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Vector calculus index notation

  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data

    prove grad(a.grad(r^-1))= -curl(a cross grad (r^-1))
    2. Relevant equations

    curl(a x b)= (b dot grad)a - (a dot grad)b +a(div b) - b(div a )

    3. The attempt at a solution
    Im trying to use index notation and get
    di (aj (grad(r^-1))j)
    =grad(r^-1) di(aj) +aj(di grad(r^-1))j

    which is obviously not right. Ive tried attacking the problem from the reverse direction and haven't had much luck there either.

    Thank you :)
     
  2. jcsd
  3. Feb 7, 2016 #2

    jedishrfu

    Staff: Mentor

    Can you provide more details? Is A a constant vector not dependent on r, theta, phi?

    Have you looked at the vector identities in spherical coordinates for curl, div and grad?

    Try to solve it without going to index notation.
     
  4. Feb 7, 2016 #3

    jedishrfu

    Staff: Mentor

    Try starting with ##\nabla( A \cdot B ) = (B \cdot \nabla ) A + (A \cdot \nabla) B + B \times (\nabla \times A) + A \times (\nabla \times B)##
     
  5. Feb 8, 2016 #4
    Hi thank you for replying!
    Sorry, a is a constant vector and r is (X,y,z)

    I have not come across this identity. Any hints on how to prove it?
     
  6. Feb 8, 2016 #5

    jedishrfu

    Staff: Mentor

    Here's a more complete list of identities:

    https://en.wikipedia.org/wiki/Vector_calculus_identities

    Off hand I don't know how to prove this particular identity but often when presented with a problem such as yours you can use the identities transform the left had side to the right hand side or vice versa.to "prove" it.
     
  7. Feb 8, 2016 #6

    PeroK

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    Time to learn some latex, I suggest!
     
  8. Feb 8, 2016 #7

    jedishrfu

    Staff: Mentor

    Some latex codes:

    http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html

    in particular look at vector operators \nabla for the del operator,\cdot for the inner-product and \times for the cross-product. This site uses mathjax for display. Mathjax looks for expressions bracketted by double # characters.
     
  9. Feb 8, 2016 #8
    I think proving the identity was the purpose of the question? I could be wrong but it's to do with index notation
     
  10. Feb 8, 2016 #9

    jedishrfu

    Staff: Mentor

    You problem has a specific function ##\nabla(1/r)## in it so its seemed to me that its not an identity but an example where you apply the identity to get the result you need to prove.

    Its true I could be wrong here. Perhaps @Mark44 or @PeroK could comment more on this.
     
  11. Feb 8, 2016 #10

    PeroK

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    I assumed it did depend on the second vector being ##\nabla(1/r)##. It must. I evaluated each side for the x-term (and then used symmetry). It's not too bad if you're careful with your differentiation.
     
  12. Feb 8, 2016 #11

    PeroK

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    I checked with the second vector being ##\nabla f## and it reduces to Laplace's equation ##\nabla ^2 f = 0##
     
  13. Feb 8, 2016 #12
    Its okay, Ive managed to do it with index notation. Thank you guys for helping though- I really appreciate it :)
     
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