Q23P - Introduction to Electrodynamics

Question

(For masochists only.) Prove product rules (ii) and (vi). Refer to Prob. 1.22 for the definition of $\left(A . \overline{V}\right)\mathbf{ }\mathit{B}$.

Step-by-Step Solution

Answer

The product rules (ii) and (vi) are proved.

Step 1: Compute the left side of product rule (ii)

To prove any rule, simplify its left and right side, and comare them with each other.

 

Let the vector A.B is defined as $A.B=A_x{B}_x+A_y{B}_y+A_z{B}_z$ and the $\nabla$ operator is defined as$\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$. The gradient of vector A.B is obtaind as

$$\begin{gathered} \left|\nabla \left(A.B\right)\right|=\left(\frac{\partial }{\partial x}{A}_x{B}_x+\frac{\partial }{\partial y}{A}_y{B}_y+\frac{\partial }{\partial z}{A}_z{B}_z\right) \\ =\frac{\partial A_x}{\partial x}{B}_x+\frac{\partial B_x}{\partial X}{A}_x+\frac{\partial A_y}{\partial x}{B}_y+\frac{\partial A_z}{\partial x}{B}_z\frac{\partial A_z}{\partial x}{B}_{z .....\left(1\right)} \end{gathered}$$ 

   

 

Compute $\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|$ in x direction.

$$\begin{gathered} {\left|A\times \left(\nabla xB\right)\right|}_x=B_y{\left(\nabla xA\right)}_z\times \left(\nabla xA\right) \\ =B_y\left(\frac{\partial A_y}{\partial x}-\frac{\partial B_x}{\partial y}\right)-B_{z }+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_x}{\partial x}\right) .....\left(2\right) \end{gathered}$$ 

$$\begin{gathered} \mathrm{Now} \mathrm{Compute} {\left|\left(\mathrm{A}.\mathrm{V}\right)\mathrm{B}\right|}_{\mathrm{x} } \mathrm{in} \mathrm{x} \mathrm{direction}. \\ {\left|\left(\mathrm{A}.\mathrm{V}\right)\mathrm{B}\right|}_{\mathrm{x} }=\left({\mathrm{A}}_{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}={\mathrm{A}}_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\right){\mathrm{B}}_{\mathrm{x}} \\ ={\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}={\mathrm{A}}_{\mathrm{z}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}} \\ \mathrm{Now} \mathrm{Compute} {\left|\left(\mathrm{B}.\mathrm{V}\right)\mathrm{A}\right|}_{\mathrm{x} } \mathrm{in} \mathrm{x} \mathrm{direction}. \end{gathered}$$

 $$\begin{gathered} \mathrm{Now} \mathrm{Compute} {\left|\left(\mathrm{A}.\mathrm{V}\right)\mathrm{B}\right|}_{\mathrm{x} } \mathrm{in} \mathrm{x} \mathrm{direction}. \\ {\left|\left(B.\mathrm{V}\right)A\right|}_{\mathrm{x} }=\left(B_{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+B_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}=B_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\right)A_{\mathrm{x}} \\ =B_{\mathrm{x}}\frac{\partial A_{\mathrm{x}}}{\partial \mathrm{x}}+B_{\mathrm{y}}\frac{\partial A_{\mathrm{x}}}{\partial \mathrm{y}}=B_{\mathrm{z}}\frac{\partial A_{\mathrm{x}}}{\partial \mathrm{z}} \end{gathered}$$               

Step 2: Simplify the calculations for the left side of product rule (ii)

$$\begin{gathered} \mathrm{Now} \mathrm{add} \mathrm{the} \mathrm{equations} \left(1\right), \left(2\right), \left(3\right) \mathrm{and} \left(4\right) \mathrm{and} \mathrm{simplify}. \\ {\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{x}}+{\left|\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)\right|}_{\mathrm{x}}+{\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{x}}+{\left|\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)\right|}_{\mathrm{x}}={\mathrm{A}}_{{\mathrm{y}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-{\mathrm{A}}_{{\mathrm{y}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}} \\ -{\mathrm{B}}_{{\mathrm{z}}_{ }}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{x}}-{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}} \\ +{\mathrm{A}}_{{\mathrm{z}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{y}}-{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{Z}} \\ ={\mathrm{A}}_{{\mathrm{y}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-{\mathrm{A}}_{\mathrm{z}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{x}}-{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{x}} \\ +{\mathrm{A}}_{{\mathrm{x}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{x}} \end{gathered}$$

Substitute${\left|\nabla \left(A.B\right)\right|}_X$ for${\mathrm{A}}_{{\mathrm{y}}_{ }}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{z}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}} +{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{X}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}+{\mathrm{B}}_{\mathrm{X}}\frac{\partial {\mathrm{A}}_{\mathrm{X}}}{\partial \mathrm{x}}$ in above simplification.

$$\begin{gathered} {\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{x}}+{\left|\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)\right|}_{\mathrm{x}}+{\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{x}}+{\left|\nabla \left(\mathrm{A}.\mathrm{B}\right)\right|}_{\mathrm{y}} \\ \mathrm{Similarly} \mathrm{we} \mathrm{can} \mathrm{write} \\ {\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{Z}}+{\left|\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)\right|}_{\mathrm{Z}}+{\left|\mathrm{A}.\left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{Z}}+{\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}\right|}_{\mathrm{y}}={\left|\nabla \left(\mathrm{A}.\mathrm{B}\right)\right|}_{\mathrm{y}} \\ {\left|\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{Z}}+{\left|\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)\right|}_{\mathrm{Z}}+{\left|\mathrm{A}.\left(\nabla \times \mathrm{B}\right)\right|}_{\mathrm{Z}}+{\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}\right|}_{\mathrm{Z}}={\left|\nabla \left(\mathrm{A}.\mathrm{B}\right)\right|}_{\mathrm{Z}} \\ \mathrm{thus} \mathrm{it} \mathrm{is} \mathrm{proved} \mathrm{that} \nabla \left(\mathrm{A}.\mathrm{B}\right)=\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)+\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)+\left(\mathrm{A}.\nabla \right)\mathrm{B}+\left(\mathrm{B}.\nabla \right)\mathrm{A} \end{gathered}$$

Step 3: Compute | A × ∇ × B | in x direction

$$\begin{gathered} \mathrm{Compute} \left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right| \mathrm{in} \mathrm{x} \mathrm{direction}. \\ {\left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right|}_{\mathrm{X}} =\frac{\partial }{\partial \mathrm{y}}{\left(\mathrm{A}\times \mathrm{B}\right)}_{\mathrm{z}}\times \frac{\partial }{\partial \mathrm{y}}{\left(\mathrm{A}\times \mathrm{B}\right)}_{\mathrm{y}} \\ =\frac{\partial }{\partial \mathrm{y}}{\left({\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}-{\mathrm{A}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{x}}\right)}_{\mathrm{z}}-\frac{\partial }{\partial \mathrm{z}}\left({\mathrm{A}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{x}}-{\mathrm{A}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{x}}\right) \\ ={\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}-{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}-{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}-{\mathrm{A}}_{\mathrm{z}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}}-{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{z}}+{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{z}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}} \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{Compute} {\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}-\left(\mathrm{A}.\nabla \right)\mathrm{B}+\mathrm{A}\left(\nabla .\mathrm{B}\right)-\mathrm{B}\left(\nabla .\mathrm{A}\right)\right|}_{\mathrm{x}} \mathrm{in} \mathrm{x} \mathrm{direction}. \\ {\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}-\left(\mathrm{A}.\nabla \right)\mathrm{B}+\mathrm{A}\left(\nabla .\mathrm{B}\right)-\mathrm{B}\left(\nabla .\mathrm{A}\right)\right|}_{\mathrm{x}}=\left(\begin{array}{c}{\mathrm{B}}_{\mathrm{x}}\frac{\partial }{\partial \mathrm{X}}+\\ {\mathrm{B}}_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+\\ {\mathrm{B}}_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}+\end{array}\right){\mathrm{A}}_{\mathrm{x}}-\left(\begin{array}{c}{\mathrm{A}}_{\mathrm{x}}\frac{\partial }{\partial \mathrm{X}}+\\ {\mathrm{A}}_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+\\ {\mathrm{A}}_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}+\end{array}\right){\mathrm{B}}_{\mathrm{x}}+\left(\begin{array}{c}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{X}}+\\ \frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+\\ \frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}}\end{array}\right){\mathrm{A}}_{\mathrm{x}}-\left(\begin{array}{c}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{X}}+\\ \frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}+\\ \frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}}+\end{array}\right){\mathrm{B}}_{\mathrm{x}} \end{gathered}$$

                                                                   $$\begin{gathered} ={\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{X}}+{\mathrm{B}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{y}}{\mathrm{B}}_{\mathrm{z}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{z}}{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{X}}{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}{\mathrm{A}}_{\mathrm{z}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}} \\ ={\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{X}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{z}}{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{X}}{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}} \\ ={\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{y}}{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}} \\ ={\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{z}}+{\mathrm{B}}_{\mathrm{x}}\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{z}}{\mathrm{A}}_{\mathrm{x}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}} \\ ={\left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right|}_{\mathrm{x}} \end{gathered}$$

 

Similarly we can write,

$$\begin{gathered} {\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}-\left(\mathrm{A}.\nabla \right)\mathrm{B}+\mathrm{A}\left(\nabla .\mathrm{B}\right)-\mathrm{B}\left(\nabla .\mathrm{A}\right)\right|}_{\mathrm{y}}={\left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right|}_{\mathrm{y}} \\ {\left|\left(\mathrm{B}.\nabla \right)\mathrm{A}-\left(\mathrm{A}.\nabla \right)\mathrm{B}+\mathrm{A}\left(\nabla .\mathrm{B}\right)-\mathrm{B}\left(\nabla .\mathrm{A}\right)\right|}_{\mathrm{z}}={\left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right|}_{\mathrm{z}} \\ \mathrm{Thus}, \left|\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)\right|=\left(\mathrm{B}.\nabla \right)\mathrm{A}-\left(\mathrm{A}.\nabla \right)\mathrm{B}+\mathrm{A}\left(\nabla .\mathrm{B}\right)-\mathrm{B}\left(\nabla .\mathrm{A}\right) \end{gathered}$$