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Normal to an ellipsoid

  1. Apr 28, 2006 #1

    I am trying to find the normal to an ellipsoid of the form

    [tex] x^2/a^2 + y^2/b^2 + z^2 = 1 [/tex]

    What I did is the following: Let

    [tex] psi = x^2/a^2 + y^2/b^2 + z^2 [/tex]

    then grad psi (and thus the normal) is:

    [tex] grad (psi) = (2x/a^2, 2y/b^2, 2z) = n [/tex]

    Could anyone tell me whether that is the correct approach to find the normal to an ellipsoid? Thanks a lot!
  2. jcsd
  3. Apr 28, 2006 #2


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    Well, I'd have to work it out to tell you for sure...

    But I have a better idea: I could try to help you figure out how to work it out to tell for sure. :smile:

    The very first thing I'd do is to see if it's consistent with things I know. Are there any ellipsoids whose normals you already know? For for a general ellipsoid, are there any points for which you already know the normal vector?

    The next thing I would do is to try and verify that it satisfies the defintion of a normal vector: do you know how to find tangent vectors to your ellipsoid? (Or even better, how to find the tangent plane?) If so, you could then test to see if your alledged normal vector is, in fact, perpendicular to the tangent vectors.
  4. Apr 29, 2006 #3
    Well, if a = b =1 then it would simply be a sphere, wouldn't it? And for a sphere the normal vector is simply the vector x of a point lying on the surface of the sphere.

    Wouldn't that be the case for an ellipsoid as well?
    i.e. a point on its surface is the normal vector at that point? I calculated the normal vector this way in spherical polar coordinates (see attached file).

    However, that is different from what a got by using the gradient function. So, which one would be right and why?

    Attached Files:

    Last edited: Apr 29, 2006
  5. Apr 29, 2006 #4
    Just a thought but couldn't you parameterise your ellipsoid using the 'usual' two variable parameterisation of a unit sphere but with an extra factor of a (or b) for the x and y components. eg. instead of writing x = sinucosv, let x = asinucosv. Similarly let y = bsinusinv and z = cosu. Using these substitutions, (x/a)^2 + (y/b)^2 + z^2 = 1 as expected.

    \Phi \left( {u,v} \right) = \left( {a\sin u\cos v,b\sin u\sin u,\cos u} \right)

    Find the derivatives of the above wrt u and v, cross them and that should be the normal if I haven't neglected anything. (I haven't checked that method, it's just something I came up with off the top of my head).
    Last edited: Apr 29, 2006
  6. Apr 29, 2006 #5


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    At a point on the surface, the normal is perpendicular to the tangent plane.
    At a point on a circle (or sphere), the tangent vectors are perpendicular to the radial position vectors (since the magnitude of the radius is constant on the circle). So, in this case, the normal and radius vectors are parallel.

    Will this be true for an ellipsoid?
    Consider these physical examples:
    http://cage.rug.ac.be/~hs/billiards/billiards.html [Broken]

    Now consider this:
    Last edited by a moderator: May 2, 2017
  7. Apr 29, 2006 #6


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    As Robphy said, your original approach is correct.
    Last edited: Apr 29, 2006
  8. Apr 30, 2006 #7
    thanks very much!! Those links are really good and helped me to see that the vetor of an arbitrary point on an ellipsoid is NOT equal to its normal at that point! :smile:

    One other quick question: the normal that I had initially given was NOT normalized.

    [tex] grad (psi) = (2x/a^2, 2y/b^2, 2z) = n [/tex]

    However, in order to normalize it will become a big mess, since one has to divide everything by

    [tex] \sqrt {4x^2/a^4 + 4y^2/b^4 + 4z^2} [/tex]

    Now, if one uses spherical polar coordinates as Benny suggested how would you normalize that and would it become any simpler? I think the problem is that I don't know how t take the grad of a vector field (is that what he or you Benny meant by crossing the derivative?)
    Last edited: Apr 30, 2006
  9. Apr 30, 2006 #8


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    This may help (normalization after changing coordinates): if the change of variables

    [tex]\hat{x}=ax, \hat{y}=by, \hat{z}=z[/tex]

    is made, then the equation becomes [itex]\hat{x}^2 + \hat{y}^2 +\hat{z}^2=1[/itex] so the gradient is then

    [tex]\vec{n}=2\left< \hat{x}, \hat{y} ,\hat{z}\right> [/tex]

    and the magnitude of the gradient is then

    [tex]\| \vec{n}\| = 2 \sqrt{ \hat{x}^2 + \hat{y}^2 +\hat{z}^2} =2\sqrt{1} = 2 [/tex]

    where the equation of the transformed solid was used to simplify above, and the unit vector in the direction of the gradient is then

    [tex]\vec{u}=\left< \hat{x}, \hat{y} ,\hat{z}\right> [/tex]

    but, hey, this may not help at all...
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