Davis Relativity Model (Debate/discussion edition)

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #30 on: October 02, 2017, 01:29:27 AM »
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No we don't. By repeating this experiment with various parameters, we see that the distance changes based upon speed initial speed, height and angle. In some cases Earth is "flat" in other cases it curves downwards, in others it curves upwards.
Again, excluding those in which it isn't flat is just dishonest special pleading.
Did you not read the definition I provided? Here it is:
Did you not read what I said, which I quoted in that post?
Excluding the ones where it is not flat is dishonest special pleading. You assert it is flat, require all space time "straight lines" have it be flat, and thus conclude it is flat.
That is just as good as simply defining Earth as flat and setting everything else of that.

I am taking the geodesics traversing the Earth's surface
But you aren't traversing Earth's surface. You are orbiting significantly above it and you are still ignoring elliptical orbits.
Ignoring that, depending on how you want to define traversing, all paths which don't orbit some other body can be considered to traverse Earth. That is they go across Earth.
Sure, they don't go all the way around Earth, but they do cross it, i.e. traverse it.

Regardless, this says nothing about the shape of Earth itself. It is merely saying your path is straight.
Earth is not on your path, as such, it is not straight.

And remember, your path is a path through space time, so even then, Earth as a whole is not on your path. It merely has a few points which are.

I'll simply ignore your points on non-traversing geodesics as they purposely take this all out of context.
No it doesn't. It simply doesn't dishonestly exclude numerous paths because they don't agree with what they want to show.

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The simplest one which allows you to discuss the shape of Earth, is one in which you consider the shape of Earth in space alone, by removing the time aspect and thus consider an object an infinite velocity.
Which you actually can't do.
Yes you can. You can see the effects of velocity and extrapolate that to infinite velocity.

due to the fact that geodesics traversing the Earth are straight.
No they aren't. Only very specific ones are.

Traversals geodesics are straight regardless of where you are standing on Earth.
No they aren't.

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It does not discuss the motion of the solar system relative to anything else, and it doesn't even have the sun fixed. As such, the entire solar system can be in motion.
Yep, you are missing the point. The motion is relative, including the entire solar system, heliocentrism assumes all objects in the solar system are rotating around the stationary sun (not considering it's relative motion outside of the system of course), that implies objective motion.
No, I'm not missing the point.
Heliocentrism assumes all objects orbit the sun, not that the sun is fixed.

All can, relative to that stationary frame, we just say, 'it moves this way relative to it'.
That is not an explanation. That is a statement.

Not in relativity, it doesn't exist
Even with relativity, it still has explanatory power, unlike a geocentric position.

i.e. assuming objective motion.
Nope, assuming relative motion.

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No it isn't. HC is used and it is translated to a GC frame.
That is the same thing as assuming geocentrism, at least in how I'm using it.
Not how I am using it.
Again, HC explains the motion, GC does not.
GC needs to explain (not state what the motion is, actually provide an explanation for it) why the sun orbits us, but all the other planets orbit the sun.


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Again, HC does not assume absolute motion.

Wikipedia:
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Heliocentrism[1] is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System.
-https://en.wikipedia.org/wiki/Heliocentrism
If that isn't 'absolute', then I don't know what is.
Then you clearly have no idea what absolute is.
Where does it indicate that the sun is fixed? It doesn't.
Instead it simply has the sun at the centre of the solar system.
Does it say the solar system is fixed? No. Even if you want to try and move to an Earth-centred view, the sun is still the centre of the solar system.
Objects still orbit the sun, not us.

The observations we derive in our frame relating to motion are relative between frames of reference, if none other exist, you can only be stationary.
No, it shows the opposite, that you can't be stationary, nothing can be as there is no absolute frame in which you can say that you are still in.

This becomes an acentric relativist explanation of the solar system and universe in the absolute sense.
No, it isn't an explanation, it is merely a way of viewing it.

You didn't address it's validity, so I don't know what you are talking about here.
Yes I did, pointing out how you dishonestly exclude loads of valid geodesics just to pretend Earth is flat, including orbital ones in elliptical orbits where the distance between Earth and the object changes.
I also pointed out that you are just looking at the distance to the point below, not the actual shape of Earth (such as distances to other points on Earth's surface).
The objects in orbit are still orbiting a sphere, not a flat surface, even though they are travelling in a "straight" path through space time.

Because c (speed of light as measured in any inertial reference frame) doesn't apply to non-inertial frames, which includes centrifugal and 'G forces'.
You mean any forces, not just centrifugal or g-forces.
What you really mean is that by being in a non-inertial frame of reference you can pretend distant objects are travelling faster than the speed of light, such as when you turn your head.

You can't meaningfully compare velocities of distant objects directly or compare vectors from distinct points without bringing them to a same origin
No, you can't meaningfully compare them at all in a GR sense because the frames for GR are inherently local.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #31 on: October 02, 2017, 01:31:00 AM »
If you want to ignore most of the quibbling, and just focus on your main point, it all boils down to this:
Provide an explanation for why we should exclude all non-circular orbits (especially given the fact we are not in such an obrit) when deciding the shape of Earth and why we should only focus on the distance between the object in orbit and the point on Earth below it rather than any other point.

If you can't do both of these, then your argument is crap.

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Master_Evar

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #32 on: October 02, 2017, 02:06:41 AM »
Another thing I want to get out there is the difference between flat and flat. In layman terms (and in most scientific fields) it means a 2D plane that isn't curved. In more spatial-mathematical terms, it means any space (2D, 3D, etc.) where two lines which are parallel will stay parallel as you traverse the lines in either direction. Thus a 3 dimensional space can be flat. What this means is that earth is both curved and flat: the surface of the earth, a 2D plane, is curved. If you traverse in parallel with a person starting from the equator going north or south, you will collide at the poles. Earth as a volume is flat however, if you just traverse in any direction regardless of the surface of the earth, so you don't follow the curved surface but go completely straight, you'll be able to stay completely parallel with someone. If you "unpacked" orbits with GR, they'd be straight lines that can be parallel. So yes, space is flat. Not a 2D-plane flat, but "parallel lines will stay parallel given no other outside influences" flat.
« Last Edit: October 02, 2017, 08:00:11 AM by Master_Evar »
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #33 on: October 02, 2017, 04:32:26 AM »
Firstly, it's important to distinguish between spacetime and space. A free-falling object will follow a 'straight' path through curved spacetime, but not, in general, through curved space (unless travelling at light-speed).

Secondly, you seem to be defining 'flat' as meaning 'the surface of a body that can be orbited'. I don't see your description being more restrictive than this. And I don't see this definition being justified.

If I read your justification correctly, it seems to relate to the fact that the surface remains spatially equidistant from a satellite in circular orbit. However the satellite's orbit is not straight in curved space, but in curved spacetime, as noted above. Further, the same logic applied to an elliptical orbit would imply an undulating earth.
As I pointed out, 4D space-time is wrapped into a continuum, an 'aether'.
I don't see the problem you are pointing out here, so I'll assume you aren't.

I'm disappointed you don't see the problem. I'll try to be clearer.

...the definition I provided... Here it is:
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A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates. A “straight line” would be a line in a constant direction in three dimensional space, or any tangent vector on the surface always touching across it would be a flat surface and therefore a flat earth. So, if the Earth is able to be traversed in a straight line, it follows under the definition of a flat plane.
I am taking the geodesics traversing the Earth's surface...

The problem is that there aren't any curved space geodesics that traverse the earth's surface. Curved space-time geodesics, yes, curved space geodesics, no.

Light follows geodesics in curved space. Other free-falling objects such as satellites don't (in general). Since light doesn't traverse the earth's surface, then by your own definition the earth is not flat in curved space.

You could attempt to argue that, because there are space-time geodesics that traverse the earth's surface, then the earth is somehow 'flat' in space-time. But what would that even mean? Space-time is populated by events, not physical surfaces.
« Last Edit: October 02, 2017, 05:55:48 AM by Copper Knickers »

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #34 on: October 07, 2017, 10:11:48 PM »
If you want to ignore most of the quibbling, and just focus on your main point, it all boils down to this:
Yes, I'll just answer your two main points on the Earth being flat with specific geodesics and heliocentrism and relativity.

As I have pointed out in my intro, the Earth's surface is defined by a parallel line traversal with the same distance to the surface.
Imagine we have a surface, and we want to determine the geometry of the surface, but couldn't see it for ourselves. So, we put a rod at a certain length on the surface and drag it across it, with a pen drawing on a paper in it's path. If it's flat, the drawn line would be straight, if it was spherical, that would draw a circle. What it requires is that the tracer is at the same height above the surface throughout it's traversal, and if it say is an undulating surface, the line traversal remaining the same distance above will record the wavy surface.
Applying this to curved space-time ('aether' as I refer to it), an equidistant geodesic reveals a straight line, and therefore gives a flat earth.
I used the term 'parallel' for this to determine our flat earth, which was to essentially mean what is said here.


I understand that heliocentrism doesn't assume a stationary sun, that's why I added "not considering it's relative motion outside of the system of course", but I guess that wasn't clear enough.
My point was that heliocentrism assumes that motion is absolute, not relative between frames of reference but rather the Earth objectively spinning and orbiting a sun as if it's in motion but the sun is mostly stationary in terms of the solar system itself. Yes, as I already pointed out, heliocentrism is most simple of a coordinate system of movements and as you said, it has great explanatory power given gravity and all the other basic phenomena in modern physics.
However, my main claim is that motion as a phenomena is something that is relative between frames of reference, it's intrinsically acentric, and that is an explanation (not just a claim), since it explains motion and center points of rotation in our universe. With this, I don't need to define a barycenter point of a solar system since I'm claiming it's relative. I don't accept heliocentrism, heliocentrism is only needed if you assume motion is absolute and the forces of physics can only work plausibly with a sun centered solar system.

I did bring up geocentrism as a chosen frame but now figure that I should abandon that explanation for better clarity and consistency, I'll think about it that way for the future.

Anyways, I just wanted to see what people thought of this, I found it and looked into it and it's the model I accept. I wanted to see some criticisms and feedback of it to see how valid it may be as an explanation. I got some and have been thinking it over. Thanks.

I'll conclude this post with a quote that points something out with a relativity universe:
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Theories can only be shown to be false, they can never be proven. Such is science and its methods. Don’t listen to the layman.

Besides, heliocentricity has no meaning in a relativistic universe. So, no need to believe it one way or another.
-https://www.quora.com/What-will-happen-if-Heliocentric-Model-is-proved-false
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #35 on: October 07, 2017, 10:14:59 PM »
Another thing I want to get out there is the difference between flat and flat. In layman terms (and in most scientific fields) it means a 2D plane that isn't curved. In more spatial-mathematical terms, it means any space (2D, 3D, etc.) where two lines which are parallel will stay parallel as you traverse the lines in either direction. Thus a 3 dimensional space can be flat. What this means is that earth is both curved and flat: the surface of the earth, a 2D plane, is curved. If you traverse in parallel with a person starting from the equator going north or south, you will collide at the poles. Earth as a volume is flat however, if you just traverse in any direction regardless of the surface of the earth, so you don't follow the curved surface but go completely straight, you'll be able to stay completely parallel with someone. If you "unpacked" orbits with GR, they'd be straight lines that can be parallel. So yes, space is flat. Not a 2D-plane flat, but "parallel lines will stay parallel given no other outside influences" flat.
Space is flat? So I reckon that you aren't claiming that Earth is flat in non-euclidean space but rather a spheroid in flat space (like Minkowski Space)?
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #36 on: October 07, 2017, 10:31:06 PM »
Firstly, it's important to distinguish between spacetime and space. A free-falling object will follow a 'straight' path through curved spacetime, but not, in general, through curved space (unless travelling at light-speed).

Secondly, you seem to be defining 'flat' as meaning 'the surface of a body that can be orbited'. I don't see your description being more restrictive than this. And I don't see this definition being justified.

If I read your justification correctly, it seems to relate to the fact that the surface remains spatially equidistant from a satellite in circular orbit. However the satellite's orbit is not straight in curved space, but in curved spacetime, as noted above. Further, the same logic applied to an elliptical orbit would imply an undulating earth.
As I pointed out, 4D space-time is wrapped into a continuum, an 'aether'.
I don't see the problem you are pointing out here, so I'll assume you aren't.

I'm disappointed you don't see the problem. I'll try to be clearer.

...the definition I provided... Here it is:
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A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates. A “straight line” would be a line in a constant direction in three dimensional space, or any tangent vector on the surface always touching across it would be a flat surface and therefore a flat earth. So, if the Earth is able to be traversed in a straight line, it follows under the definition of a flat plane.
I am taking the geodesics traversing the Earth's surface...

The problem is that there aren't any curved space geodesics that traverse the earth's surface. Curved space-time geodesics, yes, curved space geodesics, no.

Light follows geodesics in curved space. Other free-falling objects such as satellites don't (in general). Since light doesn't traverse the earth's surface, then by your own definition the earth is not flat in curved space.

You could attempt to argue that, because there are space-time geodesics that traverse the earth's surface, then the earth is somehow 'flat' in space-time. But what would that even mean? Space-time is populated by events, not physical surfaces.
4D Space-time is linked in one continuum in General Relativity, and have explicitly pointed out that the model I accept depends on this specific conception of space-time.

The geodesics don't depend on the mass, but light has a much higher velocity than any object you would be able to traverse Earth with, and so has an escape velocity that exceeds the warp of space-time of Earth, it doesn't go against the claim here at all.

That last point is closer to it. A geodesic with an equidistant traversal will be straight in curved space-time, which follows a flat earth. Space-time being populated by event isn't really relevant to this.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #37 on: October 07, 2017, 10:42:58 PM »
As I have pointed out in my intro, the Earth's surface is defined by a parallel line traversal with the same distance to the surface.
You are not doing it based upon a parallel line traversal.
You are doing it based upon the distance from an object following one out of highly specific set of space-time geodesics to the surface of Earth directly below.

This highly specific set are those which remain equidistant from the surface of Earth.
As such your argument is entirely circular. You reject all geodesics which do not result in a flat Earth to try to conclude Earth is flat. It is dishonest garbage, nothing more.

That does not make it a parallel line. That does not make the surface of Earth "straight".

Imagine we have a surface, and we want to determine the geometry of the surface, but couldn't see it for ourselves.
So ignore every other way to do it, like having the rod at a point and pivoting it to measure the distance to the surface.
The big issue with your method is that it is purely 2D. You can't get the third dimension.

So, we put a rod at a certain length on the surface and drag it across it, with a pen drawing on a paper in it's path. If it's flat, the drawn line would be straight, if it was spherical, that would draw a circle.
This method requires you to rotate the rod as well, so you aren't simply having a rod and measuring the distance to the surface.

Applying this to curved space-time ('aether' as I refer to it), an equidistant geodesic reveals a straight line, and therefore gives a flat earth.
And using this method applying it to a non-equidistant geodesic produces numerous shapes. Some, such as those for elliptical orbits, are undulating.
Others, for hyperbolic orbits, have Earth as a peak.
Others, for a non-orbital path (or an elliptical orbit that hits Earth), has it as a valley.

I understand that heliocentrism doesn't assume a stationary sun, that's why I added "not considering it's relative motion outside of the system of course", but I guess that wasn't clear enough.
My point was that heliocentrism assumes that motion is absolute
No it doesn't.

not relative between frames of reference but rather the Earth objectively spinning and orbiting a sun as if it's in motion but the sun is mostly stationary in terms of the solar system itself
No, it uses the solar system as the frame of reference.
If it was going to be absolute it would have the sun absolutely stationary (or just moving a little about an absolutely fixed point in space).

You are also completely contradicting yourself. You seem to be indicating that it has absolute motion, except this motion is relative to the solar system (i.e. not considering its relative motion outside of the system)

it has great explanatory power given gravity and all the other basic phenomena in modern physics.
And that is the key point. It has the explanatory power. GC does not.

that is an explanation (not just a claim), since it explains motion and center points of rotation in our universe.
You might want to learn what an explanation is.
How does a GC reference explain the motion of the stars?

I don't need to define a barycenter point of a solar system since I'm claiming it's relative.
It doesn't need to be absolute to be defined.

I don't accept heliocentrism, heliocentrism is only needed if you assume motion is absolute and the forces of physics can only work plausibly with a sun centered solar system.
Or if you want an actual explanation for the motion rather than just saying look it moves.

I'll conclude this post with a quote that points something out with a relativity universe:
You mean which baselessly asserts it.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #38 on: October 07, 2017, 10:45:45 PM »
A geodesic with an equidistant traversal will be straight in curved space-time.
Every geodesic is "straight" in curved space-time. Not just those which remain equidistant from an object.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #39 on: October 07, 2017, 10:58:58 PM »
Also, a bit more on parallel lines, in non-Euclidean spaces (including curved space-time), Parallel lines do not necessarily remain the same distance apart, and they can be considered to not be parallel at all bar 1 (or a few) point(s) in some cases.

For example, consider 2 circular orbits around Earth. Both starting at the equator. Both heading due north. Both at the same altitude. Both at the same velocity. But separated by 90 degrees (so one is above 0 degrees east, the other is above 90 degrees east).
These start off parallel, heading in the same direction, perpendicular to the line connecting them.
But as they continue along their orbit (their "straight" paths), they get closer together, colliding at the north pole.

But don't worry, it gets more fun.
Now consider 4 objects above Earth. All heading due north (with no change in elevation). All above the equator at 0 degrees east. All at the same velocity. But now one is in a perfectly circular orbit, while another is significantly below it, in an elliptical orbit, with the other 2 above, one in an elliptical orbit and one in a hyperbolic orbit.

All 4 of these lines are currently parallel because they are perpendicular to the line connecting them, but let them move on, and their paths are completely different.

In order to show something is straight because it remains the same distance to a straight line you need to be in flat space.
If you aren't, you could be describing a curve.

As a real life example of this, in spherical geometry (e.g. Earth's surface), a geodesic is a great circle. This makes the equator a geodesic, and thus a "straight" line. If you go north of the equator, the geodesic starts out heading due east, but then appears to curve south, cross the equator, starts curving back north, goes back to being due east, continues curving north, crosses the equator again ands starts curving south again to meet up with the original point.

However, you can easily go around Earth staying 1000 km north of the equator, that same distance. Yet that path is not a geodesic. It requires turning.

So that is yet another reason your method is fundamentally flawed. You are trying to use a concept which only works in flat spaces in non-flat space-time.

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #40 on: October 07, 2017, 11:29:23 PM »
You are not doing it based upon a parallel line traversal.
I guess on a sphere, it isn't a 'line' but it's parallel in that the traversal line is at a fixed normal distance to the curve, same with a flat earth, as I have implied here.

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You are doing it based upon the distance from an object following one out of highly specific set of space-time geodesics to the surface of Earth directly below.
This highly specific set are those which remain equidistant from the surface of Earth.
As such your argument is entirely circular. You reject all geodesics which do not result in a flat Earth to try to conclude Earth is flat.
They are the only geodesics that define the Earth's surface, because they are essentially parallel, mapping it out.

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It is dishonest garbage, nothing more.
As I have expected you'd say.

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That does not make it a parallel line. That does not make the surface of Earth "straight".
Reject it all you want, I believe I made my point pretty well whether you agree with it or not.

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So ignore every other way to do it, like having the rod at a point and pivoting it to measure the distance to the surface.
The big issue with your method is that it is purely 2D. You can't get the third dimension.
It's an illustrative example for simplicity, I'm not basing my point on it specifically as a method.

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This method requires you to rotate the rod as well, so you aren't simply having a rod and measuring the distance to the surface.
Perhaps I wasn't clear, the distance to the surface

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And using this method applying it to a non-equidistant geodesic produces numerous shapes. Some, such as those for elliptical orbits, are undulating.
Others, for hyperbolic orbits, have Earth as a peak.
Others, for a non-orbital path (or an elliptical orbit that hits Earth), has it as a valley.
Except, none of these are parallel and so are inaccurate representations of mapping Earths surface geometry.

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No it doesn't.
If you proclaim Heliocentrism in the solar system as objectively correct, then yes it does.
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No, it uses the solar system as the frame of reference.
If it was going to be absolute it would have the sun absolutely stationary (or just moving a little about an absolutely fixed point in space).
I already pointed out that it is in terms of the solar system itself, I'm not considering anything outside of the solar system here. The point is that heliocentrism assumes that the motion of Earth and the other planets is orbiting the sun in an absolute sense (objective motion), not just relative motions and picking a frame (of course you could do that AND proclaim heliocentrism, but by stating heliocentrism is objectively the motions of our solar system only means you deny that all motions are relative to frames of reference).
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You are also completely contradicting yourself. You seem to be indicating that it has absolute motion, except this motion is relative to the solar system (i.e. not considering its relative motion outside of the system)
What I mean is that in viewing the solar system as an exclusive system, heliocentrism claims that the bodies in the system are orbiting the sun (that's absolute motion to say it is orbiting the sun rather than the sun orbiting them).

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And that is the key point. It has the explanatory power. GC does not.
And I'll grant you that it does, but motion is not absolute, so Heliocentrism is not needed in a relativistic universe.

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You might want to learn what an explanation is.
How does a GC reference explain the motion of the stars?
I said acentric, there is no center, it's relative motion's between frames of reference.
GC would involve other fictitious forces and modified laws such as Mach's principle to explain a rotating universe and a stationary earth.
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It doesn't need to be absolute to be defined.
True, but if it's not absolute, then it's not needed in an explanation of the universe since it is relative to the frames of references in existence.
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Or if you want an actual explanation for the motion rather than just saying look it moves.
Yeah, the motion is relative to the distinct reference frames, I don't need a set up a coordinate system to explain how celestial motions are in this model.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #41 on: October 07, 2017, 11:53:59 PM »
Also, a bit more on parallel lines, in non-Euclidean spaces (including curved space-time), Parallel lines do not necessarily remain the same distance apart, and they can be considered to not be parallel at all bar 1 (or a few) point(s) in some cases.

For example, consider 2 circular orbits around Earth. Both starting at the equator. Both heading due north. Both at the same altitude. Both at the same velocity. But separated by 90 degrees (so one is above 0 degrees east, the other is above 90 degrees east).
These start off parallel, heading in the same direction, perpendicular to the line connecting them.
But as they continue along their orbit (their "straight" paths), they get closer together, colliding at the north pole.
Yeah, Skew lines, I don't see how they are relevant to the point being discussed here.
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But don't worry, it gets more fun.
Now consider 4 objects above Earth. All heading due north (with no change in elevation). All above the equator at 0 degrees east. All at the same velocity. But now one is in a perfectly circular orbit, while another is significantly below it, in an elliptical orbit, with the other 2 above, one in an elliptical orbit and one in a hyperbolic orbit.

All 4 of these lines are currently parallel because they are perpendicular to the line connecting them, but let them move on, and their paths are completely different.
Then doesn't that mean the geodesics they follow are not parallel?
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In order to show something is straight because it remains the same distance to a straight line you need to be in flat space.
If you aren't, you could be describing a curve.
No, you wouldn't, defining vectors for a straight line work just as fine in curved space-time as flat. The definition doesn't change in non-euclidean space-time.
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However, you can easily go around Earth staying 1000 km north of the equator, that same distance. Yet that path is not a geodesic. It requires turning.
Yeah, it'd be accelerating in that case, which the geodesic path of a satellite doesn't.
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So that is yet another reason your method is fundamentally flawed.

I haven't seen it yet, this doesn't help.
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You are trying to use a concept which only works in flat spaces in non-flat space-time.
The Earth is flat in curved (non-flat) space-time, that has been my point here.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #42 on: October 08, 2017, 01:02:37 AM »
I guess on a sphere, it isn't a 'line' but it's parallel in that the traversal line is at a fixed normal distance to the curve, same with a flat earth, as I have implied here.
Nope, it being on a sphere has nothing to do with it.


They are the only geodesics that define the Earth's surface, because they are essentially parallel, mapping it out.
This means all you are doing is defining Earth as flat.
You aren't justifying it at all. You just are just asserting it is flat and rejecting anything which would indicate it isn't.

You need to justify why you need to only use these "parallel" ones.

Reject it all you want, I believe I made my point pretty well whether you agree with it or not.
Believe whatever garbage you want. You have not made your point, unless your point was that Earth must be flat and that only valid measurement is one which results in Earth being flat.

It is pure dishonest, circular garbage. Nothing more.

It's an illustrative example for simplicity, I'm not basing my point on it specifically as a method.
Yes, it is an illustrative example, one of many ways of measuring a surface. But you demand this one particular way because it is the only way you can pretend Earth is flat using curved space-time.

So no, you are basing your point on it as a method. Sure, you don't need a literal rod, but you need this method.

If you use any other method, such as measuring the distance to all the points on the surface, you get Earth being round.

Except, none of these are parallel and so are inaccurate representations of mapping Earths surface geometry.
Who gives a shit if they are parallel? Why must they be parallel to map Earth's surface? Why can't you simply use the distance?

You are literally rejecting any measurement of the surface of Earth which doesn't show it is flat with no rational justification at all.
Is it really surprising that I would call that dishonest garbage? Is that why you expected me to say it, because you know it is dishonest garbage?

Also, all of those options do go parallel to Earth at one point. As i explained, parallel lines only stay parallel in flat space (or in a flat path in curved-space, such as along the sides in cylindrical space).

If you proclaim Heliocentrism in the solar system as objectively correct, then yes it does.
No, it doesn't. If you want to claim it requires absolute motion, PROVE IT!
You even admitted the motion does not need to be absolute and that the Solar system can be moving.

I already pointed out that it is in terms of the solar system itself, I'm not considering anything outside of the solar system here.
i.e. you are using relative motion, where it is motion relative to the solar system. You are not considering absolute motion and thus HC does not need absolute motion.
Have you refuted yourself enough yet?

The point is that heliocentrism assumes that the motion of Earth and the other planets is orbiting the sun in an absolute sense
No it doesn't.
It has the planets orbiting the sun in a relative sense and is capable of explaining that orbit.

What I mean is that in viewing the solar system as an exclusive system
i.e. as a relative frame of reference and thus not containing any absolute motion but only relative motion.

And I'll grant you that it does, but motion is not absolute, so Heliocentrism is not needed in a relativistic universe.
It is if you want an explanation.

GC would involve other fictitious forces and modified laws such as Mach's principle to explain a rotating universe and a stationary earth.
I already pointed out that Mach's principle wouldn't work (and explained why).
If you are going to try repeating the same baseless bullshit you need to go and show what is wrong with my explanation (nothing).
And if you need fictitious force while other frames do not you are admitting that this reference frame does not explain the motion, that another reference frame does, and that you need to invoke this force to take the explanation from the other reference frame to this one.

Yeah, the motion is relative to the distinct reference frames, I don't need a set up a coordinate system to explain how celestial motions are in this model.
Yes you do.
Go and set up a GC model which accurately explains the motions of the planets.

Yeah, Skew lines, I don't see how they are relevant to the point being discussed here.
No. Not skew lines. "Parallel" lines in spherical space.
They are relevant to the point because you are trying to use parallel lines to pretend something is flat, when that would only work in flat space.

Then doesn't that mean the geodesics they follow are not parallel?
No, they are parallel, as they started parallel, going perpendicular to the line joining them.

As soon as you leave Euclidean geometry, "parallel" starts to get strange, just like "straight".

No, you wouldn't, defining vectors for a straight line work just as fine in curved space-time as flat. The definition doesn't change in non-euclidean space-time.
No, the vectors change a fair bit.
You need to use vectors which represent the space and are intrinsically local. You can't simply translate it and expect the vector to remain the same.

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However, you can easily go around Earth staying 1000 km north of the equator, that same distance. Yet that path is not a geodesic. It requires turning.
Yeah, it'd be accelerating in that case, which the geodesic path of a satellite doesn't.
The satellite is the equator. Earth is the line 1000 km north of the equator.

To complete the analogy:
You are travelling along the equator, in a "straight" line.
You observe a surface (say a wall for example) 1000 km north of you.
You measure it's distance, but only to the point due north of you.
You circle the planet, going all the way around the equator, and effectively "traversing" the wall.
It has remained 1000 km away from you.

Is it straight?
NO!!!
That is because your parallel lines arguments only works in flat space.

I haven't seen it yet, this doesn't help.
Is that because you are intentionally ignoring it?

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You are trying to use a concept which only works in flat spaces in non-flat space-time.
The Earth is flat in curved (non-flat) space-time, that has been my point here.
The space is the part that is important. You need Eart to be in a flat space (in the mathematical sense where curved space time is a non-flat space).
As Earth is not in a flat space (it is in curved space-time), you cannot use your method to determine if Earth is flat or not.
« Last Edit: October 08, 2017, 01:06:51 AM by JackBlack »

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #43 on: October 08, 2017, 01:07:13 AM »
Again, a brief post highlighting the killer of your argument if you just want to try focusing on that:


An analogy:
You are travelling along the equator, in a "straight" line.
You observe a surface (say a wall for example) 1000 km north of you.
You measure it's distance, but only to the point due north of you.
You circle the planet, going all the way around the equator, and effectively "traversing" the wall.
It has remained 1000 km away from you.

Is it straight?
NO!!!
That is because your parallel lines arguments only works in flat space.

(and as a bit of an extension):
If that wall was "straight" it would cross the equator twice. The distance between it and you would change.

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AltSpace

  • Flat Earth Believer
  • 411
  • Neo-Planarist
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #44 on: October 08, 2017, 03:20:24 AM »
This means all you are doing is defining Earth as flat.
You aren't justifying it at all. You just are just asserting it is flat and rejecting anything which would indicate it isn't.
It fits the definition as provided in non-euclidean space, so I conclude it's flat.
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You need to justify why you need to only use these "parallel" ones.
Because they are the only ones that define Earth's surface vector geometry, they have to remain an equal distant from the surface like parallel lines or else your traversal line isn't adequate to define the surface.
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Yes, it is an illustrative example, one of many ways of measuring a surface. But you demand this one particular way because it is the only way you can pretend Earth is flat using curved space-time.
And it's the only way that is gonna reveal to us the shape of the surface of Earth. This is accurate way since whatever is 'parallel' in this case mirrors the surface.
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If you use any other method, such as measuring the distance to all the points on the surface, you get Earth being round.
It only shows a flat earth as a parallel geodesic is flat, and so therefore is the Earth.
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Who gives a shit if they are parallel? Why must they be parallel to map Earth's surface? Why can't you simply use the distance?
Because if you have a surface parallel to a traversal that is a straight line, it shows the surface must match that straight line and therefore be flat. That's basic common sense.
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Also, all of those options do go parallel to Earth at one point. As i explained, parallel lines only stay parallel in flat space (or in a flat path in curved-space, such as along the sides in cylindrical space).
But they don't follow parallel paths through their traversal, so they don't work.
And no, that is incorrect, 'parallel' as in points are at a fixed normal distance from a given surface applies to the curved space-time I'm talking about too, not just flat space.
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No, it doesn't. If you want to claim it requires absolute motion, PROVE IT!
You even admitted the motion does not need to be absolute and that the Solar system can be moving.
I already did, it's simple.
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Heliocentrism[1] is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System.
-https://en.wikipedia.org/wiki/Heliocentrism
By saying this is correct, you are implying that the motions of the Earth and planets revolving around the sun are objective motions.
I'm accepting that it's not absolute and I am not considering the solar system moving relative to the other galaxies, I'm leaving those out for this purpose, but you thought I wasn't at first, so I wanted to make this clear.
I already pointed out that it is in terms of the solar system itself, I'm not considering anything outside of the solar system here.
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i.e. you are using relative motion, where it is motion relative to the solar system. You are not considering absolute motion and thus HC does not need absolute motion.
Right, because Heliocentrism is not an objective phenomena at all, it's simply our frame of reference relative to the sun and other frames.
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No it doesn't.
It has the planets orbiting the sun in a relative sense and is capable of explaining that orbit.
Then you agree that Heliocentrism is not objective and simply a coordinate system with specific forces in a model?
Quote
i.e. as a relative frame of reference and thus not containing any absolute motion but only relative motion.
Then it's confirmed that there was a misunderstanding, you do accept relative motion but desire a heliocentric coordinate system for your explanatory purposes.
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It is if you want an explanation.
The explanation is that reference frames throughout the universe are in motion relative to the other, it's acentric because there is no objective center, but rather motion as observed from a frame of reference.
You are assuming that an explanation is required in terms of a set coordinate system with whatever forces and laws are required to sustain it, but I'm looking at the broader picture that is independent of our abstract systems and explaining how the universe works, all motion is relative between reference frames. I don't need a motion coordinate system to explain a non-absolute/non-objective phenomena, assuming that would mean you assume that there is objective/absolute motion.
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I already pointed out that Mach's principle wouldn't work (and explained why).
If you are going to try repeating the same baseless bullshit you need to go and show what is wrong with my explanation (nothing).
Really? I haven't seen anything other than that the Earth would supposedly rotate with the universe, and I pointed out why that was wrong.
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Yes you do.
Go and set up a GC model which accurately explains the motions of the planets.
I was thinking about that, but no. The relativity universe doesn't depend on any centric frame for motions to be accurately explained as being relative.
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They are relevant to the point because you are trying to use parallel lines to pretend something is flat, when that would only work in flat space.
Ah, I think I see your point now. So, you mean it only works in 2D space, but not 3D space since parallel lines can be pointing in completely different directions?
If so, then that is simply incorrect, the geodesic being parallel to the surface of the Earth (same distance from it across at every point) simply only requires it traverses the Earth and be parallel to the surface, the directions able to be travelled in three dimensions doesn't affect the requirements of a flat surface in curved space-time.
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No, they are parallel, as they started parallel, going perpendicular to the line joining them.
But the geodesics they travel aren't parallel, so that point doesn't matter either way.
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No, the vectors change a fair bit.
You need to use vectors which represent the space and are intrinsically local. You can't simply translate it and expect the vector to remain the same.
Now that I figured out you meant 2d space by 'flat', yes, it would change then, however, if I cut a flat 2D part of earth out that is directly below a parallel geodesic, that straight line defines a flat earth regardless of how many of these parts I can cut out in 3D space.

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The satellite is the equator. Earth is the line 1000 km north of the equator.

To complete the analogy:
You are travelling along the equator, in a "straight" line.
You observe a surface (say a wall for example) 1000 km north of you.
You measure it's distance, but only to the point due north of you.
You circle the planet, going all the way around the equator, and effectively "traversing" the wall.
It has remained 1000 km away from you.

Is it straight?
NO!!!
That is because your parallel lines arguments only works in flat space.
I fail to see the relevance and how this connects to my claim. Yes, in this case, the parallel line would curve while the other would be straight. However, with the surface of the Earth as parallel to a straight line above it, I can cut out the area the line intersects through and it fits as represented in 2D while originating in 3D space.
If I took a sphere and cut it in half and took the hemisphere I removed and cut off the top but left part of the curved surface there and both sides represented parallel lines just like in this scenario, it would necessarily have to be represented in 3D spatial coordinates. However, if, as is being done by me, I cut through a parallel line above the surface and to the surface, such could be represented in 2D like brought up with the example I had earlier. This is because one has depth curvature included with length and width, while the other (what I am using), has depth and length, but since it uses a line, no width to it. So, this doesn't even apply here.
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Is that because you are intentionally ignoring it?
No, I'm considering claims and rebuttals brought up here, and you have yet to bring any working rebuttal so far.
Quote
The space is the part that is important. You need Eart to be in a flat space (in the mathematical sense where curved space time is a non-flat space).
As Earth is not in a flat space (it is in curved space-time), you cannot use your method to determine if Earth is flat or not.
Which is false, the straight parallel lines through curved space-time traversing Earth demonstrate it's flat regardless of any direction it goes to traverse the earth in a straight line through curved space-time (it is the same thing regardless of whether separate parallel geodesics may intersect each other as all will give the same flat straight line surface of Earth).
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #45 on: October 08, 2017, 03:50:08 AM »
It fits the definition as provided in non-euclidean space, so I conclude it's flat.
You mean it fits the definition you provided to limit it to being flat.

I'm going to skip through most of the BS and just get straight to chase, demanding you answer a simple thing before moving on.
Ah, I think I see your point now. So, you mean it only works in 2D space, but not 3D space since parallel lines can be pointing in completely different directions?
No. My point is that it only works in Euclidean space (aka flat space). It does not work in non-Euclidean or non-flat space.

Euclidean space (or spaces which have Euclidean parts) are the only ones in which 2, straight, parallel lines remain the same distance apart.

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The satellite is the equator. Earth is the line 1000 km north of the equator.
To complete the analogy:
You are travelling along the equator, in a "straight" line.
You observe a surface (say a wall for example) 1000 km north of you.
You measure it's distance, but only to the point due north of you.
You circle the planet, going all the way around the equator, and effectively "traversing" the wall.
It has remained 1000 km away from you.
Is it straight?
NO!!!
That is because your parallel lines arguments only works in flat space.
I fail to see the relevance and how this connects to my claim.
It's pretty simple to see the relevance.

There is some line of "unknown" shape. (In reality we know the shape and we know it isn't straight.)
There is a reference line we can use, a "straight" line, or a geodesic in this curved space.
So, we move along the "straight" line, measuring the distance to the "unknown" line.
We find the "unknown" line remains the same distance.
According to your reasoning this means the "unknown" line, as it remains the same distance from a straight line, should be parallel to this straight line and thus straight.

The issue is that that only works in flat space.

Yes, in this case, the parallel line would curve while the other would be straight. However, with the surface of the Earth as parallel to a straight line above it
The 2 are equivalent.
The satellite, the geodesic through space time is equivalent to the equator.
The surface of Earth is equivalent to the line 1000 km north of the equator.

In order for your method to work, the line 1000 km north of the equator must be a straight line.

Do you think a line around Earth which remains 1000 km north of the equator is a straight line, or do you accept it curves?
(Or to put it another way:
Do you think a line that remains "equidistant" (that is a line connecting the 2 lines, which is perpendicular to the reference line remains the same length) from a straight line is straight, even in non-flat space, or do you accept that it can curve?)

Deal with this before trying to move on.
« Last Edit: October 08, 2017, 03:53:22 AM by JackBlack »

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #46 on: October 08, 2017, 04:25:45 AM »
As for the rest of your crap (which I will ignore any responses to until you answer the above question)

Because they are the only ones that define Earth's surface vector geometry, they have to remain an equal distant from the surface like parallel lines or else your traversal line isn't adequate to define the surface.
BULLSHIT!
Any line can be used to define it.

Lets take a simple case of a sine wave, such as one of the form y=sin(x)+10.

We can use the x axis (y=0) as a reference.
In this case, the curve alternates between 9 and 11 units above it.
We can define it quite precisely.
So no, we do not need a line to remain the same distance.

That is your pathetic excuse to ignore all the other lines which show that Earth is not flat.

You need to actually justify your claim, not just spout more crap.

And it's the only way that is gonna reveal to us the shape of the surface of Earth.
No it isn't. It is the only way you can try to justify pretending Earth is flat.
And because of that, you will reject all the others because you think if they don't show Earth is flat they must be wrong.

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If you use any other method, such as measuring the distance to all the points on the surface, you get Earth being round.
It only shows a flat earth as a parallel geodesic is flat, and so therefore is the Earth.
That doesn't address the point at all. I have also shown that that is not the case at all.

Because if you have a surface parallel to a traversal that is a straight line, it shows the surface must match that straight line and therefore be flat. That's basic common sense.
For flat-space it is. Not for curved space.
Likewise, if you have a straight line which oscillates between 2 distances, that shows it cannot match the straight line and cannot be flat (which again is only true for flat space), and likewise, that is common sense. If the distance from a straight reference line is not a simple linear relation (e.g. y=mx+b), the line can't be straight.

Quote
Also, all of those options do go parallel to Earth at one point. As i explained, parallel lines only stay parallel in flat space (or in a flat path in curved-space, such as along the sides in cylindrical space).
But they don't follow parallel paths through their traversal, so they don't work.
Again, they are parallel. There is a line going through all of them, which is perpendicular to all of them. That means they are parallel. So why don't they work?

And no, that is incorrect, 'parallel' as in points are at a fixed normal distance from a given surface applies to the curved space-time I'm talking about too, not just flat space.
So parallel as in not actually parallel and instead remaining the same distance apart. While the 2 are equivalent in flat space, they are not in non-flat spaces and in non-flat spaces, a line which remains the same distance away from a straight line will likely be curved.


I already did, it's simple.
No you didn't. It is quite difficult to prove something which is clearly false.

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Heliocentrism[1] is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System.
-https://en.wikipedia.org/wiki/Heliocentrism
Nothing in here indicates the motion is absolute rather than relative.
In order for this to be absolute the centre of the solar system needs to be fixed, absolutely.

I'm accepting that it's not absolute and I am not considering the solar system moving relative to the other galaxies, I'm leaving those out for this purpose, but you thought I wasn't at first, so I wanted to make this clear.
Again, by accepting that the centre of the solar system can move, IT MEANS IT ISN'T ABSOLUTE!!! It isn't a difficult concept to grasp.

This is basically what you are saying:
"I accept that it doesn't need absolute motion, but relative to the reference frame of the solar system, it must be absolute (i.e. relative to the solar system)."

Notice the stupidity there?

Then you agree that Heliocentrism is not objective and simply a coordinate system with specific forces in a model?
Define objective.
It seems to be objective, not based upon any sentient entity, and it is capable of explaining the motions.

but desire a heliocentric coordinate system for your explanatory purposes.
Because it can explain it.

The explanation is that reference frames throughout the universe are in motion relative to the other
This is a statement, not an explanation.
You need to provide an explanation for why the planets move.

I don't need a motion coordinate system to explain a non-absolute/non-objective phenomena, assuming that would mean you assume that there is objective/absolute motion.
Stop trying to pretend objective and absolute mean the same thing.

You need to be able to describe the interactions between various entities to explain how objects move. You cannot do that from an Earth centred frame.

Really? I haven't seen anything other than that the Earth would supposedly rotate with the universe, and I pointed out why that was wrong.
You mean you baselessly dismissed it without any rational justification, and thus you are ignoring everything that shows it to be wrong.

I was thinking about that, but no.
So you can't explain the motion of anything? Good job.

Ah, I think I see your point now. So, you mean it only works in 2D space, but not 3D space since parallel lines can be pointing in completely different directions?
No. I am saying it only works in flat space, not non-flat space. Even in 3D flat space, parallel lines point in the same direction.
In flat space a line needs 2 things to define it, a vector and a point. From that point you follow that vector (or the reverse) and you have your line. That vector is effectively the direction the line is pointing.
If you have 2 lines, which do not overlap, and have the same vector, they are parallel.

You seem to like going off on tangents to cover up your complete failure.
How about you stop setting up such pathetic strawmen and instead focus on what I have actually said?

But the geodesics they travel aren't parallel, so that point doesn't matter either way.
Yes they are.
There is a line joining them which is perpendicular to each. They do not intersect. They are parallel. They just don't have the property of parallel lines which exists in flat space, that they remain the same distance. As they are no longer in flat space the distance between these parallel lines can vary.

Now that I figured out you meant 2d space by 'flat'
No. I meant flat, i.e. Euclidean, i.e. parallel lines remain the same distance apart; there is only a single straight line through a point, parallel to another straight line; the sum of the internal angles of a triangle is 180 degrees; etc.
This is different from non-flat spaces, where instead parallel lines can converge or diverge; there can be none (depending on the definition), 1 or infinitely many "straight" lines through a point, parallel to another straight line; the sum of the internal angles of a triangle can be more or less than 180 degrees.
Stop trying to pretend I am saying something else. Start addressing what I have actually said.


I fail to see the relevance and how this connects to my claim.
That is because you don't want to admit you were wrong.
It shows how you can have a line remain equidistant from a straight line, yet not be straight.
This is because it is not in flat space.

No, I'm considering claims and rebuttals brought up here, and you have yet to bring any working rebuttal so far.
I have brought plenty, which you just dismiss/ignore.

You are yet to bring any rational, honest defence of your claims.

Which is false, the straight parallel lines through curved space-time traversing Earth demonstrate it's flat
No it doesn't.
Again, that requires the space to be flat. If you don't have flat space that entire line of reasoning is baseless. It relies upon a fact of mathematics which is only valid for flat spaces, that parallel lines always remain the same distance apart.

it is the same thing regardless of whether separate parallel geodesics may intersect each other
Yes, it is the same, by which I mean completely false. But this does help show that.
The fact that you get parallel geodesics which intersect each other shows that 2 lines do not need to remain the same distance to be parallel and thus straight.
It shows that the distance between straight lines varies, and thus you can have a line which is not straight remain the same distance from a line which is straight.

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Master_Evar

  • 3381
  • Well rounded character
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #47 on: October 08, 2017, 08:49:27 AM »
Space is flat? So I reckon that you aren't claiming that Earth is flat in non-euclidean space but rather a spheroid in flat space (like Minkowski Space)?
Yes, though technically it could be flat (well, mostly flat) in non-euclidean space as long as there are at least 3 spatial dimensions, and we would have literally no way of telling. It's a moot point.
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #48 on: October 08, 2017, 01:09:28 PM »
Space is flat? So I reckon that you aren't claiming that Earth is flat in non-euclidean space but rather a spheroid in flat space (like Minkowski Space)?
Yes, though technically it could be flat (well, mostly flat) in non-euclidean space as long as there are at least 3 spatial dimensions, and we would have literally no way of telling. It's a moot point.
I disagree.
A key part is light, or simply anything which can be deemed "straight".
If space was Euclidean, you would be able to project 2 parallel beams of light, i.e. get 2 straight lines, and have them remain the same distance apart.
With non-Euclidean space, they will converge/diverge.

Re: Davis Relativity Model (Debate/discussion edition)
« Reply #49 on: October 08, 2017, 01:43:04 PM »
The problem is that there aren't any curved space geodesics that traverse the earth's surface. Curved space-time geodesics, yes, curved space geodesics, no.

Light follows geodesics in curved space. Other free-falling objects such as satellites don't (in general). Since light doesn't traverse the earth's surface, then by your own definition the earth is not flat in curved space.

You could attempt to argue that, because there are space-time geodesics that traverse the earth's surface, then the earth is somehow 'flat' in space-time. But what would that even mean? Space-time is populated by events, not physical surfaces.
4D Space-time is linked in one continuum in General Relativity, and have explicitly pointed out that the model I accept depends on this specific conception of space-time.

The geodesics don't depend on the mass, but light has a much higher velocity than any object you would be able to traverse Earth with, and so has an escape velocity that exceeds the warp of space-time of Earth, it doesn't go against the claim here at all.

Except that light travels straight in curved space and if a surface can't be traced by light it isn't flat.

That last point is closer to it. A geodesic with an equidistant traversal will be straight in curved space-time, which follows a flat earth. Space-time being populated by event isn't really relevant to this.

That space-time is populated by events renders the idea of a surface in space-time meaningless, in my view. What does 'flat' in space-time mean to you?

Even should you manage to attach a meaning to it you are still faced with the issue that your choice of reference geodesic is actually completely arbitrary. That you choose the geodesic of a circular orbit is entirely governed by your desired result of a 'flat' earth.

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #50 on: October 08, 2017, 02:32:25 PM »
Light follows geodesics in curved space. Other free-falling objects such as satellites don't (in general).
I thought that technically it still followed geodesics in curved spacetime and to have something follow geodesics in curved space it would need to travel at infinite velocity (to remove the time component)?

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #51 on: October 08, 2017, 03:17:49 PM »
As I have had some more time, a further example (graphical now) to illustrate my point:



This can represent several things, but the specific one I am having it represent is 2D spherical geometry.

There are 4 lines here. One red, one blue, one green one purple.

The green one is straight. Yes, it doesn't look straight, but that is because it is a projection of this non-Euclidean space into Euclidean space and thus it appears distorted.
However, this projection keeps distances (and direction) along a line through the centre correct.

So, we now look at the other lines, first the red one:
This red line remains the same distance from the green line. That is, the section of a line going through the centre, measured between the red and green line, remains the same length regardless of where you are along the green line. Thus by your "reasoning", this red line is straight, as it is "parallel" to the straight green line.
Similarly, the purple line remains the same distance and thus would be "straight".
Meanwhile, this blue line, which has the distance change and even crosses the green line would be deemed to be "not straight".

Unfortunately for you, that reasoning is completely wrong.
This green line is the equator. It is "straight".
The red line is 10 degrees north of the equator. It curves.
The blue line is 10 degrees south of the equator. It also curves.
The blue line is a great circle with a declination of 10 degrees. That is, it is a great circle which goes between 10 degrees north of the equator and 10 degrees south of the equator. It is straight.

Notice how your reasoning concluded that 2 non-straight lines were straight while a straight was not.

In fact, with this geometry, there is no straight line which remains equidistant from the green line. The only distance that would work is a distance of 0, making it simply the green line. Any other line which remains the same distance from this green line will not be straight.

Now do you understand the problem with your method?

Re: Davis Relativity Model (Debate/discussion edition)
« Reply #52 on: October 09, 2017, 02:56:51 AM »
Light follows geodesics in curved space. Other free-falling objects such as satellites don't (in general).
I thought that technically it still followed geodesics in curved spacetime and to have something follow geodesics in curved space it would need to travel at infinite velocity (to remove the time component)?

Light follows geodesics in curved space-time and curved space. In that sense geodesics in space-time and space can be thought to 'coincide' at light-speed.

Of course, light-speed is a kind of 'infinite' in that it is both a maximum and, more importantly, an asymptote for many properties. E.g. time 'stops' at light-speed, etc.

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #53 on: October 09, 2017, 03:57:41 AM »
Light follows geodesics in curved space-time and curved space. In that sense geodesics in space-time and space can be thought to 'coincide' at light-speed.

Of course, light-speed is a kind of 'infinite' in that it is both a maximum and, more importantly, an asymptote for many properties. E.g. time 'stops' at light-speed, etc.
See, this part doesn't make sense to me.

Yes, the speed of light is the maximum any object we know of (which we know exists) can go. But there are hypothetical particles which can travel faster than light, and it is still quite finite.
Even when an object travels at light speed, it travels along the time axis while traversing the space and thus if the time axis is curved (such as due to a gravitational field) it's path will curve because of it, whereas a geodesic in space would not be affected by such distortions in the time axis.
So I was thinking you can observe objects travelling at various finite speeds and plot a trend where as the speed increases the effect of the curvature of time is reduced until eventually you get to the point of no effect of time at an extrapolated infinite velocity.

I was thinking that as a comparison, consider 3 spatial dimensions which is flat in 2, but curved in the other.
Where if an object starts with a vector (xv,yv,zv) at position (x0,y0,z0), then its position after some time (t) (which for the real world would instead be a parameter measuring it stepping along its path and not actually time) would be given by (x0+xv*t, y0+yv*t+(zv*t-z0)^2,z0+zv*t). This can be projected onto the XY plane to show the path through this 2D space by discarding the z axis, i.e. (x0+xv*t,y0+yv*t+(zv*t-z0)^2).

Now, if an object was moving purely through x and y (corresponding to something travelling infinitely fast such that to traverse the entire space, no time passes), such that its vector is (xv,yv,0), its path would be through (x+xv*t,y0+yv*t,z0) and through the XY plane as (x+xv*t,y0+yv*t) and thus would be straight (i.e. not curving due to the curvature of space as this XY plane has no curvature).
Instead, if an object starts only moving through z (corresponding to something with an initial velocity of 0) its path would be (x0, y0+(zv*t-z0)^2,z0+zv*t) and through the XY plane as (x0,y0+(zv*t-z0)^2), and thus its path would be curved, following the curvature of the z axis through space

Light would be somewhere in between moving quite a small amount through z as it crosses the space, but still moving through z, and thus any curvature along the z axis will effect it and thus it will not necessarily follow a straight path through x and y.

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Master_Evar

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #54 on: October 09, 2017, 05:26:32 AM »
With non-Euclidean space, they will converge/diverge.
Yes, but that is if the space is homogenously elliptic or hyperbolic. If we have the perfect mix of hyperbolic and elliptic space (so at some volumes they will converge, at some volumes diverge), we could have a flat-ish object that acts like a sphere from within the non-euclidean space.

It's actually pretty easy - all we neeed to do is to add some isotropy in one spatial dimension. Essentially, in one direction we decrease the speed of light by a factor of 100, as an example. As a result, everything moving in this dimensions will move at 1/100 of the speed of things moving in the other dimensions, with the same amount of energy, force and time. As a result we can make earth 1/100 as long in this dimension.
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

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JackBlack

  • 23739
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #55 on: October 09, 2017, 01:48:39 PM »
Yes, but that is if the space is homogenously elliptic or hyperbolic. If we have the perfect mix of hyperbolic and elliptic space (so at some volumes they will converge, at some volumes diverge), we could have a flat-ish object that acts like a sphere from within the non-euclidean space.
But there would still be regions where it converges and regions where it diverges, and it can do so anisotropically.

What that seems to be is saying space is flat with a few ripples at the small scale.

It's actually pretty easy - all we neeed to do is to add some isotropy in one spatial dimension. Essentially, in one direction we decrease the speed of light by a factor of 100
And this is still flat space. It means Earth would be an elongated round object, but still in flat space.

It also has the issue of it behaving just like normal space but where you decide to multiply all distances by 100 in that direction.

Re: Davis Relativity Model (Debate/discussion edition)
« Reply #56 on: October 09, 2017, 02:01:25 PM »
Yes, the speed of light is the maximum any object we know of (which we know exists) can go. But there are hypothetical particles which can travel faster than light, and it is still quite finite.
Even when an object travels at light speed, it travels along the time axis while traversing the space and thus if the time axis is curved (such as due to a gravitational field) it's path will curve because of it, whereas a geodesic in space would not be affected by such distortions in the time axis.
So I was thinking you can observe objects travelling at various finite speeds and plot a trend where as the speed increases the effect of the curvature of time is reduced until eventually you get to the point of no effect of time at an extrapolated infinite velocity.

I was thinking that as a comparison, consider 3 spatial dimensions which is flat in 2, but curved in the other.
Where if an object starts with a vector (xv,yv,zv) at position (x0,y0,z0), then its position after some time (t) (which for the real world would instead be a parameter measuring it stepping along its path and not actually time) would be given by (x0+xv*t, y0+yv*t+(zv*t-z0)^2,z0+zv*t). This can be projected onto the XY plane to show the path through this 2D space by discarding the z axis, i.e. (x0+xv*t,y0+yv*t+(zv*t-z0)^2).

Now, if an object was moving purely through x and y (corresponding to something travelling infinitely fast such that to traverse the entire space, no time passes), such that its vector is (xv,yv,0), its path would be through (x+xv*t,y0+yv*t,z0) and through the XY plane as (x+xv*t,y0+yv*t) and thus would be straight (i.e. not curving due to the curvature of space as this XY plane has no curvature).
Instead, if an object starts only moving through z (corresponding to something with an initial velocity of 0) its path would be (x0, y0+(zv*t-z0)^2,z0+zv*t) and through the XY plane as (x0,y0+(zv*t-z0)^2), and thus its path would be curved, following the curvature of the z axis through space

Light would be somewhere in between moving quite a small amount through z as it crosses the space, but still moving through z, and thus any curvature along the z axis will effect it and thus it will not necessarily follow a straight path through x and y.

I understand your example. My initial thought was that since time doesn't pass for an object at light-speed its space-time vector would have no time component. So in your example this would be like movement in x and y only, and the object would follow spacial curvature but not temporal.

But, I then realised I hadn't taken into account that from the point of view of the object itself, there's no spacial movement either! The timeline of an object at light-speed is just a point in space-time.

So what about the object's movement through curved space? Well, from an external viewpoint it moves but also time passes. This would allow for it to be affected by time curvature and for it not to follow a curved space geodesic. In short, you'd be right and my assertion that light follows curved space geodesics would be wrong.

I feel I need to research this a bit more though. I only come at this as an amateur.

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Master_Evar

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #57 on: October 09, 2017, 04:22:33 PM »
Yes, but that is if the space is homogenously elliptic or hyperbolic. If we have the perfect mix of hyperbolic and elliptic space (so at some volumes they will converge, at some volumes diverge), we could have a flat-ish object that acts like a sphere from within the non-euclidean space.
But there would still be regions where it converges and regions where it diverges, and it can do so anisotropically.

What that seems to be is saying space is flat with a few ripples at the small scale.
That's the point, there are regions where it converges and were it diverges. That's why a flat-ish shape can appear spherical.


It's actually pretty easy - all we neeed to do is to add some anisotropy in one spatial dimension. Essentially, in one direction we decrease the speed of light by a factor of 100
And this is still flat space. It means Earth would be an elongated round object, but still in flat space.

It also has the issue of it behaving just like normal space but where you decide to multiply all distances by 100 in that direction.
[/quote]
Yes, that's what I'm kinda getting at. We could technically say that it's actually space that in one dimension is compressed to 1/100 the size of the other dimensions (and only near earth, if we want to actually make it somewhat non-euclidean). But it doesn't matter at all for how we perceive the universe, it only matters from an outsiders perspective.
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

*

NAZA

  • 594
Re: Davis Relativity Model (Debate/discussion edition)
« Reply #58 on: October 09, 2017, 05:22:53 PM »
.
If I am to put a name on this model, it would be most accurate to deem it the “Davis Relativity Model”, named after John Davis, the American Flat Earth Society Secretary (shout-out to Davis, you are awesome)


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AltSpace

  • Flat Earth Believer
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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #59 on: October 09, 2017, 11:04:51 PM »
You mean it fits the definition you provided to limit it to being flat.
As I've said, it is the only accurate determination of the geometry of Earth's surface, and it's straight and therefore flat.
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No. My point is that it only works in Euclidean space (aka flat space). It does not work in non-Euclidean or non-flat space.

Euclidean space (or spaces which have Euclidean parts) are the only ones in which 2, straight, parallel lines remain the same distance apart.
I see what you mean here, but it doesn't rebut my point here, as I will explain.
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It's pretty simple to see the relevance.

There is some line of "unknown" shape. (In reality we know the shape and we know it isn't straight.)
There is a reference line we can use, a "straight" line, or a geodesic in this curved space.
So, we move along the "straight" line, measuring the distance to the "unknown" line.
We find the "unknown" line remains the same distance.
According to your reasoning this means the "unknown" line, as it remains the same distance from a straight line, should be parallel to this straight line and thus straight.

The issue is that that only works in flat space.
Yes, I understand, and here's why it's not relevant.

What I'm claiming here is that if you have a line cross a surface, and you draw a straight line to the surface across the traversal at an equal distance, it will reveal the geometry of the surface. The key word here is "surface", I used that in the definition.
Your example with the wall 1000 km or so north of the equator you are traversing is able to get a parallel straight line geodesic parallel to a curved line, this is because in Non-Euclidean geometry (will represent a sphere here for demonstration), the line traversing the surface is parallel to a line straight across, which also change position in a third dimension relative to the parallel lines, which means this can only be represented in a 3D coordinate system.

This is required to define the curved and straight as parallel in this specific case.
Now, in defining the surface, since it uses a line traversal, it only needs to be plotted on a 2D plane, and we can determine that in this case, the curved and straight lines being parallel doesn't hold.

This is because this isn't relying on this principle of non-euclidean geometry to determine a flat surface, so it doesn't hold and is irrelevant.
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The 2 are equivalent.
The satellite, the geodesic through space time is equivalent to the equator.
The surface of Earth is equivalent to the line 1000 km north of the equator.
No it's not, since one doesn't rely on 3D non-euclidean geometry to determine this, the only reason why the curved and straight line can be parallel is because it is all plotted in an exclusively 3D coordinate system, it can't be represented in a 2D plane, like the surface geodesics can (these lines passing straight through the ground can be represented as 2D and straight parallel lines being the same distance apart holding.
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In order for your method to work, the line 1000 km north of the equator must be a straight line.
It doesn't, there is a clear distinction.
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BULLSHIT!
Any line can be used to define it.

Lets take a simple case of a sine wave, such as one of the form y=sin(x)+10.

We can use the x axis (y=0) as a reference.
In this case, the curve alternates between 9 and 11 units above it.
We can define it quite precisely.
So no, we do not need a line to remain the same distance.

That is your pathetic excuse to ignore all the other lines which show that Earth is not flat.

You need to actually justify your claim, not just spout more crap.
No, not any line. A Sine wave can't be congruent to a surface, If you use 9 and 11 unit alterations above the x axis, it could never match a line at y=1, so can't define the surface.
This method works because the lines are parallel in a plane, so they match the surface, and reveal it as straight.
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No it isn't. It is the only way you can try to justify pretending Earth is flat.
And because of that, you will reject all the others because you think if they don't show Earth is flat they must be wrong.
It is literally the only way a line matches a surface, and it is straight and therefore a flat surface.
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That doesn't address the point at all. I have also shown that that is not the case at all.
You failed to, you went on to an irrelevant 'gotcha' with non-euclidean geometry that misses how the surface is being determined here, I hardly call that 'showing it's not the case'.
For flat-space it is. Not for curved space.
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Likewise, if you have a straight line which oscillates between 2 distances, that shows it cannot match the straight line and cannot be flat (which again is only true for flat space), and likewise, that is common sense. If the distance from a straight reference line is not a simple linear relation (e.g. y=mx+b), the line can't be straight.
But as I have said, it can only match the surface geometry it is parallel, an equal distance to match the changes in the surface directly, doing this with Earth (in non-euclidean space-time) reveals it's flat.
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Again, they are parallel. There is a line going through all of them, which is perpendicular to all of them. That means they are parallel. So why don't they work?
Because the geodesic lines aren't parallel, ones traversing Earth's surface as a whole and not just straight down through.
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So parallel as in not actually parallel and instead remaining the same distance apart. While the 2 are equivalent in flat space, they are not in non-flat spaces and in non-flat spaces, a line which remains the same distance away from a straight line will likely be curved.
I mean a line straight above a surface that can be represented in a 2D plane that are also able to be defined as a line whose points are at a fixed normal distance from a given surface, what you proclaim here doesn't apply to parallel surface geodesics.
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Nothing in here indicates the motion is absolute rather than relative.
In order for this to be absolute the centre of the solar system needs to be fixed, absolutely.
No, it can be absolute in that the planets are orbiting the sun in an absolute sense.
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Again, by accepting that the centre of the solar system can move, IT MEANS IT ISN'T ABSOLUTE!!! It isn't a difficult concept to grasp.

This is basically what you are saying:
"I accept that it doesn't need absolute motion, but relative to the reference frame of the solar system, it must be absolute (i.e. relative to the solar system)."

Notice the stupidity there?
It would be most simple to get what I mean by this statement:
The bodies in our defined solar system are orbiting the sun in an absolute sense.
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Define objective.
It seems to be objective, not based upon any sentient entity, and it is capable of explaining the motions.
Happening independent of distinct reference frames that are being used completely to determine motion.
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This is a statement, not an explanation.
You need to provide an explanation for why the planets move.
I'm just describing how they move in this case, relative to each other by means of reference frames. It's not just a statement, it's an explanation of motion with celestial bodies.
They move relative to each other because of Newton's first law in curved space-time with distinct reference frames for motion.
You don't need a coordinate system to define relative motion since relative motion is independent of any coordinate system imposed on them.
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Stop trying to pretend objective and absolute mean the same thing.

You need to be able to describe the interactions between various entities to explain how objects move. You cannot do that from an Earth centred frame.
They do mean the same thing, it's absolute in that it doesn't depend on being between what's determined between reference frames. I describe the interactions as relative motions in curved space-time. Why they are in their specific positions instead of other positions will depend mostly on the geodesics they are present in with their state of motion.
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No. I am saying it only works in flat space, not non-flat space. Even in 3D flat space, parallel lines point in the same direction.
In flat space a line needs 2 things to define it, a vector and a point. From that point you follow that vector (or the reverse) and you have your line. That vector is effectively the direction the line is pointing.
If you have 2 lines, which do not overlap, and have the same vector, they are parallel.
Exactly, and a straight line has the same vector as the Earth's surface and can be represented in a 2D plane as actually equivalent lines and not the non-euclidean 3D exclusive example you brought up which isn't relevant.
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You seem to like going off on tangents to cover up your complete failure.
How about you stop setting up such pathetic strawmen and instead focus on what I have actually said?
As I have been doing.
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Yes they are.
There is a line joining them which is perpendicular to each. They do not intersect. They are parallel. They just don't have the property of parallel lines which exists in flat space, that they remain the same distance. As they are no longer in flat space the distance between these parallel lines can vary.
Referring to this:
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But don't worry, it gets more fun.
Now consider 4 objects above Earth. All heading due north (with no change in elevation). All above the equator at 0 degrees east. All at the same velocity. But now one is in a perfectly circular orbit, while another is significantly below it, in an elliptical orbit, with the other 2 above, one in an elliptical orbit and one in a hyperbolic orbit.

All 4 of these lines are currently parallel because they are perpendicular to the line connecting them, but let them move on, and their paths are completely different.
Only, as I have shown, the paths with the surface can be represented in 2D plane, and when you do that, the lines are not parallel as they are not an equal distant apart, and when you translate the surface geodesics to a plane, they clearly aren't parallel, in fact, if you have an elliptical orbit that gets lowest to the Earth at two points that are say x distance above and a circular orbit at x distance above constantly, they will intersect, which clearly can't be parallel lines.
So, it doesn't work, only one parallel to Earth's surface, equidistant from it at all points, to match the surface. If you have an elliptical orbit, the surface and the geodesic get's closer, so under this method, an undulating surface could only be determined if a path that never got closer to Earth revealed an undulating path, but only a straight path does this in the case of Earth in curved space-time.
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No. I meant flat, i.e. Euclidean, i.e. parallel lines remain the same distance apart; there is only a single straight line through a point, parallel to another straight line; the sum of the internal angles of a triangle is 180 degrees; etc.
This is different from non-flat spaces, where instead parallel lines can converge or diverge; there can be none (depending on the definition), 1 or infinitely many "straight" lines through a point, parallel to another straight line; the sum of the internal angles of a triangle can be more or less than 180 degrees.
Stop trying to pretend I am saying something else. Start addressing what I have actually said.
Only, the moment we transfer this to the plane representation of the line cutting to the surface (the line doesn't have width so it doesn't require 3D to represent), we reveal that the straight line reveals a flat surface. Parallel lines can't converge since they are an equal distance apart constantly by definition, their vectors are the same.
I'm not pretending either, I legitimately and unintentionally misunderstood.
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That is because you don't want to admit you were wrong.
Believe it or not, I still fail to see how I'm wrong.
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It shows how you can have a line remain equidistant from a straight line, yet not be straight.
This is because it is not in flat space.
Not in this case, you have to look at why this is the case in the non-euclidean example you used, and when you have that and apply it to the surface geodesics, it is revealed that they are not connected and therefore irrelevant.
No it doesn't.
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Again, that requires the space to be flat. If you don't have flat space that entire line of reasoning is baseless. It relies upon a fact of mathematics which is only valid for flat spaces, that parallel lines always remain the same distance apart.
In my case, Earth is a flat surface since an equidistant line traverses it as straight, the non-euclidean nature of space-time isn't quite relevant in this case (other than that a straight line meets itself because it's in non-euclidean space) since the line has no width, therefore, it can be represented in 2D, and from there it is determined that parallel to this line is a flat surface.
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Yes, it is the same, by which I mean completely false. But this does help show that.
The fact that you get parallel geodesics which intersect each other shows that 2 lines do not need to remain the same distance to be parallel and thus straight.
It shows that the distance between straight lines varies, and thus you can have a line which is not straight remain the same distance from a line which is straight.
Then show that this is the case with my specific point here with a line parallel to a surface of earth (You will only need to represent it in 2D since the lines have no width, just distance from the surface and length across) and I will abandon this model.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein