Davis Relativity Model (Debate/discussion edition)

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #120 on: October 19, 2017, 02:41:21 AM »
IDK how to define "flat" in this space-time or in non-Euclidean space-times in general. My whole point was that the definition used by AltSpace/Davis was problematic because things such as this paraboloid fit it while they clearly aren't flat
Except this statement makes no sense.
If you can't define flat how can you say they clearly aren't flat?

Some people would say similar things about straight lines in non-euclidean geometry and say they clearly aren't straight, but in that space, they are.
The issue is our Euclidean representation of it.

But when it goes to a plane, it becomes more complex due to the different ways to make a plane and them no longer agreeing.


But what does left/up mean in your space-time and why are they the same direction from the line (like lines with the same vectors are the same direction from a line in Euclidean space)?
A hypothetical object starts out with a up being the direction of the normal to the plane. Forward is the direction along its trajectory.
It starts out with another straight line off to its left at some angle. It moves along its path, and keeps projecting the line along that angle.
But as it moves forward, the space itself (space-time) curves down, yet it continues moving forward, meaning the local up/down and left/right need to curve with it.

Does that make sense?

I thought we are talking about space-time which consists of events, not space-time-speed which consists of some kind of states or something.
The events are points. The connections between points are trajectories We weren't discussing points, we were discussing trajectories:
BTW, do you think that Earth's trajectory is flat in space-time?

An event in space-time would only correspond to a single point on Earth's surface at a single time. Asking if that is flat makes as much sense as asking if a point is straight/flat.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #121 on: October 19, 2017, 05:31:08 AM »
IDK how to define "flat" in this space-time or in non-Euclidean space-times in general. My whole point was that the definition used by AltSpace/Davis was problematic because things such as this paraboloid fit it while they clearly aren't flat
Except this statement makes no sense.
If you can't define flat how can you say they clearly aren't flat?

Some people would say similar things about straight lines in non-euclidean geometry and say they clearly aren't straight, but in that space, they are.
The issue is our Euclidean representation of it.

But when it goes to a plane, it becomes more complex due to the different ways to make a plane and them no longer agreeing.
Because it's exactly the same surface on exactly the same space-time, except for some distortion somewhere that doesn't have anything to do with that surface.
But what does left/up mean in your space-time and why are they the same direction from the line (like lines with the same vectors are the same direction from a line in Euclidean space)?
A hypothetical object starts out with a up being the direction of the normal to the plane. Forward is the direction along its trajectory.
It starts out with another straight line off to its left at some angle. It moves along its path, and keeps projecting the line along that angle.
But as it moves forward, the space itself (space-time) curves down, yet it continues moving forward, meaning the local up/down and left/right need to curve with it.

Does that make sense?
Yes, thank you.
I thought we are talking about space-time which consists of events, not space-time-speed which consists of some kind of states or something.
The events are points. The connections between points are trajectories We weren't discussing points, we were discussing trajectories:
BTW, do you think that Earth's trajectory is flat in space-time?

An event in space-time would only correspond to a single point on Earth's surface at a single time. Asking if that is flat makes as much sense as asking if a point is straight/flat.
I meant to ask if you think Earth's surface, as a 3D collection of points in 4D space-time is a hyperplane.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #122 on: October 19, 2017, 01:34:14 PM »
Because it's exactly the same surface on exactly the same space-time, except for some distortion somewhere that doesn't have anything to do with that surface.
Does it have nothing to do with the surface?
If so, then it likely isn't flat.
But that isn't the case here. Here we are dealing with paths that are curved due to the curvature of space-time.

If you had a surface which was a "parabolic cylinder", of the form z=-t^2, would you accept that that is flat?
This is a plane of translation, where you have parabolas due to curvature due to the t axis, which is then translated along the x axis.

This is also a case where you can construct "parallel" surfaces.
Moving along the x axis and following the surface through z and t at that new x point produces a parallel line, remaining the same x distance apart.
Moving up or down (translating the surface), such that the surface is of the form z=z0-t^2, results in a surface which remains z0 away from the original for a given x and t.

I meant to ask if you think Earth's surface, as a 3D collection of points in 4D space-time is a hyperplane.
That is dependent upon the connectivity of the points.
You can connect the points with one type of connectivity (the circular orbits) and have them be a hyperplane (assuming that definition holds). But this connectivity doesn't correspond to the actual connectivity of these points (how the points on Earth's surface move through space-time).
You can connect the points in another way (such as how they connect in reality), and that doesn't result in a hyperplane as it now curves away from the straight line.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #123 on: October 20, 2017, 02:34:56 AM »
Because it's exactly the same surface on exactly the same space-time, except for some distortion somewhere that doesn't have anything to do with that surface.
Does it have nothing to do with the surface?
Yes, it does.
If so, then it likely isn't flat.
But AltSpace's/Davis's definition says it is.
But that isn't the case here. Here we are dealing with paths that are curved due to the curvature of space-time.
On Earth, Earth curves downwards and the time curves upwards, resulting that some paths (circular orbits) remain the same. I still don't think it makes Earth flat.
If you had a surface which was a "parabolic cylinder", of the form z=-t^2, would you accept that that is flat?
This is a plane of translation, where you have parabolas due to curvature due to the t axis, which is then translated along the x axis.
In my space-time? Let me think about it. In one hand, it is flat in every point in time, but in the other, it curves in the time axis. Also, from other frames of reference it is clearly not flat even in frozen points in time. So no.
BTW, it also fits their definition, just to see how weak it is.
I meant to ask if you think Earth's surface, as a 3D collection of points in 4D space-time is a hyperplane.
That is dependent upon the connectivity of the points.
Why? There is a collection of events. Why should the way I connect them matter?
But what does left/up mean in your space-time and why are they the same direction from the line (like lines with the same vectors are the same direction from a line in Euclidean space)?
A hypothetical object starts out with a up being the direction of the normal to the plane. Forward is the direction along its trajectory.
It starts out with another straight line off to its left at some angle. It moves along its path, and keeps projecting the line along that angle.
But as it moves forward, the space itself (space-time) curves down, yet it continues moving forward, meaning the local up/down and left/right need to curve with it.

Does that make sense?
I thought about this again and I think it doesn't, Because I don't understand why front-up-back-down don't switch places (front means sending the line with more speed in your direction, back with less, and the others mean sending it in some speed in other directions too).

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #124 on: October 20, 2017, 03:28:13 AM »
But AltSpace's/Davis's definition says it is.
As I said, it lacks a key part for non-Euclidean spaces.
That has been his problem all along. He has been trying to use properties of lines/planes which only exist in Euclidean spaces. In non-Euclidean spaces it becomes more complex.

On Earth, Earth curves downwards and the time curves upwards, resulting that some paths (circular orbits) remain the same. I still don't think it makes Earth flat.
It's more the other way around. Time curves objects downwards. The pressure inside Earth forces it up. These are 2 forces are balanced resulting in the surface of Earth remaining the same, but these pressure is not part of space-time, so in space-time Earth's surface is continually curving outwards, cancelling the curvature of time curving inwards.

But for orbits, there is no extra force. They just follow a path in space-time.
Do you accept that these orbits are straight lines?

If you had a surface which was a "parabolic cylinder", of the form z=-t^2, would you accept that that is flat?
This is a plane of translation, where you have parabolas due to curvature due to the t axis, which is then translated along the x axis.
In my space-time? Let me think about it. In one hand, it is flat in every point in time, but in the other, it curves in the time axis. Also, from other frames of reference it is clearly not flat even in frozen points in time. So no.
No, as this hypothetical space time, where a straight path is curved at a rate of -t^2, akin to parabolic arcs due to gravity.
It is a non-Euclidean space, where every line on this surface at a particular x value is straight as it follows the path through t and z.

Why? There is a collection of events. Why should the way I connect them matter?
Because the way that connects them dictates how time would curve the line.
The simplest example is looking at different orbits/trajectories.
A circular orbit around Earth is a straight line. But if you tried to go faster or slower, the straight line would become an elliptical orbit or a parabolic/hyperbolic trajectory.
So if you go at the speed of an elliptical orbit, but follow the circular orbit, that circular path is no longer a straight line.

It is somewhat akin to longitudes and latitudes on Earth or more accurately in spherical geometry.
At the equator, if you go along a path around the longitudes while keeping the same latitude (e.g. east to west), you are travelling in a straight line.
However, if you try the same north or south of the equator, you will be curving. At these locations to follow a straight line you need to change your latitude as well.
So all the points in spherical geometry can be connected by straight lines, but not all the lines connecting points are straight.

I thought about this again and I think it doesn't, Because I don't understand why front-up-back-down don't switch places (front means sending the line with more speed in your direction, back with less, and the others mean sending it in some speed in other directions too).
Why would it? It is going forward at a constant speed, along a straight line, not curving.

Again, appealing to spherical geometry, consider these maps:


Say you start out at (0E, 0N), heading East. The straight path in this geometry follows 0E, so in the first map it appears as a circle, in the second, as a straight line.
Now you start with left being due north.
After 90 degrees, which way is left?
Is it still due north (which would be fully supported by the second map), or would it now be due east with you facing due south (which the first map would have as parallel in our Euclidean representation)?
It gets more complicated in 3D as you can rotate around a straight line without noticing it, switching around left/up/down/right.
It gets even more complicated with large objects in orbit which don't rotate to follow the orbit.
« Last Edit: October 20, 2017, 03:30:54 AM by JackBlack »

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #125 on: October 20, 2017, 04:06:10 AM »
But AltSpace's/Davis's definition says it is.
As I said, it lacks a key part for non-Euclidean spaces.
That has been his problem all along. He has been trying to use properties of lines/planes which only exist in Euclidean spaces. In non-Euclidean spaces it becomes more complex.
What key parts and what properties? I didn't say that a line equidistant from another straight line is straight in a general non-Euclidean space-time. I started with a curved surface in an Euclidean space-time and then added some curvature to the space-time in a place that has nothing to do with the surface.
On Earth, Earth curves downwards and the time curves upwards, resulting that some paths (circular orbits) remain the same. I still don't think it makes Earth flat.
It's more the other way around. Time curves objects downwards. The pressure inside Earth forces it up. These are 2 forces are balanced resulting in the surface of Earth remaining the same, but these pressure is not part of space-time, so in space-time Earth's surface is continually curving outwards, cancelling the curvature of time curving inwards.
But for orbits, there is no extra force. They just follow a path in space-time.
Do you accept that these orbits are straight lines?
Yes oops, I got confused.
The time curves downwards too and the orbits are the lines where the curvature of Earth in the spatial directions matches the curvature of time, making them straight.
If you had a surface which was a "parabolic cylinder", of the form z=-t^2, would you accept that that is flat?
This is a plane of translation, where you have parabolas due to curvature due to the t axis, which is then translated along the x axis.
In my space-time? Let me think about it. In one hand, it is flat in every point in time, but in the other, it curves in the time axis. Also, from other frames of reference it is clearly not flat even in frozen points in time. So no.
No, as this hypothetical space time, where a straight path is curved at a rate of -t^2, akin to parabolic arcs due to gravity.
Of course it is straight there.
Why? There is a collection of events. Why should the way I connect them matter?
Because the way that connects them dictates how time would curve the line.
The simplest example is looking at different orbits/trajectories.
A circular orbit around Earth is a straight line. But if you tried to go faster or slower, the straight line would become an elliptical orbit or a parabolic/hyperbolic trajectory.
So if you go at the speed of an elliptical orbit, but follow the circular orbit, that circular path is no longer a straight line.

It is somewhat akin to longitudes and latitudes on Earth or more accurately in spherical geometry.
At the equator, if you go along a path around the longitudes while keeping the same latitude (e.g. east to west), you are travelling in a straight line.
However, if you try the same north or south of the equator, you will be curving. At these locations to follow a straight line you need to change your latitude as well.
So all the points in spherical geometry can be connected by straight lines, but not all the lines connecting points are straight.
I know all of that, but I showed a way to construct a huperplane that equals to Earth's surface (assuming back and left don't change):
But I think it matches the 2nd way of making a flat surface – get a circular orbit as the 1st line and a circular orbit in the opposite direction as the 2nd one. In your 3D sketch you're done, but if you are in 4D space-time you'll get the equator and will need a 3rd line, such as a orbits along longitudes.
I thought about this again and I think it doesn't, Because I don't understand why front-up-back-down don't switch places (front means sending the line with more speed in your direction, back with less, and the others mean sending it in some speed in other directions too).
Why would it? It is going forward at a constant speed, along a straight line, not curving.

Again, appealing to spherical geometry, consider these maps:


Say you start out at (0E, 0N), heading East. The straight path in this geometry follows 0E, so in the first map it appears as a circle, in the second, as a straight line.
Now you start with left being due north.
After 90 degrees, which way is left?
Is it still due north (which would be fully supported by the second map), or would it now be due east with you facing due south (which the first map would have as parallel in our Euclidean representation)?
It gets more complicated in 3D as you can rotate around a straight line without noticing it, switching around left/up/down/right.
It gets even more complicated with large objects in orbit which don't rotate to follow the orbit.
Because the directions, relative to Earth, that look like the "forward" and the "upward" of the object change.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #126 on: October 20, 2017, 05:08:46 AM »
What key parts and what properties? I didn't say that a line equidistant from another straight line is straight in a general non-Euclidean space-time. I started with a curved surface in an Euclidean space-time and then added some curvature to the space-time in a place that has nothing to do with the surface.
Then how does it fit the definition?
If the surface is a hill in the region where the space is flat, then it isn't flat.

Yes oops, I got confused.
The time curves downwards too and the orbits are the lines where the curvature of Earth in the spatial directions matches the curvature of time, making them straight.
And this works in other non-Euclidean spaces as well, were lines which appear curved in Euclidean representations are actually straight.
This makes it hard to say if a surface is flat or not.

Of course it is straight there.
And is the surface flat?
If you translate the line along the x axis, you are just using straight lines.

I know all of that, but I showed a way to construct a huperplane that equals to Earth's surface (assuming back and left don't change):
But it doesn't actually match. It is turned with respect to Earth's surface. Unlike Euclidean geometry where that wouldn't matter because straight lines all curve the same (not at all), in Euclidean space it does matter. But in non-Euclidean spaces they don't all curve the same.

Because the directions, relative to Earth, that look like the "forward" and the "upward" of the object change.
Relative to our Euclidean representation.
For the object moving along Earth, it appears the same, unless you are suggesting forward becomes south.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #127 on: October 20, 2017, 07:13:54 AM »
What key parts and what properties? I didn't say that a line equidistant from another straight line is straight in a general non-Euclidean space-time. I started with a curved surface in an Euclidean space-time and then added some curvature to the space-time in a place that has nothing to do with the surface.
Then how does it fit the definition?
If the surface is a hill in the region where the space is flat, then it isn't flat.
No, the surface is still a paraboloid.
The space-time is an Euclidean space-time with a little distortion, like a hill.
Yes oops, I got confused.
The time curves downwards too and the orbits are the lines where the curvature of Earth in the spatial directions matches the curvature of time, making them straight.
And this works in other non-Euclidean spaces as well, were lines which appear curved in Euclidean representations are actually straight.
This makes it hard to say if a surface is flat or not.
I know, but in my example I took a surface that was not flat in an Euclidean space-time, and changed a zone in space-time that doesn't intersect with the surfasce, what keeps the surface not flat.
Of course it is straight there.
And is the surface flat?
If you translate the line along the x axis, you are just using straight lines.
Yes, I meant to say "flat"
I know all of that, but I showed a way to construct a huperplane that equals to Earth's surface (assuming back and left don't change):
But it doesn't actually match. It is turned with respect to Earth's surface. Unlike Euclidean geometry where that wouldn't matter because straight lines all curve the same (not at all), in Euclidean space it does matter. But in non-Euclidean spaces they don't all curve the same.
Why are you adding speed to the discussion? It doesn't matter.
There are the same points on Earth and on the hyperplane (which is a hyperplane if we assume that forward really doesn't change).
Therefore Earth's surface is the hyperplane.
Because the directions, relative to Earth, that look like the "forward" and the "upward" of the object change.
Relative to our Euclidean representation.
For the object moving along Earth, it appears the same, unless you are suggesting forward becomes south.
South remains south (if you travel along the equator).
I suggest forward becomes up, up becomes back, back becomes down and down becomes forward. It's a very simple rotation in respect to Earth. Why isn't it the right way to decide which straight lines are in the same plane?

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #128 on: October 20, 2017, 12:46:25 PM »
No, the surface is still a paraboloid.
The space-time is an Euclidean space-time with a little distortion, like a hill.
Again, if the surface is basically Euclidean with a little distortion, where that distortion is not at the paraboloid, then it isn't flat.
You can't go between any 2 points in a straight line remaining on the surface.

I know, but in my example I took a surface that was not flat in an Euclidean space-time, and changed a zone in space-time that doesn't intersect with the surfasce, what keeps the surface not flat.
The fact that the surface doesn't follow straight lines in this space which have any relation like a normal plane would (e.g. all straight lines pass through a point where they are all normal to the same vector, or they start in the same direction relative to another straight line.

Why are you adding speed to the discussion? It doesn't matter.
There are the same points on Earth and on the hyperplane (which is a hyperplane if we assume that forward really doesn't change).
Therefore Earth's surface is the hyperplane.
Again, the connectivity of points matters. The speed determines this connectivity.
It is only Euclidean spaces where the connectivity doesn't matter.
In non Euclidean spaces you can have a curved line pass through the same points as a series of straight lines which are in turn connected by a straight line.

South remains south (if you travel along the equator).
I suggest forward becomes up, up becomes back, back becomes down and down becomes forward. It's a very simple rotation in respect to Earth. Why isn't it the right way to decide which straight lines are in the same plane?
Because in the non-Euclidean space, it is simply going forwards. Why would its direction change? Why would it suddenly start going up or down?
If an object circled you while you were walking forwards would your direction change just because the object is not in the same position relative to you?

Also, the path around the equator I made into a 2D problem. There was only forwards/backwards and left/right.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #129 on: October 22, 2017, 01:27:16 AM »
No, the surface is still a paraboloid.
The space-time is an Euclidean space-time with a little distortion, like a hill.
Again, if the surface is basically Euclidean with a little distortion, where that distortion is not at the paraboloid, then it isn't flat.
You can't go between any 2 points in a straight line remaining on the surface.

I know, but in my example I took a surface that was not flat in an Euclidean space-time, and changed a zone in space-time that doesn't intersect with the surfasce, what keeps the surface not flat.
The fact that the surface doesn't follow straight lines in this space which have any relation like a normal plane would (e.g. all straight lines pass through a point where they are all normal to the same vector, or they start in the same direction relative to another straight line.
I know it isn't flat, but AltSpace's/Davis's definition says it is.
Why are you adding speed to the discussion? It doesn't matter.
There are the same points on Earth and on the hyperplane (which is a hyperplane if we assume that forward really doesn't change).
Therefore Earth's surface is the hyperplane.
Again, the connectivity of points matters. The speed determines this connectivity.
It is only Euclidean spaces where the connectivity doesn't matter.
In non Euclidean spaces you can have a curved line pass through the same points as a series of straight lines which are in turn connected by a straight line.
It can also happen in an Euclidean space.
For example, the lines (x, m, xm) for every x where m is the parameter are all connected by the x axis but form a hyperbolic paraboloid, on which there are many curved lines.
South remains south (if you travel along the equator).
I suggest forward becomes up, up becomes back, back becomes down and down becomes forward. It's a very simple rotation in respect to Earth. Why isn't it the right way to decide which straight lines are in the same plane?
Because in the non-Euclidean space, it is simply going forwards. Why would its direction change? Why would it suddenly start going up or down?
From its perspective it's not going, from Earth's perspective it clearly changes direction.
Also, the path around the equator I made into a 2D problem. There was only forwards/backwards and left/right.
But the object that is orbiting Earth isn't walking on its surface, but going on a straight line in space-time.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #130 on: October 22, 2017, 02:32:37 AM »
I know it isn't flat, but AltSpace's/Davis's definition says it is.
No it doesn't.

It can also happen in an Euclidean space.
For example, the lines (x, m, xm) for every x where m is the parameter are all connected by the x axis but form a hyperbolic paraboloid, on which there are many curved lines.
In Euclidean spaces you loose the specialness of the time axis that you have in curved space time.
You can't just ignore one of the part of the coordinates. You need to consider all three. You cannot get from any generic coordinate to another.
Also note that there are many curved lines on Euclidean planes.

From its perspective it's not going, from Earth's perspective it clearly changes direction.
Yes, from its perspective it can be considered stationary. Regardless, why should it change direction?

But the object that is orbiting Earth isn't walking on its surface, but going on a straight line in space-time.
And that object isn't Earth's surface. It is quite a complex concept due to the nature of space-time.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #131 on: October 22, 2017, 03:32:17 AM »
I know it isn't flat, but AltSpace's/Davis's definition says it is.
No it doesn't.
Yes it does.
The space-time is not Euclidean and it's possible to travel between any two spatial coordinates in a straight line.
It can also happen in an Euclidean space.
For example, the lines (x, m, xm) for every x where m is the parameter are all connected by the x axis but form a hyperbolic paraboloid, on which there are many curved lines.
In Euclidean spaces you loose the specialness of the time axis that you have in curved space time.
You can't just ignore one of the part of the coordinates. You need to consider all three. You cannot get from any generic coordinate to another.
Also note that there are many curved lines on Euclidean planes.
I described how a curved line can pass through the same points as a series of straight lines which are in turn connected by a straight line in an Euclidean space.
From its perspective it's not going, from Earth's perspective it clearly changes direction.
Yes, from its perspective it can be considered stationary. Regardless, why should it change direction?
Because if you stand on Earth you can see its direction changes.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #132 on: October 22, 2017, 03:42:02 AM »
Yes it does.
The space-time is not Euclidean and it's possible to travel between any two spatial coordinates in a straight line.
Again, the definition provided only applies to Euclidean spaces, not non-Euclidean space-time.

General definitions also make no special sense of time. As such, when it says "spatial coordinates" that would include the time coordinate.

I described how a curved line can pass through the same points as a series of straight lines which are in turn connected by a straight line in an Euclidean space.
No you didn't.
You described a surface which has numerous straight lines on it and numerous curves on it.
You did not show that these straight lines were connected. Instead you showed that the curved lines were connected by another curved line.
And you couldn't go between 2 points in a straight line.

Because if you stand on Earth you can see its direction changes.
So?
You are standing on Earth, not the object.
Again, this is an issue with non-Euclidean spaces.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #133 on: October 22, 2017, 05:20:14 AM »
Yes it does.
The space-time is not Euclidean and it's possible to travel between any two spatial coordinates in a straight line.
Again, the definition provided only applies to Euclidean spaces, not non-Euclidean space-time.

General definitions also make no special sense of time. As such, when it says "spatial coordinates" that would include the time coordinate.
OK, I didn't understand you.
So again, one of us doesn't understand the definition correctly.
Please AltSpace/Davis, clarify what is your definition.
I described how a curved line can pass through the same points as a series of straight lines which are in turn connected by a straight line in an Euclidean space.
No you didn't.
You described a surface which has numerous straight lines on it and numerous curves on it.
You did not show that these straight lines were connected. Instead you showed that the curved lines were connected by another curved line.
And you couldn't go between 2 points in a straight line.
OK, let's take the parabola (p, p, p˛). It is a curved line. It passes through one point from each line from the form (x, m, mx) – the point (x, x, x˛). These lines are connected by the x axis. If it's not what you meant please provide an example for what you meant.
Because if you stand on Earth you can see its direction changes.
So?
You are standing on Earth, not the object.
Again, this is an issue with non-Euclidean spaces.
Of course it's an issue with non-Euclidean spaces. In Euclidean spaces lines in the same direction from a line are just lines with the same vector. I asked you why lines that go forward/backward or up/down from a line are really the same direction from it in GR's space-time

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #134 on: October 22, 2017, 01:12:16 PM »
OK, I didn't understand you.
So again, one of us doesn't understand the definition correctly.
Please AltSpace/Davis, clarify what is your definition.
Again, the definition would only apply to Euclidean spaces. In non-Euclidean spaces, it gets far more complex as you can use a point and a normal and depending upon how you construct the plane would get different planes.

OK, let's take the parabola (p, p, p˛). It is a curved line. It passes through one point from each line from the form (x, m, mx) – the point (x, x, x˛). These lines are connected by the x axis. If it's not what you meant please provide an example for what you meant.
The parabola is not a straight line. You need a straight line that is translated based upon a straight line.
The simplest example is the XY plane.
Consider the line with vector<1,0,0>.
Translate this line along another line with vector <0,1,0>.
This generates a plane.

A more generic example is consider the line with vector <a1,b1,c1>.
Translate this along the line with vector <a2,b2,c2>, noting that the 2 lines are not equal.

Instead, of doing this, you translated a curved line along a straight line (assuming you are saying to translate it along the x axis, the line x, x, x^2 is neither straight nor the x axis).

I asked you why lines that go forward/backward or up/down from a line are really the same direction from it in GR's space-time
Because there is no reason for it to change direction.
From its reference frame, it is sitting there, doing nothing. Why would it change direction?

The simple 2D spherical space was an example of that.
If an object travels along a straight line around the equator, does it change direction (remember, no 3D so no up and down)? No. It continues going straight. Why should it be different in 3D (or 4D)?

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Username

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #135 on: October 22, 2017, 01:45:42 PM »
The truth of the matter is, the only thing Davis relativity has against it is that all the evidence agrees with it.
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Dinosaur Neil

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #136 on: October 22, 2017, 01:49:07 PM »
The truth of the matter is, the only thing conventional Round Earth Theory has against it is that all the evidence agrees with it.

Fixed it for you. Go forth and show where I'm wrong...  :-*
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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #137 on: October 22, 2017, 02:23:57 PM »
The truth of the matter is, the only thing Davis relativity has against it is that all the evidence agrees with it.
No, almost none of the evidence agrees with it.
Instead it has things like facts of geometry that goes directly against it, such as parallel lines not working the same in non-Eculidean spaces.

It is RET that has all the evidence agree with it.

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Dinosaur Neil

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #138 on: October 22, 2017, 04:06:46 PM »
The truth of the matter is, the only thing Davis relativity has against it is that all the evidence agrees with it.

Actually, measurements of neutrinos travelling from known sources through the earth and emerging elsewhere on the surface don't agree with it, but hey, don't let a previously-mentioned-in-this-thread problem prevent you from ignoring the inconvenient facts.

Edit: just realised that seismic measurements also conflict with your dullard idea too. In fact, anything which traverses below the crust of the earth pretty much knocks a hole in it, doesn't it? It relies on having the earth's surface be a defining boundary of your plane.
« Last Edit: October 22, 2017, 04:09:30 PM by Dinosaur Neil »
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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #139 on: October 22, 2017, 09:52:21 PM »
OK, I didn't understand you.
So again, one of us doesn't understand the definition correctly.
Please AltSpace/Davis, clarify what is your definition.
Again, the definition would only apply to Euclidean spaces. In non-Euclidean spaces, it gets far more complex as you can use a point and a normal and depending upon how you construct the plane would get different planes.
Again, I think I or you didn't understand the definition correctly, so I asked AltSpace and Davis to clarify it.
OK, let's take the parabola (p, p, p˛). It is a curved line. It passes through one point from each line from the form (x, m, mx) – the point (x, x, x˛). These lines are connected by the x axis. If it's not what you meant please provide an example for what you meant.
The parabola is not a straight line.
Yes, you asked for a curved line.
You need a straight line that is translated based upon a straight line.
You didn't say you translate the line, you said the straight lines have to be connected by a straight line.
The lines are (0, m, 0), (1, m, m), (2, m, 2m), etc. (I know it's impossible to list all of them, it's an uncountable infifnity), and they are connected by the x axis.
The simplest example is the XY plane.
Consider the line with vector<1,0,0>.
Translate this line along another line with vector <0,1,0>.
This generates a plane.

A more generic example is consider the line with vector <a1,b1,c1>.
Translate this along the line with vector <a2,b2,c2>, noting that the 2 lines are not equal.
OK, that's what I thought. My example is OK.
Instead, of doing this, you translated a curved line along a straight line (assuming you are saying to translate it along the x axis, the line x, x, x^2 is neither straight nor the x axis).
No, I described straight lines that are connected by another straight line.
I asked you why lines that go forward/backward or up/down from a line are really the same direction from it in GR's space-time
Because there is no reason for it to change direction.
From its reference frame, it is sitting there, doing nothing. Why would it change direction?
It doesn't change direction from its frame of reference, because it doesn't move from its frame of reference.
But there is a direction from its frame of reference that is "forward" in the beginning. Why can't this direction become "up", i.e. point away from Earth, after some time?
The simple 2D spherical space was an example of that.
If an object travels along a straight line around the equator, does it change direction (remember, no 3D so no up and down)? No. It continues going straight. Why should it be different in 3D (or 4D)?
Because it's not a spherical space, but GR's space-time.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #140 on: October 23, 2017, 12:56:57 AM »
Again, I think I or you didn't understand the definition correctly, so I asked AltSpace and Davis to clarify it.
And again, I do understand it and I know it only applies to Euclidean spaces.

Yes, you asked for a curved line.
No, I asked for a straight line.
Spefically I objected to your claim:
Quote
I described how a curved line can pass through the same points as a series of straight lines which are in turn connected by a straight line in an Euclidean space.
Where you indicate you had straight lines connected by a straight line, when you did not.

You didn't say you translate the line, you said the straight lines have to be connected by a straight line.
I thought that would have been quite obvious based upon previous discussion.

The simplest example is the XY plane.
A more generic example is consider the line with vector <a1,b1,c1>.
Translate this along the line with vector <a2,b2,c2>, noting that the 2 lines are not equal.
OK, that's what I thought. My example is OK.
No it isn't.

The line you are moving along is <1,0,0>
But the lines drawn from it are not the same.
In one case you have the line <0,1,0>
In another you have <0,1,1>
In another you have <0,1,2>
These are not the same lines.
So you have not translated a line of the form <a1,b1,b1> along <a2,b2,c2>

Also note that these are yet again different to the lines you previously had.
You previously had the lines passing through the points defined by (x,x,x^2), which would make it a curved line.

No, I described straight lines that are connected by another straight line.
Not initially. Initially you described a parabola which was translated along another parabola.

It doesn't change direction from its frame of reference, because it doesn't move from its frame of reference.
But there is a direction from its frame of reference that is "forward" in the beginning. Why can't this direction become "up", i.e. point away from Earth, after some time?
It can't be up because it would require it to be changing direction.
So if it can't change direction from its frame of reference, how can it change direction from its frame of reference?

Because it's not a spherical space, but GR's space-time.
Deal with the 2D spherical space first, then move on to more complex spaces.
« Last Edit: October 23, 2017, 12:59:54 AM by JackBlack »

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #141 on: October 23, 2017, 01:32:54 AM »
Again, I think I or you didn't understand the definition correctly, so I asked AltSpace and Davis to clarify it.
And again, I do understand it and I know it only applies to Euclidean spaces.
You may know another similar definition and think it's exactly the same definition while it isn't. That's why I asked for clarification.
Yes, you asked for a curved line.
No, I asked for a straight line.
Spefically I objected to your claim:
Quote
I described how a curved line can pass through the same points as a series of straight lines which are in turn connected by a straight line in an Euclidean space.
Where you indicate you had straight lines connected by a straight line, when you did not.

You didn't say you translate the line, you said the straight lines have to be connected by a straight line.
I thought that would have been quite obvious based upon previous discussion.

The simplest example is the XY plane.
A more generic example is consider the line with vector <a1,b1,c1>.
Translate this along the line with vector <a2,b2,c2>, noting that the 2 lines are not equal.
OK, that's what I thought. My example is OK.
No it isn't.

The line you are moving along is <1,0,0>
But the lines drawn from it are not the same.
In one case you have the line <0,1,0>
In another you have <0,1,1>
In another you have <0,1,2>
These are not the same lines.
So you have not translated a line of the form <a1,b1,b1> along <a2,b2,c2>
OK, but how can you translate a line in a non-Euclidean space unambiguously?
Also note that these are yet again different to the lines you previously had.
You previously had the lines passing through the points defined by (x,x,x^2), which would make it a curved line.
No, I described straight lines that are connected by another straight line.
Not initially. Initially you described a parabola which was translated along another parabola.
It's possible to create a hyperbolic paraboloid in many ways. One way is translating a parabola along another parabola, a different way is to translate a line along another line while rotating it in a specific way.
It doesn't change direction from its frame of reference, because it doesn't move from its frame of reference.
But there is a direction from its frame of reference that is "forward" in the beginning. Why can't this direction become "up", i.e. point away from Earth, after some time?
It can't be up because it would require it to be changing direction.
So if it can't change direction from its frame of reference, how can it change direction from its frame of reference?
I don't understand what that even mean to change its direction from its frame of reference. In its frame of reference it isn't moving so how does it have a direction?
Because it's not a spherical space, but GR's space-time.
Deal with the 2D spherical space first, then move on to more complex spaces.
I understand why "forward" stays consistent in spherical space, but you are yet to explain why it stays in GR's space-time.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #142 on: October 23, 2017, 01:46:35 AM »
OK, but how can you translate a line in a non-Euclidean space unambiguously?
By looking at a local, Euclidean representation at each point, as I did.

It's possible to create a hyperbolic paraboloid in many ways. One way is translating a parabola along another parabola, a different way is to translate a line along another line while rotating it in a specific way.
Yes, and that rotation means it is no longer simply translating a line along another line, and thus the surface is not flat.

I don't understand what that even mean to change its direction from its frame of reference. In its frame of reference it isn't moving so how does it have a direction?
Even though it isn't moving, it still has a direction which constitutes forwards, etc. It is still facing some direction.

I understand why "forward" stays consistent in spherical space, but you are yet to explain why it stays in GR's space-time.
Because they utilise the same principles. You have a non-Euclidean space where the object is travelling along a straight line.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #143 on: October 23, 2017, 01:51:58 AM »
OK, but how can you translate a line in a non-Euclidean space unambiguously?
By looking at a local, Euclidean representation at each point, as I did.
OK, just make sure you did it right.
It's possible to create a hyperbolic paraboloid in many ways. One way is translating a parabola along another parabola, a different way is to translate a line along another line while rotating it in a specific way.
Yes, and that rotation means it is no longer simply translating a line along another line, and thus the surface is not flat.
OK.
I don't understand what that even mean to change its direction from its frame of reference. In its frame of reference it isn't moving so how does it have a direction?
Even though it isn't moving, it still has a direction which constitutes forwards, etc. It is still facing some direction.

I understand why "forward" stays consistent in spherical space, but you are yet to explain why it stays in GR's space-time.
Because they utilise the same principles. You have a non-Euclidean space where the object is travelling along a straight line.
Whatever.

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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #144 on: October 23, 2017, 01:53:23 AM »
Again, AltSpace and Davis, when you look here please clarify your definition.

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #145 on: October 24, 2017, 06:09:00 PM »
Again, AltSpace and Davis, when you look here please clarify your definition.
The Earth's flat surface?
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.

Your parabola example completely misses the point, you are simply stating that it isn't straight (and therefore not flat) as applied to a Euclidean space assuming flat space.

If you warp the space-time field so a path forms a parabola relative to some more homogeneous space surrounding it, then you could describe such a path as inertial with a trajectory and straight.
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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #146 on: October 24, 2017, 06:20:08 PM »
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.
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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #147 on: October 24, 2017, 06:24:11 PM »
Of course, "Nowhere in that article does it claim that GR boils down to a sphere in Euclidean space-time", it doesn't attempt to.
GR cannot prove the shape of the earth. GR is, among other things, a relativistic theory of gravitation.
So it could no more prove anything about the shape of planets than could Newton's Universal Law of Gravitation.[1]
General Relativity is a geometric theory of gravity, it is definitely relevant to the Earth's geometry. We define surfaces and vectors by space-time itself, it being non-homogeneous changes things a lot.
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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #148 on: October 24, 2017, 07:18:14 PM »
But the real killer is the cyan line.
This is a line that shows a geodesic starting at a point on Earth's surface, as Earth's surface. It starts off at the same velocity as Earth's surface. Notice how it doesn't stay on Earth's surface?
It is a sub-orbital velocity. This would cause it to fall to the centre of Earth.
That is what a straight line through space-time would be for the surface of Earth.

Notice that it doesn't follow the surface of Earth?
It falls away.
So no, Earth's surface, when considered as a surface through space-time, is NOT FLAT!!!
Instead, it is constantly curving upwards/expanding. The reason it doesn't "appear" to be and it instead remains the same size is that the time axis is curving in towards the centre of Earth.
Yeah, because sub-orbital velocity is not traversing 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.

It's funny how roundies will do whatever they can to twist concepts to get their favor, they like GR and when it conflicts with their round earth faith, they forget the concept of curved space-time.
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AFanOfTruth

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #149 on: October 24, 2017, 09:53:46 PM »
Again, AltSpace and Davis, when you look here please clarify your definition.
The Earth's flat surface?
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.
Nope. You didn't answer my request.
This was your definition:
Now lets look at the definition I used:
Quote
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?

Your parabola example completely misses the point, you are simply stating that it isn't straight (and therefore not flat) as applied to a Euclidean space assuming flat space.

If you warp the space-time field so a path forms a parabola relative to some more homogeneous space surrounding it, then you could describe such a path as inertial with a trajectory and straight.
My example was in an Euclidean space-time. I can choose in which space-time my example is.