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Proof of product rule for gradients

  1. Dec 23, 2015 #1
    product rule.png
    Can someone please help me prove this product rule? I'm not accustomed to seeing the del operator used on a dot product. My understanding tells me that a dot product produces a scalar and I'm tempted to evaluate the left hand side as scalar 0 but the rule says it yields a vector. I'm very confused
     
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  3. Dec 23, 2015 #2

    blue_leaf77

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    Del operator ##\nabla## is a vector operator, following the rule for well-defined operations involving a vector and a scalar, a del operator can be multiplied by a scalar using the usual product. ##\mathbf{A}\cdot \mathbf{B}## is a scalar, but a vector (operator) ##\nabla## comes in from the left, therefore the "product" ##\nabla (\mathbf{A}\cdot \mathbf{B})## will yield a vector.
     
  4. Dec 23, 2015 #3
    So it becomes a vector because it doesn't evaluate to 0 since the del operator is used on a scalar function rather than a scalar constant?
     
  5. Dec 23, 2015 #4

    blue_leaf77

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    Well that's a matter of whether ##\mathbf{A}\cdot \mathbf{B}## is coordinate dependent or not. If this dot product turns out to be constant, then ##\nabla(\mathbf{A}\cdot \mathbf{B})=0##.
     
  6. Dec 23, 2015 #5
    A and B are meant to be interpreted as vector fields (physics texts frequently conflate vector fields with vectors, using the word "vector" to refer to both vectors and vector fields, where the exact meaning is meant to be gleaned from context): functions that assign a unique vector to each point. The dot product of two vector fields is therefore a scalar field, as it is meant to be interpreted as the function that assigns the dot product of the two vectors assigned by A and B, respectively, at each point to each point. The gradient of a scalar field is a vector field (or covector field, depending on how formal you want to get).
     
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