# Homework Help: Generalizing symmetry axis of constant-contour ellipses

1. Jun 10, 2013

### Niles

Hi

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

$$f(x, y, z) = \exp(-x^2a)\exp(-y^2b)$$
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)
$$r = (\sin \theta, \cos \phi, \sin \theta\sin\phi, \cos \theta)$$
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.

Last edited: Jun 10, 2013
2. Jun 10, 2013

### Office_Shredder

Staff Emeritus
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)

3. Jun 10, 2013

### 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
$$r = (\sin \theta \cos \phi, \sin \theta\sin\phi, \cos\theta)$$
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.

Last edited: Jun 10, 2013
4. Jun 10, 2013

### 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 $(x, y, z)\rightarrow (x+x', y+y', z+z')$