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Math methods in physics book - vector calc proof

  1. Jan 14, 2013 #1
    Hi all, I just got mary boas math methods in physics book as a supplement because I'm a physics major and I'm browsing thru the vector calculus sections and came across the del operator identity:

    nambla (V dot U) = stuff

    nambla is the del operator and "dot" is dot product...

    I'm trying to figure out how to prove this seeing as i'm very rusty on my kronecker delta, levi-civita permutation tensor, and other vector calc related identities

    any tips on the first couple steps?

    the solution is on wikipedia if you google "vector calc identities" and it appears that it involves two partial derivative product rules or something

    Anyway, I've been bored and stuck on what to do with this for a while tonight and can't figure out how to get the first few steps done that would very much refresh my brain ! many thanks to anyone who is willing to help
  2. jcsd
  3. Jan 14, 2013 #2


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    Homework Helper

    Well the natural thing to do is write

    [tex]\mathbf{\nabla(a \cdot b)=\nabla_a (a \cdot b)+\nabla_b (a \cdot b)}[/tex]

    where the subscript means the derivative is partial and only effects one vector. However partial vector derivatives are hard to interpret as they represent an arbitrary and artificial separation so we employ the identity

    [tex]\mathbf{\nabla_b (a \cdot b)=(a\times\nabla)\times b+a \, (\nabla\cdot b)=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]

    The third version being the one used in most books. This illustrates several peculiar things. We have three ways of writing the same thing, yet when switching between them we can be confused. Some object to the partial derivative formulation when the others forms are the same anyway. The fact that we prefer to use right acting operators breaks symmetry (observe the two forms are right and left hand versions of the same thing). Also notice that these rules follow from the so called baccab rule

    [tex]\mathbf{a \times (b \times c)=b \, (a \cdot c)- c \, (a \cdot b)}[/tex]

    You can write all this out with your beloved epsilons and deltas if you like. The baccab rule is then written

    [tex]\epsilon_{ijk} \epsilon^{imn}=\left|
    \delta_j^m & \delta_j^n \\
    \delta_k^m & \delta_k^n \\
    \end{array} \right|=\delta_j^m\delta_k^n-\delta_j^n\delta_k^m[/tex]
  4. Jan 14, 2013 #3


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    What is your native language? I am used to "nabla", not "nambla" but it might be a difference in language. In any case, we think of "nabla", [itex]\nabla[/itex] as the "vector differential operator",[itex](\partial/\partial x)\vec{i}+ (\partial/\partial y)\vec{j}+ (\partial/\partial z)\vec{k})[/itex]. Applied to a scalar valued function f(x) that gives the vector function [itex](\partial f/\partial x)\vec{i}+ (\partial f/\partial y)\vec{j}+ (\partial f/\partial z)\vec{k})[/itex]. If f is the result of a dot product, [itex]f= \vec{u}\cdot\vec{v}= u_xv_x+ u_yv_y+ u_zv_z[/itex] then that formula becomes [itex](\partial (u_xv_x+ u_yv_y+ u_zv_z)/\partial x)\vec{i}+ (\partial (u_xv_x+ u_yv_y+ u_zv_z)/\partial y)\vec{j}+ (\partial (u_xv_x+ u_yv_y+ u_zv_z)/\partial z)\vec{k})[/itex].

    Applying the sum and product rules to those (tedious, so I am not going to do it, but doable) gives the formula.
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