Geometric Unity

[djg]

But I must commend you for controlling your emotions enough to engage logically.

i dont. i just do the same as you do. i copy and paste it into an llm. neither of us knows if the words make any sense. my advantage is that i can let the llm loose, while you constantly have to control it to keep up the pretense. my concern for you is genuine so good luck. this will not end well.

[Strat-itis]

my concern for you is genuine so good luck. this will not end well.

[Strat-itis]

Yep, I used the wrong wording there. Model still holds tho.

[ragman1]

Don't worry, Christian was going to write to the Oxford Dictionary once to tell them their definitions were all wrong...

Those were the days

[Strat-itis]

[Christian Miller]

No man, I would never expect validation from you or this forum. I just defend myself when people are absurd to me.

Thanks for your honesty. I assumed some people would engage with the content I created on its level, but I guess I overestimated. That's kind of my fault for expecting positive discourse about it.

You must be joking? I authored all the concepts. I used AI to draft the thing quickly for the purpose of stamping authorship. I knew it would be imperfect, but all my intended concepts are there, and I can manually improve things as I move foreward.

You would have to actually engage with its logic to know that.. I'm not saying it's proven. I am saying for certain that it's quite coherent and viable. And it doesn't have the glaring contradictions that GR has.

I’m still none the wiser as to why you want me to read it.

I can’t give you a professional opinion, and even if I could you’d reject it.

Sent from my iPhone using Tapatalk

[Strat-itis]

You don't have to read or entertain it, but if you're going to invalidate it, an actual scientific rationale is required.

[Strat-itis]

Here's the summary if anyone cares or can absorb this modest amount of material.

Loading Google Docs

[pauln]

...

I am pretty sure that rotation in 5D is the same as universal 3D expansion.
Universal in the sense of each particles' expansion (acceleration of radius).
No gravity, no expansion of space - just size growth of individual particles.
Equivalence principle yields an observed gravitation and space expansion.

Strat-itis, have not looked at your paper, have to go perform; I'll look at it.

[CliffR]

I am pretty sure that rotation in 5D is the same as universal 3D expansion.

I this were true, we could take two points on a particle in 3D space, say (x,y,z) and (x+k,y,z). Before rotation, the distance between the points is k. Now let's take them into 5D. (x,y,z,0,0) and (x+k,y,z,0,0). Rotation in 5D is just a multiplication of the vector by a 5x5 rotation matrix. Then you project the result back into 3d space and measure the distance between the points. Has it changed? We already know that, for all classes of 5x5 rotation matrices where only the top 3x3 elements are occupied, the answer is no - there is no change in distance. So what do you mean here? Are you suggesting that a certain sub-class of rotations produce this expansion effect? Do you have an equation to demonstrate this?

[CliffR]

For some reason I'm still spending mental energy on this.
It's true that, if we take points in N-dimensional space, rotated them with a general N+M dimensional matrix, then project (ie remove the extra dimensions in the vectors) them back into N space, the Euclidean distance between them may have changed, depending on the nature of the higher-order matrix. But this will always result in a reduction of distance, never an expansion.
Think of the trivial case of two points close to each other on a piece of card. Let's imagine they both occupy a position at the same height on the card, but at different positions along the length. Imagine the card is flat on the table in front of you. Observe the distance between the points. Now rotate the card with the edge we labelled as 'height' remaining on the table (ie rotate into the third dimension). The distance between the two points remains the same. Now 'project' those points back on to the table - which is to say, draw straight lines from you the observer (forget about perspective) through the points on the card to the table. The distance between those points on the table is going to vary according to the amount you've rotated the card. Rotate by 90 degrees, and they become coincident. I'm afraid this is not a great physical/metaphysical insight.

[BWV]

For this model to be taken seriously it has to first be able to reproduce all of the calculations of General Relativity - so maybe the author can show how his 5D model computes the precession of Mercury as accurately as GR?

[Strat-itis]

^ Sure. I conceptually explain it as spacetime jerk on page 35 of the master doc if you want to read it. Where jerk is the rate of change of m^3/s^2 spacetime, m^3/s^3. But I'll scheme about the calculations.

I'm afraid this is not a great physical/metaphysical insight.

Good walkthrough. Well it's not necessarily a bad insight because higher spatial dimensions are not observed anywhere, so it would be speculative to introduce them.

[CliffR]

The length of a vector is defined as the square root of the sum of the square of its components. So if you choose to ignore some of the higher dimensional components, the length of the vector in the subspace is always going to be less than or equal to that of the vector in the full space. It's a basic consequence of linear algebra.

[Strat-itis]

I followed you the first time. Are you saying there are higher spatial dimensions? Just curious where you're going with it.

[pauln]

For some reason I'm still spending mental energy on this.
It's true that, if we take points in N-dimensional space, rotated them with a general N+M dimensional matrix, then project (ie remove the extra dimensions in the vectors) them back into N space, the Euclidean distance between them may have changed, depending on the nature of the higher-order matrix. But this will always result in a reduction of distance, never an expansion.

That may not be inconsistent with what I described. The expansion is not of space but of the radii of particles.
For two spheres some distance apart the distance between their centers does not change (no gravity or expansion of space), but the radius of each increases so their proximate surfaces move closer.
An observer and his measuring devices are also expanding; this produces the apparent observation that the two spheres are not expanding, but moving (accelerating) toward each other as if by gravitation.
Reversing the acceleration vector allows Sp. Relativity calculations to be done with just light propagation delay and trig in flat space because the curvature is already in the expansion of radii (equivalence principle). I think this idea may have predated someone noticing that 5D rotation would look like this in 3D.

[CliffR]

I followed you the first time. Are you saying there are higher spatial dimensions? Just curious where you're going with it.

Not going anywhere, really, just pointing out that what Paul said is a basic consequence of the way linear algebra works, not a deep insight into the structure of the universe.

I'm not saying there are higher spatial dimensions, either. Linear algebra is pure math, and can be applied to any number of multi-dimensional situations.

But for simplicity's sake, let's start with a right-angled triangle drawn on a piece of paper. The base of the triangle lies along the width of the paper, and has length 'x'. The edge going up the height of the paper is length 'y'. The hypotenuse joins the two ends of these edges. Pythogoras' theorem tells us that the length of the hypotenuse is given as the square root of x^2 + y^2. That's the length of the vector.

y
|\
| \
| \
0--- x

We could give the vertices of the triangle coordinates, starting at the bottom left and going clockwise (0,0) (0,y) (x,0). The hypotenuse is then the vector given by subtracting the coordinates of one of its points from the other. (x,0) - (0,y) = (x,-y). Length of hypotenuse = sqrt(x^2 + y^2)

This generalises to any number of dimensions. But we can think of it in 3 so as not to hurt our heads. In 3d, the length of the vector (x,y,z) = sqrt(x^2 + y^2 + z^2). For 4d, you'd add a 'w' component, and so on.

Imagine now that all the points in the plane of the piece of paper on which we drew our triangle have a z coordinate of 0. If all the triangle vertices also have 0 z coordinates, then the 3d length is the same as the 2d length, since z = 0 for the hypotenuse vector.

But if, say, the top-most vertex has a non-zero z coordinate, say z. (I hope my conflation of coordinates and lengths isn't confusing - this applies to any right-angled triangle in the plane, not just this special case of one at the origin and whose edges are alined with the coordinate frame.)

The hypotenuse is then given by (x,0,0) - (0,y,z) = (x,-y,-z). Length of hypotenuse = sqrt(x^2 + y^2 + z^2). (The squaring removes the minus signs.) Clearly, this is longer than the 2d case from before. But if we *project* this hypotenuse back on to the 2d plane, which is to say, remove the z coordinate, then we measure the shorter length.

[CliffR]

That may not be inconsistent with what I described. The expansion is not of space but of the radii of particles.
For two spheres some distance apart the distance between their centers does not change (no gravity or expansion of space), but the radius of each increases so their proximate surfaces move closer.
An observer and his measuring devices are also expanding; this produces the apparent observation that the two spheres are not expanding, but moving (accelerating) toward each other as if by gravitation.
Reversing the acceleration vector allows Sp. Relativity calculations to be done with just light propagation delay and trig in flat space because the curvature is already in the expansion of radii (equivalence principle). I think this idea may have predated someone noticing that 5D rotation would look like this in 3D.

Sorry Paul, that doesn't make any sense to me.

[Strat-itis]

Linear algebra is pure math, and can be applied to any number of multi-dimensional situations.

Thx for the writeup, I like that.