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Functions & vector fields

  1. Mar 15, 2008 #1
    Suppose that normal derivative = [itex]\nabla[/itex]g . n = dg/dn,
    then f [itex]\nabla[/itex]g . n = f dg/dn
    [I used . for dot product]

    But how is this possible?
    For f [itex]\nabla[/itex]g . n, I would interpret it as (f [itex]\nabla[/itex]g) . n
    But f dg/dn = f ([itex]\nabla[/itex]g . n) which is DIFFERENT (note the location of brackets)

    I recall that if u,v are vectors, a is a scalar constant, then (au) . v=a(u . v) = u . (av), but in our case, f is a FUNCTION!!! So this rule cannot be applied.

    f [itex]\nabla[/itex]g . n = f dg/dn seems to be suggesting that f is a scalar constant, but consider the following case:
    Let div F=[itex]\nabla[/itex] . F
    If f were really a scalar constant, then by the rule, div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g)) which is WRONG becuase we know that in general div(fG)=f div G + ([itex]\nabla[/itex]f) . G where f is function and G is vector field. But if this is wrong, then HOW can you justify that (f [itex]\nabla[/itex]g) . n = f ([itex]\nabla[/itex]g . n)?

    I seriously can't understand this...there are two seemingly contradicting ideas crashing in my mind...


    Would someone be nice enough to clear my doubts? Thanks!
     
    Last edited: Mar 15, 2008
  2. jcsd
  3. Mar 15, 2008 #2

    Hurkyl

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    Remember that two scalar fields are equal if and only if they have the same value at every point! What happens if you evaluate both [itex](f \nabla g) \cdot \vec{n}[/itex] and [itex]f (\nabla g \cdot \vec{n})[/itex] at a generic point x?
     
  4. Mar 15, 2008 #3
    At every fixed point, f would be a scalar constant, so "the rule" can be applied, right?

    But if we can treat f as a scalar constant, then by "the rule", div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g)) which is WRONG becuase we know that in general there is an identity div(fG)=f div G + ([itex]\nabla[/itex]f) . G where f is function and G is vector field. Using this identity gives a different answer.

    Note: "the rule": if u,v are vectors, a is a scalar constant, then (au) . v=a(u . v) = u . (av)
     
    Last edited: Mar 15, 2008
  5. Mar 15, 2008 #4

    Hurkyl

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    You made an error when evaluating the derivative, though. The value of the derivative of a function at a point is (usually) not the derivative of the value of the function at that point.

    In particular...

    [tex](\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0)[/tex]

    Another example, for comparison:

    [tex]f'(a) \neq f(a)'[/tex]
     
    Last edited: Mar 15, 2008
  6. Mar 15, 2008 #5
    Sorry, I don't quite get your point.
    Can you please inform me where (which equal sign) exactly I am wrong?
    div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g))

    Also, for your " f ", is it a scalar or a vector field?
     
  7. Mar 16, 2008 #6
    What's the problem? All you have done is multiply the normal derivative by f. There's no differential operation on f at all so you do not need to be concerned that it is a function rather than a scalar.
     
  8. Mar 16, 2008 #7

    nicksauce

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    Indeed. I think I am more stumped about why you are stumped than you are stumped about the problem at hand.
     
  9. Mar 16, 2008 #8

    Hurkyl

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    I was using f for scalar field, as you were. Note that these expressions would be nonsense if f were a vector field.

    This is wrong -- you cannot simply pass f through the derivative. You need to use the chain rule.
     
  10. Mar 16, 2008 #9
    You said that [tex](\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0)[/tex], but how can a vector be dotted with a scalar?



    But if u,v are vectors, a is a scalar, then (au) . v=a(u . v) = u . (av)

    Now [itex]\nabla[/itex] is a vector, f is a scalar, [itex]\nabla[/itex]g is a vector, so I just applied the above property, getting [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . [itex]\nabla[/itex]g)

    How come it works for the first case (normal derivative case), but not this one? This is the part I don't understand...

    Thanks for explaining!
     
    Last edited: Mar 16, 2008
  11. Mar 16, 2008 #10

    nicksauce

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    [tex]\nabla[/tex] not only a vector, but an operator as well. You can't just blindly apply vector rules, but you must apply derivative rules well.

    [tex] \nabla \cdot (f\nabla g) \neq f (\nabla \cdot \nabla g)[/tex]
    because it is operating on both f and g.

    [tex](f\nabla g) \cdot (\vec{n}) = f(\nabla g \cdot \vec{n})[/tex]
    because it is operating just on g.
     
    Last edited: Mar 16, 2008
  12. Mar 16, 2008 #11

    Hurkyl

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    You're right; I forgot about that example (and the example with f'). Those examples were a vector field and a function of the reals, respectively.
     
  13. Mar 16, 2008 #12
    So for the first case, we need to use the product rule, right?
     
  14. Mar 16, 2008 #13

    nicksauce

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  15. Mar 16, 2008 #14
    Sorry, but something is still not perfectly clear to me...
    Why is [tex]\nabla[/tex] operating on BOTH f and g for the first case?
    If we could treat f as a scalar, we should be able to pull it out like a constant...
     
  16. Mar 16, 2008 #15

    Hurkyl

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    The operator [itex]\nabla \cdot[/itex] is not operating on either of them. It's operating on the vector field [itex]f \nabla g[/itex].


    f is a scalar (field), but it's not a constant scalar (field). If you want to treat scalar fields like numbers, you have to distinguish between the constant scalars and the nonconstant scalars.




    The point I made earlier was about evaluating the expression at a point. If I have the scalar field g given by (u and v are vector fields)

    [tex]g = u \cdot v[/tex]

    then what is the value of g at a point a? i.e. what is g(a)?


    Now, if I instead have the scalar field

    [tex]g = \nabla \cdot u[/tex]

    then what is the value of g at a point a? i.e. what is g(a)?
     
    Last edited: Mar 16, 2008
  17. Mar 17, 2008 #16
    For the first case, g(a)=(u.v)(a), just dot u with v to get an expression and substitute in the value a?

    For the second case, g(a)=(del.u)(a) just do the dot product FIRST and then sub. in the value a? I can't see any difference... :(
     
  18. Mar 18, 2008 #17
    One more related question:

    (f . [itex]\nabla[/itex]) g

    f . ([itex]\nabla[/itex]g)

    Are these equivalent? If so, then why write in the fist form? My textbook keeps writing in the first form and I don't know why...
     
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