ScienceNostalgia101 Posted September 11, 2018 Share Posted September 11, 2018 So I was recently thinking about the similarity in formulae between spheres' volume (4*pi*r*r*r/3) and their surface area. (4*pi*r*r) Firstly, I noticed that the surface area looks like it's the derivative of volume with respect to radius... which come to think of it makes sense as the rate of change in volume at a point in time is that outer spherical shell being added times its thickness. But secondly I also noticed that the ratio of the two is r/3. As in, as if the average particle in a sphere were only a 1/3 of the way to the outside. More generally, V/A is in length units. Am I figuring this right? Does V/A represent average distance, root mean square distance, or whatever other measure of central tendency from center to outside? More generally than that, how is this extrapolated to other shapes? Does V/A represent anything in particular more generally or is its dimensionality usually meaningless? Link to comment Share on other sites More sharing options...

Endy0816 Posted September 12, 2018 Share Posted September 12, 2018 Not sure on V/SA specifically. We do see this the other way around expressed in terms of Surface Area to Volume. https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio Important in a number of areas. Can look at it in terms of an enclosing cube too. r/3 = d/6 = d^{3}/6d^{2} I'm sure someone more versed on the subject can go into ridiculous detail on this one. Link to comment Share on other sites More sharing options...

timo Posted September 13, 2018 Share Posted September 13, 2018 (edited) On 9/11/2018 at 11:18 PM, ScienceNostalgia101 said: [] I noticed that the surface area looks like it's the derivative of volume with respect to radius... which come to think of it makes sense as the rate of change in volume at a point in time is that outer spherical shell being added times its thickness. Absolutely. Adding up thin shells to create the full sphere is actually a very common technique to access the volume, i.e. [math] V(r) = \int_0^r A(r) \, dr[/math] Quote But secondly I also noticed that the ratio of the two is r/3. As in, as if the average particle in a sphere were only a 1/3 of the way to the outside. Well, picking up on the integration example: The average distance <r> of a particle from the center in a sphere of radius R is [math] <r> = \frac{1}{V(R)} \int_0^R \, r \cdot A(r) dr = \frac{1}{V(R)} \int_0^R \, r \cdot 4\pi r^2 dr = \frac{1}{\frac 43 \pi R^3} \left[ \pi r^4 \right]_0^R = \frac 34 R[/math] (modulo typos: Tex does not seem to work in preview mode ... EDIT: And apparently not in final mode. The result in the calculation above is <r> = 3/4 R). There is a general tendency that the higher the dimension, the more likely a random point in a sphere lies close to the surface. There is a famous statement in statistical physics that in a sphere with 10^23 dimension, effectively all points lie close to the surface. Quote More generally, V/A is in length units. Am I figuring this right? Does V/A represent average distance, root mean square distance, or whatever other measure of central tendency from center to outside? In all sensible definitions of volume and area I am aware of (at least in all finite-dimensional ones), volume has one more dimension of length than area. Hence, their quotient indeed has dimensions of length. I don't think the quotient itself has a direct meaning. But there are theorems like that a sphere is the shape that maximizes the V/A ratio for a fixed amount of V or A. Quote More generally than that, how is this extrapolated to other shapes? Does V/A represent anything in particular more generally or is its dimensionality usually meaningless? I already commented on the dimensionality. But I still encourage you to just play around with other shapes: Cubes are the next simple thing, I believe. Edited September 13, 2018 by timo 2 Link to comment Share on other sites More sharing options...

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