1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another mathamatician neede to check this! CURL-GRAD-DIV

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I have repeated this problem over and over again so either the question is wrong or my likely its my method thats at fault. Please can someone check this for me.

    Prove this is true:
    [tex]\nabla\times(\nabla \times F)=-\nabla^{2}F+\nabla(\nabla.F)[/tex]

    for [tex]F=x^{2}z^{3}i[/tex]


    3. The attempt at a solution

    Im getting for the LHS:

    [tex](\nabla \times F)= 3x^{2}z^{3} j[/tex]

    [tex]\nabla\times(\nabla \times F)= -6x^{2}z i + 6xz^{2}k[/tex]

    and for the right:

    [tex]-\nabla^{2}F=-2z^{3} i[/tex]

    [tex](\nabla.F)=2xz^{3}[/tex]

    [tex]\nabla(\nabla.F)=2z^{3}i + 6xz^2k[/tex]

    [tex]-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k[/tex]

    [tex]\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)[/tex]
    for [tex]F=x^{2}z^{3}i[/tex]
     
  2. jcsd
  3. May 16, 2009 #2
    isn't [tex]\nabla[/tex].[tex]\nabla[/tex] = [tex]\nabla[/tex]^2= d^2/dx^2+d^2/dy^2+etc.

    ugh the text never comes out right why is line 1 not equal to negative line 3?
     
    Last edited: May 16, 2009
  4. May 16, 2009 #3
    This is not correct.
     
  5. May 16, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    It seems you've taken the scalar Laplacian and then made a vector out of it. Make sure you use the vector Laplacian and you will get the extra term you need to proof the equality.
     
  6. May 16, 2009 #5

    Vid

    User Avatar

    This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.
     
  7. May 16, 2009 #6

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Presumably, given that we are in Cartesian coordinates, you use the fact that [tex]\nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}[/tex]
     
  8. May 17, 2009 #7
    http://en.wikipedia.org/wiki/Vector_Laplacian

    f9bc37c3b3575ca8bbc223a9438d437f.png

    [tex]A_{x}=x^{2}z^{3} [/tex]

    [tex] A_{y}=0 [/tex]

    [tex] A_{z}=0[/tex]


    [tex]\nabla^{2}A_{x}=2z^{3}[/tex]

    [tex]\Rightarrow\nabla^{2}A=2z^{3}i[/tex]

    [tex]\Rightarrow-\nabla^{2}F=-2z^{3}i[/tex]

    is this not right?
     
    Last edited by a moderator: Apr 24, 2017
  9. May 17, 2009 #8

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    This isn't correct. [tex]\nabla^2 F_x=\nabla^2x^2z^3=2z^3+6zx^2[/tex]
     
    Last edited by a moderator: Apr 24, 2017
  10. May 17, 2009 #9

    Cyosis

    User Avatar
    Homework Helper

    The normal Laplacian is given by [tex]\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}[/tex]. You already stated that [itex]A_x=x^2 z^2[/itex]. Now all you have to do is apply the normal Laplacian to A_x.
     
  11. May 17, 2009 #10
    Thanks guys, sorted it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Another mathamatician neede to check this! CURL-GRAD-DIV
  1. GRAD DIV CURL question (Replies: 3)

Loading...