Vector analysis for a variational problem

[Angelos K]

I'm reading a proof that is via variation. $$\vec {\delta A}$$ stands for a variation of the vectorpotential $$\vec{A}$$. If I understand the argument correctly [many steps are presented as one] it means:

$$(\nabla \times \vec{A})* (\nabla \times \vec{\delta A}) = (\nabla \times \nabla \times \vec {A})*\vec{\delta A}$$.

I do not believe it. If I calculate the brackets on the left hand side individually I don't get any undifferentiated $$\vec{\delta A}$$. What am I doing wrong?

You can find the "whole" calculation on page 10 of the attached file.

 

[Valerie Park]

The formula you have written is incorrect. The right hand side should be (\nabla \times \nabla \times \vec {A})*\vec{\delta A} - (\nabla \cdot \vec{\delta A})* \vec{A}. The first term on the right hand side corresponds to the calculation of the brackets on the left hand side individually, while the second term is the extra contribution that you are missing. To understand why this is true, you can consult page 10 of the attached file, which explains the entire calculation in more detail.

 

[sir-pinski]

Vector analysis for a variational problem involves using vector calculus to solve problems involving variations in vector quantities. In this case, the proof you are reading is using variations to solve a problem involving the vector potential, represented by \vec{A}. The notation \vec{\delta A} represents a variation of the vector potential.

The equation you have presented, (\nabla \times \vec{A})* (\nabla \times \vec{\delta A}) = (\nabla \times \nabla \times \vec {A})*\vec{\delta A}, is a key step in the proof. It is showing that the cross product of the curl of \vec{A} and the curl of \vec{\delta A} is equal to the curl of the curl of \vec{A} multiplied by \vec{\delta A}. This step is valid because of the properties of vector calculus, specifically the fact that the curl of a curl is equal to the gradient of the divergence minus the Laplacian.

It is possible that you are not getting the same result when calculating the brackets on the left hand side because you are not considering the properties of vector calculus. It is important to remember that vector calculus involves operations on vector quantities, not just scalar quantities. Therefore, the rules for differentiation and multiplication may not always apply in the same way. I recommend reviewing the properties and rules of vector calculus to better understand this step in the proof.