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Vector Calculus: Level curves and insulated boundaries

  1. Aug 29, 2008 #1
    Need help checking if my reasoning is sound for this.

    1. The problem statement, all variables and given/known data
    Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated.


    2. Relevant equations
    [tex]u(t,\underline{X})[/tex] is the temperature at time [tex]t[/tex] and spatial location [tex]\underline{X}[/tex]

    Let V be the region of the interest, then the at the points in the boundary of V, we have [tex]\nabla{u}\cdot\underline{n} = 0[/tex] (insulation), where [tex]\underline{n}[/tex] is the outward normal at that point.


    3. The attempt at a solution
    Consider, for a fixed time [tex]t = \alpha[/tex], the isobar [tex]u(\alpha,\underline{X}(s)) = C[/tex], where [tex]C[/tex] is an arbitrary constant and [tex]s[/tex] parametrizes a curve [tex]\partial{X}[/tex]in n-dimensional space. Its derivative for a fixed time is [tex]\displaystyle \displaystyle \frac{du}{ds} = \nabla{u}\cdot\frac{d\underline{X}(s)}{ds} = 0[/tex]. Therefore, the gradient [tex]\nabla{u}[/tex] is perpendicular to the tangent line.

    Next, consider a point [tex]X = X_{0}[/tex] in an insulated boundary [tex]\nabla{u(t,X_{0})}\cdot\underline{n} = 0[/tex] [tex]\forall t[/tex], where [tex]\underline{n}[/tex] is understood as the outward normal to the boundary at [tex]X = X_{0}[/tex], which means that the gradient [tex]\nabla{u}[/tex] at that point lies in the tangent n-1 hyperplane.

    This means that at [tex]X = X_{0}[/tex], the tangent of [tex]\partial{X}[/tex] lies normal to the n-1 tangent hyperplane, and they are thus perpendicular.
     
  2. jcsd
  3. Aug 29, 2008 #2

    HallsofIvy

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    Actually isobars are curves of constant pressure not temperature. ("bar" is from "barometric"). I'm not sure if there is a special name for constant temperature.

    If T(x,y,z) is the temperature at the point (x,y,z) then [itex]\nabla f\cdot\vec{v}[/itex] is the rate of change of T in the direction of the unit vector [itex]\vec{v}[/itex]. In particular, if [itex]\vec{v}[/itex] is tangent to a curve of constant temperature then [itex]\nabla T\cdot\vec{v}= 0[/itex] because T does not change in that direction. Combine that with the condition on an insulated boundary.
     
  4. Aug 29, 2008 #3
    Then, errr, isn't that what I wrote?
     
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