Re: Found a fully working flat earth model?
« Reply #260 on: May 01, 2022, 03:38:33 PM »
I should have written: "In reality, the same lines of longitude cannot both converge and diverge in the same direction moving  south of the equator."

Better?
But that's not what happens under RET (which I presuppose to be your definition of reality), and troolon's "model" is just a restatement of RET with no functional changes. His entire argument relies on taking RET piecemeal and throwing a layer of confusion on top of it.

Hmm.  Let me make sure I understand something:

You are saying that in the RET model, lines of longitude don’t converge as they go south and then meet at the South Pole? 

I understand they are also curving, in this model, but that’s along a different plane, and not really at issue for purposes of this discussion.

When I look at a 3D model of a globe and  at the lines of longitude south of the equator, they certainly narrow and narrow until finally converging at the South Pole.

What terminology do you suggest to be clearer in explaining how lines of longitude differ in the RET and FET (monopole) models than “converging” and “diverging” ?  (Referring to the lines as one moves south along them, to be clear).

« Last Edit: May 01, 2022, 03:56:06 PM by existoid »

*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #261 on: May 01, 2022, 07:03:49 PM »
You are saying that in the RET model, lines of longitude don’t converge as they go south and then meet at the South Pole?
No, I'm saying they don't diverge, as long as we only consider what's happening south of the equator.

What terminology do you suggest to be clearer in explaining how lines of longitude differ in the RET and FET (monopole) models than “converging” and “diverging” ?  (Referring to the lines as one moves south along them, to be clear).
Let's be extremely clear here - troolon's "model" is not an FET model in any sense of the word, and it certainly is not a monopole model. It's RET with extra steps. It is nothing more, and it is nothing less. It is just RET, stated in a way that's confusing to some. If you think it is anything other than that, you are mistaken.
« Last Edit: May 01, 2022, 07:07:03 PM by Pete Svarrior »
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume

*

Offline Clyde Frog

  • *
  • Posts: 1045
  • [kʰlaɪ̯d fɹɒg]
    • View Profile
Re: Found a fully working flat earth model?
« Reply #262 on: May 01, 2022, 11:53:38 PM »
This is what I meant when I said you two hadn't actually read his posts. It's exactly the globe, expressed in a coordinate system that looks different without being different. The fact anyone is still suggesting the lines of longitude are diverging between the equator and either one of the poles in this thing he shared with us is clear evidence that the people making said suggestion haven't bothered paying any attention, yet they are still here to broadcast their ignorance and laziness for all to see while declaring some sort of weird victory. It's pigeons and chess.
« Last Edit: May 03, 2022, 07:27:33 PM by Clyde Frog »

Re: Found a fully working flat earth model?
« Reply #263 on: May 03, 2022, 03:56:20 PM »
Quote
The map has a different distance metric. Distance is just a formula, it's up to you to choose a meaningful one.
Using the correct metric, the circumference of 80N and 80S is the same as the globe and so it's smaller than the equator

It sounds like the OP just made up his own metric.  That's not how it works.  You can't just choose whatever metric you want to use. Metric tensors transform according to specific rules and that's what determines the geometry.  When done correctly, according to the rules, the geometry of the manifold doesn't change.


*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #264 on: May 03, 2022, 04:11:36 PM »
You can't just choose whatever metric you want to use.
Of course you can. In fact, you have no other option than to do so.

Metric tensors transform according to specific rules and that's what determines the geometry.
In Euclidean spaces, sure. This is emphatically not one. Considering you've missed that, I somehow doubt you know what you're talking about.
« Last Edit: May 03, 2022, 04:24:35 PM by Pete Svarrior »
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume

Re: Found a fully working flat earth model?
« Reply #265 on: May 03, 2022, 05:52:19 PM »
You can't just choose whatever metric you want to use.
Of course you can. In fact, you have no other option than to do so.

Metric tensors transform according to specific rules and that's what determines the geometry.
In Euclidean spaces, sure. This is emphatically not one. Considering you've missed that, I somehow doubt you know what you're talking about.

Euclidean or non has nothing to do with it.  Metrics are tensors and tensors, by definition, are invariant under coordinate transformations.

Quote
A Tensor is an object that in invariant under a change of coordinate systems, with components that change according to a special set of mathematical formulae

If you transform the individual components correctly according to the transformation laws, the net result is the same metric that you started with.

If the OP has different metric for his globe model and his FE model, he didn't transform the components properly.


*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #266 on: May 03, 2022, 06:02:21 PM »
Euclidean or non has nothing to do with it.
Right, I know everything I needed to know. Before you come back to argue against RET (which is what you're currently trying to do), get a grasp of geometry.

If the OP has different metric for his globe model and his FE model
He doesn't, and it's not a FE model. Please form an understanding of what's being discussed before you explain how proudly you disagree with it.
« Last Edit: May 03, 2022, 06:08:04 PM by Pete Svarrior »
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume

Re: Found a fully working flat earth model?
« Reply #267 on: May 03, 2022, 09:51:06 PM »
Quote
Right, I know everything I needed to know. Before you come back to argue against RET (which is what you're currently trying to do), get a grasp of geometry.

 It doesn’t matter if you are talking about a Euclidean or non-Euclidean space, the metric tensor is invariant under  any  coordinate transformation.  If you are going from Euclidean to Euclidean, it is invariant.  If you are going from non-Euclidean to non-Eucldiean, it is invariant.  If you are going from Euclidean to non-Euclidean or the other way, it is invariant.

The whole point is that you can’t transform a Euclidean space to a non-Euclidean space accurately because they have different metrics.  The Euclidean metric is  the Pythagorean Theorem.  If you correctly transform to a non-Euclidean coordinate system, the metric remains the Pythagorean Theorem and distances and angles won’t make sense.  The PT doesn’t work in a non-Euclidean space.  You can’t randomly decide not to use the PT.
 
I guess you can decide that and just make up your own metric  and invent a bendy ruler to make it work.  The physics police won’t come and drag you away, but the results are meaningless.  It makes distances entirely subjective if you can arbitrarily decide to “measure differently”.  What makes his random metric the right one?  What makes his bendy ruler better than my super bendy ruler?  By the OP’s logic, I can change the defintion of a pound and claim I have lost weight. I still won’t fit into my skinny jeans though so it doesn’t mean anything.

Quote
He doesn't, and it's not a FE model. Please form an understanding of what's being discussed before you explain how proudly you disagree with it.

He specifically said they have different metrics.
Quote
The map has a different distance metric.
  And you may not consider it an FE model, but the OP does.  If you disagree with that then take it up with him.  I don’t care what the OP calls it. Or what you call it.  Either way, its not the right way to do coordinate transformation.

*

Offline Clyde Frog

  • *
  • Posts: 1045
  • [kʰlaɪ̯d fɹɒg]
    • View Profile
Re: Found a fully working flat earth model?
« Reply #268 on: May 04, 2022, 12:47:16 AM »
OP does not consider it a FE model. Try reading. If it doesn't work the first time, try again. He said more then once that it's still the globe. Other physicists reviewed it and said it's a globe (also, that's why it's "practically useless" - because using this framework is really impractical when you could use the actual globe). If you spend a few cycles pondering the implications of everything troolon described, it's obviously still a globe. But people get so hung up on fighting against a perceived enemy that it doesn't even matter what is actually being said.

Go ahead though. Tell us all how wrong it is, and how it can't possibly describe reality as we observe it, despite matching observations because it's, you know, the actual fucking globe and stars just illustrated in a way that looks scary and foreign and so, obviously, it must be bad.

*

Offline stack

  • *
  • Posts: 3583
    • View Profile
Re: Found a fully working flat earth model?
« Reply #269 on: May 04, 2022, 01:41:21 AM »
I agree. It took me a while to wrap my head around it purely because the physical manifestation presented of a globe didn’t look like a globe. But at the end of the day, and this is a vast over-simplification, mathematically, creatively, and perhaps with newly devised measuring metrics instruments, you can make almost any shape represent a globe.

Even more simplified, just look at the dozens of globe projections that exist. None of them look like a sphere.

Re: Found a fully working flat earth model?
« Reply #270 on: May 04, 2022, 04:51:41 AM »
Quote
Tell us all how wrong it is, and how it can't possibly describe reality as we observe it, despite matching observations

It only matches observations if you use a “bendy ruler”.  That’s like saying a movie perfectly reflects reality as long as you wear 3-d glasses.

Quote
OP does not consider it a FE model.

Then why did he title the thread  “Found a fully working flat earth model?” 

But like I said, I don’t care what he considers it or what you consider it.  I simply made the observation he didn’t transform the metric tensor correctly. It isn’t debatable that he did it wrong.  Tensors are invariant with coordinate transformations. Period. Full stop.

His two models, or  coordinate systems, if that makes you feel better, have different metrics. That means he didn’t transform correctly.  That’s tensor calculus 101….and basic differential geometry.  And we don’t even know if his “bendy ruler metric” is even valid. Valid metrics have defined characteristics.  You can’t just make up some random formula and call it a “metric”. 

I'll leave it to those who want to argue what the implications of those observations are.

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #271 on: May 07, 2022, 10:00:06 AM »
I believe there are still a few nuances in the way how i see this model versus how many posters view it.

Personally i don't think there has been enough emphasis on the fact that this model can be, and in fact was originally empirically derived.
2000 years ago, the Greeks made an incompleteness error: They assumed a globe and straight light and showed it was consistent with measurements in reality.
However, what they really measured was the relationship between light and earth shape. The earth shape was an assumption.

For my approach i've started over, but this time assuming a flat earth (later to be generalized to shape-agnostic)
I've started with the observation of horizon distance: ie that the formula R/cos(phi) - R   defines the relation between earth-shape and light-ray shape (basically height difference between a laser and a lake).
From this formula (and assuming a flat earth it's possible to create an explanation for day, night, seasons, .....)
In fact i believe it's possible to rederive all of physics this way without relying on the globe, at all.

At this point we have created an alternative model to the globe model and then there are 2 possibilities:
- we find a difference between both models and derive a test to see which one is correct
- both models are equivalent

In this case the latter can be proven. In history this has happened multiple times before. Think of the ptolemaic and tychonic models for planetary motion. It can be shown both are equivalent/approximations of the globe model)
In all posts so far, there has been a lot of emphasis on the equivalence with the globe model (as it's vastly easier to explain what's happening this way than having to rederive 2000 years of gnarly mathematics :)
Personally i do believe it is correct to call this a "flat earth model" (tough i admit there's a a bit of semantics involved)

Diving into this i've also discovered that the flat earth debate is a lot more nuanced than i initially thought. For example a picture of globe earth from space or a ship disappearing hull first are not actually proofs of intrinsic curvature for example and this is not something i would have guessed before this post.
And then there are of course the more philosophical questions about what it means to have 2 differently shaped models.

I also see this work as a potential bridge between RE and FE. When we see a ship disappearing hull first behind the horizon, and the globies and flatiies start warring whether it's the globe, the light or the aether that's curving, we now know they're really in agreement.
« Last Edit: May 07, 2022, 10:15:29 AM by troolon »

*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #272 on: May 07, 2022, 10:07:10 AM »
I simply made the observation he didn’t transform the metric tensor correctly.
You don't know what that means. :)

Tensors are invariant with coordinate transformations.
Assuming a Euclidean space, which this emphatically is not.
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume

Offline jimster

  • *
  • Posts: 303
    • View Profile
Re: Found a fully working flat earth model?
« Reply #273 on: June 04, 2022, 12:17:20 AM »
Troolon,

If you have oroved that the earth could be any shape, then the earth could be round. Interesting. The only thing I know all FEs to agree on is that the earth is not round. Could be north pole centric, could be south pole centric, could be bi-polar, but they all agree couldn't possibly be round. FEs will like you better if you prove it isn't round rather than that it could be round. To do that, I think you would have to prove that light rays in vacuum can't be straight and measurement is definitely broken. You will be the hero of FE.

I have a degree in math, my son is a math major, and I have reviewed the math behind your claims. Known mathematicians and discussed math all my life. Your claim might be restated as: The earth appears round in Euclidean 3-space but reality is secretly non-Euclidean (measurement is broken, as you say, and rays of light that seem to travel straight are actually bent, perhaps in different ways depending on the position of the observer). Once you assume measurement is broken and light curves in unknown ways, an infinite number of possibilities with no way to know which one.

You also did not incorporate the astronomical transform. When you apply your transform to earth, what about stars millions of light years away? Somehow they moved into a dome, but you have no math for that. Once you say the light bends and you don't know the equation, those stars could be anywhere, sun, moon, etc. The earth could be a cylinder a million miles long and an inch diameter. Meanwhile, Euclidean 3 space with RE gives us GPS, ICBM, airliner finds the airport, sextant north star latitude makes perfect sense, and the solution is singular. If light doesn't bend and measurement isn't broken, the earth is round.

Your ideas re coordinate transforms, basis, and "mathematically equivalent", and your physics claim that it follows a shape change through coordinate transformation are wrong. If you are right, you can go to the math and physics community and explain, for example, the power that coordinate transforms have to bend light. You will be famous, but you will need experiments and equations. You found a reasonable (to you, although not to a professor) of how the earth's shape is unknown. I imagine you will not share your discoveries with them, perhaps posting on TFES is the thing to do with it. To what end?

So the choice of what to do with your discoveries is up to you. You can post on TFES and get some agreement and some explanation of why you are wrong from RE. If I knew some part of science was wrong, I would go to scientists and explain in an attempt to set them straight. But I suppose if they did not agree, it would be because of conspiracy, or perhaps scientists and mathematicians are stupid. Conspiracy can really explain a lot, and you never have to have details or evidence, because its secret, so how could you know the details and the evidence is they all say the wrong RE stuff.

Still, just the idea that changing coordinates changes the shape of a geometric figure would be a starting point. Go to some mathematicians and show them how you changed a sphere into a disk by coordinate conversion. If they say "It doesn't work that way, you don't understand coordinate conversion, the figure remains the same size and shape." Perhaps all mathematicians are wrong, just the ones at UCLA, maybe you can explain, but this is their definition, so don't they get to decide?

You started with a globe where Australia's size matched real world measurements and Sigma Octantus made sense. Then you transformed the surface into a disk, where Australia was wrong size and Sigma Octanus made no sense. Then you claimed this as mathematically equivalent, and then that the physics is then equivalent.  Forget which is true, which is useful? If it pleasures you to think that under the RE appearance, non-Euclidean math means it is flat, well, I can't stop you. But the fact that you can map the points on a sphere to a disk if you don't care that the size is wrong and the light bends, but that doesn't prove the earth could be any shape. If anything, you have proved that it is round. In the infinite possibilties of shape agnostic earth, only one works with measurement and light waves traveling straight in a vacuum, Polaris and SIgma Octantus, etc, etc etc.

Does that mean anything? As I said, forget true, what is useful?
I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

*

Offline Clyde Frog

  • *
  • Posts: 1045
  • [kʰlaɪ̯d fɹɒg]
    • View Profile
Re: Found a fully working flat earth model?
« Reply #274 on: June 04, 2022, 10:59:17 PM »
It's literally a RE but ok

Re: Found a fully working flat earth model?
« Reply #275 on: June 06, 2022, 01:56:51 PM »

Tensors are invariant with coordinate transformations.
Assuming a Euclidean space, which this emphatically is not.

This is incorrect. A true tensor is invariant under any coordinate transformation. A tensor is a multilinear map from vectors and dual vectors to real numbers and does not depend on a coordinate system to be defined. Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.

*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #276 on: June 06, 2022, 04:56:21 PM »
A true tensor is invariant under any coordinate transformation.
I said nothing about coordinate transformations, merely the assumptions required for this statement to hold. Restating the statement louder is not gonna help.

Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.
Correct, but also delightfully irrelevant.
« Last Edit: June 06, 2022, 04:57:52 PM by Pete Svarrior »
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume

Offline jimster

  • *
  • Posts: 303
    • View Profile
Re: Found a fully working flat earth model?
« Reply #277 on: June 06, 2022, 06:31:06 PM »
The OP said that a sphere turns into a disk when you change coordinate systems. He said that the disk was mathematically equivalent to a sphere. A sphere is the set of points equidistant from a point. A disk is not that. There is so much wrong with the OP's ideas in basic high school geometry. Why even bring up tensors? Complexity is an FE tactic to make their idea possible by obscuring simple truth. The OP made numerous math errors ecxplained in the 14 pages above by myself and others.

Before any discussion of tensors, wouldn't it be wise to make sure the basic idea is not wrong?

Could we have a discussion of whether changing coordinate systems makes a sphere into a disk?

Does anyone else see the math errors? Crush me, show your math skills and prove to me that OP understands math correctly, coordinate systems, sphere, basis, and measurement. OP has changed all these things from math as taught in high school and college. Either all of math is wrong, or OP is wrong. OP wants to redefine math the way he needs to in order to prove the earth is flat, or at least could be flat.

The actual conclusion from the OP is that no one can know the shape of the earth because the light could bend in any direction and measurement could be broken. I am astounded that a site dedicated to the shape of the earth can end up talking about the math of tensors when the evidence of the shape of the earth is found in simple things. Only by ignoring those simple things can the earth be flat, so off to tensors and whatever.

You end up with 14 pages of people arguing esoteric math by people who mostly learned that math through a google search without a thorough background to understand correctly what those words mean. This is convenient for FEs so they can substitute their idea of what the math words mean so as to make FET possible. There are a few RE math knowledgeable people who try to convince them of the correct meaning, but that is like trying to teach a math course through a comment column to a person who does not want to learn the true math.

Troolon should go to a community college and enroll in a geometry class. Perhaps he could explain his ideas. Would be fun to see him explain how a coordinate conversion makes a sphere into a disk and yet is mathematically equivalent. Never fear, Troolon, you can save your belief system through conspiracy, just add mathemiticians to the NASA/RE conspiracy theory. Troolon correctly understands math, and the entire worldwide math community is made of a few conspirators and a bunch of stupid sheep.

Troolon right and they are all wrong? If they are right, then measurement works and light travels straight in a vacuum. If Troolon is right, measurement is broken and no one knows the shape and size of anything, not just the earth. How do you know about an object across the room when light bends and measurement doesn't work? In order to make it possible for the earth to be flat, you have to make it possible for anything to be any shape or size.

I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

Re: Found a fully working flat earth model?
« Reply #278 on: June 06, 2022, 07:32:47 PM »
A true tensor is invariant under any coordinate transformation.
I said nothing about coordinate transformations, merely the assumptions required for this statement to hold. Restating the statement louder is not gonna help.

Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.
Correct, but also delightfully irrelevant.

You didn't say, but you quoted specifically a part of someone else's post mentioning tensors are invariant under coordinate transformations, stating that this assumes Euclidian space. This is incorrect, tensors are invariant under coordinate transformations in any space they are defined, Euclidian or not. Perhaps my post wasn't clear enough

*

Offline Pete Svarrior

  • e
  • Planar Moderator
  • *****
  • Posts: 16269
  • (◕˽ ◕ ✿)
    • View Profile
Re: Found a fully working flat earth model?
« Reply #279 on: June 06, 2022, 07:43:18 PM »
Complexity is an FE tactic to make their idea possible by obscuring simple truth.
This is your twice-pagely reminder that OP is a RE'er and he's making fun of all of you. There are no "FE tactics" in place here. OP is a RE'er and he is presenting RET, unaltered. It is extremely disingenuous of you to take a RE troll and describe his jokes as "FE tactics".

This is incorrect, tensors are invariant under coordinate transformations in any space they are defined, Euclidian or not. Perhaps my post wasn't clear enough
Your post was quite clear, and my response stands. If you think repeating yourself for the fourth time will help, it will not.

Perhaps your confusion lies in the fact that it is possible to come up with a non-Euclidean space in which your assertion holds. I could have been more precise, but I wasn't - it wasn't particularly necessary given that the exact space we're discussing has been specified, and given that the incorrect assumptions about this space were also made clear.
« Last Edit: June 06, 2022, 07:59:31 PM by Pete Svarrior »
Read the FAQ before asking your question - chances are we already addressed it.
Follow the Flat Earth Society on Twitter and Facebook!

If we are not speculating then we must assume