The Holographic Universe and Pi = 4 in Kinematics!

[MarkE]

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]

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]

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"?

I already answered that directly.

You are off in the bushes again.

Just answer directly "yes" or "no" instead of writing about bushes.

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]

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]

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