Proof that the world is not flat

[syhprum1]

I have evidence that the world is flat, I see innumerable comments in the scientific press that missiles move in a parabolic path that only occurs when the Earth is flat and of infinite extent.
If the Earth was spherical they would take an elliptical path !

 

[Nugatory]

I have evidence that the world is flat, I see innumerable comments in the scientific press that missiles move in a parabolic path that only occurs when the Earth is flat and of infinite extent.
If the Earth was spherical they would take an elliptical path !

They do take an elliptical path. The parabola is a very good approximation used when the angle subtended by the path of the missile is small.

 

[fresh_42]

The ancient Greeks knew this.

Eratosthenes 200 BC

 

[mfb]

https://en.wikipedia.org/wiki/Foucault_pendulum

To me, this is the most convincing proof not only that the Earth is not flat, but that it rotates as explained in the rotating spheroid model.

For me, it is more convincing, because it can be done in a single room and does not depend on outside sources of data that may be less reliable.
If that's not a slam dunk, the repeating the experiment at several latitudes should be.

The Sagnac effect is a modern version of the experiment. It allows to measure the rotation of Earth with a parts per billion precision, and the axis of rotation can be determined with a precision of millimeters. http://www.fs.wettzell.de/LKREISEL/G/LaserGyros.html

 

[Vanadium 50]

After a few takes on this, I have the answer to Vanadium 50's question. My first intuition (finally confirmed) is that distance pairs between 4 points are sufficient to determine radius on the assumption that the points are on a 2-sphere.

Thanks. I thought about this a bit after my question and came up with this. Three points is clearly too few. Four points takes 6 numbers to specify. You have 8 coordinates of the 4 cities, but one city can be placed at (0,0) and the whole setup can be rotated around the (0,0) point, making 5 - but you also need (as PAllen said) a single number R to specify the radius, so you have 6.

Now, how many numbers do you have? Between 4 cities you have (4 x 3)/2 = 6 distances. So in principle you have enough information. I think you also do in practice, provided the 4 points are convex.

How might this work? If A, B, C and D form a convex quadrilateral, you could find the area of ABC using Brahmaguptra's formula, and the sum of the areas of ABC and BCD using Heron's formula, and note that they don't match. The spherical versions of these are called (I looked this up) Lhulier's and Cagnoli's formulas, and they also include a parameter E from which R can be extracted. It's definitely quadratic, but you want the positive solution for R, so I think it's still OK.

 

[PAllen]

Now, how many numbers do you have? Between 4 cities you have (4 x 3)/2 = 6 distances. So in principle you have enough information. I think you also do in practice, provided the 4 points are convex.

I'm not seeing how 4 non-coplanar points in [Euclidean] 3-space can fail to be convex; or that there are any 4 non-coplanar points in 3-space that you can't embed in a 2-sphere of unique radius.

 

[Vanadium 50]

'm not seeing how 4 non-coplanar points in [Euclidean] 3-space can fail to be convex

Like a chevron.

 

[PAllen]

Like a chevron.

But if it is only 4 points, when you add all the sides it still makes a convex tetrahedron.

 

[PAllen]

There is one problem with my post #30. The logic and conclusions are all fine, but the formula referenced is not actually what we want. The distances used in the formula based on the Cayley-Menger determinant would be the 'through the earth' distances, which are not the ones we want to use. These can be converted in terms of the radius we seek, and you have a complex implicit formula for radius (it being on both sides of the equation). Most likely, it is not solvable in closed form, but could readily be solved numerically.

 

[mfig]

If the Earth is flat, then why has nobody ever taken a picture of the edge?
I don't get it. Are there really people out there who honestly believe the Earth is flat?

 

[WWGD]

If the Earth is flat, then why has nobody ever taken a picture of the edge?
I don't get it. Are there really people out there who honestly believe the Earth is flat?

I think it is more of an abstract exercise in proving it, sort of assuming you are arguing with someone who legitimately believes it.

 

[PAllen]

I think it is more of an abstract exercise in proving it, sort of assuming you are arguing with someone who legitimately believes it.

Also, in terms of motivation of the original source (differential geometry lite for physicists) discussing the minimum quasi-local geometric information needed to distinguish curvature, particularly the idea pure metric (distance) quantities are enough (if you allow angle measurements, you need fewer points).

The answer is you need pairwise distances between 4 points to detect curvature of a two manifold, and 5 points for a 3 manifold. Then we had a further discussion of whether pairwise distances between 4-points allows determination of the radius of curvature of a 2-manifold on the assumption that curvature is constant. I believe I have clearly demonstrated that this is so.

 

[PAllen]

One more note on the idea of using only distances rather than angles is that in GR, the notion of rigid grids whose angles you can measure is highly problematic; even angles between light rays is non-trivial in the presence curvature. This motivated Synge's purely metric approach; the real subtlety of his argument is going from observations about curvature of Riemannian 3-manifolds from 5 points to measuring curvature of a Lorentzian 4-manifold using only round trip times times between 5 world lines; this relies on a whole series of earlier results in the book on Fermi-Walker transport, and it is very surprising (to me) that it is sufficient.

 

[mfb]

Hmm... good gravimeters can measure g with a precision of 10^-9. Let them measure the horizontal component along the line of a laser, and you should get visible deviations within a meter. They can also measure how g decreases with height - again, a meter difference is sufficient. A measurement of the radius of Earth to ~1%, done in a lab, if you know the mass distribution of the lab itself with sufficient precision to account for its effect.

 

[PAllen]

Hmm... good gravimeters can measure g with a precision of 10^-9. Let them measure the horizontal component along the line of a laser, and you should get visible deviations within a meter. They can also measure how g decreases with height - again, a meter difference is sufficient. A measurement of the radius of Earth to ~1%, done in a lab, if you know the mass distribution of the lab itself with sufficient precision to account for its effect.

But a gravimeter is providing direct information about curvature by a different means. So, having determined this, you then measure that it has predictable effects on light paths. That is interesting, but not the same as, e.g. making a series of measurements of angles between light rays and trying to deduce geometry from that. E.g. the idea that you can use sum of angles of triangle, when your triangle consists of light rays following null geodesics of unknown geometry, is an extremely non-trivial undertaking.

 

[mfb]

But a gravimeter is providing direct information about curvature by a different means.

Well, there is no GR solution that would make sense with a flat Earth, but I think if you trust GR then you see enough proofs of a round Earth anyway.
The gravimeters are a GR-independent way to see that the Earth is not flat (at least not relative to a light path).

 

[PAllen]

Well, there is no GR solution that would make sense with a flat Earth, but I think if you trust GR then you see enough proofs of a round Earth anyway.
The gravimeters are a GR-independent way to see that the Earth is not flat (at least not relative to a light path).

On further thought, I think you are not measuring curvature at all this way. The aim is to detect curvature over the 'lab size'. Curvature is divergence from SR spacetime. Change in g over the lab size has only a second order difference from the Rindler flat spacetime prediction. My understanding is that this difference has never been measured on Earth (at lab scale). [edit: Tides, of course, measure curvature over Earth as a whole. ]

Validating GR and measuring a direct 4-geometric affect of curvature are different endeavors.

 

[mfb]

I never said you would measure curvature of spacetime. You measure an inhomogeneous gravitational field.

 

[PAllen]

I think further discussion of space-time curvature measurement should be in another thread, as it is physics of GR. This thread should stay on the original topic of geometric measurement of curvature within Riemannian manifold, specifically the case of a 2-surface.

 

[mgkii]

Yeah I got it but what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"? What does it have to do with the measurement mismatch?

Hi Whitehole

I don't think anyone actually got around to giving you the answer to your specific question. What he means is if you only use 3 cities and make triangles, then you'll never see a problem; you can take a hundred cities and draw a hundred different "planar triangles" (i.e. triangles on a flat piece of paper) and the Earth's curvature will not "show up"; the curvature will simply "get lost" as each triangle will have a small error in each of its angles that makes up for the "lost" curvature as you translate the triangles from the curved Earth onto the flat planar surface of your paper.

However, if you then go back and add a forth city onto each of your planar triangles - extending each triangle into a planar quadrangle, you'll see the curvature problem come into play - if your original triangle was Paris, Berlin, Barcelona (as in the example you used), then you'll not be able to find a single point for the fourth corner that is the correct distance away from all these three cities to Rome.

 

[fresh_42]

Isn't it a lot easier? How can we have day and night at the same time in different places if Earth was flat?

 

[Hornbein]

Isn't it a lot easier? How can we have day and night at the same time in different places if Earth was flat?

Big squares blocking the Sun? Gravity bending the Sun's rays so they hit they Earth before they get to the nighttime region?

 

[WWGD]

Hi Whitehole

I don't think anyone actually got around to giving you the answer to your specific question. What he means is if you only use 3 cities and make triangles, then you'll never see a problem; you can take a hundred cities and draw a hundred different "planar triangles" (i.e. triangles on a flat piece of paper) and the Earth's curvature will not "show up"; the curvature will simply "get lost" as each triangle will have a small error in each of its angles that makes up for the "lost" curvature as you translate the triangles from the curved Earth onto the flat planar surface of your paper.

However, if you then go back and add a forth city onto each of your planar triangles - extending each triangle into a planar quadrangle, you'll see the curvature problem come into play - if your original triangle was Paris, Berlin, Barcelona (as in the example you used), then you'll not be able to find a single point for the fourth corner that is the correct distance away from all these three cities to Rome.

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

 

[WWGD]

Big squares blocking the Sun? Gravity bending the Sun's rays so they hit they Earth before they get to the nighttime region?

But how do we then switch from night to day in the dark region?

 

[Hornbein]

But how do we then switch from night to day in the dark region?

The Sun gets teleported back to its starting point? (Once the Earth is flat, anything is fair game.)

 

[mgkii]

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

Just throwing a thought on this; I could well be wrong, but I think you might need to think of this from a different starting point. Moving from a 3 dimensional Earth to a 2 dimensional plane by ignoring angles of the triangle is a form of dimensional reduction. You may be able to maintained angles by choosing a different form of reduction, and find a different type of breakaway - i.e. the value of 4 may simply be a feature of the chosen reduction.

 

[PAllen]

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

I don't know if there is a name, but the theory can be seen as follows:

You hypothesize your manifold is flat, and is embedded in a flat manifold of one higher dimension. Then, ask what is the fewest points in the higher dimensional flat manifold to make a simplex (polytope) with volume. For 2 space it is 3, 3-space it is 4, for 4-space, it is 5, etc. Then, consider e.g. 4 points in the two manifold. They would form a degenerate zero volume simplex in the embedding 3-space. This means that some expression of pairwise lengths (e.g. the Cayley-Menger determinant) that is monotonically related to simplex volume, must be zero, and this gives a criterion for flatness of the two manifold. Summary: the number of points needed determine flatness (purely by distance measures) is the minimum number to form a simplex in one higher dimension.

 

[PAllen]

Just throwing a thought on this; I could well be wrong, but I think you might need to think of this from a different starting point. Moving from a 3 dimensional Earth to a 2 dimensional plane by ignoring angles of the triangle is a form of dimensional reduction. You may be able to maintained angles by choosing a different form of reduction, and find a different type of breakaway - i.e. the value of 4 may simply be a feature of the chosen reduction.

This is correct. It has already been pointed out earlier in the thread that you need fewer points if include angle measurements.

 

[Mark Harder]

To me, this is the most convincing proof not only that the Earth is not flat, but that it rotates as explained in the rotating spheroid model.
For me, it is more convincing, because it can be done in a single room and does not depend on outside sources of data that may be less reliable.
If that's not a slam dunk, the repeating the experiment at several latitudes should be.

Nice. I think you've hit on the weakness in all the comments I've read so far. Namely, what do we mean by 'distance'? How would one know the 'distance'? By definition, the metric applicable to the Earth's surface is the contour length the great circle connecting any two cities. But this is a non-Euclidean geometry, a spherical geometry. (For one thing, the angles of a spherical triangle do not add up to 180 deg, which may play a role in our problem.) Therefore, the inconsistency between the distances defined by the spherical metric and the Euclidean distances is the result of the naive isomorphism between the 2 geometries. Perhaps there is a projection of the sphere onto the plane which preserves the distances between points on the two surfaces and the angles of triangles. I wouldn't know. But I do know what any 6th grader knows - that the Mercator projection is not the right one - all that business about Greenland being bigger than South America and so forth.

 

[WWGD]

Nice. I think you've hit on the weakness in all the comments I've read so far. Namely, what do we mean by 'distance'? How would one know the 'distance'? By definition, the metric applicable to the Earth's surface is the contour length the great circle connecting any two cities. But this is a non-Euclidean geometry, a spherical geometry. (For one thing, the angles of a spherical triangle do not add up to 180 deg, which may play a role in our problem.) Therefore, the inconsistency between the distances defined by the spherical metric and the Euclidean distances is the result of the naive isomorphism between the 2 geometries. Perhaps there is a projection of the sphere onto the plane which preserves the distances between points on the two surfaces and the angles of triangles. I wouldn't know. But I do know what any 6th grader knows - that the Mercator projection is not the right one - all that business about Greenland being bigger than South America and so forth.

Maybe one can somehow generalize the fact that any projection will produce distortions of some sort to make a general argument on the inequality/inequivalence between Euclidean and Spherical.