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Outer differentiation?

  1. Feb 19, 2004 #1
    "outer" differentiation?

    Since I didn't get a response on the "Diff. Eqns." thread, I put it here.

    Can someone give me a clear explaination of what is meant by "outer" differentiation.

    As for example, when we differentiate the vector potential, dA/dx, in order to arrive at the magnetic field, B, why would this be called an 'outer' differentiation??


    [?]
     
  2. jcsd
  3. Feb 19, 2004 #2

    turin

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    Re: "outer" differentiation?

    Probably because d/dx is not exactly what you do. You take the cross product of the gradient operator with the vector potential to get the magnetic field. The cross product is sometimes called/related to the outer product, but that's tensor jargon with which I am unfamiliar.
     
  4. Feb 20, 2004 #3
    Re: Re: "outer" differentiation?

    Thanks for the response, Turin;

    and thanks for the correction. Oc course, you are accurate in saying it is the cross product of the gradient.
    I guess then if someone says "dA/dt is the outer differentiation" I should interpret that to mean they are simply referring to it as being a cross product.??

    :smile:
     
  5. Feb 20, 2004 #4

    turin

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    Re: Re: Re: "outer" differentiation?

    I would ask them what they mean. I wouldn't know how to make sense of that.
     
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