Hypersphere Hypervolume Applications?

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Hello all,
I was curious on the practical applications of representing a sphere in four dimensions. I recently had to prove that the V=∏2R4/2. I hope I was able to format that correctly. Anyways I couldn't come up with a reason to do some beyond simply proving it for proofs sake. Perhaps modeling the effects of temperature on volume?
Thanks
 
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For 4 dimensions specifically there are certainly applications but probably the main point of the question was just to practice the integration. For n-dimensions in general there are lots of nice reasons to know the volume of a ball/sphere (which is actually the surface of a ball) - for example there are probability/analysis statements about random vectors in high dimension which draw heavily on being able to calculate the volume of various portions of the sphere.
 
Thanks for taking the time to answer my question. And yes the main point is to do just the integration. I'm taking calculus 3 currently and have already worked it out, took awhile. I was mostly just curious if an engineer would ever use such an equation. I could see the practicality of using higher dimensional functions to calculate probabilities. I may look into that, only a little, for entertainment purposes.
 
Interestingly, we used the formula for the "surface area" for the hypersphere in thermal physics! It came up when deriving the multiplicity of a ideal gas with fixed amount of energy (related to probability of each microstate of the gas).

First we just assumed there is one particle. So all possible microstates create a surface in the 6 dimensional position-momentum graph (3 component position, 3 component momentum). All possible position states are in a closed shape in the position graph of fixed volume (e.g. a box in the shape of a cube. then all points in the cube are possible position states). But in momentum graph only points on a surface of a sphere (3 dimension in the case of one particle) are allowed because of the total energy constraint. When considering more particles, the sphere in the momentum graph becomes a hypersphere and eventually in the derivation, the "surface area" of this hypersphere is required.
 
It's a bit strange how such simple object as a n-sphere is also so difficult.
V(n)=2(n+1) div 2 πn div 2 n‼-1 Rn
 

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