Question

I need help with b (picture). Are there any identities I should know to answer this? There are so many and I need help pointing out which ones.

Answer 1

Use\[
\mathbf a=\langle
a_1,a_2,a_3
\rangle\\
\mathbf b=\langle
b_1,b_2,b_3
\rangle
\]

Answer 2

In otherwords... just perform the operations with these filler variables.

Make sense @wolfe8

Answer 3

Right. I could probably work it out if I use explicit vector components, but the question asks me to use index notation, which is where I need help with.

Answer 4

Following @wio, for the first case (the simplest),
\[\sum_{i=1}^3\frac{\partial}{\partial x_i}(a_i+b_i)=\sum_{i=1}^3\frac{\partial}{\partial x_i}a_i+\sum_{i=1}^3\frac{\partial}{\partial x_i}b_i=\nabla \mathbf{A}+\nabla \mathbf{B}\]

Answer 5

Thanks John but I got the first part. I used \[(A _{i}+B _{i})_{i}=A _{ii}+B _{ii}=the things on RHS\] Sorry I don't know how to do some notations here

Answer 6

A side question: does it make a difference if there is a dot between the symbol and vectors? From what I understand, if there is none, it's a gradient. If there is a dot it's a divergent and if there's a cross it's a curl.

Answer 7

Ok, may be I would have used the notation,
\[\sum_{i=1}^{3}\frac{\partial}{\partial x_i}(\mathbf{A}+\mathbf{B})_i=\sum_{i=1}^{3}\frac{\partial {A_i}}{\partial x_i}+\sum_{i=1}^{3}\frac{\partial {B_i}}{\partial x_i}\]

Answer 8

Yes that's better than what I wrote! Thanks! I am almost completely clueless on the second part though.

Answer 9

Gradient (with or without dot) is only between the nabla symbol an a scalar function. Divergence (with or without dot) is only between the nabla symbol an a vectorial function. Curl (with the cross symbol) is only between the nabla symbol an a vectorial function.

Answer 10

Ok, for the second part, you will find usefull the Levi Civita symbol,
\[(\mathbf{A}\times\mathbf{B})_i=\varepsilon_{ijk}A_jB_k\]

Answer 11

Right. So after some lines I have in the end \[\sum_{i}^{}\frac{ \delta }{ deltax _{i} }\sum_{jk}^{}\in _{ijk} A _{j}B _{k}\] And then what do I do?

Answer 12

Oh well that is supposed to be a delta/deltax

Answer 13

Now you must use the product rule for derivatives.

Answer 14

I think I have this wrong but I get \[\sum_{ijk}^{}\frac{ \delta B _{k} }{ \delta x _{i} }A _{j}+\frac{ \delta A _{i} }{ \delta x _{i} }B _{k}\]

Answer 15

\[\sum_{i=1}^3\varepsilon_{ijk}A_jB_k=\sum_{i=1}^3\varepsilon_{ijk}\frac{\partial A_i}{\partial x_i}b_k+\sum_{i=1}^3\varepsilon_{ijk}\frac{\partial B_k}{\partial x_i}a_j\]

Answer 16

Now you must swap index. The first term need an even permutation so doesn't change its sign, but the second needs an odd permutation so changes its sign.
\[\sum_{i=1}^3\varepsilon_{ijk}A_jB_k=\sum_{i=1}^3\varepsilon_{ijk}\frac{\partial A_j}{\partial x_i}b_k+\sum_{i=1}^3\varepsilon_{ijk}\frac{\partial B_k}{\partial x_i}a_j=\\

=\sum_{i=1}^3b_k\varepsilon_{kij}\frac{\partial A_j}{\partial x_i}+\sum_{i=1}^3a_j\varepsilon_{jik}\frac{\partial B_k}{\partial x_i}=\mathbf{B}(\nabla\times\mathbf{A})-\mathbf{A}(\nabla\times\mathbf{B})\]

Answer 17

I corrected an index I missed before.

Answer 18

Oh so I wasn't supposed to just remove the Epsilon. I must be really tired because I don't see any difference between what you wrote and your correction. Oh wow thanks a bunch John! You really are a life saver.

Answer 19

;), the difference was in the first term, after the equal, I wrote A_i instead A_j. Now is ok.

Answer 20

Ohh yeah.