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    (Original post by Jammy4410)
    i just realised i did mean that 'a' was a scalar field and U was a constant vector sorry about that
    No problem, I was pretty sure you just misread it.

    (Original post by Jammy4410)
    problem 5 (based on problem 3, similar to problem 4):

    is there any similar relationship for the vector product i.e.

    is there a relationship between:

    Va x U and V x aU, and maybe even equivalent if a is a constant vector??
    I don't want to just give you the answer, but write it out in component form, and remember that the components of del have the form  \partial / \partial x , and proceed. It's just product rule and a bit of rearrangement.
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    (Original post by Rinsed)
    Not quite.

    You actually said a was a constant vector. Because a is a scalar, I assumed you meant U was a constant vector. Worth saying, that gives the right answer. Are you sure the question said a was constant?

    Plus, if a were a constant, the fact that the answer includes a differential of a would seem strange.
    ok i think i got it:

    V.(aU) = (Va).U + a(V.U)


    where
    V is del
    a is scalar field

    1) if U is a constant v.field the second term becomes zero
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    (Original post by Rinsed)
    No problem, I was pretty sure you just misread it.



    I don't want to just give you the answer, but write it out in component form, and remember that the components of del have the form  \partial / \partial x , and proceed. It's just product rule and a bit of rearrangement.
    ok i got

    V x aU = (Va) x U + a(V x U)

    where

    V is del
    a is s.field
    U is v.field

    i feel like this is wrong..

    method:

    =Eijk dj (aU)k
    =Eijk dj aUk
    =Eijk (dja)Uk + Eijk a(djUk)
    =Eijk (dja)Uk + Eijk a(djUk)
    = (Va) x U + a(V x U)
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    also, for problem 4:

    would there be anyway to get to the answer starting with

    (Va).U

    if there is a direct method, i think i still don't understand a little of what i can and cannot do
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    (Original post by Jammy4410)
    ok i got

    V x aU = (Va) x U + a(V x U)

    where

    V is del
    a is s.field
    U is v.field

    i feel like this is wrong..

    method:

    =Eijk dj (aU)k
    =Eijk dj aUk
    =Eijk (dja)Uk + Eijk a(djUk)
    =Eijk (dja)Uk + Eijk a(djUk)
    = (Va) x U + a(V x U)
    Nah I'm pretty sure that's right.

    But don't forget that if U is a constant its curl goes away.
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    (Original post by Jammy4410)
    also, for problem 4:

    would there be anyway to get to the answer starting with

    (Va).U

    if there is a direct method, i think i still don't understand a little of what i can and cannot do
    Well, yes, just apply the steps in reverse. In component form you have

     (\partial^i a) U_i

    Noting that U is a constant, it can be brought into the bracket. Then it's pretty obvious you can write it as  \nabla \cdot (a \textbf{U})

    Remember, the same rules apply here as for any old scalar maths (as long as you don't start messing around with indices) and del is just a differential operator, like d/dx.
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    (Original post by Rinsed)
    Well, yes, just apply the steps in reverse. In component form you have

     (\partial^i a) U_i

    Noting that U is a constant, it can be brought into the bracket. Then it's pretty obvious you can write it as  \nabla \cdot (a \textbf{U})

    Remember, the same rules apply here as for any old scalar maths (as long as you don't start messing around with indices) and del is just a differential operator, like d/dx.
    still a bit confused how to do it in reverse?


    problem 6:

    we know that

    a x (b x c) = b(a.c) -c(a.b)

    why is this not the case for

    V x (bxc)

    where a,b,c arbitrary vectors

    il check it in an hour, but i think i know the answer and how to prove it using index notation, just don't really get why it works

    on page 5 of:
    http://internal.physics.uwa.edu.au/~...g/CM/index.pdf
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