Davis Relativity Model (Debate/discussion edition)

Nope. You didn't answer my request.
This was your definition:

Quote from: AltSpace on September 26, 2017, 11:20:39 PM

Now lets look at the definition I used:

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it is something based on space and how it relates to the Earth. A flat plane would be defined by the ability to traverse it in a straight line between two spatial coordinates.
...

I asked you to clarify this definition.
Maybe I wasn't specific enough (although if you read my and JackBlack's discussion you could understand it). I want you to specifically clarify what spatial coordinates are – 3D coordinates relative to Earth? 4D space-time coordinates? Something else?

I gave you it, the point there was that only on a flat surface could an inertial orbit work, as it works with GR. It could be described as a tangent vector remaining tangent across earth, a straight surface.
I also pointed out that it was 4D space-time coordinates, with time being interconnected with 3D space.

My example was in an Euclidean space-time. I can choose in which space-time my example is.

Which is why it missed the point, so it doesn't apply here.

Quote from: Copper Knickers on October 19, 2017, 01:04:30 AM

You seem reluctant to acknowledge that space-time is a geometry of events rather than objects and therefore fundamentally different from space. I don't know whether this is wilful ignorance to support your 'model' or that you don't quite get it.

Anyway, this isn't a bad link if you're genuinely interested: http://www.ws5.com/spacetime/

General Relativity has space-time as an inseperable continuum, your link even supports this, they are one and the same aspect of our universe.
I am not being 'wilfully ignorant' or 'don't get it', I am simply holding to a GR conception of space and time, if you disagree with it, that's fine, but someone not accepting your conception doesn't imply any of that.

I've nowhere indicated disagreement with the GR concept of space-time. Yes, it is a continuum of space and time, but this makes its nature very different from space on its own. Space is a collection of objects separated by distances. Space-time is a collection of events separated by space-time intervals. The consequence is that a surface in space-time is not a physical surface as you seem to want it to be, but a particular geometric arrangement of events.

That is when the surface is a straight surface in 4D space-time, like a geodesic in that field. It's a quite simple concept.

No it isn't.
Surfaces aren't straight, they are flat.
And that is what you need to define.
What makes a surface flat.

You can't just say it is a flat surface.

General Relativity has space-time as an inseperable continuum, your link even supports this

Where?

they are one and the same aspect of our universe.

No. They are two related aspects of our universe.
If "they" are one and the same it would be it.
So what are you referring to? They as space and time, or it as space-time?

I am not being 'wilfully ignorant' or 'don't get it'

It has to be one of those two. Either you don't get it, or you do get it and are just rejecting it (i.e. being wilfully ignorant). There is no other alternative.

I am simply holding to a GR conception of space and time, if you disagree with it, that's fine, but someone not accepting your conception doesn't imply any of that.

No. You are holding to your misunderstanding of it.

We define surfaces and vectors by space-time itself

No. We typically define surfaces by space, not space-time. We can do the same for vectors.
We define trajectories by space-time.

Yeah, because sub-orbital velocity is not traversing

I'm not talking about traversing anything. I am talking about what the surface of Earth would do if it followed straight lines in space-time.

but is a launch like a giant parabolic arc throw that reaches space and goes back to the Earth's surface. This doesn't address the Earth's surface being a geodesic in curved space-time, which is flat.

No, it does address it, as it shows what Earth's surface would look like if it actually followed a collection of geodesics in space-time.

Earth's surface is not a geodesic in space-time, nor does it follow them.

It's funny how roundies will do whatever they can to twist concepts to get their favor, they like GR

No. What is funny is how you are projecting your own inadequacies onto others.
You are the one doing whatever you can to pretend GR magically makes Earth flat.

when it conflicts with their round earth faith, they forget the concept of curved space-time.

Again, that seems to be you. When trying to show Earth is flat, you forget that space-time is curved and instead try to use things from Euclidean geometry which do not hold in non-Euclidean geometry (at least not always).

I am not the one ignoring curved space-time, nor am I the one trying to twist anything.
Meanwhile you are yet to justify any of the BS you have spouted or refute any of my arguments.

I gave you it, the point there was that only on a flat surface could an inertial orbit work, as it works with GR.

And that doesn't work by following Earth's surface. Instead a variety of inertial orbits exist.
Some stay the same distance (in space) from Earth's surface while travelling through time at a vastly different rate to Earth's surface.
Others have the distance (in space) from Earth's surface vary.

The fact that a point on the surface of Earth moving with the surface of Earth, is not following an inertial orbit shows that Earth's surface IS NOT FLAT!!!

It could be described as a tangent vector remaining tangent across earth, a straight surface.

This doesn't describe it at all, and yet again has your baseless assertion that Earth is flat.

You need a general description which applies in any non-Euclidean space.

only on a flat surface could an inertial orbit work, as it works with GR.

Define "orbit" for any shape in a general space-time and prove this conjecture.

It could be described as a tangent vector remaining tangent across earth, a straight surface.

But it doesn't. Let's take a point on Earth's surface. A tangent to Earth on it is a hypothetical inertial trajectory of an object that starts in its location and velocity. This object will fall down immediately and its trajectory won't remain tangent to Earth. So no, Earth isn't flat.

I also pointed out that it was 4D space-time coordinates, with time being interconnected with 3D space.

So JackBlack was right. There are many pairs of spatial coordinates that the straight line between them isn't on Earth's trajectory. For example, Let's take an event on Earth, and an event in the sams time but at its opposite point. The straight line between them goes inside Earth and therefore isn't on Earth. So Earth isn't flat, again.

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My example was in an Euclidean space-time. I can choose in which space-time my example is.

Which is why it missed the point, so it doesn't apply here.

A definition for something on non-Euclidean space-times should generalize this thing on Euclidean space-times. Anyway, forget about my parabola example. I didn't understand the definition correctly.

I've nowhere indicated disagreement with the GR concept of space-time. Yes, it is a continuum of space and time, but this makes its nature very different from space on its own. Space is a collection of objects separated by distances. Space-time is a collection of events separated by space-time intervals. The consequence is that a surface in space-time is not a physical surface as you seem to want it to be, but a particular geometric arrangement of events.

And space on it's own doesn't exist. If you place a stationary object in a curved space-time region, it's path in time will warp and translate to acceleration through space. This implies space and times are variables, and are interconnected. No, objects and space are not synonymous and time is not events, but rather, GR reveals that a space-time region can be geometric with geodesic paths that are unique to that region.



Damn BlackJack, you really like to make a post focus sentence by sentence, rather than more summary responses to larger chunks, but I guess I'll have to try to work with this style:

No it isn't.
Surfaces aren't straight, they are flat.
And that is what you need to define.
What makes a surface flat.

You can't just say it is a flat surface.

The fact that a tangent vector plane touches at all points across a surface makes it flat, and this holds in curved space-time, because this follows as a geodesic in curved space-time regions.

No. They are two related aspects of our universe.
If "they" are one and the same it would be it.
So what are you referring to? They as space and time, or it as space-time?

If specific wording is so important to you, it is an 'it' since they are interconnected.
Sure, you can model it as something else, like 2 spatial dimensions and one time or even no time, but I'm accepting an interconnected space-time continuum.

No. We typically define surfaces by space, not space-time. We can do the same for vectors.
We define trajectories by space-time.

I define it by space-time, because all of the universe is affected and interconnected with a temporal dimension as part of 3D space.

I'm not talking about traversing anything. I am talking about what the surface of Earth would do if it followed straight lines in space-time.

Then could you elaborate more on the main problem with the Earth's surface having a straight line through space-time?
It seems I don't understand your point on this.

No, it does address it, as it shows what Earth's surface would look like if it actually followed a collection of geodesics in space-time.

Earth's surface is not a geodesic in space-time, nor does it follow them.

This gives me a clue on what you mean, I presume it is that these different orbital geodesics would have earth's surface as not possibly being straight since it doesn't end up following these geodesics equivalently, correct?
However, revealing these as straight geodesic paths also reveals the numerous other straight geodesics such as the Earth's surface. Earth's surface is best described as flat here in non-euclidean space-time.

And that doesn't work by following Earth's surface. Instead a variety of inertial orbits exist.
Some stay the same distance (in space) from Earth's surface while travelling through time at a vastly different rate to Earth's surface.
Others have the distance (in space) from Earth's surface vary.

The fact that a point on the surface of Earth moving with the surface of Earth, is not following an inertial orbit shows that Earth's surface IS NOT FLAT!!!

If you follow earth's surface, you still get a straight geodesic and a straight surface. The importance of the interconnected space-time components is that any time warped path directly connects to a space path, so the Earth's surface in time vs the orbital traversal in time could easily match the straight line spatial geometry and so Earth's surface is best explained as a flat surface in non-euclidean space-time. This is what simply follows from the Geometric curved space-time.

Define "orbit" for any shape in a general space-time

In this case, a path of a particular momentum in curved space-time in which the orbit matches the description of Newton's first law of motion (object either remains at rest or continues to move at a constant velocity, unless acted upon by a force).

and prove this conjecture

Define 'prove' and prove it. 

But it doesn't. Let's take a point on Earth's surface. A tangent to Earth on it is a hypothetical inertial trajectory of an object that starts in its location and velocity. This object will fall down immediately and its trajectory won't remain tangent to Earth. So no, Earth isn't flat.

But it is not an inertial orbit since it doesn't equal earth's surface with a consistent velocity.
However, what's really silly here is that you miss the point that this is Earth's surface here in curved space-time and you want to throw a random non-inertial (since it falling to the surface is not described as inertial) trajectory to dismiss earth's surface.
Also, the bold makes zero sense, please be more clear.

So JackBlack was right. There are many pairs of spatial coordinates that the straight line between them isn't on Earth's trajectory. For example, Let's take an event on Earth, and an event in the sams time but at its opposite point. The straight line between them goes inside Earth and therefore isn't on Earth. So Earth isn't flat, again.

Earth has no trajectory here, only the orbital geodesic does. What you said here was irrelevant, it is the Earth's surface as being defined by a vector in space-time, which is a straight geodesic, and leads to the conclusion of a flat earth.

I find it funny that you all act like a curved space-time region is exactly like a sphere in a flat space-time region.
John Davis pointed out this silly presumption:

Our sight would lead us to believe this might be foolish, but if space is curved (and Relativity relies on the assumption that it is) it would be silly to not question our visual representation of space since by all accounts it appears as if our observational (and theoretical) language is ill equipped to deal with description of it.

And space on it's own doesn't exist.

Nothing exists on it's own. That doesn't mean we can't analyse it on its own.

If you place a stationary object in a curved space-time region, it's path in time

i.e. you are analysing it in space-time, not just in space.

Damn BlackJack, you really like to make a post focus sentence by sentence, rather than more summary responses to larger chunks

This is because I am pointing out what is wrong with each piece. It is quite easy to make a general response to a large chunk which completely fails to address the issues raised in the large chunk. Similarly, it is also quite easy to then respond to that claiming one completely ignored what was raised.

It is much harder to do that when a single sentence (or part thereof) is focused on and shown to be wrong.

The fact that a tangent vector plane touches at all points across a surface makes it flat, and this holds in curved space-time, because this follows as a geodesic in curved space-time regions.

Define a tangent vector plane.

I showed previously that there are many ways to try and define a plane in non-euclidean 3D space-time, with the results being different for each one.
Which of the planes I provided do you think are flat, and why?

I define it by space-time, because all of the universe is affected and interconnected with a temporal dimension as part of 3D space.

And again, that would not be the surface of an object, but the trajectory. As soon as you bring in time you discuss trajectories, not just shapes.

Shapes are spatial, not temporal.

This also means the velocity becomes important as 2 objects can follow a similar path through space with one path being straight through space-time and another being curved, simply because they are travelling at different velocities.

Then could you elaborate more on the main problem with the Earth's surface having a straight line through space-time?
It seems I don't understand your point on this.

It is quite simple, if you take a point on Earth's surface (that is moving with Earth's surface) and projected it along a straight line in space-time, and did this for every point on the equator, you would end up with something like this:

The coloured (and apparently curved) surface is composed of straight lines following the Earth's surface at the equator.
The grey cylinder represents the actual path of these points through space-time.
Notice how they are not the same?
If Earth's surface was to follow a straight line through space-time it would follow the coloured path, falling towards the centre of Earth. It doesn't. Instead it follows the grey path, constantly accelerating upwards through space-time (due to the pressure inside Earth).
This shows Earth's surface is not flat in space-time.

This gives me a clue on what you mean, I presume it is that these different orbital geodesics would have earth's surface as not possibly being straight since it doesn't end up following these geodesics equivalently, correct?

No, it is the above. Earth's surface is not travelling fast enough for it to be an orbital trajectory and thus is not a geodesic in space time.

However, revealing these as straight geodesic paths also reveals the numerous other straight geodesics such as the Earth's surface. Earth's surface is best described as flat here in non-euclidean space-time.

Again, dealt with above. Earth's surface does not follow geodesics in space time.

If you follow earth's surface, you still get a straight geodesic and a straight surface.

Again, dealt with above. You don't.

The importance of the interconnected space-time components is that any time warped path directly connects to a space path

No it doesn't. (Again, dealt with above)
It warps to a trajectory. You cannot simply ignore the temporal component.
If you ignore the temporal component and analyse 2 lines purely in the spatial component you are unable to determine if it is straight or curved. This is because a curved line in space time can follow the same spatial coordinates as a straight line in space-time.

The simplest examples of this are orbits.

Consider a perfectly circular orbit at some altitude. This is a straight line through space-time and describes a circle in space.
Now consider an object following this path, but at twice the speed. This is no longer a straight line through space-time, instead it curves.
So you get 2 paths, both the same through space, but one is curved and one is straight.

Again, here is an image to demonstrate that point:

This is a surface made by lines having the vector <0,1,tk> and passing through the point <R,0,0>.
By changing tk you get completely different paths through space.

Quote from: AFanOfTruth on October 25, 2017, 02:29:35 AM

Define "orbit" for any shape in a general space-time

In this case, a path of a particular momentum in curved space-time in which the orbit matches the description of Newton's first law of motion (object either remains at rest or continues to move at a constant velocity, unless acted upon by a force).

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But it doesn't. Let's take a point on Earth's surface. A tangent to Earth on it is a hypothetical inertial trajectory of an object that starts in its location and velocity. This object will fall down immediately and its trajectory won't remain tangent to Earth. So no, Earth isn't flat.

But it is not an inertial orbit since it doesn't equal earth's surface with a consistent velocity.

Notice how you are appealing to aspects of orbit which weren't in your definition.
Why does an inertial orbit need to equal Earth's surface with a consistent velocity?

Also, your definition is wrong (at least by the common meaning of momentum)
It is a path of consistent total energy, which includes gravitational potential energy and kinetic energy, you could also have it include both angular momentum and momentum.

However, what's really silly here is that you miss the point that this is Earth's surface here in curved space-time and you want to throw a random non-inertial (since it falling to the surface is not described as inertial) trajectory to dismiss earth's surface.

Wrong again.
An "inertial" path through space-time is one which follows the curvature of space. This includes any object in free fall, even those falling to Earth. It stops being an inertial path when it hits Earth's surface and begins accelerating upwards with Earth's surface.

I find it funny that you all act like a curved space-time region is exactly like a sphere in a flat space-time region.

No we don't. We are examing Earth in both flat space and curved space-time. In neither case is the surface of Earth flat.
Earth's surface is travelling far too slowly to be following a geodesic in space-time.

John Davis pointed out this silly presumption:

You mean attempted to dismiss reality to pretend Earth is flat.

Define a tangent vector plane.

Simply a 2D plane placed on earth's surface (or line). A sphere:

In a curved space-time region, we have the vectors defining the plane as changing since space-time is not homogeneous at this point, so, the tangent vector will always be touching the equator at a consistent direction since the surface of Earth is a geodesic in this space-time region.

I showed previously that there are many ways to try and define a plane in non-euclidean 3D space-time, with the results being different for each one.
Which of the planes I provided do you think are flat, and why?

Sorry, I didn't see that.
The Earth being non-euclidean doesn't equal a plane, but it's surface has a straight path with a consistent vector always touching it's surface, which would drop away on a basic sphere, and the plane (like a flat sheet of paper) following this warp in space-time would match this surface. However, I was pretty sure only one plane type exists, which can be distorted in a universe with a space or space-time region that is uneven.

And again, that would not be the surface of an object, but the trajectory. As soon as you bring in time you discuss trajectories, not just shapes.

Shapes are spatial, not temporal.

But the Earth's surface is temporal as well, it couldn't have the form it has without the space-time component as whole, having masses accelerate into each other to form this planetary mass we call Earth. And that's the thing, we can have vectors in space-time, as defining the straight line through this space region as connected to time, and the Earth's surface is spatially flat.

Like seriously, I don't get how you people don't understand that if I take a flat plane and place it in a universe with non-homogeneous spaces and small curved regions, that the plane will start to have apparent dips and dents but remain straight, because the regions are not the same everywhere, same applies here.

It is quite simple, if you take a point on Earth's surface (that is moving with Earth's surface) and projected it along a straight line in space-time, and did this for every point on the equator, you would end up with something like this:

The coloured (and apparently curved) surface is composed of straight lines following the Earth's surface at the equator.
The grey cylinder represents the actual path of these points through space-time.
Notice how they are not the same?
If Earth's surface was to follow a straight line through space-time it would follow the coloured path, falling towards the centre of Earth. It doesn't. Instead it follows the grey path, constantly accelerating upwards through space-time (due to the pressure inside Earth).
This shows Earth's surface is not flat in space-time.

I was more looking for explanation of why this is so, giving me a picture and saying it should look a certain way on that picture but really looks a different way isn't gonna tell me much. I do notice how they are not the same, but how do you explain this picture and why would this apply to reality?
I am not sure of your reasoning with pressure inside the earth accelerating it outwards, I certainly am not making that claim, objects falling accelerate due to warped space-time, but I presume you are trying to make a point with this, and that is?
I am also not proclaiming the Earth's surface is like a moving projectile path of it's own nature through space but more like an object which has a surface equaling a straight line through space as a geodesic (and 'space' includes space-time since they are connected). An apple stationary in space can be described as moving in the fourth time dimension, which when it curves that time dimension, it accelerates in space.

No, it is the above. Earth's surface is not travelling fast enough for it to be an orbital trajectory and thus is not a geodesic in space time.

And this is one of the other things, what do you mean by Earth's surface moving like a trajectory? Didn't discuss that before.

It warps to a trajectory. You cannot simply ignore the temporal component.
If you ignore the temporal component and analyse 2 lines purely in the spatial component you are unable to determine if it is straight or curved. This is because a curved line in space time can follow the same spatial coordinates as a straight line in space-time.

Depending on the region they are in, if you launch a satellite above earth and it travels certain spatial co-ordinates, and you launch another satellite, then they cannot match without both being straight lines in space-time. If one satellite were to gain speed, it's spatial coordinate path would change.

Consider a perfectly circular orbit at some altitude. This is a straight line through space-time and describes a circle in space.
Now consider an object following this path, but at twice the speed. This is no longer a straight line through space-time, instead it curves.
So you get 2 paths, both the same through space, but one is curved and one is straight.

Again, here is an image to demonstrate that point:

This is a surface made by lines having the vector <0,1,tk> and passing through the point <R,0,0>.
By changing tk you get completely different paths through space.

We define a curved geometric line with 3D space (while connected to 4D time), a curve in time may translate to different spatial paths, but the same spatial paths will be equivalent in curvature or flatness.

Notice how you are appealing to aspects of orbit which weren't in your definition.
Why does an inertial orbit need to equal Earth's surface with a consistent velocity?

Because it's in an inertial frame of reference, and considering that this happens with 'circular orbits', with it being equidistant to the surface across the traversal.

Wrong again.
An "inertial" path through space-time is one which follows the curvature of space. This includes any object in free fall, even those falling to Earth. It stops being an inertial path when it hits Earth's surface and begins accelerating upwards with Earth's surface.

No, it is in an inertial frame of reference, so it has a constant velocity in a straight line. This can't happen on a round earth (since orbit must have acceleration by change in direction to traverse a sphere), but works on a flat earth since a straight line can traverse it's surface. A circular orbit has this property in GR.

Simply a 2D plane placed on earth's surface (or line). A sphere:

This is purely spatial, there is no temporal component. That also means that that plane is flat, and as Earth's surface doesn't lie on the plane, it isn't flat.

In a curved space-time region, we have the vectors defining the plane as changing since space-time is not homogeneous at this point, so, the tangent vector will always be touching the equator at a consistent direction since the surface of Earth is a geodesic in this space-time region.

Again, this is completely false and entirely circular.
Stop assuming that Earth's surface is flat. That is what you need to prove. Until you do, you can't use it as part of your definition of flat.

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I showed previously that there are many ways to try and define a plane in non-euclidean 3D space-time, with the results being different for each one.
Which of the planes I provided do you think are flat, and why?

Sorry, I didn't see that.

I did it here:
https://www.theflatearthsociety.org/forum/index.php?topic=72129.msg1969579#msg1969579
And then did another option which I don't agree with here:
https://www.theflatearthsociety.org/forum/index.php?topic=72129.msg1971755#msg1971755

Also, you saw the first one and responded to a part of the post here:
https://www.theflatearthsociety.org/forum/index.php?topic=72129.msg1974886#msg1974886

The Earth being non-euclidean doesn't equal a plane, but it's surface has a straight path with a consistent vector always touching it's surface

No it doesn't.
Not as it's surface. You have a theoretical orbital trajectory which would remain on Earth's surface, but goes much faster than Earth's surface and isn't Earth's surface.

However, I was pretty sure only one plane type exists, which can be distorted in a universe with a space or space-time region that is uneven.

In Euclidean spaces, yes. In non-Euclidean spaces it gets much more complicated.

But the Earth's surface is temporal as well, it couldn't have the form it has without the space-time component as whole, having masses accelerate into each other to form this planetary mass we call Earth. And that's the thing, we can have vectors in space-time, as defining the straight line through this space region as connected to time, and the Earth's surface is spatially flat.

NO IT ISN'T!!!
Stop repeating the same lie as if you saying it enough will make it true.

If you want to use the temporal component, you can't then go and ignore it.
Yes, Earth's surface does have a temporal component to its trajectory. It is moving through space with a certain speed.
But this speed is far too slow to be an orbital trajectory and thus it remaining the same distance from the centre of Earth means it isn't flat.

You can't just have one line through space time which is straight, then strip away the time component and pretend any line that goes through space like that is straight.
You need to keep both, the space and time components.

Like seriously, I don't get how you people don't understand that if I take a flat plane and place it in a universe with non-homogeneous spaces and small curved regions, that the plane will start to have apparent dips and dents but remain straight, because the regions are not the same everywhere, same applies here.

I do get that. I just understand that a plane in non-Euclidean spaces are a lot more complex and that Earth's surface is not a plane.
What I don't get is how you expect anyone to take you seriously when you keep appealing to the temporal component to establish that straight lines can curve, only to then ignore the temporal component.

These apparent dips and dents will change depending upon the vector of the lines making the plane, which includes the temporal component.
You can't just ignore it.

I was more looking for explanation of why this is so, giving me a picture and saying it should look a certain way on that picture but really looks a different way isn't gonna tell me much. I do notice how they are not the same, but how do you explain this picture and why would this apply to reality?

I thought it was already enough of an explanation, but to spell it out more:
A geodesic through space-time has apparent curvature dependent upon speed (and direction).

Considering a point on a tangential vector, that is moving perpendicularly to the line connecting it to the centre of Earth, there is one particular speed which results in a circular orbit, that is one particular speed where it remains the same distance from Earth. This vector at this point has some spatial component and some temporal component.
If you have it move at a different velocity, the ratio between these 2 components change. If it moves faster then it will be moving through less time for a given movement in space, and thus the apparent curvature is less pronounced, resulting in it moving further away from Earth. As the speed increases it goes from a circular orbit, to an elliptical orbit, growing more and more elliptical until it eventually reaches a parabolic trajectory and then goes to hyperbolic trajectories. Extending beyond the physically possible to further analyse this plane in space time (we are keeping the normal constant and merely rotating the vector) you eventually reach a line that is straight with no temporal component at all.
Going the other way, reducing speed, we go to elliptical orbits where it gets closer to the centre of Earth.

These are all straight lines through space-time.
There isn't just one.
As such, you can't just look at the spatial part, you need to look at the temporal component as well.

For the equator, with a radius of 6378.1 km, assuming g to be 9.8 m/s, the circular orbit would correspond to a speed of roughly 7900 m/s.
So you can have a straight line remain the same distance from Earth's surface, to follow the same spatial coordinates as Earth's surface. But only if it was going at 7900 m/s.
Earth's surface is only travelling at roughly 464 m/s. This is significantly below the speed required to maintain a circular orbit. Thus Earth's surface (if it was going to follow a straight line) would fall towards the centre of Earth.

The surface I showed you was composed of many of these elliptical orbits, each one for a specific point on the equator.

So if earth's surface was going to follow a straight line through space-time, it would need to contract in space as it moved through time.
Instead, it follows the grey cylinder.
As this is further away (and grows further and further) from the straight line path, this means Earth's surface is curving outwards and is not flat.

I am not sure of your reasoning with pressure inside the earth accelerating it outwards, I certainly am not making that claim, objects falling accelerate due to warped space-time, but I presume you are trying to make a point with this, and that is?

If Earth's surface was just affected by gravity (or curved space-time) it would fall, just like if you jumped off a chair you fall.
That would be because there is no force acting on you to counter gravity (and thus you remain on an "inertial" path through space-time.

But typically that isn't what happens. Instead the ground beneath you is pushing you up, and the ground beneath it is pushing it up.

This is pressure. Pressure results in a force, and the Earth's surface provides a force accelerating you upwards, resulting in you deviating from a straight line through space time.

Likewise the pressure inside Earth acts on the surface of Earth resulting in it being accelerated upwards to not follow a geodesic in space-time. As such, Earth's surface is not flat.

I am also not proclaiming the Earth's surface is like a moving projectile path of it's own nature through space but more like an object which has a surface equaling a straight line through space as a geodesic (and 'space' includes space-time since they are connected).

And you have the same problem.
You want to invoke the time component to make it curve, but then strip the time component. YOU CAN'T DO THAT!!!
Earth's surface does not have any straight lines through space-time. Nor does it have any line which equals a straight line through space-time.

Instead, there are lines where the spatial parts match while the temporal components are completely different.

In order to equal a straight line through space-time, you need to have both the spatial and temporal components match.
If you don't, they are not equal.

And this is one of the other things, what do you mean by Earth's surface moving like a trajectory? Didn't discuss that before.

As you pointed out, all objects move through time. This means the surface (or points on the surface) have some trajectory through space time. If you don't want to discuss it as trajectories you lose the temporal component and can no longer appeal to time resulting in apparent curvature through space.

Depending on the region they are in, if you launch a satellite above earth and it travels certain spatial co-ordinates, and you launch another satellite, then they cannot match without both being straight lines in space-time. If one satellite were to gain speed, it's spatial coordinate path would change.

They cannot match without both following the exact same path, which means they need to have the same space-time coordinates.
However, you can have them offset by a simple rotation.

But yes, if one changes speed, its spatial coordinates change and it follows a different straight line.
Or to put it another way, UNLESS THE SPEED MATCHES THE PATHS DO NOT MATCH!!!!

And guess what? the speed of Earth's surface does not match that for an orbit and thus it doesn't match a straight line.

We define a curved geometric line with 3D space (while connected to 4D time), a curve in time may translate to different spatial paths, but the same spatial paths will be equivalent in curvature or flatness.

I just showed how that is pure bullshit.
Again, you cannot strip the time part from it.
The motion through time is what dictates the curvature through space.
There is only one specific path through space-time which results in a spatial path that is straight. Other paths through space-time that follow that spatial path will be curved in space-time.
As you said before, the speed needs to match.

Because it's in an inertial frame of reference, and considering that this happens with 'circular orbits', with it being equidistant to the surface across the traversal.

Who cares? They are not the only inertial orbits, you also have elliptical orbits, which are not equidistant to the surface (in space), and removing the orbit part you also have parabolic and hyperbolic trajectories.

These are inertial paths.

Also, circular orbits do not maintain a constant velocity. They maintain a constant speed. There is a difference.
The direction of their motion changes.

No, it is in an inertial frame of reference, so it has a constant velocity in a straight line.

That then raises the question of what constitutes a constant velocity through space-time.
According to GR, all free-falling frames are inertial. That includes ones where the reference frame changes velocity relative to other reference frames.
And regardless, as soon as there is curvature of the path, the velocity changes.

This can't happen on a round earth (since orbit must have acceleration by change in direction to traverse a sphere), but works on a flat earth since a straight line can traverse it's surface. A circular orbit has this property in GR.

No it doesn't. All orbits in GR are the same, the velocity changes from external references frames, but the local reference frame in orbit is inertial. A circular orbit is not special in GR.

Digging this up in order to not derail a different thread:

There was a thread where someone took a pretty good look at it, but I can't seem to find it now. I'll provide you with the base thought experiment that defines the model here. The more formal definitions I'm working on. I was originally looking at redefining the euclidean axioms to fit the model in a new way, but have since changed my direction.


First we start with a hypothetical planet; let us call it terra. Above this hypothetical planet we have a 'perfectly orbiting satellite.'  By this I mean one which has a constant altitude at any point in its trajectory. Let us say this height = a.

Secondly, we recognize Newton's laws, presented here in common language -
a) Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
b) The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. F=ma
c) For every action there is an equal and opposite reaction.

Third we recognize the equivalence principle:
In short, gravitational pull is shown as an observational illusion hinging on our naive view of our frame of reference being inertial. Gravity is actually revealed as an inertial force (also known as a fictitious force).

Now we can build from here.

An object in motion tends to stay in motion and in the direction it is in motion. We can certainly say that the object in orbit that it feels no experimentally verifiable difference in force or pseudo-force - which is equivalent to saying it is experimentally not accelerating (and thus not changing direction or speed.) This leads us to believe the orbiting object is then traveling in a straight line because it does not feel acceleration and all force interactions we might say exist on it are pseudoforces. Thus, this orbiting object is traveling a straight line through curved space - or its geodesic.

Next let us consider all such possible orbits at height = a.  It is simple to see each such object would form its own geodesic. Combining these geodesics we can create geodesic surface which is finite and closed and like the geodesics forming it formed of straight lines - or in other words flat. Additionally, at any point we can see that it is at a constant height to the surface (terra) it is surrounding - a.

From here we could do some work to show that the surface it is surrounding is indeed also a geodesic surface by considering the paths of non-orbiting objects and how they differ from their ideal geodesics if they were not gravitationally affected, however for brevity I will say that it is acceptable that since a flat surface bounds terra, terra itself is also flat.

What you describe reads to me quite like general relativity, except that in GR, free-falling orbits describe geodesics through spacetime, not space itself. Before diving into an argument I have a few questions (sorry if they have been answered already, I am not in the mood to dig through several pages of threads):
- Do you accept the mathematical basics of GR for your model (meaning differential geometry, especially the covariant derivative defined by the Levi-Civita connection)?
- Do you accept that spacetime is a 4-dimensional manifold, that photons describe light-like geodesics and all matter with mass>0 describes timelike geodesics in that spacetime?
- Do you accept that the metric of spacetime is defined by the presence of matter, pressure, and energy?
- Do you accept the Einstein field equations?