Confusion about properties of the del operator

[bcjochim07]

Homework Statement

I was reading some notes about the del operator, and they make the statement

∇ x (Ua) = U(∇ x a) + (∇U) x a.

However, I disagree with this because it seems to me that in the right hand side of the equation for the second term, the ∇ is operating on a, since a appears after it.

I also worked on some other examples to see if understood the properties of del. For instance, for an arbitrary vector function A and position vector r,

(A dot ∇)r = A and

(A x ∇) dot r = 0.

Both of these examples seem to confirm that ∇ operates on what appears after it (crossed, multiplied, or dotted, but not added or subtracted).

If I expand the right hand side of the above equation, I don't get the left hand side.

I think that the statement should read
∇ x (Ua) = U(∇ x a) + -a x (∇U)

Could somebody help clear my confusion? Thanks.

Homework Equations

The Attempt at a Solution

 

[lurflurf]

I would probably write it the way you do, but botth forms are correct.
recall
axb=-bxa

 

[quantumdude]

Homework Statement

I was reading some notes about the del operator, and they make the statement

∇ x (Ua) = U(∇ x a) + (∇U) x a.

However, I disagree with this because it seems to me that in the right hand side of the equation for the second term, the ∇ is operating on a, since a appears after it.

No, this goes back to the order of operations that you learned in arithmetic. Compute what's in the parentheses first, which is the vector $\nabla U$. Then cross that vector with the vector $\vec{a}$. In any case it would make no sense for $\nabla$ to act on $\vec{a}$. Gradients act on scalars, not vectors.

I also worked on some other examples to see if understood the properties of del. For instance, for an arbitrary vector function A and position vector r,

(A dot ∇)r = A and

(A x ∇) dot r = 0.

Both of these examples seem to confirm that ∇ operates on what appears after it (crossed, multiplied, or dotted, but not added or subtracted).

It does act on $\vec{r}$. That's because $\vec{A}\cdot\nabla$ and $(\vec{A}\times\nabla )\cdot$ are operators, not vectors.

If I expand the right hand side of the above equation, I don't get the left hand side.

I do get it to work out. So perhaps you should post what you've done so we can examine it.

 

[lanedance]

The del operator is a differential operator, compare the first equation you gave with the ordinary product rule with a(x,y,z) and U(x,y,z) as U is scalar it can be taken to either side of the cross product without affecting the product.

So the equations you have written at the top & bottom are actually equivalent as you have changed the order of the cross product

have a look at this
http://en.wikipedia.org/wiki/Del

 

[bcjochim07]

Tom,

Thinking back to (A x ∇) dot r = 0.
(A x ∇) is a vector isn't it?

and ∇ is a vector and an operator.

Could you please clarify what you mean? Thanks.

When I expand (∇U) x a, I am doing it like this
(U daz/dy + az dU/dz - U day/dz - ay dU/dz)i + ...

 

[quantumdude]

I wouldn't call $\nabla$ a vector. When I say "vector" (in normal 3d space), I mean an element of $\mathbb{R}^3$, which $\nabla$ certainly is not. Rather, $\nabla$ is an operator on $\mathbb{R}^3$. And so is $(\vec{A}\times\nabla )\cdot$.

 

[lurflurf]

∇ is a vector operator
it is not a vector
treating it as if it is a vector will lead to much confusion
for example b.(axb)=0
b.(∇xb) need not be 0

 

[bcjochim07]

I think I'm starting to understand what you are saying. ∇U is the operation on a scalar function which yields a vector function (gradient), so that really can't operate on anything. However, when you dot or cross ∇ with another vector, that does operate on the terms after it. Right?

 

[lurflurf]

It is true that the usual ∇ is a right acting operator. Another symbor should be used otherwise. It is very dangerous (as in wrong) to think of say crossing ∇ with a vector, it is much better to think of
∇()
∇.()
∇x()
as three different operators
then there are things like
a.∇()
(ax∇).()
(ax∇)x()
(ax∇)()