Del operator problem driving me nutty

In summary, Afken says to first use the bac-cab rule to each of the cross products, and then add the two resulting equations together. He then says he removes a factor of 2 from the sum with no explanation. I am not sure how to arrange the indices to make the position of the k's match so that I can apply the identity later.
  • #1
go_ducks
24
0
I have several problems like these, I am learning about Levi-Civita and Kroenecker delta so those may be relevant, however, what I am really looking for is further discussion or good books/online tutorials that are specific to usage of the del operator.

Homework Statement



I'm actually going to give you two problems because they're so well related anyway.

1. (this is p46 of Arfken & Weber 6th edition Mathematical Methods - its an example!)

[tex]\nabla[/tex](A.B) = (B.[tex]\nabla[/tex])A + (A.[tex]\nabla[/tex])B + Bx([tex]\nabla[/tex] x A) + A x ([tex]\nabla[/tex] x B)2.

[tex]\nabla[/tex](A.B) = (A x [tex]\nabla[/tex]) x B + (B x [tex]\nabla[/tex]) x A + A([tex]\nabla[/tex] . B) + B([tex]\nabla[/tex] . A)

Homework Equations



eijkeilm = delta jl delta km = delta jm delta kl
Ax(BxC) = B(A.C) - (C.A)B

The Attempt at a Solution



K.. I just lost my post. But basically I got (immediately):

[tex]\nabla[/tex](A.B) = A.[tex]\nabla[/tex]B + B.[tex]\nabla[/tex]A

and I do not know how they came up with these cross products from there. For instance, B.[tex]\nabla[/tex]A came from Bi dA/dxi (these should be partial derivatives, curly d's, to be clear). Am I supposed to show (B.[tex]\nabla[/tex])A + Bx([tex]\nabla[/tex] x A) gives same answer in components as that? Eyeballing it seemed like it might, but when I tried it, I got Bx([tex]\nabla[/tex] x A) = 0.

Specifically, if you look at the writeup in Afken (p46), he cheats. He starts by applying the bac-cab rule to each of the cross products on the RHS, using the commutative property to move around the del operator. He then says he adds two resulting equations together, which normally I'd say, ok Afken, you're pulling it out of the hat but that seems fine, except - in this case it looks like a factor of 2 has been removed from the sum with no explanation. I'm sorry for not typing that out more clearly but the last time I did , the forum chewed it up. Hopefully someone has the book and can verify that mistake or explain why he's correct?.
 
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  • #2
Try starting with the cross products first. Using the Levi-Civita symbol do you know how to re-write the following?

[tex]\vec{B}X(\nabla X \vec{A})[/tex]
 
  • #3
Sorry I left to go to Borders to look for a book. Came back empty handed. Anyway I tried that before. Let me know what I do wrong here then.

B x ([tex]\nabla[/tex] x A)

let C = [tex]\nabla[/tex] X A => [B x C]i = eijk BjCk

now [[tex]\nabla[/tex] X A]i = eimn [tex]\nabla[/tex] m An

=> [B x ([tex]\nabla[/tex] x A) ]i = eijkBj ekmn [tex]\nabla[/tex] m An

...here's where I'm confused a bit. I am summing del X A over "k" ? A classmate told me that I can just drop the k, leaving the second epislon symbol as eimn. However I really feel like it has to replace the i after reading my notes. I know if I am wrong here the rest is stuffed, but I don't think I am.

=> [B x ([tex]\nabla[/tex] x A) ]i = eijk ekmn Bj [tex]\nabla[/tex] m An

at this point I say [tex]\nabla[/tex] is acting on one thing so I'm happy about that. So what I want is to swap the indices to make the position of k's match, to use identity later.

=> [B x ([tex]\nabla[/tex] x A) ]i = eijk emnk Bj [tex]\nabla[/tex] m An (2 swaps, so no sign change)

next thing to do is, eijk emnk = delta I am delta jn - delta in delta jm

=> [B x ([tex]\nabla[/tex] x A) ]i = (delta I am delta jn - delta in delta jm) Bj [tex]\nabla[/tex] m An

=> [B x ([tex]\nabla[/tex] x A) ]i = delta I am delta jn Bj [tex]\nabla[/tex] m An - delta in delta jm Bj [tex]\nabla[/tex] m An

next apply fact m=i, n=j to return anything from the first term, and n=i, m=j from the second term

=> [B x ([tex]\nabla[/tex] x A) ]i = delta ii delta jj Bj [tex]\nabla[/tex] i Aj - delta ii delta jj Bj [tex]\nabla[/tex] j Ai

=> [B x ([tex]\nabla[/tex] x A) ]i = Bj [tex]\nabla[/tex] i Aj - Bj [tex]\nabla[/tex] j Ai

but now I am not sure how to arrange. This is a guess based on the position of i, but honestly I'm not sure what to do with it:

=> [B x ([tex]\nabla[/tex] x A) ]i = B x ([tex]\nabla[/tex] x A) - (B . [tex]\nabla[/tex]) x A
 
  • #4
You know, I guess my suggestion isn't very obvious. Let me try to help a bit.

[tex]\vec{B}X(\nabla X \vec{A})=\epsilon_{ijk}B_{i}(\nabla X \vec{A})_{j}\hat{x}_{k}[/tex]

[tex]\vec{A}X(\nabla X \vec{B})=\epsilon_{ijk}A_{i}(\nabla X \vec{B})_{j}\hat{x}_{k}[/tex]
 
  • #5
I will have to take a break because my head is hurting, but thank you, I had forgotten all about the x-hat. What a horrible mistake. I will see how this goes after my frontal lobe stops pulsating.
 
  • #6
Ok, tried BX(del X A)

first of all a forethought, I expected result to look like the BAC-CAB rule, Ax(BxC) = B(A.C) - (C.A)B ...here, A->B, B->del, C->A, implying Bx(delxA) = del(B.A) - (A.B)del . Not that it makes sense to move the operator around like that...

So anyway here's my results (did some more reading and learned the x isn't necessary):

bx(del x a)
= eijk bj ekmn Dm an (take D to mean the curly D, for partial deriv, while below take d as delta.)
= eijk ekmn bj Dm an
= ekij ekmn bj Dm an
= (dimdjn - dindjm) bj Dm an
= dimdjn bj Dm an - dindjmbj Dm an
= diidjj bj Di aj - diidjjbj Dj ai (take m=i, n=j for first term, etc.)
= bj Di aj - bj Dj ai

I'm stuck here. I don't know how the indices relate to one another. It looks like del(A.B) - (B.del)A ...? I guess this looks like a part of the equation but is it right? Also this is exactly what I got before you gave another hint.

So it looks like I am claiming:

BX(del X A) = del(A.B) - (B.del)A ... del(A.B) = (B.del)A + BX(del X A)
AX(del X B) = del(A.B) - (A.del)B ... del(A.B) = (A.del)B + AX(del X B)

This is the same as the Arfken result but it has the same problem as his. That is, when you add them together, it should give you 2 del (A.B), so you have to divide by 2, and clearly that wouldn't be the identity. Unless there is some reason I can drop the 2 which I'm not seeing. I have my textbook example doing the same thing and I would like to know what is the argument.
 
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  • #7
go_ducks said:
Ok, tried BX(del X A)

first of all a forethought, I expected result to look like the BAC-CAB rule, Ax(BxC) = B(A.C) - (C.A)B ...here, A->B, B->del, C->A, implying Bx(delxA) = del(B.A) - (A.B)del . Not that it makes sense to move the operator around like that...

So anyway here's my results (did some more reading and learned the x isn't necessary):

bx(del x a)
= eijk bj ekmn Dm an (take D to mean the curly D, for partial deriv, while below take d as delta.)
= eijk ekmn bj Dm an
= ekij ekmn bj Dm an
= (dimdjn - dindjm) bj Dm an
= dimdjn bj Dm an - dindjmbj Dm an
= diidjj bj Di aj - diidjjbj Dj ai (take m=i, n=j for first term, etc.)
= bj Di aj - bj Dj ai

I think you may be missing a cross term (I'm no expert at this and I'm having trouble relating our solutions) when you add them. Here is the way I have seen this problem worked out.

[tex]\vec{B}X(\nabla X \vec{A})+\vec{A}X(\nabla X \vec{B})=\epsilon_{ijk}B_{i}(\nabla X \vec{A})_{j}\hat{x}_{k}+\epsilon_{ijk}A_{i}(\nabla X \vec{B})_{j}\hat{x}_{k}[/tex]

[tex]\epsilon_{ijk}B_{i}(\nabla X \vec{A})_{j}\hat{x}_{k}+\epsilon_{ijk}A_{i}(\nabla X \vec{B})_{j}\hat{x}_{k}=(\epsilon_{ijk}B_{i}\epsilon_{jlm}\partial_{l}A_{m}+\epsilon_{ijk}A_{i}\epsilon_{jlm}\partial_{l}B_{m})\hat{x}_{k}[/tex]

[tex](\epsilon_{ijk}B_{i}\epsilon_{jlm}\partial_{l}A_{m}+\epsilon_{ijk}A_{i}\epsilon_{jlm}\partial_{l}B_{m})\hat{x}_{k}=\epsilon_{ijk}\epsilon_{jlm}(B_{i}\partial_{l}A_{m}+A_{i}\partial_{l}B_{m})\hat{x}_{k}[/tex]

[tex]\epsilon_{ijk}\epsilon_{jlm}(B_{i}\partial_{l}A_{m}+A_{i}\partial_{l}B_{m})\hat{x}_{k}=-(\delta_{il}\delta_{km}-\delta_{im}\delta_{kl})(B_{i}\partial_{l}A_{m}+A_{i}\partial_{l}B{m})\hat{x}_{k}[/tex]

[tex]-(\delta_{il}\delta_{km}-\delta_{im}\delta_{kl})(B_{i}\partial_{l}A_{m}+A_{i}\partial_{l}B{m})\hat{x}_{k}=(-B_{i}\partial_{i}A_{k}-A_{i}\partial_{i}B_{k}+B_{i}\partial_{k}A_{i}+A_{i}\partial_{k}B_{i})\hat{x}_{k}[/tex]

[tex](-B_{i}\partial_{i}A_{k}-A_{i}\partial_{i}B_{k}+B_{i}\partial_{k}A_{i}+A_{i}\partial_{k}B_{i})\hat{x}_{k}=-(B_{i}\partial_{i})(A_{k}\hat{x}_{k})-(A_{i}\partial_{i})(B_{k}\hat{x}_{k})+\partial_{k}(B_{i}A_{i})\hat{x}_{k}[/tex]

[tex]-(B_{i}\partial_{i})(A_{k}\hat{x}_{k})-(A_{i}\partial_{i})(B_{k}\hat{x}_{k})+\partial_{k}(B_{i}A_{i})\hat{x}_{k}=-(\vec{B}\cdot\nabla)\vec{A}-(\vec{A}\cdot\nabla)\vec{B}+\nabla(\vec{B}\cdot\vec{A})[/tex]

I'm also no expert on LaTeX formatting. I hope that is clear.
 
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  • #8
Well its equivalent until the last step, where you combined your BiDkAi and AidkBi together to form dk(BiAi) which looks like use of the product rule.

At the end I didnt do that, I claimed "bj Di aj" = "del(A.B)" I think that is the issue. Because its not true... its one half of the gradient of the vector dot product. When I add up the two sides it will be possible to combine each piece to form del(A.B)

Well that is nice but it still leaves me troubled on Afken's version. He seems to be giving bad information...
 

Related to Del operator problem driving me nutty

1. What is the "Del operator" problem in science?

The "Del operator" problem, also known as the "divergence problem", is a mathematical issue that arises when trying to calculate the divergence of a vector field using the del operator. It is often encountered in fluid mechanics, electromagnetics, and other areas of physics.

2. Why is the "Del operator" problem important to solve?

The "Del operator" problem is important to solve because it is a fundamental issue in many fields of science and engineering. It can impact the accuracy and reliability of calculations and simulations, leading to incorrect results and potentially costly mistakes.

3. What causes the "Del operator" problem?

The "Del operator" problem is caused by the non-uniqueness of the solution to the divergence equation. This means that there are infinitely many possible solutions, making it difficult to determine the correct answer.

4. How do scientists and mathematicians approach solving the "Del operator" problem?

Scientists and mathematicians have developed various techniques to address the "Del operator" problem, such as regularizing the vector field, using finite difference approximations, and applying boundary conditions. These methods help to narrow down the possible solutions and find a more accurate answer.

5. Are there any current solutions to the "Del operator" problem?

While the "Del operator" problem remains a challenging issue, there are ongoing efforts to improve and refine existing methods for solving it. Some researchers are also exploring new approaches, such as using machine learning algorithms, to tackle this problem in a more efficient and accurate manner.

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