Are subatomic partices n-dimensional hyperspheres or another n-dimensional shape?

Question

This guy derives the formula for the volume of an n-dimensional hypersphere, which can also be found here on the Wikipedia page, and he shows that it is equal to this equation: $V_n = \frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2} + 1)}$ perhaps better illustrated by this table:

Anyway, this shows that (assuming the radius is always 1) the volume will approach 0 as the dimension goes up very quickly because there is essentially a factorial function on the bottom, and that quickly outpaces the exponentiated π.

WolframAlpha http://www4a.wolframalpha.com/Calculate/MSP/MSP6651f24a06537b2c4750000518ea98863add5i7?MSPStoreType=image/gif&s=43&w=300.&h=194.&cdf=RangeControl

$\lim_{n \to +\infty} V_n = \frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2} + 1)} = 0$

I don't have any proof that this has anything to do with subatomic particles, but I think it's interesting that here we have shapes with higher dimensions, but incredibly small volumes. Meaning, an n-sphere where n is a large number can have a large radius and many degrees of freedom for rotation and movement that increase with n according to the triangular numbers:

Wikipedia:

In general, the number of euler angles in dimension D is quadratic in D; since any one rotation consists of choosing two dimensions to rotate between, the total number of rotations available in dimension $D$ is $N_{rot}=\binom{D}{2}=D(D-1)/2$, which for $D=2,3,4$ yields $N_{rot}=1,3,6$.

I feel like this might explain things like the spin of an electron being one half because higher dimensional rotations can take more than 360 degrees to complete before you see the same shape again.

Answer 1

If one uses the sphere volume as $\pi^{n/2}/\frac n2 ! $ and then apply stirling's approximation to $n! = (n/e)^n$, then one gets $ (2.\pi.e/n)^{n/2}$ as the approximation for the volume of an n-sphere. This then equates to a cube of edge $\sqrt{2\pi e/n}$.

Of course, one can use by way of limits, that even a cube of side $1/2$ must 'go to zero', even though it is clear that it is roughly the same size as the original cube.

But it offers no hope of explaining the nature of sub-atomic particles. A sphere is not a circle, it has extent in 3d, and the only way that you might suppose that it lives in 2d, is to imagine that that 2d space is trapped between sheets of glass, so to speak.

Moreover, a sphere of N dimensions, trapped into a 3D space, will appear no different to ordinary 3d spheres. So even though the size of a 19D sphere of unit radius is less than a unit-edge 19-d cube, the thing trapped in 3d, will produce a 3d space that's identical to a 3d sphere.

While there is a model that supposes 11D, the sheets of glass are placed too close for the size of subatomic particles to be realised in that space. The order postulated in the model is the size of Planck's length, which is about 15 orders of magnitude smaller than any real particle.

The quote from the Wikipedia gives something different, which really does not give any thing different to more than 360 degrees. It's more of nature of a phase-space, the composite rotations of an object in 2d, 3d, and 4d might be represented by a point in 1, 3, or 6 dimensional space.

Imagine you had a thing in 2d space. It just basically rotates, and the nature of this rotation can be controlled by a point on a line. You move right if you want clockwise, and left if you want anticlockwise, the distance is the speed of rotation.

In 3D, the rotation-space is 3d too. A sphere can turn in any direction, but you can use the right-hand-rule to determine a 'north pole'. When you plot a point in 3dpsce, you are passing a ray through the north pole, and the length of the ray is the speed it rotates at.

In 4D, it is possible to have a 'double rotation', different speeds in WX and YZ spaces, for example. It's a little more complex, but you get in the realms of left-clifford parallels and right-clifford parallels: in essence, if WX=YZ, the you have one set of parallels, if WX=-YZ, you have another set.

You can easily prove that all points go around a common centre in a circle, if you are familiar with complex numbers. Put $X = w+xi$ and $Y= y+zi$. Now, you have all lines passing through the point (0,0) are of the form $aX=Y$. If you multiply both sides by $e^{i\omega t}$, then the (complex) gradient does not change, but over a cycle of $t$

Since the gradient does not change, one can map the argand diagram onto a sphere by way of placing a sphere at centre (0,0,1). A ray from any (x,y) to (0,0,2) will cross this sphere at a particular point. The polar point represents a gradient perpendicular to the plane (ie the $Y$ axis in the previous example).

Since complex numbers only demonstrate a left-turning, there is a matching right-turning set. All rotations in 4D are represented by a sum of a left-rotation and a right-rotation, of potentially different intensities. The left-rotation represents a 3d space, the right rotation also a 3d space, and the product makes a 6d space.

In none of these examples, do we admit more than 360 degrees to a circle. We merely admit the possibilites of multiple circles rotating at the same time.

Unlike 2d and 3d space, the rotation-space for 4d is 'granular', in that a point anywhere in the 6d space, will by the law of equipartition of energy of the modes, will tend to migrate to one of the two orthogonal 3d spaces representing 'clifford-rotations'.