Vector Analysis: Help needed

  • Thread starter Hassan2
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  • #1
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Dear all,

I have two vector fields [itex] \vec{B}[/itex] and [itex]\vec{A}[/itex] related by:

[itex] \vec{B}=\nabla \times \vec{A}[/itex]

How can I simplify the following term:

[itex]\frac{\partial }{\partial \vec{A}} B^{2}[/itex]

where [itex]\frac{\partial }{\partial \vec{A}}=(\frac{\partial }{\partial A_{x}} \frac{\partial }{\partial A_{y}} \frac{\partial }{\partial A_{z}} )[/itex]

I would also like to know what are this kind of derivatives ( derivatives with respect to a vector field) called.

Thanks.
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132
Dear all,

I have two vector fields [itex] \vec{B}[/itex] and [itex]\vec{A}[/itex] related by:

[itex] \vec{B}=\nabla \times \vec{A}[/itex]

How can I simplify the following term:

[itex]\frac{\partial }{\partial \vec{A}} B^{2}[/itex]

where [itex]\frac{\partial }{\partial \vec{A}}=(\frac{\partial }{\partial A_{x}} \frac{\partial }{\partial A_{y}} \frac{\partial }{\partial A_{z}} )[/itex]

I would also like to know what are this kind of derivatives ( derivatives with respect to a vector field) called.

Thanks.

Hey Hassan2.

Try expanding out the cross product of del and A first.

Also when you say the vector derivative, are the elements of each vector mapped to the same corresponding element in the other? In other words if A = [x0,y0,z0] and B = [x1,y1,z1] then is x0 = f(x1), y0 = g(y1) and z0 = h(z1) (and the components are completely orthogonal)?

If this is the case, you will be able to expand del X A using the determinant formulation and simplify terms depending on how you define your elements of your vector (even if they are more general than above).
 
  • #3
426
5
The elements of the vectors are NOT mapped correspondingly. In fact the first equation is the definition of B, thus, the components are intertwined.

I couldn't simplify it by expanding the curl.It results in partial derivatives of second order multiplied by partial derivatives of first order.

Thanks.
 

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