Davis Relativity Model (Debate/discussion edition)

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #60 on: October 09, 2017, 11:13:34 PM »
Except that light travels straight in curved space and if a surface can't be traced by light it isn't flat.
Geodesic's aren't defined by light. Update your knowledge by reading this useful post: https://physics.stackexchange.com/questions/20069/what-does-it-mean-for-objects-to-follow-the-curvature-of-space
Geodesics are independent of velocity to attain them.
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That space-time is populated by events renders the idea of a surface in space-time meaningless, in my view.

I accept that there is a 4D space-time continuum that can be altered according to General Relativity, that's what I use.
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What does 'flat' in space-time mean to you?
Surface parallel to a straight line, of the same vectors.
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Even should you manage to attach a meaning to it you are still faced with the issue that your choice of reference geodesic is actually completely arbitrary. That you choose the geodesic of a circular orbit is entirely governed by your desired result of a 'flat' earth.
Only one is parallel to the Earth's surface, a straight line, which has a flat surface.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #61 on: October 09, 2017, 11:56:44 PM »
Space is flat? So I reckon that you aren't claiming that Earth is flat in non-euclidean space but rather a spheroid in flat space (like Minkowski Space)?
Yes, though technically it could be flat (well, mostly flat) in non-euclidean space as long as there are at least 3 spatial dimensions, and we would have literally no way of telling. It's a moot point.
How is it a moot point in determining if Earth is flat? If I can trace a parallel line to the surface that can be represented in a 2D plane (since the lines have no width of course) and it's straight, then isn't that a way of telling?
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #62 on: October 10, 2017, 01:28:55 AM »
As I've said, it is the only accurate determination of the geometry of Earth's surface, and it's straight and therefore flat.
Yes, you have said that repeatedly, but you are yet to justify it. Until you do, your argument is just a pile of crap.

I see what you mean here, but it doesn't rebut my point here, as I will explain.
Yes it does, as the example I provided clearly does.

Address that example as it uses your method and gets a completely false result.

What I'm claiming here is that if you have a line cross a surface, and you draw a straight line to the surface across the traversal at an equal distance, it will reveal the geometry of the surface. The key word here is "surface", I used that in the definition.
Your example with the wall 1000 km or so north of the equator you are traversing is able to get a parallel straight line geodesic parallel to a curved line, this is because in Non-Euclidean geometry (will represent a sphere here for demonstration), the line traversing the surface is parallel to a line straight across, which also change position in a third dimension relative to the parallel lines, which means this can only be represented in a 3D coordinate system.
No. It can be represented in a 2D non-Euclidean system.
You can also represent it in a 3D Euclidean system noting that it is a surface.
The same applies to space-time, which can be represented as a 4D surface in 5D.

Again, here is a 2D representation, noting that it is non-Euclidean space and thus the line, while it appears curved, is actually straight.


The blue and green lines are straight, the red and purple lines curve.

This is required to define the curved and straight as parallel in this specific case.
No, not just this specific case.
In all cases dealing with non-Euclidean geometry.

This is because this isn't relying on this principle of non-euclidean geometry to determine a flat surface, so it doesn't hold and is irrelevant.
No, you are trying to define a surface in non-Euclidean geometry. You can do the same with the wall example.
Rather than accepting reality of the wall is (a curved line), you can instead try to determine the surface (you can't say you are using to determine a flat surface as that is what you are trying to prove. If you need to start with the asserted that Earth is flat you have made your entire argument useless).

Using your method you will conclude that the red and purple circles are straight. In reality, they are not.

No it's not, since one doesn't rely on 3D non-euclidean geometry to determine this
Neither rely on 3D. Both are 2D problems. (alternatively you can state yours is 4D, mine is 2D)
Yours uses radius and longitude (or fraction around orbit).
Mine uses latitude and longitude, where latitude can be treated as the radius.

Both are allegedly trying to map surfaces in non-euclidean space.

the only reason why the curved and straight line can be parallel is because it is all plotted in an exclusively 3D coordinate system, it can't be represented in a 2D plane, like the surface geodesics can (these lines passing straight through the ground can be represented as 2D and straight parallel lines being the same distance apart holding.
Again, pure bullshit.
I represented it in a 2D plane. Straight "parallel" lines were not the same distance apart.
Instead a straight line remained the same distance from 2 non-straight lines.

It doesn't, there is a clear distinction.
No there isn't.
Again, both work exactly the same.

You have a reference line. This line is straight.
You have some other line which you are trying to determine if it is straight.
You traverse your reference line, measuring the distance to the other line (perpendicular to your reference).
The other line remains the same distance during the entire traversal.
Thus you conclude the other line is straight as well.


This describes both situations.

(Again, I'm ignoring the rest of your bullshit until you deal with this).
You failed to, you went on to an irrelevant 'gotcha' with non-euclidean geometry that misses how the surface is being determined here, I hardly call that 'showing it's not the case'.
No. I went on a highly relevant "gotcha", with non-Euclidean geometry, and attempting to determine a "surface" just like you were, showing that it does not work.
So no, that is showing it is not the case.

For flat-space it is. Not for curved space.
Yes, for flat space it is the case. 2 straight, parallel lines remain equidistant. This is not the case for curved space (or curved space-time).


But as I have said, it can only match the surface geometry it is parallel, an equal distance to match the changes in the surface directly, doing this with Earth (in non-euclidean space-time) reveals it's flat.
Similarly, doing it with a line which remains 10 degrees north of the equator of a sphere in spherical geometry shows that that is straight. But in reality, it is curved.
So no, doing this doesn't reveal anything except your dishonesty and unwillingness to admit error.
This doesn't work.

Exactly, and a straight line has the same vector as the Earth's surface and can be represented in a 2D plane as actually equivalent lines and not the non-euclidean 3D exclusive example you brought up which isn't relevant.
Is your space-time Euclidean or non-Euclidean.
Because this argument seems to be indicating your space-time is flat, going completely against GR, meaning orbits are no longer geodesics, and which would show that Earth is round.

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You seem to like going off on tangents to cover up your complete failure.
How about you stop setting up such pathetic strawmen and instead focus on what I have actually said?
As I have been doing.
No you haven't. You have been continually dismissing them as irrelevant without any rational justification.

Only, as I have shown, the paths with the surface can be represented in 2D plane, and when you do that, the lines are not parallel as they are not an equal distant apart
And here you go off on circular reasoning yet again.
That is entirely the point, they are parallel yet don't remain the same distance apart because it is not in flat space.
Until you justify this baseless claim of yours, where you use something which only applies in Euclidean space, the rest of your argument is pure bullshit.
You also need to address where I showed this is not the case.

Only, the moment we transfer this to the plane representation of the line cutting to the surface (the line doesn't have width so it doesn't require 3D to represent), we reveal that the straight line reveals a flat surface. Parallel lines can't converge since they are an equal distance apart constantly by definition
Nope. Exactly the opposite. When we transfer this non-Euclidean space to a Euclidean representation, the straight lines appear to curve and parallel lines can converge or diverge.

They only remain the same distance apart by definition in Euclidean space.
Outside of Euclidean space that axiom does not hold.
Instead it is replaced by 1 of 2 other options for parallel lines.
The first one is the simplest which is simply any 2 lines in a plane which do not intersect.
In this case, spherical geometry has no straight parallel lines; Euclidean geometry has a single straight line passing through a point, which is parallel to another straight line; and hyperbolic geometry has infinitely many.

Perhaps this will help you, this is for hyperbolic space:


The other option is using one of the methods of creating a straight parallel line through a point:
You take a straight line (the reference one). You draw another straight line (the construction line) perpendicular to it (technically at any angle), which passes through the point, and then draw a straight line through the point perpendicular (or at a supplementary angle) to the construction line. This line will be parallel to the first.
The issue is that it relies upon key points on the lines, as if you move along the lines, this supplementarity of the lines doesn't always hold.

their vectors are the same.
No they aren't.
The vectors in non-Euclidean geometries are significantly more complex than that.

Believe it or not, I still fail to see how I'm wrong.
It's quite simple, the red and purple lines remain the same distance from the "straight" green line, yet are not straight.
Thus, the method you are trying to use is leading to a false conclusion. As such, it is flawed (i.e. WRONG).
Is that simple enough?

In order for you to be correct, the red and purple lines need to be straight.

Not in this case, you have to look at why this is the case in the non-euclidean example you used
I already know why. It is because the argument you are using ONLY APPLIES TO EUCLIDEAN SPACE!!!!
It does not apply to non-Euclidean spaces.

Instead of trying to look at why it shouldn't apply, try to figure out when it should and what conditions are required for it.

when you have that and apply it to the surface geodesics, it is revealed that they are not connected and therefore irrelevant.
Again, PURE BULLSHIT!!!
Your geodesics are in non-Euclidean space. As such, you have the same issues. It is connected and thus is relevant.

In my case, Earth is a flat surface since an equidistant line traverses it as straight
And in my case, the red and purple lines (which are not straight) were deemed to be straight as an equidistant line traverses it as straight.

the non-euclidean nature of space-time isn't quite relevant in this case (other than that a straight line meets itself because it's in non-euclidean space)
No, it is relavent as it means the parallel lines do not remain the same distance apart.

since the line has no width
It doesn't matter.
The "width" corresponds to the distance between the line and the surface of Earth.

Then show that this is the case with my specific point here with a line parallel to a surface of earth (You will only need to represent it in 2D since the lines have no width, just distance from the surface and length across) and I will abandon this model.
That would require you to first accept that you are wrong and that your method does not work.
You are basically asking me to show your method is wrong, without using anything other than your method and the thing you are applying it to.
In what world does that sound reasonable?

I'm taking the more honest/rational approach. Using your method in a situation where there is agreement on what the facts are (that the red and purple lines are curved), and thus showing your method does not reach the correct conclusion and thus is wrong.

How is it a moot point in determining if Earth is flat? If I can trace a parallel line to the surface that can be represented in a 2D plane (since the lines have no width of course) and it's straight, then isn't that a way of telling?
No it isn't, because you aren't tracing a parallel line, you are tracing a line which is equidistant, but more importantly, because Earth isn't in flat space-time.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #63 on: October 10, 2017, 01:46:33 AM »
And again, to be nice, a short version:
It can be represented in 2D, including my version, the width in yours which you claim is missing is the distance between Earth and the line.
Here is the representation of my case in 2D (which also works for your case but gets more complicated due to the time axis, and elliptical orbits not being the same):


Here is your method:
You have a reference line that is straight (the green line).
You have another line which you would like to determine if it is straight (the red or purple line).
You start at some point along the green line, and measure the distance to the other line.
You then move along the green line, measuring the distance distance to the other line.
You note the distance remains the same.
Thus you conclude that the line is straight.

That is your method. You note the distance remains the same and use that to conclude it is straight.
But, this is wrong. In reality the red and purple lines are curved and the blue line is straight.

You can also stick in an elliptical orbit which goes between the red and green line. Because it is an orbit, it is straight. But it doesn't remain the same distance.


Your method only works in Euclidean space.

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Master_Evar

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #64 on: October 10, 2017, 01:54:03 AM »
How is it a moot point in determining if Earth is flat? If I can trace a parallel line to the surface that can be represented in a 2D plane (since the lines have no width of course) and it's straight, then isn't that a way of telling?

Can you elaborate? Trace a parallel line from where to the surface, and what is this line tracking? The picture doesn't help, too many lines.
Math is the language of the universe.

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

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

When in doubt; sources!

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #65 on: October 10, 2017, 02:32:31 AM »
How is it a moot point in determining if Earth is flat? If I can trace a parallel line to the surface that can be represented in a 2D plane (since the lines have no width of course) and it's straight, then isn't that a way of telling?

Can you elaborate? Trace a parallel line from where to the surface, and what is this line tracking? The picture doesn't help, too many lines.
I think it does help to some extent. It helps to show why he is wrong.
You have the orbital path. This is a geodesic in space-time, and thus is "straight".
Below it, you have the path of a ball. This is also a geodesic in space-time and thus is "straight".
But the vertical distance between these 2 paths varies as you go across the picture. Thus the distance between 2 straight lines (which are parallel in this hyperbolic space) (at least it has some properties of hyperbolic space) varies.
Meanwhile you have a non-gravitationally affected path. This path thus does not follow a geodesic in space-time (if it did it would be the path of the ball) and thus it is not straight..
This line does remain the same distance from the orbital path.
This shows that just because a line remains the same distance from a straight line does not mean it is straight.

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Master_Evar

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #66 on: October 10, 2017, 06:13:08 AM »
I think it does help to some extent. It helps to show why he is wrong.
You have the orbital path. This is a geodesic in space-time, and thus is "straight".
Below it, you have the path of a ball. This is also a geodesic in space-time and thus is "straight".
But the vertical distance between these 2 paths varies as you go across the picture. Thus the distance between 2 straight lines (which are parallel in this hyperbolic space) (at least it has some properties of hyperbolic space) varies.
Meanwhile you have a non-gravitationally affected path. This path thus does not follow a geodesic in space-time (if it did it would be the path of the ball) and thus it is not straight..
This line does remain the same distance from the orbital path.
This shows that just because a line remains the same distance from a straight line does not mean it is straight.
I see.
Math is the language of the universe.

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

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

When in doubt; sources!

Re: Davis Relativity Model (Debate/discussion edition)
« Reply #67 on: October 11, 2017, 02:45:50 AM »
Except that light travels straight in curved space and if a surface can't be traced by light it isn't flat.
Geodesic's aren't defined by light. Update your knowledge by reading this useful post: https://physics.stackexchange.com/questions/20069/what-does-it-mean-for-objects-to-follow-the-curvature-of-space
Geodesics are independent of velocity to attain them.
Your link (or at least the replies to the question therein) are clear that they relate to curved space-time. I was writing about curved space.

Nevertheless, JackBlack has pointed out that I may be not wholly accurate in my claim. However, within the scale of the earth, light won't deviate from curved space geodesics by very much, and not remotely by enough to detract from my point. Which is that the earth is not flat in curved space.

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That space-time is populated by events renders the idea of a surface in space-time meaningless, in my view.

I accept that there is a 4D space-time continuum that can be altered according to General Relativity, that's what I use.
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What does 'flat' in space-time mean to you?
Surface parallel to a straight line, of the same vectors.

So what does a 'surface' in space-time mean?

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #68 on: October 11, 2017, 04:34:16 AM »
So what does a 'surface' in space-time mean?
I had been working on something along those lines.
If I could figure out simple parametric based solutions (where time is the parameter) for an elliptic and hyperbolic orbit, and an "orbit" that goes straight down I would be further along.

But this is what I have so far:

This is 2D space+time, the vertical axis is time, and the other 2 are spatial.

All of these lines are "straight" in curved space time.
The 2 helixes correspond to circular orbits.
The 3 other lines correspond to paths purely through space (i.e. not moving through time).
I was hoping to be able to add in some other options as well.

Anyway, as this shows, surfaces in space-time are quite complex, and you will get different results depending upon how you make a surface.

In Euclidean geometry, where there is a single plane normal to a line through a point, you can use at least 2 techniques to make a plane and always get the same plane:
You can start with the point and normal, and draw many many straight lines through the point normal to the normal, this would be akin to a plane of rotation as you are effectively rotating a straight line about the point.
Or you can start with the point and a normal, draw a single straight line through the point normal to the normal, then draw multiple lines normal to both the previous 2 lines, this would be akin to a plane of translation as you are effectively translating a line along a line that passes through the point. You can also have the translated line at different angles. Technically, the lines don't even need to be straight, as long as both lines remain normal to the normal.

In non-Euclidean space-time, this technique no longer works. Firstly, as it is 4D, the surface is a 3-space (at least it would be if the space was Euclidean), but lets ignore that for now and deal with 2D space+time. In this case, as it is 3D, a "surface" should be 2D.
As I can't easily plot the surface of other orbits, I am stuck with just those kinds of lines and thus I can't do a plane of rotation. But needless to say, it is somewhat complex and intersects itself repeatedly (as the 2 helixes would), including hyperbolic, parabolic and elliptic orbits between the red line and the others, and including elliptical and sub-orbital trajectories past the others. (these 3 lines at the point of intersection share a common normal of [+x,0,0]

But I have shown parts of a plane of translation.
The purple line is a result of translating the red line along the blue line. This would produce a winding sheet.
But you can also translate it in other directions, such as to get the yellow line.

Even more fun, you can translate it through the spatial dimensions and just get a flat plane, and in this flat plane, the surface of Earth (the grey cylinder), is a circle.

I may try it with angle as a parameter later.
« Last Edit: October 11, 2017, 05:04:12 AM by JackBlack »

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #69 on: October 12, 2017, 07:03:45 AM »
There is no such thing as the Ferrari effect.
:P :P Miracle of miracles! For about the first time I agree with one statement from Sandokhan. :P :P
I doubt it will happen again!

I only worry that rab may have received the maximum safe recommended life time dosage of ioning << spell checker suggests: :D ironing  :D>> radiation.  Did you even take your potasium pills?
What is :P ioning radiation? Do you mean ironing radiation? In case you didn't know there are radiation free irons available.
I have a good supply here and I can forward a couple if you'll just send you to bank account details to rabinoz@radiation_free_ions.co.ng
;D happy to oblige. ;D

Have you been taking "potassium pills"? What an idiot!
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Potassium can cause stomach upset, nausea, diarrhea, vomiting, intestinal gas, and other side effects. Too much potassium is UNSAFE and can cause feelings of burning or tingling, generalized weakness, paralysis, listlessness, dizziness, mental confusion, low blood pressure, irregular heart rhythm, and death.
Those symptoms seem familiar?
I guess that mental confusion you continually suffer from is due to the "potassium pills".
You probably mean "bromide pills".
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Potassium bromide (KBr) is a salt, widely used as an anticonvulsant and a sedative in the late 19th and early 20th centuries, with over-the-counter use extending to 1975 in the US. Its action is due to the bromide ion (sodium bromide is equally effective).
But that's so old fashioned and not usually recommended these days.


But you think that the Ferrari Effect is real? Duh!
Leo Ferrari was a real joker and it looks as though you and John Davis haven't caught on yet.
You did study the video carefully? so that you will understand:
          "light curved by mass makes the earth look a sphere from space" at 14:31 - gee, that sounds like the Ferrari Effect! Would a YouTube video ever mislead you  :D?
          "Earth Moves up and down"  at 19:50 - but I thought that the earth was stationary! Oh, well live and learn :D.
          "Space/non-Space"  at 20:30
          "Cows in Antarctica" at 22:35 - must be real hardy cows, they sure breed 'em tough down there!
       


Ferrari was no joker; he was a world renowned academic. As far as his flat earth stuff goes, it seems as if you wish to poison the well. Just because he proposed an idea in jest (I don't believe he did; he seems to have proposed it as a philosophical interest), does not mean the idea is bunk.

He also had many views that differ from the real society. This does not mean we cannot, as Rowbotham suggests, make useful his work.

As far as antarctic cows - could he not be referencing Penguins?
So long and thanks for all the fish

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #70 on: October 12, 2017, 05:46:18 PM »
No. It can be represented in a 2D non-Euclidean system.
You can also represent it in a 3D Euclidean system noting that it is a surface.
The same applies to space-time, which can be represented as a 4D surface in 5D.
It all boils down to this, so I may as well sum it up here.

Demonstrate that it applies with a traversal parallel to the surface, showing that the line traversing parallel to the surface would not show the surface is flat with a straight line.
You tried to show that this isn't true by pointing to the wall example, I am saying this doesn't apply specifically to the example of a traversal line being equidistant from the surface.

And to the larger point, I am claiming that surface itself is a flat surface in curved space-time, it would be under this scenario.

Now, to your illustration, with what you previously said:
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This can represent several things, but the specific one I am having it represent is 2D spherical geometry.

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

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

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

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

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

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

Now do you understand the problem with your method?
This is not my method. You take lines that are in different areas on the surface, with only two crossing through the center (the blue and green). With what I am saying, you got one straight line traversing above the equator, straight lines draw down to the equator that are of equal length throughout the entire traversal, and the equator is therefore straight itself and so a flat surface. You can do this with any straight line geodesics parallel to the surface.
Now, if you care to take it down, demonstrate how your 'wall' example showing straight lines parallel to curved lines applies to this specifically and renders it useless.

Do you not agree that a surface can be flat in non-euclidean space-time?
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #71 on: October 12, 2017, 05:49:37 PM »
How is it a moot point in determining if Earth is flat? If I can trace a parallel line to the surface that can be represented in a 2D plane (since the lines have no width of course) and it's straight, then isn't that a way of telling?

Can you elaborate? Trace a parallel line from where to the surface, and what is this line tracking? The picture doesn't help, too many lines.
You see the 'Orbit path' line and 'the earth' line? The orbit traversal path and the surface being two straight parallel lines.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #72 on: October 12, 2017, 05:54:08 PM »
I think it does help to some extent. It helps to show why he is wrong.
You have the orbital path. This is a geodesic in space-time, and thus is "straight".
Below it, you have the path of a ball. This is also a geodesic in space-time and thus is "straight".
But the vertical distance between these 2 paths varies as you go across the picture. Thus the distance between 2 straight lines (which are parallel in this hyperbolic space) (at least it has some properties of hyperbolic space) varies.
Meanwhile you have a non-gravitationally affected path. This path thus does not follow a geodesic in space-time (if it did it would be the path of the ball) and thus it is not straight..
This line does remain the same distance from the orbital path.
This shows that just because a line remains the same distance from a straight line does not mean it is straight.
Both are straight since they have a consistent tangent vector touching at every point, just one is in non-euclidean space-time and the other in flat euclidean space-time, they are equivalent in that they are both straight lines, it's just that the space they are within is non-homogeneous relative to the other. Also, of course, the non-euclidean space-time has the straight line meet itself.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #73 on: October 12, 2017, 06:09:51 PM »
Your link (or at least the replies to the question therein) are clear that they relate to curved space-time. I was writing about curved space.
There is no distinction, they are both one and the same continuum.
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Nevertheless, JackBlack has pointed out that I may be not wholly accurate in my claim. However, within the scale of the earth, light won't deviate from curved space geodesics by very much, and not remotely by enough to detract from my point. Which is that the earth is not flat in curved space.
*space-time

The photon sphere concept points out how wrong this is:
Where photons (light) can be in orbit: r = 3GM/c˛
r is the Schwarzschild radius (radius of the event horizon), G is the gravitational constant, M is the mass (black hole mass in this case), and c is the speed of light in a vacuum.
With Earth:
(3*6.674×10^-11*5.9726x10^24)/(299792458*299792458) =1.330545 cm.
The Earth would need to have a less than 1.33 cm radius for light orbit.
A mass like the ISS or a human could follow geodesics, but light goes beyond escape velocity on Earth. So, you can hardly use that reasoning as 'close enough'.

Quote
So what does a 'surface' in space-time mean?
It means the upper most layer of the Earth, with land and water mainly in this case (not exactly the atmospheric strata). If you want a specific defining case, then imagine the Earth being covered completely in water, water serves as an excellent level defining the surface, it follows the path across, and that would represent the surface.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #74 on: October 13, 2017, 12:09:44 AM »
Isn't it funny when the roundies are wrong? All of a sudden 'oh it just boils down to a round earth'. This is a fancy pants way of saying 'yeah it matches our experimental evidence, but I still believe it's round.' Downright silly. Clearly, we are both equally wrong. But I suppose that's relative. From flat earth to quasar.
So long and thanks for all the fish

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rabinoz

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #75 on: October 13, 2017, 12:26:16 AM »
But you think that the Ferrari Effect is real? Duh!
Leo Ferrari was a real joker and it looks as though you and John Davis haven't caught on yet.
You did study the video carefully? so that you will understand:
          "light curved by mass makes the earth look a sphere from space" at 14:31 - gee, that sounds like the Ferrari Effect! Would a YouTube video ever mislead you?
          "Earth Moves up and down"  at 19:50 - but I thought that the earth was stationary! Oh, well live and learn!
          "Space/non-Space"  at 20:30
          "Cows in Antarctica" at 22:35 - must be real hardy cows, they sure breed 'em tough down there!
Ferrari was no joker;
After seeing that video, along with the embedded clues you expect us to believe that Our Leo Ferrari was no joker'.
What with "Earth Moves up and down", "Space/non-Space" and "Cows in Antarctica"!

Quote from: John Davis
he was a world renowned academic.
Yes, I quite agree that Leo Ferrari "was a world-renowned academic" and one of his interests was the type of education that we are giving our children.
He, along with Neil deGraffe Tyson believed that many school-children are force-fed facts, making education more like an indoctrination. He put it like this:
Quote from: Neil deGrasse Tyson
This is a deep failure of our educational system [...] There's something deeper going on in our society that somehow enables people to believe they're making cogent arguments when they're not.
You know what I think it is - the way we teach science is:

You’re just some empty vessel and we pour this science into you and then you regurgitate in on an exam –
Whereas science is a way of thinking; it’s a way of understanding and probing the operations of nature
He's bothered that modern science teaching is close to indoctrination and not really teaching what science is and how to think.

The video

Flat Earth - In search of the edge (1990 Documentary)
is just one of Leo Ferrari's educational tools to assist the students in understanding the shape of the earth.

Quote from: John Davis
As far as his flat earth stuff goes, it seems as if you wish to poison the well.
Not I, Leo Ferrari did that quite intentionally with the clues in that video!
Quote from: John Davis
Just because he proposed an idea in jest (I don't believe he did; he seems to have proposed it as a philosophical interest), does not mean the idea is bunk.
He also had many views that differ from the real society. This does not mean we cannot, as Rowbotham suggests, make useful his work.
As far as antarctic cows - could he not be referencing Penguins?

Penguins?
Yup, sure look like heavily bio-engineered penguins, bred as a food source!.

In search of the edge of the flat earth 1990 All parts at 22:35
But on second thoughts,
it might be just possible that our joker, Leo Ferrari, might be planting clues that the video is not to be taken seriously.
« Last Edit: June 29, 2018, 04:57:09 AM by rabinoz »

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #76 on: October 13, 2017, 12:36:21 AM »
The orbit of earth is a flat line. This is explained in the aforementioned and linked post. What do you see an inertial path travelling? Some sort of path that is changing direction?

If a moon flies around, it surely isn't round.
« Last Edit: October 13, 2017, 12:39:20 AM by John Davis »
So long and thanks for all the fish

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #77 on: October 13, 2017, 01:07:13 AM »
Demonstrate that it applies with a traversal parallel to the surface, showing that the line traversing parallel to the surface would not show the surface is flat with a straight line.
I did. You ignored it (technically you dismissed it as not-relevant)
Again, here it is:

The green line is straight. This is a great circle in spherical geometry.
The red and purple lines remain the same distance (your definition of parallel), yet both curve.
Meanwhile, another straight line, the blue line, has the distance vary.

As such, you have a straight line "parallel" to a non-straight line.
As such, your method does not work.

I am saying this doesn't apply specifically to the example of a traversal line being equidistant from the surface.
Yes, you are saying it, but you are yet to back it up in any way.

And to the larger point, I am claiming that surface itself is a flat surface in curved space-time, it would be under this scenario.
The only surface that makes sense for an object is one that is at a specific time point. This makes it purely spatial, and thus allows you to describe the surface of the object. This results in Earth being round.
As soon as you include any temporal aspect you are no longer describing an object and instead are describing a trajectory.

This is not my method.
If it isn't your method, point out what part is not, or clearly describe your method, in point form like that.

You take lines that are in different areas on the surface, with only two crossing through the center (the blue and green).
Exactly as you do, taking lines in different areas on a space-time surface.
Unless you are trying to say that the distance between your hypothetical orbit and the surface of Earth is always 0?

With what I am saying, you got one straight line traversing above the equator, straight lines draw down to the equator that are of equal length throughout the entire traversal, and the equator is therefore straight itself and so a flat surface. You can do this with any straight line geodesics parallel to the surface.
And that is what I did.
You have a straight line traversing about the pole (or 10 degrees north (I think, it may be different in this example, but the argument still holds)). Straight lines drawn down (well up) to 10 degrees north remain equal length throughout the entire traversal, thus 10 degrees north is therefore straight itself.

So notice how you can't actually do this in non-Euclidean geometry?

Now, if you care to take it down, demonstrate how your 'wall' example showing straight lines parallel to curved lines applies to this specifically and renders it useless.
I have. Now if you care to take it down, demonstrate how my wall example is wrong.
Do so quite explicitly, showing exactly where I have went wrong, as I have a line which remains equidistant from a straight line, not being straight.
That is exactly what you have.
Both examples (for what we are considering anyway) is 2D. You have the dimensions (r,theta), where r is the distance above Earth and theta is the angle around the orbit. (technically you also have a time dimension, but I'm ignoring that for now).
I have the dimensions (phi,theta), where phi is the angular distance from the north pole and theta is the angular distance around the space.

Do you not agree that a surface can be flat in non-euclidean space-time?
Yes, although a surface in space-time is quite a complex thing as is any surface in non-euclidean space due to different surfaces produced by different methods, primarily because the parallel postulate no longer holds. That is you no longer have a single unique, parallel line through a point.

You see the 'Orbit path' line and 'the earth' line? The orbit traversal path and the surface being two straight parallel lines.
No. The 'Orbit path' and the 'Path of our ball' are 2 straight lines.

the other in flat euclidean space-time
No. Both are in non-Eculidean space-time.

There is no distinction, they are both one and the same continuum.
No, there is a very significant distinction.
Space does not have the temporal aspect. It does not have the curvature due to time.
It allows you to describe the shape of objects rather than trajectories.

*space-time
The photon sphere concept points out how wrong this is:
He was discussing space, not space time.
Orbits are trajectories, not objects and thus are in space-time, not space.

A mass like the ISS or a human could follow geodesics, but light goes beyond escape velocity on Earth. So, you can hardly use that reasoning as 'close enough'.
No he can. Light is not significantly deflected by Earth's gravity. This allows light to be used as a fairly accurate measurement for the curvature of space itself, as its trajectory through spacetime is not significantly curved by time.

It means the upper most layer of the Earth
Not specifically for Earth, in general.

water serves as an excellent level defining the surface, it follows the path across, and that would represent the surface.
But as it is not in free fall it does not represent a straight line through space-time.

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #78 on: October 13, 2017, 01:10:15 AM »
Isn't it funny when the roundies are wrong? All of a sudden 'oh it just boils down to a round earth'. This is a fancy pants way of saying 'yeah it matches our experimental evidence, but I still believe it's round.' Downright silly. Clearly, we are both equally wrong. But I suppose that's relative. From flat earth to quasar.
And when are the roundies wrong?
Do you mean above when light following straight paths through space and space-time were confused? Big deal. It doesn't make Earth flat.

All evidence indicates Earth is round.
The only evidence indicating Earth is flat is ambiguous and also indicates Earth is round.
So when that is presented, it is a case of saying "so what?" You have evidence that can't distinguish between a flat and round Earth, that isn't going to make me think Earth is flat.

What do you see an inertial path travelling?
In GR, inertial reference frames are infinitesimally small (aka "local"). This is because as soon as it is any real size, the curvature of space-time results in the frame distorting and thus not being inertial.

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #79 on: October 13, 2017, 01:21:05 AM »
Isn't it funny when the roundies are wrong? All of a sudden 'oh it just boils down to a round earth'. This is a fancy pants way of saying 'yeah it matches our experimental evidence, but I still believe it's round.' Downright silly. Clearly, we are both equally wrong. But I suppose that's relative. From flat earth to quasar.
And when are the roundies wrong?
Do you mean above when light following straight paths through space and space-time were confused? Big deal. It doesn't make Earth flat.

All evidence indicates Earth is round.
The only evidence indicating Earth is flat is ambiguous and also indicates Earth is round.
So when that is presented, it is a case of saying "so what?" You have evidence that can't distinguish between a flat and round Earth, that isn't going to make me think Earth is flat.

What do you see an inertial path travelling?
In GR, inertial reference frames are infinitesimally small (aka "local"). This is because as soon as it is any real size, the curvature of space-time results in the frame distorting and thus not being inertial.
Think of any advance ever since the advent of the round earth model. That is where roundies were 'wrong.'.

Yes, and the universe is made up of infinitesimally small local frames. And they all agree - the earth is flat.
So long and thanks for all the fish

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #80 on: October 13, 2017, 01:33:18 AM »
Think of any advance ever since the advent of the round earth model. That is where roundies were 'wrong.'.
No, it isn't.
Sliced bread didn't mean they were wrong.
You also need to learn the difference between being wrong and having an approximate answer.
Having Earth being an oblate spheroid doesn't mean they were wrong. Earth still isn't flat.

Yes, and the universe is made up of infinitesimally small local frames. And they all agree - the earth is flat.
No. The vast majority of them don't have Earth in them.
The ones which do only have an infinitesimally small part of Earth and thus are able to make no statement regarding its shape.

So none of them indicate Earth is flat.

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #81 on: October 13, 2017, 01:40:06 AM »
Think of any advance ever since the advent of the round earth model. That is where roundies were 'wrong.'.
No, it isn't.
Sliced bread didn't mean they were wrong.
You also need to learn the difference between being wrong and having an approximate answer.
Having Earth being an oblate spheroid doesn't mean they were wrong. Earth still isn't flat.

Yes, and the universe is made up of infinitesimally small local frames. And they all agree - the earth is flat.
No. The vast majority of them don't have Earth in them.
The ones which do only have an infinitesimally small part of Earth and thus are able to make no statement regarding its shape.

So none of them indicate Earth is flat.
They all do have the earth in them. Also, they all agree both that the earth is flat, in such that any body with an orbiting body must be flat.
So long and thanks for all the fish

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #82 on: October 13, 2017, 01:43:00 AM »
To the other point, I didn't say that because they are wrong the earth is flat. I said they were wrong. They were. Slice bread was not wrong - unsliced bread was. To use your inaccurate example. In reality, I like both kinds of bread. I prefer flat bread, I suppose.
So long and thanks for all the fish

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #83 on: October 13, 2017, 01:56:31 AM »
They all do have the earth in them. Also, they all agree both that the earth is flat, in such that any body with an orbiting body must be flat.
Repeating the same BS won't make it true.

Do you understand what infinitesimally small means?
It means they have no spatial extent.
So how does a point light years away have Earth in it?

And no, this thread has shown that being able to have an object orbit you does not mean you are flat.

Objects can orbit Earth without Earth being flat.
Just like they can be in hyperbolic "orbits" without Earth being massively convex and they can be in sub orbital trajectories without Earth being concave and they can be in elliptical orbits without Earth being undulating.

What you (and alt-space) are trying to claim, only works if space-time is Euclidean, which would mean no orbits.
To the other point, I didn't say that because they are wrong the earth is flat. I said they were wrong. They were. Slice bread was not wrong - unsliced bread was. To use your inaccurate example. In reality, I like both kinds of bread. I prefer flat bread, I suppose.
There is nothing wrong with unsliced bread.

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rabinoz

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #84 on: October 13, 2017, 02:03:01 AM »
The orbit of earth is a flat line. This is explained in the aforementioned and linked post. What do you see an inertial path travelling? Some sort of path that is changing direction?
Just who are you replying to?
But, you describe "The orbit of earth is a flat line".
I would far prefer that to be stated as "The orbit of earth is a geodesic in spacetime" and
there is absolutely reason that a geodesic in spacetime cannot be periodic in the spacelike component of spacetime.

In actual orbital motion under GR, however, a perfectly periodic orbit in the spacelike component of spacetime is the exception.
This very slight precession the cause of the very slight discrepancies between the orbits of the planets (especially that of Mercury) predicted by Newtonian Gravitation and those predicted by GR, and as observed.

Quote from: John Davis
If a moon flies around, it surely isn't round.
What does that mean?
The orbit of the moon is simply another geodesic in spacetime, why is that a problem to you?
Just as the orbits of the artificial satellites are geodesics in spacetime.

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rabinoz

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #85 on: October 13, 2017, 05:11:00 AM »
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Another important part of the model here is geocentricity. Flat Earth models are generally geocentric. In this model, the universe is considered as geocentric as well. In relativity, only the motion of two material bodies relative to each other can be physically detected, but the motions taking place are not absolute, they are relative.
This is a point where you start to depart from Einstein's GR.
You state that "In relativity, only the motion of two material bodies relative to each other can be physically detected".
If you take "the motion of two material bodies" to include all motions, including rotation then you are describing Mach's Principle and not Einstein's GR.

This distinction caused Einstein considerable problems in the early days of his "Pathway to General Relativity".
He, at first, tried to build a theory of gravitation that included  Mach's Principle  but was unable to.

Though Isaac Newton preceded Ernst Mach by almost 300 years, he considered many of the same thoughts.

The nett result of this is that there is no simple relativity of angular motion and there are certainly situations in which it leads to ridiculous results - more on this later.
But as background to this, you should read Einstein's Pathway to General Relativity, John D. Norton.
And the section Relativity of Inertia ("Mach's Principle").

Quote from: AltSpace
So, relative to our frame of reference, geocentrism is equally valid to heliocentrism since either can explain the relative motions, to say one is false and the other true implies that the physical motions taking place are absolute, but they are in fact not absolute according to relativity. In fact, on Earth, we take geocentrism as a valid framework since we are observing bodies in motion around us. Telescopes track and move with celestial bodies, it is just as if it is us as stationary and the celestial bodies rotating around us. It turns out to be not only a valid framework to assume geocentrism, but also a more relevant one, since us as observers are stationary relative to the Earth.
This is a situation where strict adherence to Mach's Principle leads to impossible situations.
JackBlack has already alluded to the impossibility of a stationary earth and the so-called fixed stars orbiting earth,
showing how it leads to those stars having to travel many times the velocity of light.

There is much current material written on this topic. Just do a search on "Mach's principle versus Einstein's General Relativity".

I also have big questions about your definition of a flat earth, but I'll have to collect my thoughts a bit on that one.

What makes me so critical of your whole premise is that proper application of Einstein's General Relativity does not cjange th e shape of the earth at all, just a minute change to its size.
So, your hypothesis is not consistent with a solution to Einstein's General Relativity.

There is absolutely nothing in Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space.

By the way, the fact that GR reduces exactly to Newton's Laws of Motion and Gravitation is no accident.
Some of Einstein's earlier attempts did not reduce to Newton's Laws in the low speed, low mass situation, and were discarded.

All that stuff is in the reference I gave you.
« Last Edit: October 13, 2017, 05:50:26 PM by rabinoz »

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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #86 on: October 13, 2017, 11:49:57 AM »
I did. You ignored it (technically you dismissed it as not-relevant)
Again, here it is:

The green line is straight. This is a great circle in spherical geometry.
The red and purple lines remain the same distance (your definition of parallel), yet both curve.
Meanwhile, another straight line, the blue line, has the distance vary.

As such, you have a straight line "parallel" to a non-straight line.
As such, your method does not work.
I already pointed out that you are misrepresenting what I'm saying here, this isn't what I'm talking about.
Quote
Yes, you are saying it, but you are yet to back it up in any way.
It's up to you to justify that it does apply to the surface of the Earth, you made the claim that it does, I already tried explaining why it doesn't.
Quote
The only surface that makes sense for an object is one that is at a specific time point. This makes it purely spatial, and thus allows you to describe the surface of the object. This results in Earth being round.
As soon as you include any temporal aspect you are no longer describing an object and instead are describing a trajectory.
Except 4D space-time are one continuum, all objects, whether at 0 velocity or accelerating are temporally in connection with space-time, this is why
Quote
If it isn't your method, point out what part is not, or clearly describe your method, in point form like that.
Which I did.
Quote
Exactly as you do, taking lines in different areas on a space-time surface.
Unless you are trying to say that the distance between your hypothetical orbit and the surface of Earth is always 0?
No, I am saying that this straight line geodesic matches Earth's surface, describing a flat surface.
Quote
You have a straight line traversing about the pole (or 10 degrees north (I think, it may be different in this example, but the argument still holds)). Straight lines drawn down (well up) to 10 degrees north remain equal length throughout the entire traversal, thus 10 degrees north is therefore straight itself.

So notice how you can't actually do this in non-Euclidean geometry?
From what I read, you took one line traversing the equator, another traversing across with a declination of 10 degrees. These both pass through the center and are therefore straight. The other lines cross 10 degrees north of the equator, and another 10 degrees south of the equator. This was to show that this method would result in describing two non-straight lines as straight (since they remain equidistant from the straight green line), while it concludes that the blue line is not straight.
This is not what I'm pointing out. I'm saying that you have a line, the equator, straight across, and you have a geodesic parallel to it straight above, what a satellite may follow, and these are both straight lines, the surface of the Earth is flat in warped space-time. You are instead comparing other lines that cross the Earth and using that as your example rather than getting to what I'm saying with the surface being equal to a geodesic vector, being flat.
Quote
I have. Now if you care to take it down, demonstrate how my wall example is wrong.
IT DOESN'T DEAL WITH MY POINT!
Stop pretending it does and deal with it or at least admit you can't falsify it.

Quote
Yes, although a surface in space-time is quite a complex thing as is any surface in non-euclidean space due to different surfaces produced by different methods, primarily because the parallel postulate no longer holds. That is you no longer have a single unique, parallel line through a point.
So, the Earth's surface cannot match a straight line in Non-euclidean space-time whatsoever?
Quote
No. Both are in non-Eculidean space-time.
Not a non-gravitationally affected path, that wouldn't be in curved space-time.
Quote
No, there is a very significant distinction.
Space does not have the temporal aspect. It does not have the curvature due to time.
It allows you to describe the shape of objects rather than trajectories.
But it does. Earth warps the way it does into a non-euclidean mass with a gravitational field due to curved space-time.
Quote
He was discussing space, not space time.
Orbits are trajectories, not objects and thus are in space-time, not space.
Like I said, Space and time are one in the same as a continuum.
Quote
No he can. Light is not significantly deflected by Earth's gravity. This allows light to be used as a fairly accurate measurement for the curvature of space itself, as its trajectory through spacetime is not significantly curved by time.
But light can't necessarily follow a mass's geodesics unless it's mass can allow for it, clearly doesn't for Earth.
Quote
But as it is not in free fall it does not represent a straight line through space-time.
Alright, let's suppose you have a straight geodesic line above Earth, a satellite could hypothetically follow it. Then you raise the water level to meet that Geodesic. It matches that geodesic, it equals a flat surface.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
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AltSpace

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #87 on: October 13, 2017, 12:27:28 PM »
This is a point where you start to depart from Einstein's GR.
You state that "In relativity, only the motion of two material bodies relative to each other can be physically detected".
If you take "the motion of two material bodies" to include all motions, including rotation then you are describing Mach's Principle and not i]Einstein's GR[/i].
They aren't at odds, Mach's principle simply states that that inertia is not absolute, so acceleration in respect to other masses in the universe. This would mean the configuration of matter would determine inertia, and also that there isn't another source.
Frame dragging confirms the main postulate here.
There would need to be a matching of an observer in a non-rotating frame examining a rotating frame and the observer in the rotating frame considering the effect of the universe rotating around it. This does indeed match at the right scales. But I guess it would be problematic in respect to 'whole universe'.
I don't think you can so readily dismiss inertia being not absolute, GR doesn't conflict with this.
Quote
The nett result of this is that there is no simple relativity of angular motion and there are certainly situations in which it leads to ridiculous results - more on this later.
Alright, think of it this way. If observer A is in a rotating room with respect to observer B, which is observing the rotating room. Could the analysis of observer A considering a rotating shell of the universe dragging for inertial forces match observer B's analysis of the independent acceleration of that rotating room? If so, then we can postulate that inertia may not be intrinsically absolute.
Quote
This is a situstion where strict adherence to Mach's Principle leads to impossible situations.
JackBlack has already alluded to the impossibility of a stationary earth and the so-called fixed stars orbiting earth,
showing how it leads to those stars having to travel many times the velocity of light.
Only, that's debunked by the fact that c (speed of light) as applied to inertial frames doesn't apply to all accelerating frames. The Sagnac Effect clearly puts this to shame.
Quote
What makes me so critical of your whole premise is that proper application of Einstein's General Relativity does not cjange th e shape of the earth ay all, just a minute change to its size.
Does not change that it's flat, yes.
GR directly depends on Gravity being the geometry of space-time itself, you can't say it has a 'minute change' in Earth's geometry, a round earth simply doesn't match it.
Quote
There is absolutely nothing in Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space.
Curved space-time does, it's flat in non-euclidean space-time.
Quote
By the way, the fact that GR reduces exactly to Newton's Laws of Motion and Gravitation is no accident.
No it doesn't. Newton's gravitation failed to predict the 4D continuum that GR champions, it certainly matches it in many ways, but they are distinct representations.
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

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JackBlack

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #88 on: October 13, 2017, 03:02:54 PM »
I already pointed out that you are misrepresenting what I'm saying here, this isn't what I'm talking about.
No. You baselessly asserted it. You are yet to justify it.
In my example, my 2D example, you have a line which remains the same distance from a straight line yet is not straight, showing your method does not work.

It's up to you to justify that it does apply to the surface of the Earth, you made the claim that it does, I already tried explaining why it doesn't.
No, it is up to you to justify that your method works in non-Euclidean space. You made the claim it does, I showed it does not. I also explained why your attempt at an explanation was wrong.

Except 4D space-time are one continuum, all objects, whether at 0 velocity or accelerating are temporally in connection with space-time, this is why
Yes, 4D space-time is a continuum. That doesn't mean it makes sense to describe the surface of an object by appealing to space-time.
Just like you wouldn't try discussing the cross sectional area of an object using 3D space, and instead rely upon the cross section which is intrinsically 2D.

If you use the time axis, that is a trajectory, not an object.

Which I did.
No you didn't. You just dismissed it saying it isn't your method.
Exactly which point was wrong? There were several steps. Here they are again, numbered for you:
1 - You have a reference line that is straight.
2 - You have another line which you would like to determine if it is straight.
3 - You start at some point along the straight line, and measure the distance to the other line.
4 - You then move along the green line, measuring the distance distance to the other line.
5 - You note the distance remains the same.
6 - Thus you conclude that the line is straight.

What part of this is wrong? What part of it isn't your method?
Here it is more specific for you:
1 - You have a straight line through space-time, an orbit.
2 - You have another line you wish to determine if it is straight, the surface of Earth.
3 - You start at some point along the orbit, and measure the distance to the surface of Earth.
4 - You then move along the orbit, measuring the distance distance to the surface of Earth.
5 - You note the distance remains the same.
6 - Thus you conclude that the surface is straight.

As far as I can tell, that is your method, and thus it should also work for me example.

No, I am saying that this straight line geodesic matches Earth's surface, describing a flat surface.
No. This straight line geodesic is a trajectory through space time, and thus does not describe the surface of Earth (as a shape).
Also, as it passes through a different region of space-time, with different curvature, it does not match the surface of Earth.

From what I read, you took one line traversing the equator, another traversing across with a declination of 10 degrees. These both pass through the center and are therefore straight. The other lines cross 10 degrees north of the equator, and another 10 degrees south of the equator. This was to show that this method would result in describing two non-straight lines as straight (since they remain equidistant from the straight green line), while it concludes that the blue line is not straight.
Yep, pretty much. You have 2 straight lines in non-Euclidean space (2D, not 3D) which cross twice and 2 non-straight lines which remain the same distance from one of these straight lines.

This is not what I'm pointing out. I'm saying that you have a line, the equator, straight across, and you have a geodesic parallel to it straight above
And apart from the curvature being different, that is what I am describing.
You have the straight line (the great circle about the equator).
Directly "above" it, you have the line 10 degrees south, and directly "below" you have the line 10 degrees north.
According to your reasoning, these are both straight.

Stop trying to turn my 2D non-Euclidean problem into a 3D problem.


You are instead comparing other lines that cross the Earth and using that as your example rather than getting to what I'm saying with the surface being equal to a geodesic vector, being flat.
I am comparing a line remaining the same distance from a straight in non-Euclidean space-time, to a line remaining the same distance to a straight line in non-Euclidean space.
They are basically the same thing.

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I have. Now if you care to take it down, demonstrate how my wall example is wrong.
IT DOESN'T DEAL WITH MY POINT!
Stop pretending it does and deal with it or at least admit you can't falsify it.
Yes it does. It shows that remaining the same distance from a straight line does not mean a line is straight in non-Euclidean spaces.
Stop pretending it doesn't and deal with it, or admit your method is pure BS.

So, the Earth's surface cannot match a straight line in Non-euclidean space-time whatsoever?
No. Earth's surface is in space, not space-time.
The trajectory of a hypothetical object can follow Earth's surface in non-Euclidean space time, but that isn't Earth's surface itself.

A straight line through non-Euclidean space-time is either a trajectory and thus cannot describe the shape of an object, or it has no temporal component and thus would not proceed through time to stay with an orbit.

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No. Both are in non-Eculidean space-time.
Not a non-gravitationally affected path, that wouldn't be in curved space-time.
That is right, it would be in Euclidean space, not space time. This means you can't compare it to a trajectory like that of an orbit. you need to remain at the same point in time.
The easiest way to do this is to see what happens as the speed increases eventually to an infinitely velocity where a hypothetical object is capable of traversing Earth in a single instant in time.
When you do, you find that Earth's surface does not remain the same distance. Instead, it curves.

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No, there is a very significant distinction.
Space does not have the temporal aspect. It does not have the curvature due to time.
It allows you to describe the shape of objects rather than trajectories.
But it does. Earth warps the way it does into a non-euclidean mass with a gravitational field due to curved space-time.
No. Space itself does not have the temporal aspect, only space-time does.
Gravity results in curvature along the time axis, not the spatial axes.

Like I said, Space and time are one in the same as a continuum.
And in intersection of space time with a purely spatial space to produce space is not space-time.

What you are doing now is akin to saying a line and a plane are the same and that there is no distinction between them.

But light can't necessarily follow a mass's geodesics unless it's mass can allow for it, clearly doesn't for Earth.
Again, we are discussing the shape of Earth, not the trajectory of a hypothetical mass through space-time.

Alright, let's suppose you have a straight geodesic line above Earth, a satellite could hypothetically follow it. Then you raise the water level to meet that Geodesic. It matches that geodesic, it equals a flat surface.
No it doesn't. Not until you go to a geostationary orbit. But then it only matches at the equator.
For all the other points it is going too slowly.
Remember, these orbits are not path's through space, they are paths through space-time.
You can't simply ignore the temporal aspect with things like this.


The Sagnac Effect clearly puts this to shame.
Yes, by showing Earth is rotating.

GR directly depends on Gravity being the geometry of space-time itself, you can't say it has a 'minute change' in Earth's geometry, a round earth simply doesn't match it.
A round Earth matches it quite well.

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There is absolutely nothing in Einstein's General Relativity that justifies what has been labelled "The Ferrari Effect - a flat earth appearing to be a Globe when seen from space.
Curved space-time does, it's flat in non-euclidean space-time.
No it doesn't. This is because the entire principle of it requires Euclidean spaces and does not work in non-Euclidean spaces.

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Re: Davis Relativity Model (Debate/discussion edition)
« Reply #89 on: October 13, 2017, 04:54:03 PM »
No. You baselessly asserted it. You are yet to justify it.
In my example, my 2D example, you have a line which remains the same distance from a straight line yet is not straight, showing your method does not work.
But that is not dealing with my point. The Geodesics are straight above the surface, equating the surface to a straight line, as flat.
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No, it is up to you to justify that your method works in non-Euclidean space. You made the claim it does, I showed it does not. I also explained why your attempt at an explanation was wrong.
You didn't show it was anything near wrong, you ignored it and instead use lines along varying longitudes and declination's rather than the surface of the Earth itself as directly across from a geodesic, matching it.
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Yes, 4D space-time is a continuum. That doesn't mean it makes sense to describe the surface of an object by appealing to space-time.
Just like you wouldn't try discussing the cross sectional area of an object using 3D space, and instead rely upon the cross section which is intrinsically 2D.
You would need to in this case, since the surface of the Earth is in a unification as a mass by 4D space-time itself, you can't remove time in the description.
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If you use the time axis, that is a trajectory, not an object.
The object is 'moving' through time, even if it were to be at 0 velocity through 3D spatial coordinates, so it defines an object as well.
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Exactly which point was wrong? There were several steps. Here they are again, numbered for you:
1 - You have a reference line that is straight.
2 - You have another line which you would like to determine if it is straight.
3 - You start at some point along the straight line, and measure the distance to the other line.
4 - You then move along the green line, measuring the distance distance to the other line.
5 - You note the distance remains the same.
6 - Thus you conclude that the line is straight.
What you did wrong was taken a geodesic and compared to something not part of the surface of the Earth it crosses through. I think it's because you know it's straight and choose to ignore it. I really find it hard to believe that you are simply 'missing the point'.
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No. This straight line geodesic is a trajectory through space time, and thus does not describe the surface of Earth (as a shape).
Also, as it passes through a different region of space-time, with different curvature, it does not match the surface of Earth.
This trajectory is following a geodesic that matches the Earth's straight surface.
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And apart from the curvature being different, that is what I am describing.
You have the straight line (the great circle about the equator).
Directly "above" it, you have the line 10 degrees south, and directly "below" you have the line 10 degrees north.
According to your reasoning, these are both straight.
But guess what? These aren't directly through the center like my example has it.
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I am comparing a line remaining the same distance from a straight in non-Euclidean space-time, to a line remaining the same distance to a straight line in non-Euclidean space.
They are basically the same thing.
They are not since your example misses the lines being so that straight connections cross through the same area through the Earth.
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Stop pretending it doesn't and deal with it, or admit your method is pure BS.
Again: Stop pretending it does and deal with it or at least admit you can't falsify it.
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No. Earth's surface is in space, not space-time.
The trajectory of a hypothetical object can follow Earth's surface in non-Euclidean space time, but that isn't Earth's surface itself.
Earth's surface is in space-time itself as well, just like any other object. By your reasoning, an apple hanging on a tree is not in time since it is stationary. Only, the apple hanging to the tree is resisting the curve of space-time and ages overtime.
You simply can't ignore the 4D space-time continuum with Earth.
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A straight line through non-Euclidean space-time is either a trajectory and thus cannot describe the shape of an object, or it has no temporal component and thus would not proceed through time to stay with an orbit.
This trajectory follows the same geodesic that matches the Earth's surface as flat. I know you are trying so hard to deny it, but it simply won't work to deny the 4D space-time continuum as formulating the accurate Geometry of Earth.
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That is right, it would be in Euclidean space, not space time.

No, that would be in the same 4D space-time continuum we all are in, just that one isn't affected by curved space-time.
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This means you can't compare it to a trajectory like that of an orbit. you need to remain at the same point in time
The Earth's surface does not remain at the same point in time.
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The easiest way to do this is to see what happens as the speed increases eventually to an infinitely velocity where a hypothetical object is capable of traversing Earth in a single instant in time.
When you do, you find that Earth's surface does not remain the same distance. Instead, it curves.
Unfortunantly for you, time exists, and defines the Geometry of the Earth as well, so your hilarious roundie excuses won't work here for GR.
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No. Space itself does not have the temporal aspect, only space-time does.
Gravity results in curvature along the time axis, not the spatial axes.
And the spatial axes too surely, as they are both linked, so an object at 0 velocity will start accelerating with translating curved 4D temporal coordinates to 3D spatial coordinates.
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And in intersection of space time with a purely spatial space to produce space is not space-time.

What you are doing now is akin to saying a line and a plane are the same and that there is no distinction between them.
Straight lines are all parallel to the surface across Earth, it seems the surface would be represented that of a flat plane.
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Again, we are discussing the shape of Earth, not the trajectory of a hypothetical mass through space-time.
These geodesics are quite independent of the masses that follow them.
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No it doesn't. Not until you go to a geostationary orbit. But then it only matches at the equator.
For all the other points it is going too slowly.
Remember, these orbits are not path's through space, they are paths through space-time.
You can't simply ignore the temporal aspect with things like this.
So is the Earth's surface, it's a geodesic itself in space-time.
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A round Earth matches it quite well.
No, the surface is flat in Non-euclidean space-time, the round earth fails to match it.
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No it doesn't. This is because the entire principle of it requires Euclidean spaces and does not work in non-Euclidean spaces.
However, the cross-section through the Earth has straight lines all the way through, which includes the Earth's surface. This clearly shows it's flat. You have yet to address that.
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