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Curl of a gradient

  1. Aug 26, 2013 #1
    Let ##v(x,y)## be function of (x,y) and not z.
    [tex]\nabla v=\hat x \frac{\partial v}{\partial x}+\hat y \frac{\partial v}{\partial y}[/tex]
    [tex]\nabla \times \nabla v=\left|\begin{array} \;\hat x & \hat y & \hat z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & 0 \end{array}\right|=\hat x\left(-\frac{\partial^2 v}{\partial y\partial z}\right)-\hat y\left(-\frac{\partial^2 v}{\partial x\partial z}\right) +\hat z\left(\frac{\partial^2 v}{\partial y\partial x}-\frac{\partial^2 v}{\partial x\partial y} \right) =0 [/tex]

    What did I do wrong?
  2. jcsd
  3. Aug 26, 2013 #2


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    What makes you think that's wrong? The curl of the gradient of a smooth scalar field ##f:\mathbb{R}^{3}\rightarrow \mathbb{R}## always vanishes: ##(\nabla \times \nabla f)^{i} = \epsilon^{ijk}\partial_{j}\partial_{k}f = 0##.
  4. Aug 26, 2013 #3
    I did not know that!!!

  5. Aug 26, 2013 #4


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    What is going on here is the use of Clairaut's theorem, which says that with enough continuity the mixed partial derivatives in any order give the same value. That's why, for example, ##v_{xy} = v_{yx}##.
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