Del Operator


by newbie101
Tags: operator
newbie101
newbie101 is offline
#1
Apr20-05, 05:33 AM
P: 16
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
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jdavel
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#2
Apr20-05, 09:05 AM
P: 618
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?
arildno
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#3
Apr20-05, 10:32 AM
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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..

newbie101
newbie101 is offline
#4
Apr20-05, 12:22 PM
P: 16

Del Operator


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






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
jtbell
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#5
Apr20-05, 02:01 PM
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Quote Quote by newbie101
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"
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).
jdavel
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#6
Apr20-05, 02:22 PM
P: 618
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?
newbie101
newbie101 is offline
#7
Apr20-05, 02:47 PM
P: 16
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
jdavel
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#8
Apr20-05, 03:15 PM
P: 618
Quote Quote by newbie101

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
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.
newbie101
newbie101 is offline
#9
Apr20-05, 09:59 PM
P: 16
Yes the source is distributed. Thanks jdavel


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