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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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5 Answers 5

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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, 2014 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, 2014 at 4:50
  • $\begingroup$ i know what is product rule, but we are finding triple product $\endgroup$
    – Holy cow
    Jun 27, 2014 at 4:53
  • $\begingroup$ what you said, gives the right answer but not the satisfaction $\endgroup$
    – Holy cow
    Jun 27, 2014 at 4:54
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    $\begingroup$ 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, 2014 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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  • $\begingroup$ This is the way I would have suggested! (+1) $\endgroup$
    – Mark Viola
    May 21 at 17:59
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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, 2014 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, 2014 at 13:04
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Derivation of the Vector Triple Product Using the Vector Calculus Chain Rule

The Vector Triple Product is Given from Wolfram Math World as: $$ \overset{\rightarrow}{\boldsymbol A} \times \left(\overset{\rightarrow}{\boldsymbol B}\times \overset{\rightarrow}{\boldsymbol C} \right) = \overset{\rightarrow}{\boldsymbol B} \left( \overset{\rightarrow}{\boldsymbol A}\bullet \overset{\rightarrow}{\boldsymbol C} \right) - \overset{\rightarrow}{\boldsymbol C} \left( \overset{\rightarrow}{\boldsymbol A}\bullet \overset{\rightarrow}{\boldsymbol B} \right) \tag{Eq. 1}$$ But this triple vector product does not have any differentiation in it, so it still needs application of the chain rule to be correct when a differentiation is involved, with $\overset{\rightarrow}\nabla$ for instance. So now, the vector chain rule needs to be applied. Just like when $f$ and $g$ are commutative, that $\frac{\partial}{\partial x} \left(f\right)\left(g\right)= \left(f\right) \frac{\partial}{\partial x} \left(g\right)+ \left(g\right) \frac{\partial}{\partial x} \left(f\right)$, the term inside $\left(\vec{\nabla}\bullet\vec{C}\right)$ are also swapped to apply the vector chain rule as also $\left(\vec{C}\bullet\vec{\nabla}\right)$, to either have one of $\vec{B}$ or $\vec{C}$ constant as the derivative is applied to the remaining variable that is not constant as in the Equation below. Because the ordinary triple cross product has two terms, when the chain rule is applied to two quantities $\vec{A}$ and $\vec{B}$ to each of the two quantities from the ordinary vector cross product multiplication formula, it results in four terms as follows:

$$ \begin{align} \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol B}\times \overset{\rightarrow}{\boldsymbol C} \right) &= \left.\left( \overset{\rightarrow}{\boldsymbol B} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol C} \right) \right)\right|_\text{B is constant} + \left.\left(\left( \overset{\rightarrow}{\boldsymbol C}\bullet \overset{\rightarrow}{\boldsymbol \nabla}\right) \overset{\rightarrow}{\boldsymbol B}\right)\right|_\text{C is constant}\\ &- \left.\left(\overset{\rightarrow}{\boldsymbol C} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol B} \right)\right)\right|_\text{C is constant} - \left. \left(\left(\overset{\rightarrow}{\boldsymbol B} \bullet \overset{\rightarrow}{\boldsymbol \nabla} \right) \overset{\rightarrow}{\boldsymbol C} \right)\right|_\text{B is constant} \end{align}$$ $$\boxed{ \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol B}\times \overset{\rightarrow}{\boldsymbol C} \right) =\overset{\rightarrow}{\boldsymbol B} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol C} \right) +\left(\vec{\boldsymbol C}\bullet\vec{\boldsymbol\nabla}\right) \vec{B} - \overset{\rightarrow}{\boldsymbol C} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol B} \right) -\left(\vec{\boldsymbol B}\bullet\vec{\boldsymbol \nabla}\right) \vec{\boldsymbol C} }$$ $$\tag{Eqs. 2}$$ To understand terms like $\vec{C} \bullet \vec{\nabla}$ it is a derivative in the direction of $\vec{C}$ first multiplied by the magnitude of $C \hat{c}$ being $C$. Some of the other answers mentioned the directional derivative but forgot to mention that the magnitude of $\vec{C}$ needs to be multiplied first.

For confirmation of the proof above, see the reference "Curvilinear Coordinate Systems" quoted here:

Dell Cross A Cross B formula

An Application

Here is an example in Physics calculating the Cross product of the $\overset{\rightarrow}{\boldsymbol J}\text{ }$ Current Density for a Spherical shell of radius $r$ with a charge $Q$, in a shell element thickness $(\partial r)$ in Spherical Coordinates from Wolfram Math World:

Spherical Coordinates from Wolfram Math World

$$ \overset{\rightarrow}{\boldsymbol J} =\frac{Q}{Area}\overset{\rightarrow}{\boldsymbol v} =\frac{Q}{Area}\overset{\rightarrow}{\boldsymbol v} =\frac{Q*\text{Length}}{Volume}\overset{\rightarrow}{\boldsymbol v} =\frac{Q*r*\sin(\phi)\partial \theta}{4\pi r^2 \partial r} \left(\overset{\rightarrow}{\boldsymbol \omega} \times \overset{\rightarrow}{\boldsymbol r}\right) $$ $$ \overset{\rightarrow}{\boldsymbol J} =\frac{Q*r*\sin(\phi)\partial \theta}{4\pi r^2 \partial r} \left(\overset{\rightarrow}{\boldsymbol \omega} \times \overset{\rightarrow}{\boldsymbol r}\right) =\frac{Q*\sin(\phi)\partial \theta}{4\pi* \partial r} \left(\overset{\rightarrow}{\boldsymbol \omega} \times \boldsymbol{ \hat r }\right) $$ $$ \overset{\rightarrow}{\boldsymbol \nabla} \times \overset{\rightarrow}{\boldsymbol J} \text{ }(\partial r) (r \partial \phi)\text{ }= \overset{\rightarrow}{\boldsymbol \nabla} \times \frac{Q}{4\pi r^2}\overset{\rightarrow}{\boldsymbol \omega} \times \overset{\rightarrow}{\boldsymbol r} \text{ }(\partial r) (r \partial \phi)\text{ } =\frac{Q}{4\pi}(\partial r)(\partial \phi) \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol \omega}\times \hat{\boldsymbol r} \right) $$ $$ \overset{\rightarrow}{\boldsymbol \nabla} \times \overset{\rightarrow}{\boldsymbol J}(\partial r)(r \partial \phi) =\frac{Q*(\partial \theta) (\partial \phi) }{4\pi} \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol \omega} \times \overset{\rightarrow} {\boldsymbol{r}}\sin(\phi)\right) $$ $$ \tag{Eqs. 1} $$ So in this case, the curl triple product formula is needed for $ \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol \omega}\times \hat{\boldsymbol r} \sin(\phi) \right) $ , but it is generic for other vector variables with different names than $\overset{\rightarrow}{\boldsymbol \omega}$ and $\hat{\boldsymbol r}$.

Another Application: Deriving the Cross Product in General Coordinates

To determine the curl $\vec{\nabla}\times\left(\vec{D}\right) =\vec{\nabla}\times\left(\vec{B}\times\vec{C}\right)$ for suitably constructed vectors $\vec{B}$ and $\vec{C}$ (such that $\vec{D}=\vec{B}\times\vec{C}= D\left(\hat{B}\times\hat{C}\right)$ along with the fairly obvious vector identity (which is just an application of the Calculus Chain Rule) $\vec{\nabla}\times\left(f\vec{a}\right)= f\left(\vec{\nabla}\times\left(\vec{a}\right)\right)+ (\vec{\nabla} f) \times\left(\vec{a}\right)$ quoted below:

del cross a scalar times f

From "Divergence in curvilinear coordinates", the formula for Divergence $\overset{\rightarrow} {\boldsymbol \nabla} \bullet \overset{\rightarrow} {\boldsymbol E}$ can be determined (Quoting the Answer from TurlocTheRed) as follows:

By definition: $\vec{\nabla} \cdot \vec{E}=\lim_{\Delta V\to 0 }\frac{\int \vec{E}\cdot \hat{n}dA}{\Delta V} =\lim_{\Delta V\to 0} \frac{\int E_in_ih_jh_kdx_jdx_k +\int E_jn_jh_ih_kdx_idx_k+ \int E_kn_kh_jh_idx_jdx_i}{\int h_ih_jh_kdx_idx_jdx_k}$ $$= \frac{1}{h_ih_jh_k}\left(\frac{\partial(E_ih_jh_k) }{\partial x_i }+\frac{\partial(E_jh_ih_k) }{\partial x_j }+ \frac{\partial(E_kh_jh_i) }{\partial x_k } \right)$$

The reason to start with the Divergence in general curvilinear coordinates for $\vec{\nabla}\bullet \vec{E}$ is that its formula is very similar to the formula for the Divergence in Cartesian Coordinates, just with a few additional scaling parameters.

Not to mention, but also in Equation 2:

$$\boxed{ \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol D} \right)=\\ \overset{\rightarrow}{\boldsymbol \nabla} \times \left(\overset{\rightarrow}{\boldsymbol B}\times \overset{\rightarrow}{\boldsymbol C} \right) =\overset{\rightarrow}{\boldsymbol B} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol C} \right) +\left(\vec{\boldsymbol C}\bullet\vec{\boldsymbol\nabla}\right) \vec{B} - \overset{\rightarrow}{\boldsymbol C} \left( \overset{\rightarrow}{\boldsymbol \nabla}\bullet \overset{\rightarrow}{\boldsymbol B} \right) -\left(\vec{\boldsymbol B}\bullet\vec{\boldsymbol \nabla}\right) \vec{\boldsymbol C} }$$ $$\tag{Eq. 2}$$

Calculation in Curvilinear coordinates of $\vec{\nabla}\times\vec{D}$ is still in the works. But the point is that one can start with a divergence formula that looks very similar to $\vec{\nabla}\times\vec{E}$ in Cartesian Coordinates and from there to derive the Cross Product in General Coordinates.

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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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