Generalizing symmetry axis of constant-contour ellipses

[Niles]

Hi

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

<br /> f(x, y, z) = \exp(-x^2a)\exp(-y^2b)<br />
Here a\neq b are both positive, real constants. The axis of these ellipses is along z. 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)
<br /> r = (\sin \theta, \cos \phi, \sin \theta\sin\phi, \cos \theta)<br />
If a=b=1 the task would be easy: In this case we can write d^2 = x^2 + y^2, and generalize this such that it gives the distance between the point/coordinate (x,y,z) and the above line (p, r). But when a\neq b I can't write d^2 like that. What can I do in this general case?

Note that this question is a generalization of this thread, where the case a=b=1 was treated. Thanks in advance for hints/help.

 

[Office_Shredder]

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: (x,y,z)= \alpha w_1 + \beta w_2+\gamma v then in the exponential we have
e^{-\alpha^2 a -\beta^2 b }.

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

 

[Niles]

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
<br /> r = (\sin \theta \cos \phi, \sin \theta\sin\phi, \cos\theta)<br />
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 \alpha,\beta, sinply by shifting x, y and z. So v is simply given by the unit vector in spherical coordinates for now, r.

 

[Niles]

I figured it out, w1 and w2 are of course just the other spherical components, besides r. So now I know what \alpha and \beta are in terms of (x, y, z). If I want to shift them by a constant amount, I just shift (x, y, z)\rightarrow (x+x&#039;, y+y&#039;, z+z&#039;)