Question

One vector calculus problem

Answer 1

@Michele_Laino

Answer 2

is the symbol "\(*\)" for vector product?

Answer 3

yeah I couldn't find the cross product symbol

Answer 4

ok! Here are your steps, please wait a moment...

Answer 5

I check your identity on one generic component, say the \(i-\)th component
please wait...

Answer 6

ok..

Answer 7

here are my answer:

Answer 8

I cannot understand, not used to this notation.

Answer 9

I have used the ricci symbol \(\Large \epsilon_{ijk}\) and this identity:
\[\Large {\varepsilon _{kij}}{\varepsilon _{klm}} = {\delta _{il}}{\delta _{jm}} - {\delta _{im}}{\delta _{jl}}\]

Answer 10

\[\Large \begin{gathered}
{\left( {f \times \left( {\nabla \times f} \right)} \right)_i} = {\varepsilon _{ijk}}{f_j}{\left( {\nabla \times f} \right)_k} = \hfill \\
\hfill \\
= {\varepsilon _{ijk}}{f_j}{\varepsilon _{klm}}{\partial _l}{f_m} = \hfill \\
\hfill \\
= {\varepsilon _{ijk}}{\varepsilon _{klm}}{f_j}\left( {{\partial _l}{f_m}} \right) = \hfill \\
\hfill \\
= {\varepsilon _{kij}}{\varepsilon _{klm}}{f_j}\left( {{\partial _l}{f_m}} \right) = \hfill \\
\hfill \\
= \left( {{\delta _{il}}{\delta _{jm}} - {\delta _{im}}{\delta _{jl}}} \right){f_j}\left( {{\partial _l}{f_m}} \right) = \hfill \\
\hfill \\
= {\delta _{il}}{\delta _{jm}}{f_j}\left( {{\partial _l}{f_m}} \right) - {\delta _{im}}{\delta _{jl}}{f_j}\left( {{\partial _l}{f_m}} \right) = \hfill \\
\hfill \\
= {f_j}\left( {{\partial _i}{f_j}} \right) - \left( {{f_j}{\partial _j}} \right){f_i} = \hfill \\
\hfill \\
= \frac{1}{2}{\partial _i}\left( {{f_j}{f_j}} \right) - \left( {{f_j}{\partial _j}} \right){f_i} \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]

Answer 11

oh ricci calculus, I haven;t started it yet.. I started tensor calculus about a week or so..

Answer 12

with such method I'm able to check all the vector identities

Answer 13

It is the only method that I know

Answer 14

OK..:) then I am writing it..But first I have to learn it

Answer 15

It is not difficult. Please keep in mind that for the product vector between two vextors \(a,b\), I can write this:
\[\huge {\left( {{\mathbf{a}} \times {\mathbf{b}}} \right)_i} = {\varepsilon _{kij}}{a_i}{b_j}\]
where sums over repeated indices \(i,j\) are implicit

Answer 16

oops..I have made a typo
\[\huge {\left( {{\mathbf{a}} \times {\mathbf{b}}} \right)_k} = {\varepsilon _{kij}}{a_i}{b_j}\]

Answer 17

of course, indices \(i,\,j,\,k\) runs from \(1\) to \(3\)

Answer 18

run*

Answer 19

Oh I see the notation. I never did a vector calculus problem using tensors... its easy though

Answer 20

yes!

Answer 21

@IrishBoy123 do you have other method? I mean vector calculus methods?

Answer 22

michele is the maestro.


my crude offering is from the bac-cab rule
\(\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})\)

\(\mathbf{f}\times (\mathbf{\nabla}\times \mathbf{f}) = \mathbf{\nabla}(\mathbf{f}\cdot\mathbf{f}) - \mathbf{f}(\mathbf{\nabla}\cdot\mathbf{f})\)

michele might know where the \(\frac{1}{2}\) comes in

Answer 23

\[\epsilon _{ijk}=\epsilon _{kij}\
??

Answer 24

@Michele_Laino

Answer 25

yes! since I have made this permutations:
\[\huge ijk \to ikj \to kij\]
and for each permutation (better is swapping) sign changes from \(+1\) to \(-1\) and then from \(-1\) to \(+1\)

Answer 26

ok.. and have you written

Answer 27

please note that I have written this:
\[\huge {\nabla _i} = {\partial _i} = \frac{\partial }{{\partial {x_i}}}\]

Answer 28

ok.. and just one more thing, what does
\[(a \times b)_{i}\]
mean,,, i mean are we operating it on i vector?

Answer 29

yes! it is the \(i-\)th component of the vector:
\[\huge {{\mathbf{a}} \times {\mathbf{b}}}\]

Answer 30

where, of course, \(i\) stands for \(1,2\) or \(3\)

Answer 31

or \(x,y\) or \(z\)

Answer 32

ok.. thanks.:).. I learnt a new method today, though I will be able to apply it after I learn Ricci calculus,.

Answer 33

yes! Please keep in mind that it is not a difficult method

Answer 34

yeah,
\[\epsilon _{ijk}\]
is the generalized Kronecker delta, right?

Answer 35

yes!