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### Author Topic: The Holographic Universe and Pi = 4 in Kinematics!  (Read 226323 times)

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #75 on: May 19, 2014, 01:45:59 PM »
The path length isn't changing as it converges, and we can visually see this, so this is the exact path length of the circumference.  The "better" approximation you speak of simply doesn't exist as you wrongly assert.

Gravock
Indeed the Manhattan path length does not change.  Neither does it converge with the circumferential path length.  The outline formed by the inner square vertices converges with the circumference.  Those points are not the path.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #76 on: May 19, 2014, 01:57:44 PM »
Indeed the Manhattan path length does not change.  Neither does it converge with the circumferential path length.  The outline formed by the inner square vertices converges with the circumference.  Those points are not the path.

Again, we can visually see in each successive squaring method the number of inner square vertices are also exponentially increasing around the circumference, and at the planck length, the inner square vertices will be at all points on the rectilinear circumference of the circle itself.

Gravock

#### verpies

• Hero Member
• Posts: 3480
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #77 on: May 19, 2014, 02:02:03 PM »
Really?  What is that difference?  What qualifies a "physical circle" to be a circle and what properties may it have that are different than an "abstract circle"?

You are off in the bushes again.

Slay those men of straw.  I have stated clearly that Pi is defined as the ratio of a circle's circumference to its diameter,
That's not what I am disagreeing with.
You have also stated that if Pi so defined is not 3.14 then the circle fails to have the property of the circle that you are familiar with and because of that is not a real circle.
You also disqualified a physical circle as a circle because physical circle is not a two-dimensional figure.  These two are not Straw Men.

I asked you whether that circle in the article you quoted qualified as a circle and if not then what is it in your opinion.   I still have not received a direct answer.

and the mutual claim you make with GravityLock that the ratio is numerically equal to four is patently false.
Prove it for kinematic circles.

"Those coordinates" are not the same as what?
"Not the same" as in "not identical".  Not the same time coordinate for each point.

A plane has two axes.  A circle is a construct of plane geometry.  Be sure that your "physical circle" conforms to those requirements.Again, a circle is a construct of plane geometry.  There are only two dimensions.  If you cannot draw it on a piece of paper then it isn't plane geometry.
In abstract geometry - yes. All of the points belonging to an abstract geometric figure exist at the same instance in time, so the time can be disregarded.
In physics time cannot be disregarded and a physical circle is not a strictly 2D object.

The paper's premise is utter and total BS.
Did you pay attention which limit is reached first?

#### verpies

• Hero Member
• Posts: 3480
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #78 on: May 19, 2014, 02:09:15 PM »
Indeed the Manhattan path length does not change.  Neither does it converge with the circumferential path length.  The outline formed by the inner square vertices converges with the circumference.  Those points are not the path.
That is the correct analysis for abstract timeless circles only.  It should be added that the Manhattan area converges to the area of the abstract circle, too.

@Gravityblock
MarkE is correct that the Manhattan path does not converge with the circumference of an abstract timeless circle at the limit.  In an abstract circle, the chords (or hypotenuses) converge with the circle/arc - not the catheti of the right triangles.  Please remember that when you discuss this issue with him or he will eat you alive.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #79 on: May 19, 2014, 02:27:37 PM »
That is the correct analysis for abstract timeless circles only.  It should be added that the Manhattan area converges to the area of the abstract circle, too.

@Gravityblock
MarkE is correct that the Manhattan path does not converge with the circumference of an abstract timeless circle at the limit.  In an abstract circle, the chords (or hypotenuses) converge with the circle/arc - not the catheti of the right triangle.  Please remember that when you discuss this issue with him or he will eat you alive.

Matter doesn't move in a continuous motion, it moves in discrete jumps at the planck length.  When the squaring method reaches the planck length, the inner square vertices will be at all points on the rectilinear circumference of the circle itself, which is not continuous and is made of discrete jumps.  The Manhattan path does correctly simulate the time variable in real circles at the planck length!

Edited for better clarification.

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #80 on: May 19, 2014, 03:03:12 PM »
Again, we can visually see in each successive squaring method the number of inner square vertices are also exponentially increasing around the circumference, and at the planck length, the inner square vertices will be at all points on the rectilinear circumference of the circle itself.

Gravock
That is both true and irrelevant.  Since we can connect the inner vertices with chords, as the number of chords tends towards a large number, such as 1/lP, all the vertices being on the circumference, the chords will closely approximate the path of the circumference.  If they did not, then the area approximation established by those vertices would not match the area of the circle.  Ergo the path formed by taking a straight line between those vertices closely approximates the true circumferential path length.  Since that path length converges towards ~3.1415953 and not 4.0, the claim that the sum of 1/lP diversions from the circumference and back matches the path length of the circumference fails.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #81 on: May 19, 2014, 03:13:49 PM »
That is both true and irrelevant.  Since we can connect the inner vertices with chords, as the number of chords tends towards a large number, such as 1/lP, all the vertices being on the circumference, the chords will closely approximate the path of the circumference.  If they did not, then the area approximation established by those vertices would not match the area of the circle.  Ergo the path formed by taking a straight line between those vertices closely approximates the true circumferential path length.  Since that path length converges towards ~3.1415953 and not 4.0, the claim that the sum of 1/lP diversions from the circumference and back matches the path length of the circumference fails.

No, the inner vertices at the planck length can not be connected with chords in a real circle with a time variable.  By connecting the inner vertices at the planck length with chords, then you are saying matter moves in a continuous motion and not in discrete jumps.

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #82 on: May 19, 2014, 03:22:15 PM »
I have yet to see any such description.  If you think you have provided one kindly point at the post that provides it.
Quote

That's not what I am disagreeing with.
You have also stated that if Pi so defined is not 3.14 then the circle fails to have the property of the circle that you are familiar with and because of that is not a real circle.
You also disqualified a physical circle as a circle because physical circle is not a two-dimensional figure.  These two are not Straw Men.
Once again you misstate what I have said.  Pi is a defined relationship.  The value of that relationship has been reliably approximated to eight decimal digits as: ~3.1415953.  I have said that a geometric object that purports to have a circumference to diameter ratio that does not conform within the measured and/or calculated error bands to a correspondingly precise expression of Pi, then the the object fails to demonstrate a basic property of circles.  This tortured idea that an object that fails to demonstrate a circumference to diameter ratio in close accord with a reliably approximated value of Pi is completely silly.  How long do you two intend to keep this nonsense up?

Circles are plane geometry objects.  They exist in two dimensions.  That is not my doing.  That is the accepted definition.
I have yet to see a definition of a "physical circle" from you that allows for the special pleadings that you make that such an object has a time component or any other property distinct from the plane geometry object known as a circle.
Quote

I asked you whether that circle in the article you quoted qualified as a circle and if not then what is it in your opinion.   I still have not received a direct answer.
Even the form of the question is silly.
Quote
Prove it for kinematic circles.
What is a kinematic circle?
Quote
"Not the same" as in "not identical".  Not the same time coordinate for each point.
Time is irrelevant to plane geometry.
Quote
In abstract geometry - yes. All of the points belonging to an abstract geometric figure exist at the same instance in time, so the time can be disregarded.
Time never entered.  It is not a matter of disregarding something that is not significant. Time plays no part.
Quote
In physics time cannot be disregarded and a physical circle is not a strictly 2D object.
Kindly provide an academic link that identifies and describes one of these time dependent "physical circles".

Did you pay attention which limit is reached first?
[/quote]Apparently among other things, Mr. Mathis did not.  If you wish to attempt to show by deriving the limit of the chord slope that the limit is other than zero, feel free to show your math.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #83 on: May 19, 2014, 03:25:15 PM »
No, the inner vertices at the planck length can not be connected with chords in a real circle with a time variable.  By connecting the inner vertices at the planck length with chords, then you are saying matter moves in a continuous motion and not in discrete jumps.

Gravock
A plane geometry object has nothing to do with time.  Plane geometry is a field of mathematics.  If you want to play:  "You can't do XYZ in the physical world" then you are already stuck with much bigger problems than whether chords can be mapped between presumed squares with presumed vertices at static positions in space.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #84 on: May 19, 2014, 03:26:24 PM »
Matter doesn't move in a continuous motion, it moves in discrete jumps at the planck length.  When the squaring method reaches the planck length, the inner square vertices will be at all points on the rectilinear circumference of the circle itself, which is not continuous and is made of discrete jumps.  The Manhattan path does correctly simulate the time variable in real circles at the planck length!

Edited for better clarification.

Gravock
That's rubbish.  The Manhattan path has no time element to it.  It is plane geometry.  How long are you going to insist on this silly game?

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #85 on: May 19, 2014, 03:44:16 PM »
A plane geometry object has nothing to do with time.  Plane geometry is a field of mathematics.  If you want to play:  "You can't do XYZ in the physical world" then you are already stuck with much bigger problems than whether chords can be mapped between presumed squares with presumed vertices at static positions in space.

Plane geometry must also obey the laws of physics when dealing with the physical world.  In other words, if you want to use geometry to analyse the motion of a physical object, then a circle should be drawn in discrete jumps to represent the discrete jumps in the physical world.  If the physical world says matter moves in discrete jumps at the planck scale, then the geometry must correctly represent this.  If not, then you are using geometry to break the laws of physics.  Now, if you're analysing an abstract circle with no time variable, then you are free to show a continuous motion by connecting the inner vertices with chords to show a continuous motion.  However, this is not the case for a real circle with a time variable.

Gravock

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #86 on: May 19, 2014, 03:58:18 PM »
That's rubbish.  The Manhattan path has no time element to it.  It is plane geometry.  How long are you going to insist on this silly game?

The Manhattan path does have a time element to it when we are at the planck scale.  The planck time is the amount of time it takes light to move one planck length.  Since we can't go beyond the planck length in the Manhattan path without breaking the laws of physics, then the planck length correctly represents the time as points in those discrete jumps

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #87 on: May 19, 2014, 04:21:52 PM »
Plane geometry must also obey the laws of physics when dealing with the physical world.  In other words, if you want to use geometry to analyse the motion of a physical object, then a circle should be drawn in discrete jumps to represent the discrete jumps in the physical world.  If the physical world says matter moves in discrete jumps at the planck scale, then the geometry must correctly represent this.  If not, then you are using geometry to break the laws of physics.  Now, if you're analysing an abstract circle with no time variable, then you are free to show a continuous motion by connecting the inner vertices with chords to show a continuous motion.  However, this is not the case for a real circle with a time variable.

Gravock
That is some really tortured boot strapping.  You have as far as I know been arguing that Pi = 4 based on the plane geometry of a circle.  The behaviors of a circle do not change because of how you might want to apply a circle in a model or an experiment.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #88 on: May 19, 2014, 04:22:51 PM »
The Manhattan path does have a time element to it when we are at the planck scale.  The planck time is the amount of time it takes light to move one planck length.  Since we can't go beyond the planck length in the Manhattan path without breaking the laws of physics, then the planck length correctly represents the time as points in those discrete jumps

Gravock
Two dimensional geometry has no time element.

#### verpies

• Hero Member
• Posts: 3480
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #89 on: May 19, 2014, 04:30:02 PM »
I have yet to see any such description.
If you think you have provided one kindly point at the post that provides it.
The definition was in this post.
"A circle is a set of points on a spatial plane equidistant from the center of the circle. The difference between an abstract and physical circle is whether these points have time coordinates or not.  Physical circles do and those coordinates are not the same" - Later you even asked me what the phrase "not the same" referred to.

Once again you misstate what I have said.  Pi is a defined relationship
No I do not.  You are plainly stating that a circle is not a circle if the relationship of circumference to diameter is not ~3.1415953 while failing to define the circumference of a physical circle and conflating it with the circumference of an abstract circle.
By doing it you are letting ~3.1415953 define the circle instead of letting the Circle define the ratio between its circumference and diameter.  Such reversal makes a conclusion out of the premise.
I agree with you you that Pi=c/d but I disagree with you how c & d are measured physically.

The value of that relationship has been reliably approximated to eight decimal digits as: ~3.1415953.  I have said that a geometric object that purports to have a circumference to diameter ratio that does not conform within the measured and/or calculated error bands to a correspondingly precise expression of Pi, then the the object fails to demonstrate a basic property of circles.
This is the reversal. You are attempting to prove that circle is not a circle because it does not conform to your expected ratio of circumference to diameter.

This tortured idea that an object that fails to demonstrate a circumference to diameter ratio in close accord with a reliably approximated value of Pi is completely silly.  How long do you two intend to keep this nonsense up?
Silly is only your insistence on conflating the circumference of an abstract circle to a circumference of a physical circle and using the same measuring processes for both.
I will keep it up a long time if you won't answer my questions directly.

Circles are plane geometry objects.  They exist in two dimensions.  That is not my doing.  That is the accepted definition.
This is true of abstract circles only.
What is "that" in the article you quoted ?