News for hackers. Stuff that matters.

A comment on If you put your hand in the Large Hadron Collider.... Parent comment

Re:If you put your hand in the Large Hadron Collider... by demallien on Saturday September 25, 2010 @04:19AM (#1726581)
No, it isn't universal, even in our universe. If you go and draw a circle on the surface of the Earth, with a 1km diameter, the ratio of the diameter to the circumference will not be pi, due to the curvature of the Earth's surface. I suppose that you could say that I wasn't measuring the right diameter in this case, and that the true diameter passes underneath the surface of the Earth. But there is an similar deformation introduced by the curvature of spacetime caused by the Earth's mass, so you still won't get pi as a result when measuring the ratio between the circumference and the diameter.

That said, I would very much like to understand why c = 2 x pi x r actually works. pi itself can be derived from pure mathematics, knowing the relationship e^(i x pi) = -1. So why a circle should have such a relationship with this constant is to me a fascinating question.

[ Parent | on HN ]
Re:If you put your hand in the Large Hadron Collider... by derefr on Saturday September 25, 2010 @04:38AM (#1726610)
> Why [should] a circle should have such a relationship with this constant?

Because the "circles" we talk about in mathematics are found in Euclidean space—which is defined to have simple metric properties. Our own space is non-Euclidean.

http://en.wikipedia.org/wiki/Euclidean_space

[ Parent | on HN ]
Re:If you put your hand in the Large Hadron Collider... by demallien on Saturday September 25, 2010 @06:01AM (#1726680)
Umm, yes, I understand that c = 2 x pi x r only holds in Euclidean space (indeed, my original post gave two examples of circles in non-Euclidean spaces, to illustrate my point), but my question was more about why it holds at all. That is, if you consider pi as being a constant that you can derive from e (which is itself just a power series), and i (again, just a mathematical construct), it seems quite stunning to me that this constant (pi) should have such strong ties of Euclidean space. It implies a deeper connection between Euclidean Space, e, and i. Why should that be so?
[ Parent | on HN ]
Re:If you put your hand in the Large Hadron Collider... by derefr on Saturday September 25, 2010 @07:05AM (#1726735)
1. Taking e to a complex power (using θi) walks you through the complex plane at a given angle θ. (That is, f(x) "draws" a continuous curve, where the "pen" has a constant rotational torque of θi.)

2. When θ = pi, f(x) is periodic (the angle is such that f(x) intersects itself or "loops.")

3. The set of complex numbers is isomorphic to a 2D Euclidean plane. Therefore, a walk at angle i*pi through the set of complex numbers is isomorphic to a circle.

[ Parent | on HN ]