Does the gradient operator apply only to g in a product F(x'y'z')g(x,y,z)?

[yungman]

If $\vec F(x'y'z')$ is function of $(x'y'z')$. $\nabla$ is operator on $(x,y,z)$.

So:
$$\nabla\left[\vec F(x'y'z') g(x,y,z)\right]=(\vec F(x'y'z') \nabla g(x,y,z)$$
or
$$\nabla(\vec F g)=\vec F \nabla g$$

Am I correct?

 

[ehild]

Is g(x,y,z) a scalar function?

ehild

 

[vanhees71]

Such tasks are most easily solved using the (Euclidean) Ricci calculus. Your question would read
$$\vec{\nabla} \vec{F} g(\vec{x})=\partial_i(F_i g)=F_i \partial_i g=\vec{F} \cdot \vec{\nabla} g,$$
where $\vec{F}=\text{const}$, i.e., independent of $\vec{x}$.

 

[HallsofIvy]

If $\vec F(x'y'z')$ is function of $(x'y'z')$. $\nabla$ is operator on $(x,y,z)$.

So:
$$\nabla\left[\vec F(x'y'z') g(x,y,z)\right]=(\vec F(x'y'z') \nabla g(x,y,z)$$
or
$$\nabla(\vec F g)=\vec F \nabla g$$

Am I correct?

Yes, if x', y', z' and x, y, z are independent variables, that is correct.

 

[yungman]

Is g(x,y,z) a scalar function?

ehild

Yes.

 

[yungman]

Thanks everyone, I just need to confirm this. I am not a math major, I just encountered all sort of math issues when I am studying antenna theory. This and Electromagnetics really put vector calculus through the ringer! I am sure I'll be posting many more of these kind of stupid questions.

Thanks

 

[Ray Vickson]

If $\vec F(x'y'z')$ is function of $(x'y'z')$. $\nabla$ is operator on $(x,y,z)$.

So:
$$\nabla\left[\vec F(x'y'z') g(x,y,z)\right]=(\vec F(x'y'z') \nabla g(x,y,z)$$
or
$$\nabla(\vec F g)=\vec F \nabla g$$

Am I correct?

You would only be correct if you define what you mean by the product $\vec{A}\vec{B}$ for vectors $A = \vec{F}(x',y',z')$ and $\vec{B} = \vec{\nabla}g(x,y,z)$. Presumably, you mean the outer product, which gives a 3x3 matrix with $\vec{A}\vec{B}_{i,j} = a_i b_j.$ (Of course, $\vec{F}$ could be a vector or other than 3 dimensions, so the matrix could be non-square.)

 

[yungman]

You would only be correct if you define what you mean by the product $\vec{A}\vec{B}$ for vectors $A = \vec{F}(x',y',z')$ and $\vec{B} = \vec{\nabla}g(x,y,z)$. Presumably, you mean the outer product, which gives a 3x3 matrix with $\vec{A}\vec{B}_{i,j} = a_i b_j.$ (Of course, $\vec{F}$ could be a vector or other than 3 dimensions, so the matrix could be non-square.)

g is only a scalar function of (x,y,z).

 

[lurflurf]

^Yes but F and ∇g are vectors so F∇g is the dyad product of F and ∇g.

 

[yungman]

^Yes but F and ∇g are vectors so F∇g is the dyad product of F and ∇g.

But my original question is
$$\nabla(\vec F g)=\vec F \nabla g$$

It's only after the gradient that it become a vector $\nabla g$, not before.

That actually raise a funny question. If the result is $\vec F \nabla g$, what is this? A vector multiplying a vector?

 

[Dick]

But my original question is
$$\nabla(\vec F g)=\vec F \nabla g$$

It's only after the gradient that it become a vector $\nabla g$, not before.

That actually raise a funny question. If the result is $\vec F \nabla g$, what is this? A vector multiplying a vector?

It's a tensor product. It's an object with two indices.

 

[Ray Vickson]

But my original question is
$$\nabla(\vec F g)=\vec F \nabla g$$

It's only after the gradient that it become a vector $\nabla g$, not before.

That actually raise a funny question. If the result is $\vec F \nabla g$, what is this? A vector multiplying a vector?

Yes, we all know that g is a scalar function.

However, the question is whether or not YOU realize the issues. These are: (1) what do you mean by asking for the gradient of a vector function---that is, what do you mean by $\vec{\nabla}\vec{G}?$; and (2) what do you mean by the product of two vectors that you wrote, namely, $\vec{A} \vec{B},$ where $\vec{A} = \vec{F}(x',y',z')$ and $\vec{B} = \vec{\nabla} g(x,y,z)?$