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Proving vector calculus identities using summation notation

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{∂x_{i}}{∂x_{j}}[/itex] = δ[itex]_{ij}[/itex]

    2. Relevant equations

    [itex]\vec{r}[/itex] = x[itex]_{i}[/itex]e[itex]_{i}[/itex]

    3. The attempt at a solution
    [itex]\frac{∂x_{i}}{∂x_{j}}[/itex] = 1 iff i=j

    δ[itex]_{ij}[/itex] = 1 iff i=j

    therefore

    [itex]\frac{∂x_{i}}{∂x_{j}}[/itex] = δ[itex]_{ij}[/itex]

    1. The problem statement, all variables and given/known data

    r[itex]^{2}[/itex] = x[itex]_{k}[/itex]x[itex]_{k}[/itex]

    2. Relevant equations

    [itex]\vec{r}[/itex] = x[itex]_{k}[/itex]e[itex]_{k}[/itex]
    [itex]\vec{r}[/itex] = x[itex]_{j}[/itex]e[itex]_{j}[/itex]


    3. The attempt at a solution

    r[itex]^{2}[/itex] = x[itex]_{k}[/itex]e[itex]_{k}[/itex][itex]\bullet[/itex]x[itex]_{j}[/itex]e[itex]_{j}[/itex]

    = e[itex]_{k}[/itex]e[itex]_{j}[/itex]x[itex]_{k}[/itex]x[itex]_{j}[/itex]

    e[itex]_{k}[/itex]e[itex]_{j}[/itex] = δ[itex]_{jk}[/itex] = 1 iff j=k

    r[itex]^{2}[/itex] = x[itex]_{k}[/itex]x[itex]_{k}[/itex] iff j=k


    1. The problem statement, all variables and given/known data

    ([itex]\nabla[/itex]r[itex]^{2}[/itex])[itex]_{j}[/itex]= [itex]\frac{∂}{∂x_{j}}[/itex]([itex]x_{l}[/itex][itex]x_{l}[/itex])= 2x[itex]_{j}[/itex]

    2. Relevant equations

    r[itex]^{2}[/itex] = [itex]x_{k}[/itex][itex]x_{k}[/itex]

    3. The attempt at a solution

    pretty confused by now, so far I've guessed my way through.

    is the j index communative? if so where did the l index come from. there is only 2 l's so maybe they cancel and you get x[itex]^{2}[/itex] which differentiation gives 2x. and it inherits the i index from the d/dxi.
     
  2. jcsd
  3. Mar 15, 2012 #2

    tiny-tim

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    hi lostminty! :smile:

    (try using the X2 button just above the Reply box :wink:)
    fine :smile:
    no, the whole point of knowing that ∂xi/∂xj = ejek = δij

    is that this notation enables you to avoid using those "iffs" …

    ejekxjxk

    = δjkxjxk

    = xkxk :wink:

    (and try the third one again)
     
  4. Mar 15, 2012 #3
    Ok, sounds good.

    so

    3. Attempt at solving

    ([itex]\nabla[/itex]r[itex]^{2}[/itex])[itex]_{j}=[/itex]
    [itex]\frac{∂r^{2}}{∂x_{j}}=[/itex]

    [itex]\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})= [/itex]

    [itex]\frac{∂}{∂x_{j}}(x_{l}x_{l}) =[/itex]

    [itex]\frac{∂}{∂x_{j}}(x^{2}) = [/itex]

    [itex]2x_{j}[/itex]
     
  5. Mar 15, 2012 #4

    tiny-tim

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    hi lostminty! :smile:
    ok so far :smile:

    (though you could miss out the line with δ, it's not necessary)
    no, you need to turn [itex]\frac{∂}{∂x_{j}}(x_{l})[/itex] into a δ :wink:
     
  6. Mar 15, 2012 #5

    HallsofIvy

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    What does it equal if [itex]i\ne j[/itex]? Don't you think you should say that?

     
  7. Mar 15, 2012 #6

    hmmm


    so

    [itex]\frac{∂}{∂x}(x_{l}x_{l})_{j} =[/itex]


    [itex]\frac{∂}{∂x}(x_{l}^{2})_{j} =[/itex]


    [itex]\frac{∂}{∂x}x^{2}_{l}δ_{jl} =[/itex]

    if l=j [itex]δ_{jl}=1[/itex] else [itex]δ_{jl}=0[/itex]

    [itex]\frac{∂}{∂x}x^{2}_{j} =[/itex]

    [itex]2x_{j}[/itex]
     
  8. Mar 15, 2012 #7

    tiny-tim

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    but that doesn't mean anything! :redface:
     
  9. Mar 15, 2012 #8
    Still struggling with the concept. So I can use a kronecker delta to have the condition of index l being j...

    [itex]\frac{∂}{∂x}(δ_{jl}x_{l}x_{l})_{j} =[/itex]

    ( [itex]δ_{jl}=1[/itex] if l=j else 0)

    [itex]\frac{∂}{∂x}x_{j}x_{j} =[/itex]

    [itex]2x_{j}[/itex]
     
  10. Mar 16, 2012 #9

    tiny-tim

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    hi lostminty! :smile:

    (just got up :zzz:)

    you need a push-start :wink:

    ∂/∂xj (xlxl)

    = (∂/∂xj xl) xl + xl (∂/∂xj xl)

    = 2(∂/∂xj xl) xl

    carry on from there :smile:
     
  11. Mar 16, 2012 #10
    That makes sense! So you do it that way because you can't have x2 instead you do product rule

    = 2(∂ xl/∂xj) xl

    = δjlxl

    if l=j

    = 2xj
     
  12. Mar 16, 2012 #11
    well I'll assume thats close to right and move onto the next problem which seems to make sense

    1. The problem statement, all variables and given/known data

    [itex]\nabla\cdot \vec{r}= [/itex]δii=3

    2. Relevant equations


    ∂xi/∂xjij


    3. The attempt at a solution

    [itex]\nabla\cdot \vec{r}=\nabla_{i}r_{i} [/itex]

    = ∂/∂xiri

    = ∂xi/∂xi = δii

    =[itex]\sum[/itex] between i=1 and 3 of δii
    =[itex]\sum[/itex] between i=1 and 3 of 1
    = 1+ 1 + 1
    =3
     
  13. Mar 17, 2012 #12

    tiny-tim

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    hi lostminty! :wink:
    yes :smile:
    that "if" line is unnecessary and wrong

    stop putting "ifs" into your proofs! (what is it with you and "if"? :rolleyes: do you keep saying "if" in ordinary speech?)

    the δ takes care of that!

    (and you missed a "2" :wink:)
    yes that's fine :smile:

    (though i'd be inclined to shorten the ending to just δii = 3 or δii = trace(δ) = 3)
     
  14. Mar 20, 2012 #13
    This is another problem I'm a bit stuck on with similar content

    1. The problem statement, all variables and given/known data

    F is a constant vector field. hence [itex]\nabla\cdot[/itex] F = 0

    this means there is a vector potential F = [itex]\nabla\times[/itex] A

    also [itex]\nabla\times[/itex] F = 0

    this means there is a scalar potential F = [itex]\nabla[/itex] ∅

    verify

    ∅ = F [itex]\cdot[/itex] r

    and

    A = 1/2F x r


    2. Relevant equations


    3. The attempt at a solution

    ∅ = F [itex]\cdot[/itex] r

    = Firi

    F = [itex]\nabla[/itex] ∅

    = [itex]\nabla[/itex] (Firi)

    = ∂/∂xk (Firi)

    = (∂Fi/∂xk)ri + (∂ri/∂xk)Fi


    since [itex]\nabla\cdot[/itex] F = 0

    (∂Fi/∂xk)ri = 0

    and

    ∂xi/∂xj = δij = 1 when j=i

    (∂ri/∂xk)Fi = δikFiei = Fiei = F
     
    Last edited: Mar 20, 2012
  15. Mar 21, 2012 #14
    F = [itex]\nabla\times[/itex] A = [itex]\nabla\times[/itex]1/2F x r

    1/2F x r = 1/2εijkFjrk

    F= [itex]\nabla\times[/itex] 1/2εijkFjrk

    = εimn∂/∂xi(1/2εijkFjrk)m

    =∂/∂ximjδnknjδmk)(1/2Fjrk)m

    =(1/2)∂/∂xi(Fmrn - Fnrm)m

    =?
     
  16. Mar 21, 2012 #15

    tiny-tim

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    hi lostminty!
    no, leave out "F = " … that's what you're trying to prove!

    no, you need to write either

    [itex](\nabla[/itex] (Firi))_k

    = ∂/∂xk (Firi)​

    or

    [itex]\nabla[/itex] (Firi)

    = ∂/∂xk (Firi) e_k​
    fine :smile:
    no, that doesn't follow at all, does it? :redface:

    ((∂Fi/∂xk) has nothing to do with [itex]\nabla\cdot[/itex] F)

    you need the stronger condition, that F is constant

    you're doing it again!! :rolleyes:

    using "when" is the same as using "if"!!

    just write "∂xi/∂xj = δij"
    where did those e's suddenly come from?

    if you were going to use them, they should have been there from the beginning :wink:

    (i'll look at the other one later)
     
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