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Math Help - Vector Analysis identities proof

  1. #1
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    Vector Analysis identities proof

    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.
    Last edited by tsang; August 28th 2010 at 03:18 AM. Reason: solved second part
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  2. #2
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    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}<br />
0, &\text{if any index is equal to any other index}\\<br />
+1, &\text{if}\;i, j, k\;\text{form an {\it even} permutation of}\;1,2,3\\<br />
-1, &\text{if}\;i, j, k\;\text{form an {\it odd} permutation of}\;1,2,3<br />
\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))?
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  3. #3
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    Quote Originally Posted by tsang View Post
    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.
    Last edited by HallsofIvy; August 28th 2010 at 09:39 AM.
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    Quote Originally Posted by Ackbeet View Post
    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.
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  5. #5
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    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]<br />
=\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?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    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]<br />
=\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?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    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.
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  8. #8
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    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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