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Verify Laplace equation in rectangular coordinates.

  1. Jul 11, 2011 #1
    Just want to verify Laplace equation in rectangular coordinates that:

    [tex] \nabla ^2 \vec E = 0 [/tex]

    [tex]\Rightarrow\; \nabla^2 \vec E = \left ( \frac {\partial^2}{\partial x^2} +\frac {\partial^2}{\partial y^2} +\frac {\partial^2}{\partial z^2} \right ) ( \hat x E_x +\hat y E_y + \hat z E_z) = 0 [/tex]

    [tex]\hbox {(1)}\;\Rightarrow \;\frac {\partial^2 \vec E}{\partial x^2} = 0,\;\frac {\partial^2 \vec E}{\partial y^2} = 0 \;\hbox { and } \frac {\partial^2 \vec E}{\partial z^2} = 0 [/tex]

    And

    [tex]\hbox {(2)}\; \nabla ^2 \vec E = \nabla^2_{xy}\vec E + \frac {\partial^2 \vec E}{\partial z^2} = 0 \;\hbox { where }\; \nabla^2_{xy}\vec E = \left ( \frac {\partial^2}{\partial x^2} +\frac {\partial^2}{\partial y^2} \right ) ( \hat x E_x +\hat y E_y + \hat z E_z) [/tex]
     
    Last edited: Jul 12, 2011
  2. jcsd
  3. Jul 11, 2011 #2

    Pengwuino

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    Gold Member

    (1) is not right (which implies (2) is not what you're looking for). Are you sure you don't mean

    [tex]\nabla^2 \phi = 0[/tex]

    as this is the typical potential you work with in classical physics.

    If you mean what you wrote, what that tells you is that each component of this vector satisfies Laplace's equation; that is,

    [tex]{{\partial^2 E_x}\over{\partial x^2}} + {{\partial^2 E_x}\over{\partial y^2}} + {{\partial^2 E_x}\over{\partial z^2}} = 0[/tex]

    and so on for each component
     
  4. Jul 12, 2011 #3
    Thanks for the reply. I don't mean [itex]\nabla^2 \phi[/itex]. I was referring to Laplacian of a vector where:

    [tex] \nabla^2 \vec E = \left ( \frac {\partial^2}{\partial x^2} +\frac {\partial^2}{\partial y^2} +\frac {\partial^2}{\partial z^2} \right ) ( \hat x E_x +\hat y E_y + \hat z E_z) = 0 [/tex]

    So You say [itex] \nabla^2 \vec E = 0 \;\hbox { don't mean }\;\;\frac {\partial^2 \vec E}{\partial x^2} = 0,\;\frac {\partial^2 \vec E}{\partial y^2} = 0 \;\hbox { and } \frac {\partial^2 \vec E}{\partial z^2} = 0 [/itex]

    The second question is totally independent to the first question, even I got the first one wrong, that has no bearing on the second question.

    What I meant in the second question is I can split the [itex] \nabla^2_{xyz} \;\hbox { into Laplacian in x and y plus Laplacian in z } \; \Rightarrow\;\nabla ^2_{xyz} \vec E = \nabla^2_{xy}\vec E + \frac {\partial^2 \vec E}{\partial z^2}[/itex]

    Thanks

    Alan
     
  5. Jul 12, 2011 #4

    Pengwuino

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    Gold Member

    I see, so yah, you can split the Laplacian into parts like that. However, again, what i said in the first part still matters. You'll be doing the Laplacian on each individual component, but it can be split up like that.
     
  6. Jul 12, 2011 #5
    Thanks
     
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