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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.
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.
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
You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality
It’s certainly an interesting thought experiment, but agreed, it’s not a model, let alone a ‘fully working’ model.
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
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
Mathematically it behaves like a euclidean globe.
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.
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: https://en.wikipedia.org/wiki/Isometry.
It's basically any transformation that preserves distance
My coordinates are in (lat, long). So my distances are calculated with haversine, and will thus give the same result as on your globe.
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.
Straight is only defined the way you think it is in an orthonormal basis.
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.
Since the curvature of the sphere does not vanish, it CANNOT BE LOCALLY ISOMETRICALLY MAPPED TO THE EUCLIDEAN PLANE
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.
At what point have i tried to "flatten" the sphere?
The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.
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).
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.
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
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
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.
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.
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.
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.
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.https://www.infoplease.com/math-science/mathematics/geometry/geometry-isometries#:~:text=There%20are%20many%20ways%20to,also%20known%20as%20rigid%20motion.
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.
I don't agree pythagorean distances are the only distance metric. In our discussion we're more interested in arclength along a greatcircle distances.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.
But fine, let's go with straight distances through the earth.
These are not arbitrary formulas. This is the way you're supposed to do coord transformsSome 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.
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.http://wiki.gis.com/wiki/index.php/Metric_space
How will you tell?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.
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.
10.2 Parallel Transport in a Curved Space, D=2https://www.av8n.com/physics/geodesics.htm
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.
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
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.
And this brings us back to my original point: physics works regardless of shape.
flat earth can not work with cartesian distancesThis 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.
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 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.
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.
The intrinsic curvature of my space is 1/R²How 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.
Imagine i have 2D cartesian coordinates (0,0) and (4, 4). Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²What “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.
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.
There is no test that can distinguish the flat earth representation from the globe representation
