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Sketching the Level Surface of a Scalar Field

  1. May 1, 2014 #1
    Question: For the scalar field [itex]\Phi = x^{2} + y^{2} - z^{2} -1[/itex], sketch the level surface [itex] \Phi = 0 [/itex]. (It's advised that in order to sketch the surface, [itex]\Phi[/itex] should be written in cylindrical polar coordinates, and then to use [itex]\Phi = 0[/itex] to find [itex] z [/itex] as a function of the radial coordinate [itex]\rho[/itex]).

    I've done as advised, and have found:

    [itex] \Phi = \rho^{2}cos^{2}(\phi) + \rho^{2}sin^{2}(\phi) - z^{2} - 1 [/itex]

    and therefore

    [itex] z = \sqrt{\rho^{2} - 1} [/itex]

    However, i don't actually know how this is supposed to enable me to construct the sketch. My instinct was to rewrite the expression for [itex]\Phi[/itex] in Cartesian coordinate form as [itex] z(x,y) [/itex] i.e. [itex] z = \sqrt{x^{2} + y^{2} - 1} [/itex], and to plot this using a suitable program. I think i must have a deep misunderstanding of what the scalar field expression actually represents.

    Any advise would be appreciated.
     
  2. jcsd
  3. May 1, 2014 #2

    LCKurtz

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    If I were plotting this, I would ignore that hint. You are asked to graph the surface ##x^2+y^2-z^2=1##. I would just plot a few traces, beginning with the coordinate planes. What is the curve in the ##xy## plane (##z=0##)? The ##xz## plane (##y=0##)? The ##yz## plane (##x=0##)? If that doesn't give you an idea of the shape, add a few traces for ##z = \pm c## for a few constants.

    [Edit - Added]: In both of your equations with a square root, you should have ##z=\pm\sqrt{...}##. The surface is symmetric in ##z##. Also, with respect to what the scalar field represents, think of ##\Phi(x,y,z)## as representing the temperature at ##(x,y,z)##. The level surface ##\Phi = 0## represents the surface in space where the temperature is ##0##.
     
    Last edited: May 1, 2014
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