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Generalizing symmetry axis of constant-contour ellipses

  1. Jun 10, 2013 #1
    Hi

    I am looking at the contours of the following function f, which trace out an ellipse:

    [tex]
    f(x, y, z) = \exp(-x^2a)\exp(-y^2b)
    [/tex]
    Here [itex]a\neq b[/itex] are both positive, real constants. The axis of these ellipses is along [itex]z[/itex]. Now, I am wondering how to generalize the function f such that the symmetry axis of these elliptical contours lies along an arbitrary vector defined by some line that goes through the point p=(x0, y0, z0) and has directional vector r (in usual spherical coordinates)
    [tex]
    r = (\sin \theta, \cos \phi, \sin \theta\sin\phi, \cos \theta)
    [/tex]
    If [itex]a=b=1[/itex] the task would be easy: In this case we can write [itex]d^2 = x^2 + y^2[/itex], and generalize this such that it gives the distance between the point/coordinate (x,y,z) and the above line (p, r). But when [itex]a\neq b[/itex] I can't write [itex]d^2[/itex] like that. What can I do in this general case?

    Note that this question is a generalization of this thread, where the case [itex]a=b=1[/itex] was treated. Thanks in advance for hints/help.
     
    Last edited: Jun 10, 2013
  2. jcsd
  3. Jun 10, 2013 #2

    Office_Shredder

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    Rather than try to do a change of coordinates directly, I think it helps to build intuition to do it backwards. In the exponential, we want a function whose value is 0 if you are lying on a line given by a specific vector v (presumably a unit vector but it doesn't really affect the problem), and furthermore we pick two vectors w1 and w2 orthogonal to v and having unit length such that if our position (x,y,z) is: [itex](x,y,z)= \alpha w_1 + \beta w_2+\gamma v[/itex] then in the exponential we have
    [tex] e^{-\alpha^2 a -\beta^2 b } [/tex].

    Now we're done basically! The function f(x,y,z) is defined in two steps:
    1) Write [itex] (x,y,z) = \alpha w_1 + \beta w_2 + \gamma v [/itex]
    2) Define
    [tex] f(x,y,z) = e^{-\alpha^2 a - \beta^2 b } [/tex]
    To write this in a single formula all you need to do is solve for [itex]\alpha[/itex] and [itex] \beta[/itex] in terms of (x,y,z) which can be done by inverting a 3x3 matrix (assuming you know w1, w2 and v)
     
  4. Jun 10, 2013 #3
    Hi Office_Shredder

    Thanks for that. You are right, doing it backwards like this is more intuitive. So it reduces to finding the unit vectors w1, w2 and v for a given symmetry axis. I write the axis generally, so it passes through some point P=(x', y', z') with unit direction vector
    [tex]
    r = (\sin \theta \cos \phi, \sin \theta\sin\phi, \cos\theta)
    [/tex]
    So the axis is given by L(t) = P + rt. I am working in units of meters so P is in meters, just like the product rt (r is unitless). This line is v. I am a little uncertain of this, because I need a unit vector, whereas this is basically a parametrized line.

    Now I need to find w1 and w2, parallel to the two axes of the ellipse. Can I get a hint to how I can determine these?

    Thanks for getting me started.


    EDIT: I realized that I don't need to take into account P at the current moment. I can always do that afterwards, when I have found [itex]\alpha,\beta[/itex], sinply by shifting [itex]x, y[/itex] and [itex]z[/itex]. So v is simply given by the unit vector in spherical coordinates for now, r.
     
    Last edited: Jun 10, 2013
  5. Jun 10, 2013 #4
    I figured it out, w1 and w2 are of course just the other spherical components, besides r. So now I know what [itex]\alpha[/itex] and [itex]\beta[/itex] are in terms of [itex](x, y, z)[/itex]. If I want to shift them by a constant amount, I just shift [itex](x, y, z)\rightarrow (x+x', y+y', z+z')[/itex]
     
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