Vector Analysis identities proof

**tsang** · August 26th 2010, 06:08 AM

Hi dear friends, I have a question and I cannot do any of them! I'm so frustrated and I fell into bad mental block now.

Please help me, I feel so helpless this moment. Please help me and show me some details. I will really appreciate for that.

Question: F and G both are vector fields. Prove identity div (F cross G)=G dot (curl F)-F dot (curl G)

So this is the identity which says "divergence of vector field A cross B = A dot (curl B)-B dot (curl A)"

Please help me, thanks a lot, I still cannot get it. I'm trying to prove by definition, so I just expand everything, but I end up quite messy.

**Ackbeet** · August 26th 2010, 09:37 AM

I would prove this using the Levi-Civita density symbol for cross products. It looks like $\epsilon_{ijk},$ and it's defined this way:

$\displaystyle{\epsilon_{ijk}=\begin{cases} 0, &\text{if any index is equal to any other index}\\ +1, &\text{if}\;i, j, k\;\text{form an {\it even} permutation of}\;1,2,3\\ -1, &\text{if}\;i, j, k\;\text{form an {\it odd} permutation of}\;1,2,3 \end{cases}.}$

An even permutation has an even number of exchanges of position of two symbols. Cyclic permutations (for example, $123\to 231\to 312$ ) are always even. Thus

$\epsilon_{122}=\epsilon_{313}=\epsilon_{211}=0,\qu ad\text{etc.}$

$\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=+1$

$\epsilon_{132}=\epsilon_{213}=\epsilon_{321}=-1$

- from Classical Dynamics of Particles and Systems, 4th Ed., p. 25-26, by Marion and Thornton.

The cool thing about the Levi-Civita symbol is that you can write cross-products with it explicitly (or at least components of it explicitly):

$\displaystyle{\left(\vec{A}\times\vec{B}\right)_{i }=\sum_{j,k}\epsilon_{ijk}A_{j}B_{k}}.$

So, getting to your first question, I can write the cross product as

$\displaystyle{\left(\vec{F}\times\vec{G}\right)_{i }=\sum_{j,k}\epsilon_{ijk}F_{j}G_{k}}.$

Then I get to say that

$\displaystyle{\nabla\cdot(\vec{F}\times\vec{G})=\s um_{i,j,k}\epsilon_{ijk}\,\frac{\partial}{\partial x_{i}}\left(F_{j}G_{k}\right)=\sum_{i,j,k}\epsilon _{ijk}\,\left(F_{j}\frac{\partial}{\partial x_{i}}G_{k}+G_{k}\frac{\partial}{\partial x_{i}}F_{j}\right)}.$

The rest of the proof consists of massaging this last expression to equal your original RHS.

As for your part 2, do you mean that you need to show

$\nabla\cdot((f\,\nabla g)\times(\nabla h))=(\nabla f)\cdot((\nabla g)\times(\nabla h))?$

**HallsofIvy** · August 26th 2010, 11:13 AM

Originally Posted by tsang

Hi dear friends, I have two questions and I cannot do any of them! I'm so frustrated and I fell into bad mental block now.

Please help me, I feel so helpless this moment. Please help me and show me some details. I will really appreciate for that.

Question 1: F and G both are vector fields. Prove identity div dot (F cross G)=G dot (curl F)-F dot (curl G)

No reason in the world why you can't just "slog it out". Let F= ai+ bj+ ck and G= di+ ej+ fk. Then F cross G is equal to $(bf- ce)i- (af- cd)j+ (ae- bd)k$ . div(F cross B) then is div((bf- ce)i- (af-cd)j+ (ae- bd)k= b_xf+ bf_x- c_xe- ce_x+ a_yf+ af_y- c_ye- ce_y+ a_ze+ ae_z- b_zd- bd_z. (NOT "div dot ...". "div ... " is "del dot ...")

Do the right side the same way and see if you get the same thing.

Question 2: f(x,y,z), g(x,y,z), and h(x,y,z) be any C2 scalar functions. Using the basic identities (write down which identities) of vector analysis, prove that
div dot (f div g cross div h)= div f dot (div g cross div h)

Forgive me for not using LaTex Code, I'm still quite bad with Latex this moment. Very sorry for that.

**tsang** · August 27th 2010, 07:26 PM

Originally Posted by Ackbeet

I would prove this using the Levi-Civita density symbol for cross products. It looks like $\epsilon_{ijk},$ and it's defined this way:

- from Classical Dynamics of Particles and Systems, 4th Ed., p. 25-26, by Marion and Thornton.

The cool thing about the Levi-Civita symbol is that you can write cross-products with it explicitly (or at least components of it explicitly):

$\displaystyle{\left(\vec{A}\times\vec{B}\right)_{i }=\sum_{j,k}\epsilon_{ijk}A_{j}B_{k}}.$

So, getting to your first question, I can write the cross product as

$\displaystyle{\left(\vec{F}\times\vec{G}\right)_{i }=\sum_{j,k}\epsilon_{ijk}F_{j}G_{k}}.$

Then I get to say that

$\displaystyle{\nabla\cdot(\vec{F}\times\vec{G})=\s um_{i,j,k}\epsilon_{ijk}\,\frac{\partial}{\partial x_{i}}\left(F_{j}G_{k}\right)=\sum_{i,j,k}\epsilon _{ijk}\,\left(F_{j}\frac{\partial}{\partial x_{i}}G_{k}+G_{k}\frac{\partial}{\partial x_{i}}F_{j}\right)}.$

The rest of the proof consists of massaging this last expression to equal your original RHS.

As for your part 2, do you mean that you need to show

$\nabla\cdot((f\,\nabla g)\times(\nabla h))=(\nabla f)\cdot((\nabla g)\times(\nabla h))?$

Thank you, the question 2 is as you typed, and I still cannot work out it, can you help me with it please? Thanks a lot.

**Ackbeet** · August 28th 2010, 07:42 AM

Ok, so we can write the LHS as follows:

$\displaystyle{\sum_{i,j,k}\epsilon_{ijk}\frac{\par tial}{\partial x_{i}}\left[(f\nabla g)_{j}(\nabla h)_{k}\right] =\sum_{i,j,k}\epsilon_{ijk}\frac{\partial}{\partia l x_{i}}\left[\left(f\frac{\partial}{\partial x_{j}} g\right)\left(\frac{\partial}{\partial x_{k}} h\right)\right]}.$

At this point, I would probably use the product rule here and then show that it equals the RHS, which looks like this:

$\displaystyle{\sum_{i,j,k}\epsilon_{ijk}\left(\fra c{\partial}{\partial x_{i}}\,f\right)\left(\frac{\partial}{\partial x_{j}}\,g\right)\left(\frac{\partial}{\partial x_{k}}\,h\right)}.$

Can you continue?

**tsang** · August 28th 2010, 05:52 PM

Originally Posted by Ackbeet

Ok, so we can write the LHS as follows:

$\displaystyle{\sum_{i,j,k}\epsilon_{ijk}\frac{\par tial}{\partial x_{i}}\left[(f\nabla g)_{j}(\nabla h)_{k}\right] =\sum_{i,j,k}\epsilon_{ijk}\frac{\partial}{\partia l x_{i}}\left[\left(f\frac{\partial}{\partial x_{j}} g\right)\left(\frac{\partial}{\partial x_{k}} h\right)\right]}.$

At this point, I would probably use the product rule here and then show that it equals the RHS, which looks like this:

$\displaystyle{\sum_{i,j,k}\epsilon_{ijk}\left(\fra c{\partial}{\partial x_{i}}\,f\right)\left(\frac{\partial}{\partial x_{j}}\,g\right)\left(\frac{\partial}{\partial x_{k}}\,h\right)}.$

Can you continue?

Thank you for your time. Thanks for helping me everytime.
I actually get a bit dizzy for the way you do this question, I can feel that you are using some high level way to solve it. Maybe a bit ahead of my current level? I assume?
I was just trying to use definition to expand everything, and I ended up with 12 terms for LHS and 12 terms RHS, but they do not really match with each other, so I cannot end up that they are equal. I just feel so helpless as I've been trying for days and still cannot solve it. Are you able to help me with the expantion please?

**tsang** · August 29th 2010, 05:46 AM

Originally Posted by HallsofIvy

No reason in the world why you can't just "slog it out". Let F= ai+ bj+ ck and G= di+ ej+ fk. Then F cross G is equal to $(bf- ce)i- (af- cd)j+ (ae- bd)k$ . div(F cross B) then is div((bf- ce)i- (af-cd)j+ (ae- bd)k= b_xf+ bf_x- c_xe- ce_x+ a_yf+ af_y- c_ye- ce_y+ a_ze+ ae_z- b_zd- bd_z. (NOT "div dot ...". "div ... " is "del dot ...")

Do the right side the same way and see if you get the same thing.

I have been keeping trying, and the RHS I just cannot end up with same things, I always have different partial derivatives. For example, LHS has b_xf+ bf_x, nut RHS is b_yf+bf_y, they always have different base. I don't know what's going on and I'm really frustrated now, since I've been trying it for many days. Please help me.

**Ackbeet** · August 30th 2010, 02:11 AM

Ok, here's another way: use other vector calculus identities. You'll need at least the following list:

$\nabla\cdot(A\times B)=B\cdot(\nabla\times A)-A\cdot(\nabla\times B).$

$\nabla\times(\nabla\phi)=\vec{0}.$

$\nabla\times(\psi A)=\psi(\nabla\times A)+(\nabla\psi)\times A.$

Finally, you'll need

$A\cdot(B\times C)=B\cdot(C\times A)=C\cdot(A\times B).$

Together, those identities will give you what you need. Can you start on the LHS of the original?

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