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What have I done?

  1. Nov 9, 2011 #1

    ShayanJ

    User Avatar
    Gold Member

    I know my question is strange(and maybe stupid) but I'm really curious about it.I once tried to calculate the gradient of the magnitude of the gradient of a scalar field which for its x coordinate,I found:

    [itex]
    \left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} =
    \frac {\frac { \partial ^ {2} \varphi } {\partial x ^ {2} } }
    {\sqrt { 1+ (\frac{\frac {\partial \varphi} {\partial y} } {\frac {\partial \varphi } {\partial x} } )^{2} } }

    [/itex]

    then I tried to write it in a simpler form and it just came into my mind that I can integrate it and write it as the derivative of a function so I made substitutions below and treated [itex] \frac {\partial \varphi} {\partial y} [/itex] as constant:

    [itex]\frac{\frac {\partial \varphi } {\partial y} }{\frac {\partial \varphi } {\partial x} }=\tan{\theta}
    [/itex]
    [itex]
    \frac{\partial ^{2} \varphi }{\partial x ^{2} }=- \frac {\partial \varphi } {\partial y} \csc ^{2}{\theta} d \theta
    [/itex]

    so I arrived at:

    [itex]

    \int \left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} =
    \frac {\partial \varphi} {\partial y} \csc {\theta}

    [/itex]

    I reverted and got the following:

    [itex]

    \int \left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} =
    \left| \nabla \varphi \right|

    [/itex]

    Now comes my question:
    When I check the things I've done,I just can tell bullsh*t.But the result tells that was really an inegration but I really don't understand how can that be an integration.

    Sorry for such a mess and thanks in advance
     
    Last edited: Nov 9, 2011
  2. jcsd
  3. Nov 9, 2011 #2
    I'm not really sure what you did (I'm not really good at that stuff), but I don't think you have to treat [itex]\frac {\partial \varphi} {\partial y}[/itex] as constant since [itex]\theta[/itex] can be a variable (I'm not really sure if I'm right, but I think it works that way).
     
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