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Del Operator

  1. Apr 20, 2005 #1
    Hello All,

    May I know what is the difference between
    1) Del operator with respect for field point
    2) Del operator with respect to source point

    thanks
    newbie
     
  2. jcsd
  3. Apr 20, 2005 #2
    newbie,

    Not sure what you mean.

    Del is an operation on a scalar that gives a vector (namely, the gradient of the scalar)

    What are the "field" and "source" that you're talking about?
     
  4. Apr 20, 2005 #3

    arildno

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    Science Advisor
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    Dearly Missed

    I am quite certain that he is confused by the convention of regarding the divergence of a source potential as a multiple of dirac's delta function.

    However, only newbie knows for sure..
     
  5. Apr 20, 2005 #4
    Hi All,

    thanks for helping.. let me explain
    i'm reading this text on the derivation of helmholtz theorem

    let me just quote directly from the book

    [​IMG]

    [​IMG]


    Page 2 top half
    " In Equations (A-2) through (A-5), the operator 'del-f' differentiates with respect to field point rf, while the operator 'del-s' differentiates with repect to the source point rs"

    May I know the difference between the operators here.


    Page 2 bottom half
    " From Equation (A-1) since F(rs) is a function of the source point alone, but "del-f" differentiates with respect to the field point.... "

    Well apparently we can move F(rs) out of the lapacian here. Please help explain how this is possible

    thanks again
    newbie101
     
    Last edited: Apr 20, 2005
  6. Apr 20, 2005 #5

    jtbell

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    Staff: Mentor

    That means that, for example,

    [tex]\nabla_f V = \frac {\partial V}{\partial x_f} \hat {\bold i} + \frac {\partial V}{\partial y_f} \hat {\bold j} + \frac {\partial V}{\partial z_f} \hat {\bold k} [/tex]

    whereas

    [tex]\nabla_s V = \frac {\partial V}{\partial x_s} \hat {\bold i} + \frac {\partial V}{\partial y_s} \hat {\bold j} + \frac {\partial V}{\partial z_s} \hat {\bold k} [/tex]

    where V is some function of [itex]x_f[/itex], [itex]y_f[/itex], [itex]z_f[/itex], [itex]x_s[/itex], [itex]y_s[/itex], and [itex]z_s[/itex] (that is, depends on both the field coordinates and the source coordiates).
     
  7. Apr 20, 2005 #6
    jtbell,

    So your vector V is analogous to the Green's function G(rs,rf) since it's a function of both rs and rf. But since F(rs) is a function only of rs, it doesn't vary with rf, so when derivatives are taken wrt rf, F acts like a constant.

    newbie, does that help at all, or am I missing your point entirely?
     
  8. Apr 20, 2005 #7
    jtbell & jdavel,

    yes it does explain everything if vector V here is a function of both (x,y,z) at field point and (x,y,z) at source point.... which should be the case

    since the E field at a point would depend on both
    1) where the field point is as well as
    2) where the source is


    however, im still not understanding the partial derivative here ... i mean how is dV/dXf different from dV/dXs ... arent there only 3 axis here X,Y,Z so the gradient whould still be the same wouldnt it ???

    thanks again all
    newbie101

    ** if necessary, i can scan more pages **

    BTW the book is "Numerical Computation of Electric and Magnetic Fields" by Charles W Stelle
     
  9. Apr 20, 2005 #8
    newbie, When you say "the E field at a point would depend on....where the source is" it sounds like you think the source is located at a single point. That's not true here; the source is distributed over the entire volume.
     
  10. Apr 20, 2005 #9
    Yes the source is distributed. Thanks jdavel
     
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