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Confusion about properties of the del operator

  • Thread starter bcjochim07
  • Start date
1. 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.



2. Homework Equations



3. The Attempt at a Solution
 

lurflurf

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

Tom Mattson

Staff Emeritus
Science Advisor
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1. 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 [itex]\nabla U[/itex]. Then cross that vector with the vector [itex]\vec{a}[/itex]. In any case it would make no sense for [itex]\nabla[/itex] to act on [itex]\vec{a}[/itex]. 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 [itex]\vec{r}[/itex]. That's because [itex]\vec{A}\cdot\nabla[/itex] and [itex](\vec{A}\times\nabla )\cdot[/itex] 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

Homework Helper
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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
 
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 + ...
 
Last edited:

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
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I wouldn't call [itex]\nabla[/itex] a vector. When I say "vector" (in normal 3d space), I mean an element of [itex]\mathbb{R}^3[/itex], which [itex]\nabla[/itex] certainly is not. Rather, [itex]\nabla[/itex] is an operator on [itex]\mathbb{R}^3[/itex]. And so is [itex](\vec{A}\times\nabla )\cdot[/itex].
 

lurflurf

Homework Helper
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∇ 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
 
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

Homework Helper
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122
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∇)()
 

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