91Ó°ÊÓ

Verify the following identities in which \(f\) and \(g\) are arbitrary differentiable scalar functions of position, and \(\mathbf{F}\) and \(\mathbf{G}\) are arbitrary differentiable vector functions of position. (a) \(\nabla(f g)=f \nabla g+g \nabla f\). (b) \(\nabla(\mathbf{F} \cdot \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}+(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F} \times(\nabla \times \mathbf{G})+\mathbf{G} \times\) \((\nabla \times \mathbf{F})\) (c) \(\nabla \cdot(f \mathbf{F})=f \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla f\) (d) \(\nabla \cdot(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G})\) (e) \(\nabla \times(f \mathbf{F})=f \nabla \times \mathbf{F}+(\nabla f) \times \mathbf{F}\) (f) \(\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})-\mathbf{G}(\nabla \cdot \mathbf{F})\) (g) \(\nabla \times(\nabla \times \mathbf{F})=\nabla(\nabla \cdot \mathbf{F})-\nabla^{2} \mathbf{F}\)

Here is a "proof" that there is no such thing as magnetism. One of Maxwell's equations tells us that $$ \nabla \cdot \mathbf{B}=0 $$ where \(\mathbf{B}\) is any magnetic field. Then using the divergence theorem, we find $$ \iint_{S} \mathbf{B} \cdot \hat{\mathbf{n}} d S=\iiint_{V} \nabla \cdot \mathbf{B} d V=0 $$ Because B has zero divergence, we know (see Problem III-24) there exists a vector function, call it \(\mathbf{A}\), such that $$ \mathbf{B}=\nabla \times \mathbf{A} $$ Combining these last two equations, we get $$ \iint_{s} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Next we apply Stokes' theorem and the preceding result to find $$ \oint_{C} \mathbf{A} \cdot \hat{\mathbf{t}} d s=\iint_{S} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Thus we have shown that the circulation of \(\mathbf{A}\) is path independent. It follows that we can write \(\mathbf{A}=\nabla \psi\), where \(\psi\) is some scalar function. Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that $$ \mathbf{B}=\nabla \times \nabla \psi=0 $$ that is, all magnetic fields are zero! Where did we go wrong? [Taken from G. Arfken, Amer. J. Phys., 27, 526 (1959).]