Functions & vector fields

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Main Question or Discussion Point

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!
 
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Answers and Replies

  • #2
Hurkyl
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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.
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?
 
  • #3
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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?
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)
 
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  • #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]
 
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  • #5
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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]
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?
 
  • #6
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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.
 
  • #7
nicksauce
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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.
Indeed. I think I am more stumped about why you are stumped than you are stumped about the problem at hand.
 
  • #8
Hurkyl
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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?
I was using f for scalar field, as you were. Note that these expressions would be nonsense if f were a vector field.

[itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g))
This is wrong -- you cannot simply pass f through the derivative. You need to use the chain rule.
 
  • #9
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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.
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?



This is wrong -- you cannot simply pass f through the derivative. You need to use the chain rule.
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!
 
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  • #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.
 
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  • #11
Hurkyl
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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?
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.
 
  • #12
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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.
So for the first case, we need to use the product rule, right?
 
  • #13
nicksauce
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Yes.
 
  • #14
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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.
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...
 
  • #15
Hurkyl
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Why is [tex]\nabla[/tex] operating on BOTH f and g for the first case?
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].


If we could treat f as a scalar, we should be able to pull it out like a constant...
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)?
 
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  • #16
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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)?
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... :(
 
  • #17
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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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