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For the sake of clarity: Do you think the OP is suggesting they are showing us a Euclidean space FE in this thread?
Could you point me to an example of a tensor that IS NOT INVARIANT under coordinate transformation?I don't know how many times I need to say that coordinate transforms are completely irrelevant here, but I'll try once more. If you keep rambling about this, I'll just ignore you.
Tensors are invariant with coordinate transformations.Assuming a Euclidean space, which this emphatically is not.
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 enoughYour post was quite clear, and my response stands. If you think repeating yourself for the fourth time will help, it will not.
[/b]Perhaps your confusion lies in the fact that it is possible to come up with a non-Euclidean space in which your assertion holds.[/b] 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.
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.
Tensors are invariant with coordinate transformations.Assuming a Euclidean space, which this emphatically is not.
You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality.Yes, but a flat surface, by definition, has an euclidian metric. That is, the euclidian metric gives the shortest distance between 2 points, not haversine.
Intrinsically my model behaves like a globe. Extrinsically it's flat. When we look at a model, we look at it from the outside so extrinsically.
So i would argue it's possible to create a flat earth representation, that mathematically behaves just like a globe would.
- What does the earth look like in an orthonormal reference frame/euclidian geometryThis question is a bit strange. We have mentioned this before, but curvature does not care about coordinates. I don't understand how you can look at a surface in "euclidian geometry". Its properties are not dependent on the coordinate system you use. And by the way, a (lon,lat) system of coordinates on a sphere is also orthonormal, as are polar coordinates in a plane, cylindrical coordinates, ...
- Can we model physics on a flat earth?You have yet to show that since your model is not flat, unless in the sense described above.
- What is the real shape of the planet / what is the correct coordinate system?These questions are completely unrelated, and there's no correct coordinate system. You can choose whatever you want, but as we said before, the properties of the surface such as curvature are not coordinate dependent.
https://www.education.vic.gov.au/school/teachers/teachingresources/discipline/science/continuum/Pages/scimodels.aspxQuoteA scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted. The models scientists create need to be consistent with our observations, inferences and current explanations. However, scientific models are not created to be factual statements about the world.
It's... not a model?
The extrinsic curvature is only defined when you embed a submanifold in a higher dimensional one. So for us it does not matter, since we can only experience 3 space dimensions; if the earth has zero extrinsic curvature for some beings that can experience 4D space, whatever. But even they will agree that for us mere 3D beings, earth is curved. So your model is curved (where it matters), not flat.Also, this is wrong. As it has been mentioned, curvature is invariant over a change of coordinates. If it's non-zero, there is no possible change of coordinates that can turn it to zero globally.You're quite right the intrinsic curvature is indeed curved. Mathematically it's a curved space.
Doesn't matter the axis or coordinate basis. If your model has non-zero intrinsic curvature, it isn't flat
But i believe extrinsic curvature to be zero (i'm sure Rog will have a 70 page proof or something about this
When you look at a model of the world, you look at the model from the outside. So i believe it's correct to say it's a flat representation of physics.
For me personally i can even look at the world and experience it as the flat representation.
For example when i see a picture of a ship with a missing keel, i can very often imagine it's the light that's curving.
Or when i see a globe from space, i can superimpose the model and tell myself: ah yes, that's a flat disc distorted by curvy light.
Some people seem to feel more comfortable looking at the world as flat, and now science can explain the world in their view if they want to...
The really impressive thing is how patient OP has been in explaining it in 10 pages of posts.
He had me worried for a minute talking about teleporting, but it was just a lack of a better word to describe some of the artefacts in looking at things after the coordinate transformation in the model
A disc in an orthonormal basis, with orthonormal distances, will have a curvature of 0.Also, this is wrong. As it has been mentioned, curvature is invariant over a change of coordinates. If it's non-zero, there is no possible change of coordinates that can turn it to zero globally.
A disc in celestial coords, with spherical distances, will have the curvature of a sphere. (as shown)
The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.Doesn't matter the axis or coordinate basis. If your model has non-zero intrinsic curvature, it isn't flat
No, curvature does not depend on the choice of coordinates. You could define curvature without ever mentioning coordinates.QuoteAnd to finish it, if you calculate the curvature of your model to be different from 0, you don't have a flat earth. I hope this is clearOnly in an orthonormal reference frame. Shapes only make sense if you have defined a basis.
We have presented two differently shaped models of the same physics.You don't have two different shapes. You said it yourself - take the globe, change coordinates, this is what you get. This cannot alter the "shape" you were using, no matter what coordinates you try to use. If your curvature was non-zero to start with, you can't make it zero (globally) just by a change in coordinate system.
They can't be differentiated by any test.
Therefore it's impossible to know the shape of the earth.
We now have 2 models: one with a flat earth, and one with a globe.No, still only a globe represented differently than usual
How do we know which one is correct? We device a test and check which model matches the results.They give you the same result because they are the same model. Try to explain the sun/moon going below the horizon for an observer, but for others is still visible high in the sky. Or the fact that constellations keep their shape throughout the night, or the sun keeping the same size all day, or earth being seen as a globe from space. Of course you could chalk it up to "earth is flat, therefore some weird thing happens". But then you're not devising a test to know the shape, you're assuming the shape and then fitting everything else to that. Or, again, you invoke some handwavy-not-supported physics for that.
Only problem, both models are identical and always give the same result. There's no way to tell which one is right.
Ergo, we can't measure the true shape of the earth.
As for inventing physics: please tell me why light travels straight in a globe world?Light travels straight provided there are no ways to refract, deflect or interact with light. We see they go straight (as in follow a geodesic) because this is what is observed and measured, no matter what the shape of the earth is. The sun is not the only source of light we have, you know. In this case, theory and experiment agree. However, bendy light on a flat earth is a complete ad hoc hypothesis, as it starts with the earth being flat and then it has to explain sunsets, sunrises, etc.
My guess is because it matches observations. Same thing with curvy light on a flat planet.
The two models are just one. The explanation remains the same.
Only if you're assuming an orthonormal basis. In an orthonormal basis Australia is broken.I'm not assuming any basis. If the earth is flat, any map of it is a projection of a disk onto a smaller disk; no distortion happens. That means that if I measured the width of Australia to be 3000km (random number, doesn't matter for the argument) and 3000 pixels in a map, the scale is 1km/pixel everywhere. Therefore, measuring 3000 pixels anywhere else on the map means that distance is also 3000km.
The flat world i'm presenting has the equator and the NS-line as axis and coordinates are expressed in degrees (lat/long along these axis, just like a celestial coordinates)
How will you tell?They are a single model, and it was not me who said it, it was you. You took a globe earth, changed coordinates and projected it in a disk. Everything else was made to fit that disk based on the observations made on a globe earth.
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.
I don't get what all the fuss is about, really, unless I've missed something.Based on your reactin, you have probably totally understood.
From a physics viewpoint this model has nothing new to offer.
However in the context of the flat-earth-"debate", it offers a few interesting tidbits:
- There exists a working flat-earth model. People have been looking for this for over 100 years
- The true shape of the earth can never be known. It can be a globe, or flat or a pyramid. There's no way to measure it. It seems people do not always realize this.
But yes, you have probably understood everything correctly and your reaction is very typical for a physics graduate.
But as a general answer, my 3D model is globe physics, just rendered differently. If you take a globe, express it in in celestial coords (lat, long, distance), and then draw latitude on a straight instead of a radial axis, you'll get a flat-earth-universe. Physics works with celestial coords, and so it behaves like a globe, it just looks flat. So if a wave works on a globe, it also works on the flat renderering. The only difference is the shape of one axis.The OP him/herself said that the model is globe earth rendered on a flat disk, with whatever it takes to make the shape of the sun's illumination match the observations, and calling the boundary of the disk a "mathematical artefact". It is not flat earth. The title of the thread is a bit disingenuous.
The milky way is composed mostly of dark matter. If anything, ordinary matter should follow, not the other way around. And it actually does, the density profiles for both baryonic and dark matter are similar (DM halo extends way beyond the visible disk for reasons we have already discussed).There's no gap anywhere, DM is present. It's just that the sun is at the outer edge of the galaxy and therefore the density is lower (for both kinds, compared to regions closer to the center).You got it, DM has no efficient way of losing energy, that's why it's more diffuse. I still don't understand why the fixation with earth; if anything it would be attracted to the sun in the solar system, not the earth (or earth-moon). Besides that, there are millions of other stars way more massive than the sun. Given the collisionless nature of DM, it would most likely follow the denser regions. And models of structure formation predict that there are subhalos within the halo, exactly what you are saying. And we see that the distribution is not totally uniform. But, as we both agreed that DM is collisionless, the substructures formed are larger than that of ordinary matter.The same point works for the Sun too, added mass is even less likely there given it's primarily said to be composed of the lightest elements (and the only way to do away with that is to lose a gaseous Sun altogether, which RET can't do without a lot of revision) so there's no room to remove any as dark matter.
The (possible) reason that there's more DM in the center it's because the halo was formed by gravitational collapse. It grew later on by mergers or pure accretion.
I agree dark matter would form larger haloes with this model, but it's when you get within that that there starts to be an issue. The dark matter that aggregated in areas that affect us, I don't agree that those haloes would be stable. You use the example of one collapsing at galactic center, but what about elsewhere? All the subhalos seem as though they should end up attracted to other massive bodies, and vice versa; for that to instead leave a gap with only one type of matter is headscratching. Sure, regular matter loses energy faster but that's a lot of time for their mutual gravitational pulls to not interact.
You got it, DM has no efficient way of losing energy, that's why it's more diffuse. I still don't understand why the fixation with earth; if anything it would be attracted to the sun in the solar system, not the earth (or earth-moon). Besides that, there are millions of other stars way more massive than the sun. Given the collisionless nature of DM, it would most likely follow the denser regions. And models of structure formation predict that there are subhalos within the halo, exactly what you are saying. And we see that the distribution is not totally uniform. But, as we both agreed that DM is collisionless, the substructures formed are larger than that of ordinary matter.
Dark matter has no limitations on its movement, I agree with that.
My issue still has to be with the effect of gravity on dark matter though. It would not need to form any actual structure, particularly if it collisionless. Take the RE moon and its orbit around the Earth's core; why not have dark matter do the same within the Earth? You could have tonnes of the stuff in the same location. That's a halo, the same structure you're talking about, just less diffuse.
The only reason I can see for it to be as diffuse as you've said is that it lacks an efficient way to lose energy, though they overlaps with the discussion I've had with ScienceThat; it needs to lose energy if it's to be present at all to affect the galaxy, especially galactic center.
I may be misunderstanding what you're saying, but it still doesn't quite seem to work.
The problem is that regular matter would be subject to the same forces; as dark matter was drawn to those fluctuations, it would be too. Sure, it may be less efficient if it occurs in that sweet spot where matter's moved beyond the subatomic, but how could a galaxy form with that halo dragging everything outwards?First, the bold part is wrong. A hollow uniform sphere has zero gravitational force anywhere inside, not just the middle.
I'm aware of the fact that an equal gravitational pull from all directions has net force zero, but that's only going to be the case for a hollow perfect sphere, and even then only for objects right in the middle. No kind of orbit could form around anything when the dust cloud would just be pulled outwards.
A key tenet of RET is dark matter; without such an entity the whole model falls apart. And further, it is true that dark matter is supported by evidence when the world is viewed from the RE perspective, the only way to make sense of the motion of planets and stars (supposedly due to gravity) is by recourse to these dark bodies.Dark matter only is only needed to explain the motion of stars within galaxies. There's absolutely no need for dark matter to explain anything in the solar system, and therefore dark matter is not a key tenet for round earth. If dark matter does not exist in any form and we actually need to revise general relativity, the new theory would have to have GR as its limit when applied to solar systems, much in the same way GR tends to newtonian gravity if we are dealing with small enough velocities and weak gravitational fields.
The earliest reference I can find to dark matter as a vague concept is 1884, though this was very slight.
There are some estimates of dark matter density in the solar system. Even if you take the upper limits, the amount of dark matter on earth would be around 10-18 times the total mass, completely negligible and undetectable. But why didn't DM clump together as ordinary matter, you might ask?
So where is its impact on calculations of the Earth's mass? RET does have excuses, but none of them can explain why it is dark matter fails to be attracted to centers of mass like planets. The moon doesn't simply stop orbiting the Earth just because the Sun or Galactic Center exist. If dark matter exists, it should be drawn to stars, moons, planets, according to RET.
This whole discussion is basically a discussion on semantics but you're wrong. In usual terminology, linear acceleration is any change in velocity, which is a vectorial quantity. Therefore, any rotational movement will have linear acceleration.QuoteThe term linear has nothing to do with the movementThis part is just wrong.
Did you maybe forget that the object in a perfectly circular orbit is still accelerating, at a right angle to its instantaneous velocity, toward the center of the orbit? If it weren't, it would fly off (standard pedantic disclaimer on FoRs) and not be in orbit any longer.No, and that's why I said that any object in uniform circular motion will have a non zero linear acceleration.
Someone correct me if I'm wrong, but as far as I know linear acceleration is just the force divided by the mass, and it's the second derivative of the position vector w.r.t. time. Since there's only one force acting on the ISS and its velocity vector changes with time, it follows that it does have linear acceleration. The term linear has nothing to do with the movement, it's just to differentiate it from angular acceleration, which can be defined as the torque divided by the moment of inertia, or the derivative of the angular velocity w.r.t. time. Any object in uniform circular motion has a non-zero linear acceleration and zero angular acceleration.ur denseNo. The ISS rotates about the RE. Why do you ask? Do you think an object can't rotate about the RE and have linear acceleration?No REer has suggested describing rotational motion as simple linear motion (I think that what you meant by "linearly".).
Your whole schpiel has been that the ISS has linear acceleration relative to the earth. Do you mean to say that its motion relative to the earth isn't rotational?
Think hard about that. Your previous response indicates their is a severe gap in your critical thinking.
I can see why Thork dropped out of this conversation, you people just refuse to read the posts and assume they say something they don't. Angular acceleration is what has been talked about this entire time. Go back through and read the posts and you'll see that.
The ISS doesn't accelerate. Its in low earth orbit at a constant speed of around 17000mph To have a force of acceleration in one direction and an opposing force that cancels that out in the opposite direction does not mean one is accelerating. If I get in my car and accelerate to 150mph before the wind resistance is so great my car stops accelerating doesn't mean I am still accelerating. It means I am now at 150mph and that is that as the forces balanced.He says VERY clearly that the ISS doesn't accelerate and that the forces on it are balanced. If he was talking about angular acceleration, he should have added the word angular, which he did not. Therefore it'safe to assume he was talking in the more general sense of acceleration.
You are confusing angular velocity with angular acceleration.
https://en.wikipedia.org/wiki/Angular_velocity
