Vector calculus identities proof using suffix notation

  • Thread starter U.Renko
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  • #1
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I must become good at this ASAP.

Homework Statement



prove [itex] \vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \vec{b} \cdot(\vec\nabla\times\vec{a}) - \vec{a}\cdot(\vec\nabla\times\vec{b})[/itex]

Homework Equations


[itex] \vec a \times \vec b = \epsilon_{ijk}\vec a_j \vec b_k [/itex]
[itex] \vec\nabla\cdot = \Large\frac{\partial}{\partial x_i} [/itex]
summation over [itex] i [/itex]

The Attempt at a Solution



I don't know where to start. I'm sure it must involve some product rule.
but I'm not 100% sure whether or not [itex] \vec\nabla\cdot(\vec a \times \vec b) = (\vec b \times\vec\nabla\cdot\vec a) + (\vec a \times \vec\nabla\cdot\vec b) [/itex] (or something resembling it)

....
....Right now I have no decent book with me and searching on the internet has done more harm than good.

If that identity correct I might (probably) be able to do the rest.
 

Answers and Replies

  • #2
Matterwave
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Look at the types of quantities you are using in your attempt at a solution. What type of quantity is ##\nabla\cdot \vec{a}##? Does it make sense to take a cross product of a vector with this quantity?

As for how to prove the necessary assertion. You can always go back and use components! The brute force method is always doable (and it's not so hard for this problem).

For a more elegant method of proof, none immediately come to mind. Perhaps with some more thought, one can find a more elegant proof.
 
  • #3
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Look at the types of quantities you are using in your attempt at a solution. What type of quantity is ##\nabla\cdot \vec{a}##? Does it make sense to take a cross product of a vector with this quantity?

Nope. it doesn't make sense.
what was I thinking...
 
  • #4
Ray Vickson
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Nope. it doesn't make sense.
what was I thinking...

Letting ##\partial_i \equiv \partial / \partial x_i##, we have
[tex] \vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \partial_i (\epsilon_{ijk} a_j b_k)
= \epsilon_{ijk} \partial_i (a_j b_k)[/tex]
(using the Einstein summation convention). Now use the differentiation product rule, and simplify.
 
  • #5
lurflurf
Homework Helper
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$$\vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \vec{b} \cdot(\vec\nabla\times\vec{a}) - \vec{a}\cdot(\vec\nabla\times\vec{b})=(\vec{b} \times\vec\nabla)\cdot\vec{a} - (\vec{a} \times\vec\nabla)\cdot\vec{b}=[\vec{b} \vec\nabla\vec{a}]-[\vec{a} \vec\nabla\vec{b}]$$
are several notations for the same thing

Though I personally like to include the parentheses some people do not. The justification for omitting the parentheses is that only one interpretation is possible. It is silly to make the nonsensical interpretation.

The problem you have is with parity of a permutation. Each switch in order changes the sign.
(123)=-(132)=(312)=-(213)=(231)=-(321)
we see that
$$\vec{b} \cdot(\vec\nabla\times\vec{a})$$
corresponds to (312) and thus has a positive sign
and
$$\vec{a}\cdot(\vec\nabla\times\vec{b})$$
corresponds to (213) and thus has a negative sign
 
  • #6
76
26
Show that $$(1){~~~~~~~~~~~~~~~~~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b){~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}$$ Using suffix notation, we get for the left hand side of eq. (1) $$(2){~~~~~~~~~~~~~~~~~~~~~}[~\vec \nabla\cdot(\vec a \times \vec b)~]_i =\begin{cases} \begin{align}&~\partial_i (\vec a \times \vec b)_i = \partial_i (πœ€_{ijk} a_j b_k) = πœ€_{ijk} \partial_i (a_j b_k) \nonumber \\ & {~~~~~} πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~]~\dots\nonumber\end{align} \end{cases}$$For the right hand side of eq. (1), we obtain $$(3){~~~~~~~~~~~~}[~\vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b)~]_i =\begin{cases} \begin{align}& ~b_i ( πœ€_{ijk} \partial_j a_k) - a_i ( πœ€_{ijk} \partial_j b_k) \nonumber \\ &~πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] \nonumber\end{align} \end{cases}$$According to eqs. (1), (2), and (3) $$(4){~~~~~~~~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] = πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~]{~~~~~~~~}$$ For the left hand side of eq. (4), we get after summing over ##k##
##{~~~~~~}(5){~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~πœ€_{ij1} [~b_1 (\partial_i a_j) + a_j ( \partial_i b_1)~]\nonumber \\ & + πœ€_{ij2} [~b_2 (\partial_i a_j) + a_j ( \partial_i b_2)~]\nonumber \\ & + πœ€_{ij3} [~b_3 (\partial_i a_j) + a_j ( \partial_i b_3)~] \nonumber\end{align}\end{cases}##
Now ... 1st line of (5) ##j\neq1##, ##j## = 2, 3 ... 2nd line ##j\neq2##, ##j## = 1, 3 ... 3rd line ##j\neq3##, ##j## = 1, 2
##{~~~~~~}(6){~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~πœ€_{i21} [~b_1 (\partial_i a_2) + a_2 ( \partial_i b_1)~] \nonumber \\ &{~~~~~}+ πœ€_{i31} [~b_1 (\partial_i a_3) + a_3 ( \partial_i b_1)\nonumber \\ & + πœ€_{i12} [~b_2 (\partial_i a_1) + a_1 ( \partial_i b_2)~] \nonumber \\ &{~~~~~} + πœ€_{i32} [~b_2 (\partial_i a_3) + a_3 ( \partial_i b_2)~] \nonumber \\ & + πœ€_{i13} [~b_3 (\partial_i a_1) + a_1 ( \partial_i b_3)~] \nonumber \\ &{~~~~~} + πœ€_{i23} [~b_3 (\partial_i a_2) + a_2 ( \partial_i b_3)~] \nonumber\end{align}\end{cases}##
Summing over ##i## as in the preceding, we find that
##{~~~~~~~~~~~~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~πœ€_{321} [~b_1 (\partial_3 a_2) + a_2 ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ πœ€_{231} [~b_1 (\partial_2 a_3) + a_3 ( \partial_2 b_1)\nonumber \\ & + πœ€_{312} [~b_2 (\partial_3 a_1) + a_1 ( \partial_3 b_2)~] \nonumber \\ &{~~~~~} + πœ€_{132} [~b_2 (\partial_1 a_3) + a_3 ( \partial_1 b_2)~] \nonumber \\ & + πœ€_{213} [~b_3 (\partial_2 a_1) + a_1 ( \partial_2 b_3)~] \nonumber \\ &{~~~~~} + πœ€_{123} [~b_3 (\partial_1 a_2) + a_2 ( \partial_1 b_3)~] \nonumber\end{align}\end{cases}##
Using the fact that ##~πœ€_{ijk} =\begin{cases}\begin{align} &~0~\text {if any index is equal to} \nonumber \\&~\text {any other index}~\dots\nonumber \\ &+1~\text {if}~i ,j, k~\text {form an} \nonumber \\ &~\text {even permutation (cylic}\nonumber \\ &~\text {permutation) of 1, 2, 3}~\dots\nonumber
\\ &-1~\text {if}~i ,j, k~\text {form an odd}\nonumber \\ &~\text {permutation of 1, 2, 3}~\dots\nonumber \end{align}\end{cases}##
##{~~~~~~~~~~~~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~(-1) [~b_1 (\partial_3 a_2) + a_2 ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ (+1) [~b_1 (\partial_2 a_3) + a_3 ( \partial_2 b_1)~] \nonumber \\ & + (+1) [~b_2 (\partial_3 a_1) + a_1 ( \partial_3 b_2)~] \nonumber \\ &{~~~~~} + (-1) [~b_2 (\partial_1 a_3) + a_3 ( \partial_1 b_2)~] \nonumber \\ & + (-1) [~b_3 (\partial_2 a_1) + a_1 ( \partial_2 b_3)~] \nonumber \\ &{~~~~~} +(+1) [~b_3 (\partial_1 a_2) + a_2 ( \partial_1 b_3)~] \nonumber\end{align}\end{cases}##
##{~~~~~~~~~~~~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~ - b_1 (\partial_3 a_2)^* - a_2 ( \partial_3 b_1){\uparrow\uparrow} \nonumber \\ &{~~~~~}+ b_1 (\partial_2 a_3)^* + a_3 ( \partial_2 b_1){\uparrow\uparrow\uparrow}\nonumber \\ & + b_2 (\partial_3 a_1)^{**} + a_1 ( \partial_3 b_2)^{\uparrow} \nonumber \\ &{~~~~~} - b_2 (\partial_1 a_3)^{**} - a_3 ( \partial_1 b_2){\uparrow\uparrow\uparrow} \nonumber \\ & - b_3 (\partial_2 a_1)^{***} - a_1 ( \partial_2 b_3)^{\uparrow} \nonumber \\ &{~~~~~} + b_3 (\partial_1 a_2)^{***} + a_2 ( \partial_1 b_3)^{\uparrow\uparrow} \nonumber \end{align}\end{cases}##
##{~~~~~~}(7){~~~~~~~~~~}πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~b_1 [~(\partial_2 a_3) - (\partial_3 a_2)~] \nonumber \\ &{~~~~~} + b_2 [~(\partial_3 a_1) - (\partial_1 a_3)~] \nonumber \\ & + b_3 [~(\partial_1 a_2) - (\partial_2 a_1)~] \nonumber \\ &{~~~~~}+ a_1 [~( \partial_3 b_2) - (\partial_2 b_3)~] \nonumber \\ & + a_2 [~( \partial_1 b_3) - ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ a_3 [~( \partial_2b_1) - ( \partial_1 b_2)~] \nonumber\end{align}\end{cases}##
Going back to the right side of eq. (4), we get after summing over ##k##:
$$πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~πœ€_{ij1} [~b_i (\partial_j a_1)~] - πœ€_{ij1} [~a_i (\partial_j b_1)~]\nonumber \\ &~+ πœ€_{ij2} [~b_i (\partial_j a_2)~] - πœ€_{ij2} [~a_i (\partial_j b_2)~]\nonumber \\ &~+ πœ€_{ij3} [~b_i (\partial_j a_3)~] - πœ€_{ij3} [~a_i (\partial_j b_3)~]\nonumber\end{align}\end{cases}$$ Summing over ##j##:
##πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~πœ€_{i21} [~b_i (\partial_2 a_1)~] - πœ€_{i21} [~a_i (\partial_2 b_1)~]\nonumber \\ &~+ πœ€_{i31} [~b_i (\partial_3 a_1)~] - πœ€_{i31} [~a_i (\partial_3 b_1)~]\nonumber \\ &~+ πœ€_{i12} [~b_i (\partial_1 a_2)~] - πœ€_{i12} [~a_i (\partial_1 b_2)~]\nonumber \\ &~+ πœ€_{i32} [~b_i (\partial_3 a_2)~] - πœ€_{i32} [~a_i (\partial_3 b_2)~]\nonumber \\ &~+ πœ€_{i13} [~b_i (\partial_1 a_3)~] - πœ€_{i13} [~a_i (\partial_1 b_3)~]\nonumber \\ &~+ πœ€_{i23} [~b_i (\partial_2 a_3)~] - πœ€_{i23} [~a_i (\partial_2 b_3)~]\nonumber\end{align}\end{cases}##
Summing over ##i##:
##πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~πœ€_{321} [~b_3 (\partial_2 a_1)~] - πœ€_{321} [~a_3 (\partial_2 b_1)~]\nonumber \\ &~+ πœ€_{231} [~b_2 (\partial_3 a_1)~] - πœ€_{231} [~a_2 (\partial_3 b_1)~]\nonumber \\ &~+ πœ€_{312} [~b_3 (\partial_1 a_2)~] - πœ€_{312} [~a_3 (\partial_1 b_2)~]\nonumber \\ &~+ πœ€_{132} [~b_1 (\partial_3 a_2)~] - πœ€_{132} [~a_1 (\partial_3 b_2)~]\nonumber \\ &~+ πœ€_{213} [~b_2 (\partial_1 a_3)~] - πœ€_{213} [~a_2 (\partial_1 b_3)~]\nonumber \\ &~+ πœ€_{123} [~b_1 (\partial_2 a_3)~] - πœ€_{123} [~a_1 (\partial_2 b_3)~]\nonumber\end{align}\end{cases}##
##πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~(-1) b_3 (\partial_2 a_1) - (-1) a_3 (\partial_2 b_1) \nonumber \\ &~+ (+1) b_2 (\partial_3 a_1) - (+1) a_2 (\partial_3 b_1) \nonumber \\ &~+ (+1) b_3 (\partial_1 a_2) - (+1) a_3 (\partial_1 b_2) \nonumber \\ &~+ (-1) b_1 (\partial_3 a_2) - (-1) a_1 (\partial_3 b_2) \nonumber \\ &~+ (-1) b_2 (\partial_1 a_3) - (-1) a_2 (\partial_1 b_3) \nonumber \\ &~+ (+1) b_1 (\partial_2 a_3) - (+1) a_1 (\partial_2 b_3) \nonumber\end{align}\end{cases}##
##πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~- b_3 (\partial_2 a_1)^{***} + a_3 (\partial_2 b_1)^{\uparrow\uparrow \uparrow} \nonumber \\ &~+ b_2 (\partial_3 a_1)^{**} - a_2 (\partial_3 b_1)^{\uparrow\uparrow} \nonumber \\ &~+ b_3 (\partial_1 a_2)^{***} - a_3 (\partial_1 b_2){\uparrow\uparrow \uparrow} \nonumber \\ &~- b_1 (\partial_3 a_2)^* + a_1 (\partial_3 b_2)^{\uparrow} \nonumber \\ &~- b_2 (\partial_1 a_3)^{**} + a_2 (\partial_1 b_3)^{\uparrow\uparrow} \nonumber \\ &~+ b_1 (\partial_2 a_3)^* - a_1 (\partial_2 b_3)^{\uparrow} \nonumber\end{align}\end{cases}##
##(8){~~~~~}πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~b_1 [~(\partial_2 a_3)- (\partial_3 a_2)~] \nonumber \\
&~+ b_2 [~(\partial_3 a_1) - (\partial_1 a_3)~] \nonumber \\
&~+ b_3 [~(\partial_1 a_2) - (\partial_2 a_1)~] \nonumber \\
&~+ a_1 [~(\partial_3 b_2) - (\partial_2 b_3)~] \nonumber \\
&~+ a_2 [~(\partial_1 b_3) - (\partial_3 b_1)~] \nonumber \\ &~+ a_3 [~(\partial_2 b_1) - (\partial_1 b_2)~] \nonumber\end{align}\end{cases}##
Comparing eqs. (7) and (8), we find that $$πœ€_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] = πœ€_{ijk} [~b_i (\partial_j a_k)~] - πœ€_{ijk} [~a_i (\partial_j b_k)~]$$ $$ \Rightarrow{~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b)$$
 

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