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Need help checking if my reasoning is sound for this.
Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated.
[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.
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.
Homework Statement
Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated.
Homework 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.
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.