Zangwill, problem 10.19 - A Matching condition for the vector potential A

In summary, the jump discontinuity of the normal derivative of the coulomb gauge vector at a surface with a current density can be expressed as the cross product of the normal vector and the difference between the magnetic fields on either side of the surface. By substituting the triple cross product identity, rearranging the terms, and making sure the nabla has something to operate on, one can arrive at the expression: ##\hat{n}\times(\vec{\nabla}\times \vec{A})=\vec{\nabla}(\hat{n}\cdot \vec{A})-(\hat{n}\cdot \vec{\nabla} )\vec{A} ## which can be used to solve the given problem.
  • #1
BearShark
15
0

Homework Statement


Show that the normal derivative of the coulomb gauge vector suffers a jump discontinuity at a surface endowed with a current density [tex] K(\vec r_s ) [/tex]

Homework Equations


The vector potential A is given by:
[tex] A=\frac{\mu_0}{4\pi}\int{\frac{J(x')}{|x-x'|}d^3x} [/tex]

The magnetic field has a jump in the parallel component given by:
[tex]\mu_0K(\vec r_s )=\hat{n}\times(\vec{B_2}-\vec{B_1})[/tex]

The Attempt at a Solution


Starting with the expression for the jump discontinuity, one can substitute the relation:
[tex]B=\nabla\times{A}[/tex]
to get:
[tex]\mu_0K(\vec r_s )=\hat{n}\times(\nabla\times{A_2}-\nabla\times{A_1})[/tex]

From here I hoped I can distribute the n - hat to get:

[tex]\mu_0K(\vec r_s )=\hat{n}\times{\nabla\times{A_2}}-\hat{n}\times{\nabla\times{A_1}}[/tex]

However, I noticed most vector calculus identities refer to identities where the nabla is the first element, and trying to use the general identities for vectors lead me to have nablas that do not operate on anything. Am I in the right direction? I will appreciate advice on how to tackle this problem.

Thanks!
 
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  • #2
I suppose you are referring to the "BAC - CAB" rule for the triple cross product, namely
##\vec{A}\times(\vec{B}\times \vec {C})=\vec{B}(\vec{A}\cdot \vec {C})-\vec{C}(\vec{A}\cdot \vec {B} ) ##
Although order of operation matters on the left side, it does not on the right side. The vectors forming the dot products on the right side commute and so do the dot products themselves with the vectors that multiply them. So substitute
##\hat{n} = \vec{A}##, ##\vec{\nabla} = \vec{B}## and ##\vec{A}=\vec{C}##. Then you make a first pass and write
##\hat{n}\times(\vec{\nabla}\times \vec{A})=\vec{\nabla}(\hat{n}\cdot \vec{A})-\vec{A}(\hat{n}\cdot \vec{\nabla} ) ##
Finally you make sure the "nabla" has something to operate on and rearrange the second term on the right to get
##\hat{n}\times(\vec{\nabla}\times \vec{A})=\vec{\nabla}(\hat{n}\cdot \vec{A})-(\hat{n}\cdot \vec{\nabla} )\vec{A} ##
 

Related to Zangwill, problem 10.19 - A Matching condition for the vector potential A

1. What is the Zangwill problem 10.19?

The Zangwill problem 10.19 is a mathematical problem in the field of electromagnetism that deals with a specific condition for the vector potential A. It is named after its creator, physicist Mark Zangwill.

2. What is the matching condition for the vector potential A?

The matching condition for the vector potential A is a condition that ensures that the vector potential is continuous across a boundary between two regions with different electromagnetic properties. It is necessary for the solution of Maxwell's equations to be physically meaningful.

3. Why is the matching condition important in electromagnetism?

The matching condition is important because it ensures that the vector potential, which is a key component of the electromagnetic field, is continuous and well-defined in different regions. This allows for a consistent and accurate description of electromagnetic phenomena.

4. What are some applications of the matching condition for the vector potential A?

The matching condition for the vector potential A is used in various applications of electromagnetism, such as in the modeling of electromagnetic wave propagation, the design of antennas, and the study of electromagnetic scattering.

5. How is the matching condition for the vector potential A solved?

The matching condition for the vector potential A is typically solved by applying boundary conditions to the equations of Maxwell's equations. This involves setting the tangential components of the vector potential equal on both sides of the boundary and solving for the unknown coefficients. In some cases, numerical methods may also be used to find a solution.

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