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Differential Geometry Question

Cowboy-BebopCowboy-Bebop Registered User regular
edited November 2010 in Help / Advice Forum
While going over some information about Differential Geometry I learned last year, a friend and I hit a snag. We were reviewing principal and Gaussian curvatures of surfaces when the issue arose. I understand that for a cylinder, the Gaussian curvature is zero because in one principal direction the curvature will be zero. The other principal direction will have constant positive or negative curvature, depending on choice of normal. So along one direction the curve given by intersecting the a normal plane with the cylinder will be straight and in one direction it will be a circle. Similarly for a sphere, but in this case every curve created by the intersection of a normal plane will be a circle.

I don't understand why, for a Monkey Saddle or a parabolic sheet (z = y^2 for instance), the origin will have both principal curvatures zero and thus be a planar point. Obviously the math behind this assertion proves this but I'm trying to get an intuitive understanding.

Specifically with the parabolic sheet, in the direction along the y-axis, the curve will clearly have zero curvature as the intersection of the parabolic sheet with the normal plane is again a straight line. When you look at the normal plane intersecting in any other direction, the curve appears to have non-zero curvature. Is there something that I'm missing? This is also true for the Monkey Saddle, but in all directions.

The conclusion that we came to is that since you're measuring the curvature at the origin, somehow in any direction it must always be zero specifically at that point. This was unconvincing though. Is my intuition getting in the way? Does anyone have any insights?

Cowboy-Bebop on

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    rabidrabbitsrabidrabbits Registered User regular
    edited November 2010
    It's been a while, but I'm pretty sure the principal curvatures are nonzero at those points and suspect you're doing your math wrong.

    Correct me if I'm mistaken, because I don't remember justifications for everything and will be making some assumptions:

    z=y^2 has principal axises of y=constant and x=constant for any point. The principal curvature for x=x_0 is 0, but for y=y_0, it should be 2/(1+4y^2)^(3/2), which is never zero, so you have no planar points.

    And I don't know what a Monkey Saddle is, but the standard Saddle equation is z = y^2 - x^2, which have principal curvatures of 2 and -2 at the origin, but I think they may have planar points on the |x|=|y|>0 lines (worth checking).

    If my math is wrong, could you supply some equations for me to check my work?

    rabidrabbits on
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    Cowboy-BebopCowboy-Bebop Registered User regular
    edited November 2010
    The monkey saddle is this surface - http://en.wikipedia.org/wiki/Monkey_saddle It almost seems arbitrarily constructed to have certain geometric properties, i.e. a planar point at the origin. The equation is z = x^3 - 3xy^2

    In my Springer text, they gave two examples of surfaces with planar points (principal curvatures both zero). One was the above monkey saddle and the other was actually z = y^4. I just generalized to parabolic sheet because the properties should be similar, correct?

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    rabidrabbitsrabidrabbits Registered User regular
    edited November 2010
    Similar, yes, but not in any way important to differential geometry. The 2nd derivative at the point is incredibly important. The principal curvatures for z=y^4 are 0 and 12(y^2)/(1+16y^6)^(3/2), which happens to be 0 at y=0.

    The monkey saddle has a planar point in the center because, if you put it in cylindrical coordinates, is z = r^3 * Re(e^(i3theta)). So for all fixed theta, you have z = r^3 * constant, which means the curvature is c*6r/(1+9r^4)^(3/2), which is 0 at r=0.

    You've probably done this math already, but that helps me at least grasp where you're at so we (and other readers) are on the same page.

    Now, the reason these look *like* there is curvature when the math tells you there is none, is that the curvature only really cares about the 2nd derivative. Why? Because every plane can have a first derivative, but all have 2nd derivatives of zero. You can take any point, shift the equation so the point is now the origin, then rotate the equation so that the 0th and 1st order power series approximation (locally, at least) are z = 0. So the power series approximation is going to look like z = f(theta)r^2 +g(theta)r^3 +..., and this f(theta) happens to be the mathematical curvature at that point in the theta direction.

    This "planar point" concept does not mean that there is a tiny fragment of a plane around this point, but rather if you took a *small enough* area around this point, it would resemble a plane enough that nobody would care too much about the difference. As opposed to a point with a nonzero curvature in some direction, which makes it different from a plane in exactly the first place someone would look if they were comparing equations.

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    Cowboy-BebopCowboy-Bebop Registered User regular
    edited November 2010
    Thanks a lot, I think that helps and I do feel like it was an instance of my intuition getting in the way of what the math was telling me, similar to other subjects (damn you analysis). It does seem strange for a 4th degree parabolic sheet to have a planar point to me though because isn't it essentially a steeper parabola?

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    Cowboy-BebopCowboy-Bebop Registered User regular
    edited November 2010
    Oh wait I kind of see it now, because y = x^4 between [0,1] is more level than a parabola. Is this notion correct?

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    rabidrabbitsrabidrabbits Registered User regular
    edited November 2010
    You should train yourself to be looking at open areas such as (-1,1) for instance, because

    1) they are the bread and butter of topology and analysis, and

    2) and more importantly for this problem, you don't want your point to be on a boundary. If there is a direction with no wiggle room, it has indefinite curvature in that direction.

    But yeah, that's pretty much it.

    edit: Looking at descriptions in http://en.wikipedia.org/wiki/Curvature there is a much more intuitive notion for curvature in a direction. If you were to take a differentially continuous path along your surface intersecting with your point, say c(t) where c(0) = p, then look at the points c(-h), c(0), c(h), for h very small. These three points will define a circle. If you take lim(h=0), the circle defined will have radius 1/k, where k is the curvature in the direction of your path c(t).

    So when you're looking at flatness, don't just look at "is this more flat than that", but instead imagine a circle defined by very near points along a path. If this circle approaches infinite size as the points get closer together, then the curvature is 0.

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    Cowboy-BebopCowboy-Bebop Registered User regular
    edited November 2010
    Oh yeah I didn't actually mean to include the boundary on that interval.

    Thanks! That notion of curvature seems more appropriate.

    Cowboy-Bebop on
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