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∇)()