I'm not sure where to start on this proof...also, my book didn't give any clarification as to what $f$ and $\textbf{F}$ are. Usually $F$ is a scalar function, and $\textbf{f}$ is the vector field $f_1\textbf{i}+f_2\textbf{j}+f_3\textbf{k}$. But in this question the $\textbf{F}$ is a vector field because it's bold? Also, can you take the divergence of a scalar function? I thought it was just vector fields...
Thanks in advance.
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$\begingroup$$f=f(x,y,z)$ is a scalar function, and $\mathbf{F}=(F_1,F_2,F_3)$ is a vector field, where each component is a scalar function of three variables ($x,y,z$).
We can multiply vectors by scalars to get a new vector field: $$f\mathbf{F}=(fF_1,fF_2,fF_3)$$ This leads to $$\nabla\cdot (f\mathbf{F})=\frac{\partial }{\partial x}fF_1+\frac{\partial }{\partial y}fF_2+\frac{\partial }{\partial z}fF_3$$ Using the product rule, $$\nabla\cdot (f\mathbf{F})=(f_xF_1+fF_{1x})+\cdots=f(F_{1x}+F_{2y}+F_{3z})+(F_1f_x+F_2f_y+F_3f_z)$$ Now we re-write: $$f(\nabla\cdot\mathbf{F})+\mathbf{F}\cdot(\nabla f)$$ and the equation is proved. A good way to remember this formula (and others) is to think in terms of a generalized "product rule": using the "nabla" ($\nabla$) operator in a loose sense, we can write $$\nabla\cdot(f\mathbf{F})=(\nabla f)\cdot\mathbf{F}+f\cdot(\nabla\cdot \mathbf{F})$$ Now just figure out which dots represent scalar products, and which dot products!
$\endgroup$ 3 $\begingroup$$$ \mbox{div}(f F) = \partial_i(f F_i) = (\partial_i f)F_i +f \partial_i F_i = \mbox{grad} f \cdot F + f \, \mbox{div} F $$
Note I used these abbreviations:
- Einstein summation convention: if an index shows up twice, one sums over it $a_i b_i := \sum_{i=1}^n a_i b_i$
- $\partial_k := \partial / \partial x_k$
Other example
$$ \mbox{div} \, \mbox{rot} A = \partial_i (\epsilon_{ijk} \partial_j A_k) = \epsilon_{ijk} \partial_i \partial_j A_k = -\epsilon_{jik} \partial_i \partial_j A_k = -\epsilon_{jik} \partial_j \partial_i A_k = 0 $$
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