I have a copy of a recent 3rd year calculus exam. There's a question about curves right off the bat, which is: If \( C(t)\;=\; (c_1(t),\;c_2(t),\;c_3(t)),\; a \le t \le b \) is a curve in three dimensions how is its length defined? And: what does it mean to parametrise a curve by arc length s, and how does this lead to the notion of curvature? This has something to do with fitting circles on the curve with a known radius, doesn't it? Wikipedia says something about the triangle inequality becoming an equality for points p and q on the curve with an arc a(p,q) between, which I don't really have an intuitive grasp of.
Length: \(L=\int_a^b|C'(t)|dt\) To parametrise a curve by arc length s means that with a function \(\phi(t)=\int_a^t|C'(t)|dt\) to parameterize C by \(D(s)=C(\phi^{-1}(s))\) now s is in [0,L]. Note that \(|D'(s)|=|C'(\phi^{-1}(s))[\phi^{-1}]'(s)|=1\) so the "speed" with respect to s is 1. If the direction of D'(s) is constant then the curve will be a straight line. Since the length of D'(s) is constant, now the information left in D''(s) is essentially only how the curve deviates from being straight.
K, if I have this at all, L is the length of a curve in terms of small chord-lengths that correspond to arc-lengths. So you can say that taking the integral, as the chord-lengths tend to zero, 'straightens' the curve. Or at least, L is the equivalent linear distance of a curve. So the 'notion' of curvature depends on a pointwise relation to 1/r over the length L, and r depends on fitting circles to the curve in a form of interpolation? However this means retaining both the chord and arc distances, or the notion of the distances tending to zero 'in the limit'?
Hmm. Wikipedia goes at this curvature thing from the usual multiple aspects. But an hyperbola, say, is a one-dimensional curve in the plane. If you make a surface of revolution by rotating the hyperbola about the semimajor or semiminor axis, that sounds like a one-dimensional curve in three dimensions. Anyways, the next part of the question asks you to parametrise by arc length s the curve: \( C(t)\;=\; (r cos t, r sin t, ct),\; 0 \leq t \leq 2 \pi \) Applying the solution \(L=\int_a^b|C'(t)|dt\), \(\phi(t)=\int_a^t|C'(t)|dt\) means taking the derivative: \( C'(t)\;=\; (-r sin t, r cos t, c) \), and we want an equation for the arc length so that: D(s) is the curve C parametrised (that is, it takes as argument) the inverse of the function \( \phi (s) \). But \( t \) can be for instance, the tangent to the curve at (x,y), or an infinitesimal rotation of a one-dimensional curve, such as in a surface or solid of revolution, like say an hyperboloid sheet? Please Register or Log in to view the hidden image!
What I've been trying to relate intuitively with the curvature question is that of a conserved quantity, and symmetry, the latter being some geometrical property of general curves (as in the hyperboloid of revolution). So if in the case of the above, the conserved quantity is radial distance--which every vertical planar slice through the sheet has a different value of--then the symmetry is 'broken' by the horizontal curvature (the hyperbola lying along a horizontal slice). Or if each point on the sheet has two straight lines through it that intersect both perimeters, the symmetry is the intersection of two planes through the sheet, and the conserved quantity is hyperbolic curvature. It all depends on how you 'angle' the required planes of intersection. Note this is about me trying to get a handle on the phrase "globally hyperbolic spacetime". --in which the mathematical notion of inverse radial distance is extended to the notion of inverse function of arc/chord distance, and I suppose the circle limit (of curve interpolation).
Tsk. Silly me. Each point has two straight lines through it, so both lines lie on the same plane through the sheet. But if instead you align any plane so it intersects a single (skew) line then such a plane can meet the surface at multiple angles, or you can have a single plane slicing the sheet which intersects the two lines through a given point, and two more planes each intersecting one of the same lines, the latter slices having arbitrary angles at the surface and so they will intersect arbitrary lines elsewhere on the sheet. So there's a planar symmetry which breaks if you use the latter method of slicing. This is easier to visualise, I think, with this figure: Please Register or Log in to view the hidden image!
It is taught in freshman year in college. For derivation in two dimensions see here. Three dimensions is easy to generalize. Not even close, there is no correlation.All this is taught in algebraic geometry, in freshman year.
Hmm. Maybe someone can explain this: Please Register or Log in to view the hidden image! So is Newtonian mechanics. I have another 2nd year paper with a question about free-falling bodies in it and the answer is in a 1st year physics book. Go figure.
You need to retake differential geometry in general with accent on the Frenet formulas in particular. See here for a crash course. Spend some time studying this, it will answer all your questions. In return, perhaps you could stop trolling the thread on covariance? Thank you.
So you're saying the diagram is bullshit? Or are you just trolling? Wait, I'm getting a beam from somewhere. You aren't the someone who can explain why a circle is fitted to the curve in the diagram, or what that means.
The extrinsic curvature of a curve in a plane is, when non-zero, is adequately described by a circle tangent to the curve with matching curvature. Such a circle is called the osculating (kissing) circle. The extrinsic curvature of \(y=f(x)\) is \(\kappa(x) = \frac{f''(x)}{(1 + (f'(x))^2)^{\frac{3}{2}}}\). For a circle of radius r, we have \(f(x) = \sqrt{r^2 - x^2}\), \(f'(x) = \frac{-x}{\sqrt{r^2-x^2}}\) and \(f''(x) = \frac{-r^2}{(r^2-x^2)^{\frac{3}{2}}}\). Therefore, for a circle of radius r, \(\kappa(x) = \frac{f''(x)}{(1 + (f'(x))^2)^{\frac{3}{2}}} = \frac{\frac{-r^2}{(r^2-x^2)^{\frac{3}{2}}}}{\left(1 + \left( \frac{-x}{\sqrt{r^2-x^2}} \right)^2 \right)^{\frac{3}{2}}} = \frac{\frac{-r^2}{(r^2-x^2)^{\frac{3}{2}}}}{\left(\frac{r^2-x^2}{r^2-x^2} + \frac{x^2}{r^2-x^2} \right)^{\frac{3}{2}}} = \frac{\frac{-r^2}{(r^2-x^2)^{\frac{3}{2}}}}{\left(\frac{r^2}{r^2-x^2} \right)^{\frac{3}{2}}} = \frac{-r^2}{r^3} = - \frac{1}{r}\) which makes sense. But in three or more dimensions, curvature is not the end of the story, because a surface in three-dimensions can curve in two different directions at once. http://mathworld.wolfram.com/Curvature.html http://mathworld.wolfram.com/OsculatingCircle.html http://mathworld.wolfram.com/OsculatingCurves.html
I think Tach's memory is failing him at a critical point of this discussion, viz: Famous last words and all that.
