How do functions and vector fields interact with normal derivatives?

[kingwinner]

Suppose that normal derivative = \nablag . n = dg/dn,
then f \nablag . n = f dg/dn
[I used . for dot product]

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

 

[Hurkyl]

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 (f \nabla g) \cdot \vec{n} and f (\nabla g \cdot \vec{n}) at a generic point x?

 

[kingwinner]

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 (f \nabla g) \cdot \vec{n} and f (\nabla g \cdot \vec{n}) 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 \nablag) = \nabla . (f \nablag) = f (\nabla . (\nablag)) which is WRONG becuase we know that in general there is an identity div(fG)=f div G + (\nablaf) . 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)

 

[Hurkyl]

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...

(\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0)

Another example, for comparison:

f'(a) \neq f(a)'

 

[kingwinner]

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...

(\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0)

Another example, for comparison:

f'(a) \neq f(a)'

Sorry, I don't quite get your point.
Can you please inform me where (which equal sign) exactly I am wrong?
div (f \nablag) = \nabla . (f \nablag) = f (\nabla . (\nablag))

Also, for your " f ", is it a scalar or a vector field?

 

[dhris]

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.

 

[nicksauce]

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.

 

[Hurkyl]

Sorry, I don't quite get your point.
Can you please inform me where (which equal sign) exactly I am wrong?
div (f \nablag) = \nabla . (f \nablag) = f (\nabla . (\nablag))

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.

\nabla . (f \nablag) = f (\nabla . (\nablag))

This is wrong -- you cannot simply pass f through the derivative. You need to use the chain rule.

 

[kingwinner]

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 (\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0), 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 \nabla is a vector, f is a scalar, \nablag is a vector, so I just applied the above property, getting \nabla . (f \nablag) = f (\nabla . \nablag)

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!

 

[nicksauce]

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

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

(f\nabla g) \cdot (\vec{n}) = f(\nabla g \cdot \vec{n})
because it is operating just on g.

 

[Hurkyl]

You said that (\nabla \cdot f)(a) \neq \nabla \cdot f(a) \quad (= 0), 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.

 

[kingwinner]

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

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

(f\nabla g) \cdot (\vec{n}) = f(\nabla g \cdot \vec{n})
because it is operating just on g.

So for the first case, we need to use the product rule, right?

 

[nicksauce]

Yes.

 

[kingwinner]

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

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

(f\nabla g) \cdot (\vec{n}) = f(\nabla g \cdot \vec{n})
because it is operating just on g.

Sorry, but something is still not perfectly clear to me...
Why is \nabla 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...

 

[Hurkyl]

Why is \nabla operating on BOTH f and g for the first case?

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

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)

g = u \cdot v

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

g = \nabla \cdot u

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

 

[kingwinner]

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



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)

g = u \cdot v

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

g = \nabla \cdot u

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... :(

 

[kingwinner]

One more related question:

(f . \nabla) g

f . (\nablag)

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...