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Hypersphere Hypervolume Applications?

  1. Nov 19, 2013 #1
    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?
    Last edited: Nov 19, 2013
  2. jcsd
  3. Nov 19, 2013 #2


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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.
  4. Nov 20, 2013 #3
    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.
  5. Nov 20, 2013 #4
    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.
  6. Nov 20, 2013 #5
    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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