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Comment by evanb | Hacker News Reader

This is a nice analog but unfortunately I think it breaks down in a way the "π" calculation does not.

In the article 2π(d) = the ratio of the circumference to the radius. This is dimensionless, in the sense that the circumference and the radius are both lengths (measured in meters, or whatever), so 2π(d) is really just a number.

But the (hyper)volumes you're talking about depend on dimension, which is exactly why you say "hyper". In 2 dimensions the volume is the area, πr^2, which has dimensions L^2 [measured in m^2 or whatever]. But in 3 dimensions the volume is 4/3 πr^3, which has dimensions L^3. The 5 dimensional (hyper)volume has dimensions L^5, and so on.

So, "comparing" these to find out which is bigger and which smaller is not really meaningful---just like you shouldn't ask which is the bigger mass: a meter or a second? Neither is, they aren't masses.

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Yeah, though they are somewhat similar. We could perhaps say the volume of the unit ball is (in some sense) "larger" than the area of the unit disk, because both are measured relative to a radius of 1. So the area of the disk is fewer "units" than the volume of the ball.

Anyway, it seems independently interesting that this value peaks for the 5-ball, or the ~5.2569-ball. The non-fractional difference between fractional dimension of peak hyper volume and peak hyper surface area seems also interesting. (I assume there is some trivial explanation for this though.)

No it doesn't work that way because the units of hyper-volume and length are different.

However, once you take appropriate roots of hypervolume to get same units you can safely compare. Or the otherway round take appropriate powers of length to get same units as hypervolume.

With that it would probably mean the units monotonically approach 0, rather than first increasing and then decreasing. At least for whole dimensions. I'm not sure about monotonicity with fractional dimensions.

I agree; the fair comparison is the nth root of the hypervolume in n dimensions, (V(n))^(1/n). This monotonically decreases from n=0, which shows the counterintuitive point that people often want to make anyway: an n-sphere takes up less and less of an n-hypercube. The peak at n~=5 is illusory.

Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.

Imagine you want to compare sizes of video game levels, where some are 2D, made from pixels, and some are 3D, made from voxels. You could stipulate that n pixels are equivalent to sqrt(n) voxels. But you could also stipulate that n pixels are equivalent to n voxels. I don't think either is more correct than the other.

Which is bigger: a meter, a hectare, or a liter?

A hectare

What if the property with 1 hectare of area were 0.1m x 10^5 m? The meter is bigger than the field is wide.

Or, more to the point, suppose someone came with a different system of units and said: look, one of our standard lengths is 10^-6 of one of yours, but one of our standard areas is defined just like yours: 1 ha = (100m)^2 and (1 of our areas) = (100 of our lengths)^2.

In other words, their standard length = 10^-6m; their area = 10^-12 ha = 10^-8 m^2.

Is their standard area bigger or smaller than their standard length?

Notice that in proportion the lengths and areas are the same.

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