# Homework Help: Is there any difference between these 2 terms?

1. Jun 22, 2015

### null void

1. The problem statement, all variables and given/known data
Is there any difference between these 2 terms, if yes how are they different?
\begin{align}(\nabla\cdot\nabla\vec A)\vec B &= (\nabla\vec A\cdot\nabla)\vec B\end{align}

2. Relevant equations
From what i know about dot product, it is commutative, so does this property apply here?

2. Jun 22, 2015

### RUber

In the left equation, both operators apply to A, transferring one onto B requires that some additional conditions are met. Think something like integrals by parts.
Check out Green's theorem for a very common application. https://en.wikipedia.org/wiki/Green's_identities.
Commutative dot products work for matrices,
$A \cdot B = B \cdot A$ but not so clearly for operators.
In one dimension, you have something that says $x''y = x'y'$. There are lots of functions x and y for which this is not true, and only certain functions will satisfy this.

3. Jun 22, 2015

### Incand

There's nothing wrong writing it that way in your case. The thing is you can't threat $\nabla$ as a normal vector. A lot of those properties are indeed true but you have to use the definition of grad, div and curl to prove it (and in some cases you will get the wrong thing from threating it as a vector i.e. it's a good memory rule but you can't prove anything with it).
So in your case you have to use $\nabla A = \left( \frac{\partial A}{\partial x} , \frac{\partial A}{\partial y},\frac{\partial A}{\partial z}\right)$ and you get the same result as if you use $\nabla = \left( \frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\right)$

4. Jun 22, 2015

### null void

so for the right side, it is something like this:
\begin{align}(\nabla\vec A\cdot\nabla)\vec B &= (\frac{dA_x}{dx}\frac{d}{dx}+\frac{dA_y}{dy}\frac{d}{dy}+\frac{dA_z}{dz}\frac{d}{dz})\vec B\\ &= <(\frac{dA_x}{dx}\frac{d}{dx}+\frac{dA_y}{dy}\frac{d}{dy}+\frac{dA_z}{dz}\frac{d}{dz}) B_x,\\(\frac{dA_x}{dx}\frac{d}{dx}+\frac{dA_y}{dy}\frac{d}{dy}+\frac{dA_z}{dz}\frac{d}{dz}) B_y,\\(\frac{dA_x}{dx}\frac{d}{dx}+\frac{dA_y}{dy}\frac{d}{dy}+\frac{dA_z}{dz}\frac{d}{dz}) B_z,> \end{align}
or it is
\begin{align}(\nabla\vec A\cdot\nabla)\vec B &= <\frac{dA_x}{dx}\frac{dB_x}{dx},\frac{dA_y}{dy}\frac{dB_y}{dy},\frac{dA_z}{dz}\frac{dB_z}{dz}> \end{align}
by the way, any idea what kind of operator can be used to align the lines without "=" character?

Last edited: Jun 22, 2015
5. Jun 22, 2015

### Incand

I'm afraid I may be misleading you here, I suspect RUber is more knowledgeable than I am since all I studied myself is vector calculus.
Check out the last part of the article here https://en.wikipedia.org/?title=Del to see when you can treat it as a vector and when you can't. From what I can read there the identity is actually false.
But generally when you prove an identity like this you have to do exactly what you just did (which is correct btw).

6. Jun 22, 2015

### null void

from what i see in the wiki page on "Del", i think the right operation is this one

7. Jun 22, 2015

### RUber

On the reference that Incand posted above, it clearly showed that you can't just swap out which vector the del is applied to using standard rules for commutativity.
"A counterexample that relies on del's failure to commute:

"
In your original post you have $(\Delta A)B =$something else. What is your goal with this problem? Do you clearly need to spell out the right hand side, or is it enough to show that the differential $\nabla$ doesn't commute like a vector?

8. Jun 22, 2015

### RUber

Back to the 1D example, you have a relationship like
$\frac{\partial ^2 A}{\partial x^2} B = \frac{\partial A}{\partial x} \frac{\partial A}{\partial B}$
This works fine for a counterexample in 3D too, just let $A = (A_x(x),0,0), B = (B_x(x),0,0)$
If they are the same, then their integrals would be the same, right?
$\int_\mathbb{R} \frac{\partial ^2 A}{\partial x^2} B dx= \int_\mathbb{R} \frac{\partial A}{\partial x} \frac{\partial B}{\partial x} dx$
Integrating by parts:
$\left. \frac{\partial A}{\partial x} B\right|_{-\infty}^\infty - \int_\mathbb{R} \frac{\partial A}{\partial x} \frac{\partial B}{\partial x} dx = \int_\mathbb{R} \frac{\partial A}{\partial x} \frac{\partial B}{\partial x} dx$
Even if we assume that these vectors are zero at infinity, you still have
$- \int_\mathbb{R} \frac{\partial A}{\partial x} \frac{\partial B}{\partial x} dx = \int_\mathbb{R} \frac{\partial A}{\partial x} \frac{\partial B}{\partial x} dx$
Which looks like -a = a, implying that the whole integral must be zero for this to hold true...
In short, No. They are not the same.

9. Jun 23, 2015

### null void

Sorry, I just notice there are some naming convention problem in my equations, A is a scalar, bacause $\nabla$ can't multiply with a scalar
$(\nabla A \cdot\nabla)\vec B$

Now I know that the del is not commutative, but I want to conform if this is the right way to evaluate that expression, i have typed it out before, but there is some mistake in the equation.
\begin{align} \text{let A be a function of x,y,z} \\ \text{let} \vec B &= <{B_1, B_2, B_3}>, \text{elements in vectors B are also a function of x,y,z} \\ (\nabla A \cdot\nabla)\vec B &= (\frac{\partial A}{\partial x}\frac{\partial }{\partial x}+\frac{\partial A}{\partial y}\frac{\partial }{\partial y}+\frac{\partial A}{\partial z}\frac{\partial }{\partial z})\vec B \\ &= <{(\frac{\partial A}{\partial x}\frac{\partial B_1}{\partial x}+\frac{\partial A}{\partial y}\frac{\partial B_1}{\partial y}+\frac{\partial A}{\partial z}\frac{\partial B_1}{\partial z})},\\ {(\frac{\partial A}{\partial x}\frac{\partial B_2}{\partial x}+\frac{\partial A}{\partial y}\frac{\partial B_2}{\partial y}+\frac{\partial A}{\partial z}\frac{\partial B_2}{\partial z})}, \\{(\frac{\partial A}{\partial x}\frac{\partial B_3}{\partial x}+\frac{\partial A}{\partial y}\frac{\partial B_3}{\partial y}+\frac{\partial A}{\partial z}\frac{\partial B_3}{\partial z})}>\end{align}

Last edited: Jun 23, 2015
10. Jun 23, 2015

### RUber

That seems right.

11. Jun 23, 2015

### null void

I think I get the things I want here, thank a bunch guys.