Q17Mathematics I

In advanced vector calculus, establishing complex identities requires breaking them down into their fundamental operator definitions. We must mathematically prove that the divergence of the cross product of two vectors exactly equals $\vec{b}\cdot (\nabla \times \vec{a})-\vec{a}\cdot (\nabla \times \vec{b})$.

Step 1: Applying the Divergence Operator

We define the divergence operator $\nabla$ as the vector summation $\sum \hat{i}\frac{\partial }{\partial x}$. Applying this to the cross product gives:

We must strictly apply the vector derivative product rule to the cross product term:

We distribute the dot product linearly across the two resulting terms:

Step 2: Exploiting the Scalar Triple Product

Each summation term now forms a scalar triple product, $\vec{u}\cdot (\vec{v}\times \vec{w})$. A fundamental mathematical property of the scalar triple product is that circular shifts of the vectors leave the volume invariant: $\vec{u}\cdot (\vec{v}\times \vec{w})=\vec{w}\cdot (\vec{u}\times \vec{v})$. Reversing a cross product introduces a strict negative sign: $=-\vec{v}\cdot (\vec{u}\times \vec{w})$.

Applying these rigid rules to our terms:

• First term (circular shift): $\hat{i}\cdot \left(\frac{\partial \vec{a}}{\partial x}\times \vec{b}\right)=\vec{b}\cdot \left(\hat{i}\times \frac{\partial \vec{a}}{\partial x}\right)$
• Second term (sign reversal): $\hat{i}\cdot \left(\vec{a}\times \frac{\partial \vec{b}}{\partial x}\right)=-\vec{a}\cdot \left(\hat{i}\times \frac{\partial \vec{b}}{\partial x}\right)$

Step 3: Finalizing the Curl Definition

Substituting these rigorously transformed terms back into our main summation:

By mathematical definition, the curl of a vector $\vec{v}$ is exactly $\nabla \times \vec{v}=\sum \hat{i}\times \frac{\partial \vec{v}}{\partial x}$. We recognize these exact summation structures and finalize the proof unequivocally: