Rotation of Separation Vector
📂Mathematical Physics Rotation of Separation Vector Equation ∇ × 2 = 0
\nabla \times \dfrac{\crH }{\cR ^2} = \mathbf{0}
∇ × 2 = 0
Explanation There is nothing particularly special about this formula. It emerges in the process of calculating the divergence of a magnetic field, and its calculation is not straightforward, hence the separate explanation.
Proof If we refer to = ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ \bcR=(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}} = ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^
∣ ∣ = = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2
| \bcR |=\cR=\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}
∣ ∣ = = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2
= = ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2
\crH = \dfrac{ \bcR } { \cR}=\dfrac{(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}}
= = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^
2 = 1 ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2
\dfrac{\crH}{\cR^2}=\dfrac{1}{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}\dfrac{(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}}
2 = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 1 ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^
For the sake of simplicity, let’s say
2 = A x x ^ + A y y ^ + A z z ^
\dfrac{\crH }{\cR ^2} = A_{x} \hat{\mathbf{x}} + A_{y} \hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}}
2 = A x x ^ + A y y ^ + A z z ^
Then,
∇ × 2 = ∣ x ^ y ^ z ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z A x A y A z ∣ = ( ∂ ∂ y A z − ∂ ∂ z A y ) x ^ + ( ∂ ∂ z A x − ∂ ∂ x A z ) y ^ + ( ∂ ∂ x A y − ∂ ∂ y A x ) z ^
\begin{align*}
\nabla \times \dfrac{ \crH}{\cR^2} &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}
\\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z}
\\ A_{x} & A_{y} & A_{z} \end{vmatrix}
\\ &= \left( \dfrac{ \partial}{\partial y} A_{z} -\dfrac{\partial }{\partial z}A_{y} \right) \hat{\mathbf{x}} + \left( \dfrac{ \partial}{\partial z} A_{x} -\dfrac{\partial }{\partial x}A_{z} \right) \hat{\mathbf{y}} +\left( \dfrac{ \partial}{\partial x} A_{y} -\dfrac{\partial }{\partial y}A_{x} \right) \hat{\mathbf{z}}
\end{align*}
∇ × 2 = x ^ ∂ x ∂ A x y ^ ∂ y ∂ A y z ^ ∂ z ∂ A z = ( ∂ y ∂ A z − ∂ z ∂ A y ) x ^ + ( ∂ z ∂ A x − ∂ x ∂ A z ) y ^ + ( ∂ x ∂ A y − ∂ y ∂ A x ) z ^
Let’s first calculate the x ^ \hat{\mathbf{x}} x ^
∂ ∂ y A z − ∂ ∂ z A y = ∂ ∂ y [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 3 2 ( z − z ′ ) ] − ∂ ∂ z [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 3 2 ( y − y ′ ) ] = − 3 2 [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 5 2 ( z − z ′ ) ] ⋅ 2 ( y − y ′ ) + 3 2 [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 5 2 ( y − y ′ ) ] ⋅ 2 ( z − z ′ ) = − 3 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 5 2 ( z − z ′ ) ( y − y ′ ) + 3 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 5 2 ( y − y ′ ) ( z − z ′ ) = 0
\begin{align*}
& \dfrac{\partial }{\partial y} A_{z} - \dfrac{\partial}{\partial z}A_{y}
\\ =&\ \dfrac{\partial }{\partial y} \left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{3}{2}}(z-z^{\prime}) \right]
\\ &- \dfrac{\partial }{\partial z} \left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{3}{2}}(y-y^{\prime}) \right]
\\ =&\ -\dfrac{3}{2}\left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(z-z^{\prime}) \right]\cdot 2(y-y^{\prime})
\\ & +\dfrac{3}{2}\left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(y-y^{\prime}) \right]\cdot 2(z-z^{\prime})
\\ =&\ -3 \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(z-z^{\prime})(y-y^{\prime})
\\ & +3 \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(y-y^{\prime})(z-z^{\prime})
\\ =&\ 0
\end{align*}
= = = = ∂ y ∂ A z − ∂ z ∂ A y ∂ y ∂ [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 3 ( z − z ′ ) ] − ∂ z ∂ [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 3 ( y − y ′ ) ] − 2 3 [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 5 ( z − z ′ ) ] ⋅ 2 ( y − y ′ ) + 2 3 [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 5 ( y − y ′ ) ] ⋅ 2 ( z − z ′ ) − 3 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 5 ( z − z ′ ) ( y − y ′ ) + 3 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 5 ( y − y ′ ) ( z − z ′ ) 0
By using the same method, the y ^ \hat{\mathbf{y}} y ^ z ^ \hat{\mathbf{z}} z ^ 0 0 0
∴ ∇ × 2 = 0
\therefore \nabla \times \dfrac{ \crH}{\cR^2} = \mathbf{0}
∴ ∇ × 2 = 0
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