differential operator identity

Discussion in 'Math' started by Starhowl, Dec 7, 2013.

  1. Starhowl

    Thread Starter New Member

    Dec 3, 2013
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    Can someone please explain to me why this identity is valid? Especially how the author gets to this:
    \nabla_q \times \frac{M_c}{|r_{qa}|} = -M \times \nabla_q \frac{1}{|r_{qa}|}.

    I assume that c stands for 'constant'..?
     
    Last edited: Dec 7, 2013
  2. WBahn

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    Mar 31, 2012
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    How can it possibly be valid if there is an Mc on one side and an M on the other?

    Unless you have an already-defined relationship between the two.

    Are you sure we have all the information?
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Is this about magnetization?

    If so you need to specify the units of M since engineers and physicists use different units and this affects formulae.

    Is your author calculating total magnetization from induced plus permanent, hence the M_sub_c, or is the M_sub_c unit M in a particular direction and M the total over a whole shell ?

    As WBahn says we need more information.
     
  4. Starhowl

    Thread Starter New Member

    Dec 3, 2013
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    I'm sure I provided all the information; as far as I do understand this is a general valid relation where the vector could be also of the name 'u' or 'v'. Furthermore, as far as I can see, the index 'c' indicates 'constant', or, respectively 'held constant'.
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I'm sure you will find better responses if you cooperate with others.

    WBahn, in particular is a very skilled teacher in these subjects, so if he asks for more detail, you can be quite sure yours is sparse.

    So put your question in context and tell us what you are studying,

    What is the book ( a page reference could be helpful since others may know it),

    What area of your subject is the book developing, it certainly doesn't quote formulae in complete isolation.
     
  6. WBahn

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    Mar 31, 2012
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    But think about this. If M could be u and Mc could be v, than can't I change u one one side without changing v on the other? That means that the each side must fundamentally be equal to the same constant value independent of the values of M or Mc. Does that make sense?
     
  7. Starhowl

    Thread Starter New Member

    Dec 3, 2013
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    No, I never meant to be M like u and Mc like v.
     
  8. WBahn

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    Mar 31, 2012
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    Well, then what did you mean?

    The same argument still applies. If there is no fixed relationship between M and Mc, the what prevents one from changing and the other from staying the same? If that happens, how can the two sides remain equal?
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    The OP deleted a post, the reason given was "problem solved". It would have been nice if he had said it out loud.
     
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