor and also all other fields defined on the manifold. In case when metric, torsion and other tensor fields transform accordingly to coordinates the way tensor fields should transform, we have an coordinate transformation and manifold and all ''geometry'' (and physics) is preserved. In case metric and torsion transforms accordingly but other physical quantities does not, you are probably changing physics but preserving manifold. In case torsion or metric transformed differently than they should based on tensor transformation laws, you have defined new manifold.

Source .
Translation: Transforming the metric tensor correctly is critical to preserving the physics and geometry. It isn’t just about the physics working in any coordinate system.  Unless you transformed the metric tensor based on the tensor transformation laws, you have changed the geometry of the underlying structure.

When you correctly transform the metric when transforming from a spherical to a flat coordinate system, you get distortion, as explained here.
The procedure employed above works in general. To transform the metric from coordinates (t,x,y,z)(t,x,y,z) to new coordinates (t′,x′,y′,z′)(t′,x′,y′,z′), we obtain the unprimed coordinates in terms of the primed ones, take differentials on both sides, and eliminate t,...,dt,...t,...,dt,... in favor of t′,...dt′,...t′,...dt′,... in the expression for ds2ds2. We’ll see in section 9.2, that this is an example of a more general transformation law for tensors, mathematical objects that generalize vectors and covectors in the same way that matrices generalize row and column vectors. .
You broke physics by making up your own rules for transforming the metric tensor. If you don't have any distortion in your model, you didn't transform the metric correctly and have changed the geometry.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 04, 2022, 01:57:07 PM »
The intrinsic curvature of my space is 1/R²
How did you calculate that?
If one of the principal curvatures is zero: κ1κ2 = 0, the Gaussian curvature is zero and the surface is said to have a parabolic point.
The Gaussian curvature is the product of the two principal curvatures Κ = κ1κ2
Where are your principle curvatures?  It doesn’t look to me like there are any curvatures.

Imagine i have 2D cartesian coordinates (0,0) and (4, 4).  Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²
In polar coordinates these coordinates would be (0°, 0) en (45°, 4).  If i plug these numbers into the distance formula, i get total gibberish. When you do a coord transform, you must update all formulas. That is what i mean with a new distance metric. It's the old one with compensation for the coord transform.
What “formula” did you use?  Coordinate systems use metric tensors to fix distances and different coordinate systems have a defined metric tensor that should be used. 

There is no test that can distinguish the flat earth representation from the globe representation

You are contradicting yourself.  You claim your model has intrinsic curvature, but, by definition, intrinsic curvature can be tested for and observed from “the inside”

Flat Earth Theory / Re: Experiment to Distinguish FE from RE
« on: February 03, 2022, 04:13:19 AM »
A parallel transport can distinguish curved space from flat.  Its just moving a vector, keeping it parallel to itself, along a closed path.  If the vector has to change directions to complete the path, then the surface is curved.

You might have to use some ingenuity on the actual execution, but I'm sure it could be figured out.  Maybe a missile or rocket of some sort that is programmed to maintain a constant orientation along its trajectory.

The same concept was used to measure the curvature of spacetime using a gyroscope.  The animation is from a link describing the experiment.

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