Volumes of n-Spheres: Intuition & Calculations

[martix]

I'm trying to wrap my head around how higher dimensions work. Not over some academic pursuit, merely curiosity(which started with me discovering this neat recurrence relation).

Given that the surface area of a sphere is an object of 1 dimension lower.
So on a 3-sphere, the surface area is a plane and is measured in say m².
Then for a 4-sphere, the surface area is a 3-dimensional sphere with a volume measured in m³.

Then I thought a bit about volumes of higher dimensional spheres before realizing hypervolume doesn't mean the same as normal cubic volume, throwing any intuition as what it is(let alone how big it is) out the window.

However, in a 4-sphere, one can imagine taking slices at equal steps and obtaining a multitude of concentric 4-spheres, each of which has a "surface area" with a regular 3-dimensional volume.

So if you define a step, using this method you get a meaningful answer for "What is a the 3d volume of a 4-sphere with radius r?"

Say,
S(step) = 0.2u (as in units)
V4 = A4(u) + A4(0.8u) + A4(0.6u) + A4(0.4u) + A4(0.2u)
Or ~41.22u³ if my calculations are correct.

Does this even make sense? If no, why not?

 

[BvU]

Hi,

How do you get 41.22 ?

What do you get with steps of 0.1 ? 0.01 ? 0.001 ?

Would it work for the 2d circumference of a 3d sphere ?

Why not simply differentiate ${\pi^2\over 2}R^4$ to get $2\pi^2 R^3$ ?

 

[martix]

1. Via the recurrence relation. I translated it to python, just to check if it works, then in C#, and ran it for all those values. I confirmed its correctness with normal 3d spheres.
2. 0.1 -> V ~ 66.67; 0.01 -> V ~ 539.63; 0.001 -> V ~ 5276.96 (also, actually, with proper rounding and a step of 0.2 it's V ~ 44.23)
3. Not sure what you mean. The 2d part of a 3d sphere is a surface, not a scalar(circumference). But I don't see why not - you'd get a bunch of concentric normal spheres, each with a surface area, which you can sum to get a surface area of the sphere based on the step you chose. It's how I came up with the idea originally.
4. I did not make the connection... I'm still not sure how it fits. But with that formula, a 4d unit sphere has cubic volume of ~19.74 u³, though I have no idea how it fits anywhere.

 

[BvU]

1. if you don't show the steps in detail we have no idea what you do.
2. something that depends on a stepsize isn't sensible
3. Sorry I meant 2d area of 3d volume (same thing as 3d volume of 4d sphere but one step down)
4.

though I have no idea how it fits anywhere

If you do it right you get the volume of the shell of the 4d hypervolume .

 

[martix]

1. I guess I could have been clearer in expressing them.
2. Not sensible as in pointless or doesn't make logical sense?

 

[WWGD]

I'm trying to wrap my head around how higher dimensions work. Not over some academic pursuit, merely curiosity(which started with me discovering this neat recurrence relation).

Given that the surface area of a sphere is an object of 1 dimension lower.
So on a 3-sphere, the surface area is a plane and is measured in say m².
Then for a 4-sphere, the surface area is a 3-dimensional sphere with a volume measured in m³.

Then I thought a bit about volumes of higher dimensional spheres before realizing hypervolume doesn't mean the same as normal cubic volume, throwing any intuition as what it is(let alone how big it is) out the window.

However, in a 4-sphere, one can imagine taking slices at equal steps and obtaining a multitude of concentric 4-spheres, each of which has a "surface area" with a regular 3-dimensional volume.

So if you define a step, using this method you get a meaningful answer for "What is a the 3d volume of a 4-sphere with radius r?"

Say,
S(step) = 0.2u (as in units)
V4 = A4(u) + A4(0.8u) + A4(0.6u) + A4(0.4u) + A4(0.2u)
Or ~41.22u³ if my calculations are correct.

Does this even make sense? If no, why not?

Sorry to nitpick:
. But in 4 dimensions, the boundary being one dimension lower does not mean it is a plane, nor even planar ( embeddable or contained-within a plane) .

 

[chiro]

Hey martix.

Have you thought about using differential geometry to do this?

What you can do is find a way to go from a n-sphere to an n-box and do standard integration in so many dimensions.

The other alternative [which is probably going to be more feasible] is to use the wedge product definition of area.

A differential geometry book should contain the basic ideas in terms of wedge products and norms / inner products on those.

 

[martix]

No, I hadn't thought of that. It's a little beyond me ATM. In a year maybe.

Also, evidently I was wrong.
The correct answer my initial series of calculations should be ~35.53 u³
V = A(u, 3) + A(0.8u, 3) + A(0.6u, 3) + A(0.4u, 3) + A(0.2u, 3)
(I was using A(u, 4), which is the 4-dimensional "surface area" of a 5-sphere.)

 

[SlowThinker]

Also, evidently I was wrong.

You don't seem to realize where the real problem is.
If you perform your slicing process on a sphere, you get a 2D "area". Can you imagine where the slices are located in the sphere, and what, if any, relation your result has to the surface and volume of a 3D sphere?