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I want to prove the following identity

$$\text{curl } \left(\textbf{F}\times \textbf{G}\right) = \textbf{F}\text{ div}\textbf{ G}- \textbf{G}\text{ div}\textbf{ F}+ \left(\textbf{G}\cdot \nabla \right)\textbf{F}- \left(\textbf{F}\cdot \nabla \right)\textbf{G}$$

But I do not know how! Also, what does $\textbf{F}\cdot \nabla $ mean, isn't it the divergence of $\textbf{F}$!

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You only need two things to prove this. First, the BAC-CAB rule:

$$A \times (B \times C) = B(A \cdot C) - C(A \cdot B)$$

And the product rule. Let $\dot \nabla \times (\dot F \times G)$ mean "differentiate $F$ only; pretend $G$ is constant here". So the product rule would read

$$\nabla \times (F \times G) = \dot \nabla \times (\dot F \times G) + \dot \nabla \times (F \times \dot G)$$

Now, apply the BAC-CAB rule. I'll do this for just one term for brevity:

$$\dot \nabla \times (\dot F \times G) = \dot F (\dot \nabla \cdot G) - G(\dot \nabla \cdot \dot F)$$

Now, here's where the dots become important: since $G$ is not differentiated in this whole equation, $\dot \nabla \cdot G$ is a directional derivative, conventionally written $G \cdot \nabla$. Indeed, we have

$$\dot F(\dot \nabla \cdot G) = (G \cdot \nabla) F$$

On the other hand, the $G(\dot \nabla \cdot \dot F)$ term can just drop the dots to get something that looks like a divergence:

$$G (\dot \nabla \cdot \dot F) = G(\nabla \cdot F)$$

Carry out the same expansion for the $\dot \nabla \times (F \cdot \dot G)$ term, and you're done.

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    $\begingroup$ i dont get why should i use the product rule, why is BAC-CAB rule not enough. in the end its just a cross product, right? $\endgroup$ – Holy cow Jun 27 '14 at 4:44
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    $\begingroup$ Both $F$ and $G$ are being differentiated here. You have to use the product rule at some point. You do not have the option not to. You could use the product rule second if you prefer, after the BAC-CAB rule, but you can't choose not to use the product rule here. That would be like making the mistake $\frac{d}{dx} [f(x) g(x) ]= g(x) \frac{df}{dx}$. It's...wrong. $\endgroup$ – Muphrid Jun 27 '14 at 4:50
  • $\begingroup$ i know what is product rule, but we are finding triple product $\endgroup$ – Holy cow Jun 27 '14 at 4:53
  • $\begingroup$ what you said, gives the right answer but not the satisfaction $\endgroup$ – Holy cow Jun 27 '14 at 4:54
  • $\begingroup$ It merely sounds to me that you're unfamiliar with vector calculus versions of the product rule, but they are no more exotic than the single-variable version and follow directly from that version (which can be proved by breaking into components, if you insist). The overdot notation I used here is just a convenient way of not having to write out components while still invoking the product rule. When you differentiate a product in single-variable calculus, you use a product rule. When you differentiate a product of vectors, there is a vector extension of the product rule. Seems sensible to me. $\endgroup$ – Muphrid Jun 27 '14 at 4:57
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Here is a simple proof using index notation and BAC-CAB identity.

$$\begin{align} \nabla \times \left( {{\bf{A}} \times {\bf{B}}} \right) &= {{\bf{e}}_i} \times {\partial _i}\left( {{A_j}{{\bf{e}}_j} \times {B_k}{{\bf{e}}_k}} \right)\\ &= {\partial _i}\left( {{A_j}{B_k}} \right){{\bf{e}}_i} \times \left( {{{\bf{e}}_j} \times {{\bf{e}}_k}} \right)\\ &= \left( {{\partial _i}{A_j}{B_k} + {A_j}{\partial _i}{B_k}} \right)\left( {\left( {{{\bf{e}}_i} \cdot {{\bf{e}}_k}} \right){{\bf{e}}_j} - \left( {{{\bf{e}}_i} \cdot {{\bf{e}}_j}} \right){{\bf{e}}_k}} \right)\\ &= \left( {{\partial _i}{A_j}{B_k} + {A_j}{\partial _i}{B_k}} \right)\left( {{\delta _{ik}}{{\bf{e}}_j} - {\delta _{ij}}{{\bf{e}}_k}} \right)\\ &= {\partial _i}{A_j}{B_i}{{\bf{e}}_j} - {\partial _i}{A_i}{B_k}{{\bf{e}}_k} + {A_j}{\partial _i}{B_i}{{\bf{e}}_j} - {A_i}{\partial _i}{B_k}{{\bf{e}}_k}\\ &= {\bf{B}} \cdot \nabla {\bf{A}} - \left( {\nabla \cdot {\bf{A}}} \right){\bf{B}} + \left( {\nabla \cdot {\bf{B}}} \right){\bf{A}} - {\bf{A}} \cdot \nabla {\bf{B}} \end{align}$$

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The divergence is $\nabla\cdot\mathbf{F}$ whereas $(\mathbf{F}\cdot\nabla)$ is another way of writing the directional derivative operator. In component notation we have

$$(\mathbf{F}\cdot\nabla) = \sum_{\alpha=1}^dF_\alpha\frac{\partial}{\partial x_\alpha}$$

which when applied to each component ($\beta$) of $\mathbf{G}$ gives

$$\left((\mathbf{F}\cdot\nabla)\mathbf G\right)_\beta = \sum_{\alpha=1}^dF_\alpha\frac{\partial G_\beta}{\partial x_\alpha} $$

which is the same as if we consider $\mathbf{F}\cdot(\nabla\otimes\mathbf{G})$ where $\nabla\otimes\mathbf{G}$ is

$$\left(\nabla\otimes\mathbf{G}\right)_{\alpha \beta}= \frac{\partial G_\beta}{\partial x_\alpha}$$

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    $\begingroup$ is there an easy way to prove the main problem i posted $\endgroup$ – Holy cow Jun 16 '14 at 12:59
  • $\begingroup$ I would start with one of the sides, write it in component notation and then use rearrangement, expansion, etc to get it into the form of the the other side. Though you will end up with lots of terms and it might take a while. If you want to know how to do that, perhaps post it as another question. $\endgroup$ – Dan Jun 16 '14 at 13:04
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It's easiest to see by writing it out in components:

[ ( ∇ ⋅ a ) b ]i = ( ∂x ax + ∂y ay + ∂z az )bi = ( ∂x ax )bi + ( ∂y ay )bi + ( ∂z az )bi

Whereas

[(∇⋅a)b]i=(∂x ax +∂y ay +∂z az)bi =(∂x ax)bi +(∂y ay)bi +(∂z az)bi

and clearly these are not the same. So while a ⋅ b = b ⋅ a a⋅b=b⋅a holds when a and b are really vectors, it is not necessarily true when one of them is a vector operator. This is one of the cases where the convenience of considering ∇ ∇ as a vector satisfying all the rules for vectors does not apply.

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