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Proving properties of curl

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The curl satisfies

    (A) curl(f+g) = curl(f) + curl(g)

    (B) if h is real values, then curl(hf) = hcurl(f) + h'·f

    (C) if f is C2, then curl(gradf) = 0

    Show that (B) holds.

    2. The attempt at a solution
    I'm not quite sure how to interpret the "h is real valued" part. Does it mean that, as is, h is a scalar quantity and its derivative is a vector?

    Any help is appreciated.

    - - - -
    So far I've been reviewing the definition of curl.
    I let f = f(x, y, z) = (f1(x, y, z), f2(x, y, z), f3(x, y, z))

    hf = (hf1(x, y, z), hf2(x, y, z), hf3(x, y, z))

    curl(hf) = (D2(hf3) - D3(hf2), D3(hf1) - D1(hf3), D1(hf2) - D2(hf1)),

    but I don't see where this route is taking me.
     
    Last edited: Jan 25, 2012
  2. jcsd
  3. Jan 25, 2012 #2
    Probably means h is a real valued function. Something like
    [itex]h(x_1,x_2...)\vec{f}(x_1,x_2...)[/itex]
    so the values of h are scalars, but it is not necessarily constant.
     
  4. Jan 25, 2012 #3

    lanedance

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    yeah I would probably take it as real scalar function of position h = h(x,y,z), multiplying anything else by a vector does not make sense unless it is a vector product
     
  5. Jan 25, 2012 #4

    lanedance

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    also the fact that h' is in a dot product means it is probably the gradient of h
     
  6. Jan 25, 2012 #5
    I looked at the problems ahead, and there's a similar one involving divergence, where I have to show that

    div(hf) = hdiv(f) + gradh·f,

    so I'm not sure if h' is the gradient in the curl question.
     
  7. Jan 25, 2012 #6

    Dick

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    In my reference the h'.f term is the cross product of grad(h) with f.
     
  8. Jan 25, 2012 #7
    I ended up continuing with process and was able to simplify the expression to

    curl(hf) = h(D2f3 - D3f2, D3f1 - D1f3, D1f2 - D2f1) + ((D2h)f3 - D3h)f2, (D3h)f1 - D1h)f3, (D1h)f2 - D2h)f1)

    curl(hf) = h curl(f) + ((D2h)f3 - (D3h)f2, (D3h)f1 - (D1h)f3, (D1h)f2 - (D2h)f1)

    Is there any way to simplify the rest of this expression?
     
  9. Jan 25, 2012 #8

    Dick

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    The rest of it looks like grad(h) x f. Where 'x' is the vector cross product.
     
  10. Jan 25, 2012 #9
    I suppose there's no way of expressing "grad(h) x f" as "h' f", is there?
     
  11. Jan 25, 2012 #10

    Dick

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    I have no idea what "h' f" is supposed to mean. It's no standard notation I've ever seen. I wouldn't torture yourself over it. You've got the vector formula right.
     
  12. Jan 25, 2012 #11
    Good enough, I suppose. Thanks everyone for the help.
     
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