### Re: can you help with the maths behind an explanation of lightdilutinginasphere

Date: Sun Sep 2 20:12:50 2007
Posted By: Joe Fitzsimons, Grad student, Quantum and Nanotechnology Theory Group, Department of Materials, Oxford University
Area of science: Physics
ID: 1188340364.Ph
Message:

Good question.

In 3D space, the volume of a sphere is prpportional to the radius of the sphere cubed and the surface area is proportional to the radius squared.

Volume in N-dimensional space has dimension of length to the power of N (m^N if the length is in meters). Since a sphere is symmetric, the total volume of a hypersphere (which is what we call a sphere in more than three dimensions) is proportional to the radius to the power of N (i.e. V=C(R^N) for some constant C).

Now volume is always the integral of the surface area over the radius (or the extra dimension for arbitrary shapes), so to get the surface area all we need to do is differentiate the volume with respect to R. This gives A= (C*N)(R^(N-1)), so the area is proportional to the radius to the power of (N-1).

The light from a light bulb in N-dimensional space would spread out as the surface of an N-hypersphere, and so would be diluted over an area proportional to R^(N-1).

I hope this helps.

Here are a few places with more detailed mathematical treatments:
[1] http://en.wikipedia.org/wiki/Hypersphere
[2] http://mathworld.wolfram.com/Hypersphere.html
[3] http://mathworld.wolfram.com/SurfaceArea.html

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