Vector Operators: Grad, Div and Curl

[MatinSAR]

Homework Statement

Identify true and false.

Relevant Equations

Vector calculus.

Can someone tell me if i am wrong ?

 

[kuruman]

Numbers 2, 3 & 4 are ambiguous without parentheses. Also, please state why you think each one is True or False. You might be right for the wrong reason.

 

[MatinSAR]

Numbers 2, 3 & 4 are ambiguous without parentheses. Also, please state why you think each one is True or False. You might be right for the wrong reason.

True according to BAC-CAB

True because in scalar triple product we can change "dot" and "cross"

False. If nabla was only a vector it could be right. But here it should take the derivate of both A and B vectors.

This is also wrong with same reason. If nabla was only a vector it could be right. But here it should take the derivate of both A and B vectors.Sorry that I can't use latex right now.

 

[haruspex]

This is also wrong with same reason. If nabla was only a vector it could be right. But here it should take the derivate of both A and B vectors.

Should it? I don’t see why.
But is there a more obvious reason it is wrong?

 

[kuruman]

True because in scalar triple product we can change "dot" and "cross"

Are you sure? Note that on the left-hand side $\nabla$ operates on both $\vec A$ and $\vec B.$ On the right-hand side it does not.

 

[Orodruin]

True according to BAC-CAB

False. On the LHS the derivative acts on both A and B and in the RHS you suddenly skip the derivative of one field in each term.

Generally you seem to be treating $\nabla$ as if it were a vector. It is not. It is a differential operator.

 

[MatinSAR]

Should it? I don’t see why.
But is there a more obvious reason it is wrong?

Sorry but I cannot find out which one you are talking about. 3rd or 4th? Or both of them?

Are you sure? Note that on the left-hand side $\nabla$ operates on both $\vec A$ and $\vec B.$ On the right-hand side it does not.

Agree. I did not notice it.

False. On the LHS the derivative acts on both A and B and in the RHS you suddenly skip the derivative of one field in each term.

Generally you seem to be treating $\nabla$ as if it were a vector. It is not. It is a differential operator.

Yes. So I can't use triple dot product or triple cross product rules when I have a differentiel operator. Am I right?

Thanks everyone for their help and time.

 

[MatinSAR]

Should it? I don’t see why.
But is there a more obvious reason it is wrong?

3rd: In LHS differential operator acts on both vector fields. In RHS it doesn't act on any field.

 

[haruspex]

3rd: In LHS differential operator acts on both vector fields. In RHS it doesn't act on any field.

As I thought was clear from the text I quoted in post #4, it was re the fourth case.
A good starting point would be to consider the difference between $\frac{d}{dx}(f(x)g(x))$ and $(\frac{d}{dx}f(x))g(x)$.

 

[MatinSAR]

As I thought was clear from the text I quoted in post #4, it was re the fourth case.
A good starting point would be to consider the difference between $\frac{d}{dx}(f(x)g(x))$ and $(\frac{d}{dx}f(x))g(x)$.

I see. In first one d/dx acts on both f and g. In second one d/dx acts only on f. So they are not the same.

For using triple dot product or triple cross product rules when we have a differential operator, Should we always check it in this way?!

 

[Orodruin]

Yes. So I can't use triple dot product or triple cross product rules when I have a differentiel operator. Am I right?

You can, to some extent, but you must make sure that you respect what the differential operator acts on. There is a way of doing this formally using vector identities and the Leibniz rule. Using a notation such that it is clear what the differential operator acts on, you can do any manipulations of the vector structure as long as you end up with a proper expression in the end. For example, take $\nabla\cdot(\vec A \times \vec B)$. Using underline on a vector to denote that it is what the nabla should act on, this can be written
$$
\nabla\cdot(\vec A \times \vec B)
=
\nabla\cdot(\underline{\vec A} \times \vec B)
+
\nabla\cdot({\vec A} \times \underline{\vec B})
$$
where we have applied the Leibniz rule. You can now perform any manipuations you wish on the vector structure with the aim of having the vector the operator acts on as the only thing to the right of it. In particular, here we would do
$$
\nabla\cdot(\underline{\vec A} \times \vec B)
=
\vec B \cdot (\nabla \times \underline{\vec A})
$$
for the first term by cyclic permutation of the triple product. Once on this form the underline can be removed as it is clear what the differential operator acts on.

 

[haruspex]

I see. In first one d/dx acts on both f and g. In second one d/dx acts only on f. So they are not the same.

For using triple dot product or triple cross product rules when we have a differential operator, Should we always check it in this way?!

I'm not offering it as the key to answering all cases, just pointing out that in case 4 of post #3 (just noticed you changed the order from post #1, which may lead to confusion) your reasoning was incorrect. The differentiation only applies to A on both sides.
Conversely, in post #3 case 2, it applies to A and B on the left but only to B.

Note that in case 3 the RHS is still an operator.

 

[MatinSAR]

@kuruman @haruspex @Orodruin I think I find out my mistakes. Thanks for your time.

 

[PeroK]

@kuruman @haruspex @Orodruin I think I find out my mistakes. Thanks for your time.

I'm not convinced by your general approach. If something is true, you need a proof. If something is false, you either need to prove algebraically that the expressions are different. Or, finding a single counterexample is enough. For example, for the first identity you could let $\vec B = \vec e_x$ (the unit vector in the x-direction). Then the identity is manifestly false.

 

[MatinSAR]

I'm not convinced by your general approach. If something is true, you need a proof.

Actually It was not a homework and because of this I did not try to prove them. I wanted to know if I can use "triple dot product or triple cross product" rules without any change when I have a differential operator. Now I know I cannot. Thanks.

 

[kuruman]

Now I know I cannot.

Don't be so sure. A long time ago I came across "Feynman's vector calculus trick" (I think somewhere in one of the Feynman lectures books) where he says that you can as long as you follow a simple rule, "Don't leave any operator without something to operate on." An example of violation of this rule is the fourth equation in your problem that has a dangling $\nabla.$ Operator $\nabla$ always must have something to operate on and only to its right.

When you use Feynman's trick, it helps to introduce subscripts indicating which vector $\nabla$ operates on. In post #11 @Orodruin underlines the vectors to achieve the same goal but I prefer subscripts. Furthermore, keep in mind that if you have a cross product and you flip the order of $\vec A$ and $\vec B$, you may have to introduce a negative sign.

Example 1: $\vec{\nabla}(\vec A \cdot \vec B)$
Start by introducing the subscripts. $$\vec{\nabla}(\vec A \cdot \vec B) = \vec{\nabla}_A(\vec A \cdot \vec B)+\vec{\nabla}_B(\vec A \cdot \vec B) \tag{1}.$$ According to the triple product rule (BAC-CAB)
$\vec a \times(\vec b\times \vec c)=\vec b(\vec a \cdot \vec c)-\vec c(\vec a \cdot \vec b)\implies \vec b(\vec a \cdot \vec c)=\vec a \times(\vec b\times \vec c)+\vec b(\vec a \cdot \vec c)$

We now treat all symbols as vectors and identify
$\vec b \equiv \vec {\nabla}_B~;~~\vec a \equiv \vec A~;~~\vec c \equiv \vec B$
Then the second term in equation (1) becomes
$$\vec{\nabla}_B(\vec A \cdot \vec B)=\vec A \times( {\nabla}_B\times \vec B)+\vec B(\vec A \cdot \vec{\nabla}_B).\tag{2}$$The second term in equation (2) has a dangling $\nabla_B$. However, $(\vec A \cdot \vec{\nabla}_B)$ is treated as a scalar product which means that we can move $\vec B$ to the right. $$\vec{\nabla}_B(\vec A \cdot \vec B)=\vec A \times( {\nabla}_B\times \vec B)+(\vec A \cdot \vec{\nabla}_B)\vec B.\tag{3}$$ Now we note that there are no cross products in equation (1) which means that we can swab $A$ and $B$ without changing anything. So we write immediately $$\vec{\nabla}_A(\vec B \cdot \vec A)=\vec B \times( {\nabla}_A\times \vec A)+(\vec B \cdot \vec{\nabla}_A)\vec A.\tag{4}$$ We now add equations (3) and (4), and drop the subscripts because all $\nabla$s are satisfied. The final result is $$
\vec{\nabla}(\vec A \cdot \vec B)=\vec B \times( {\nabla}\times \vec A)+(\vec B \cdot \vec{\nabla})\vec A
+\vec A \times( {\nabla}\times \vec B)+(\vec A \cdot \vec{\nabla})\vec B$$ Note that when you swap symbols $A$ and $B$ in the above equation, you get on each side the same expression, as you should.

Example 2: $\vec{\nabla}\times(\vec A \times \vec B)$
Start by introducing the subscripts and then use the BAC-CAB rule. $$\begin{align}& \vec{\nabla}\times(\vec A \times \vec B) = \vec{\nabla_A}\times(\vec A \times \vec B)+\vec{\nabla_B}\times(\vec A \times \vec B) \nonumber \\ & =
\vec A(\vec{\nabla_A}\cdot \vec B)-\vec B(\vec{\nabla_A}\cdot \vec A)+\vec A(\vec{\nabla_B}\cdot \vec B)-\vec B(\vec{\nabla_B}\cdot \vec A).
\nonumber\end{align}$$ At this point you need to be careful. Look at the first term $\vec A(\vec{\nabla_A}\cdot \vec B)$. You want $\vec A$ to the right of $\vec{\nabla_A}$ and $\vec B$ out of the way. First you move $\vec A$ past the "scalar" in the parentheses to get $(\vec{\nabla_A}\cdot \vec B)\vec A$. Then you swap the "vectors" between the parentheses in the dot product, but when you do this, you must change the sign as demanded by the cross product in the original expression. The same applies to the fourth term. Thus, when you do the swaps in the first and fourth terms and drop subscripts, you get $$\vec{\nabla}\times(\vec A \times \vec B) =
(\vec B \cdot \vec {\nabla})\vec A-\vec B(\vec{\nabla}\cdot \vec A)+\vec A(\vec{\nabla}\cdot \vec B)-(\vec A\cdot \vec{\nabla})\vec B$$Note that when you swap symbols $A$ and $B$ in the above equation, you get on each side an overall negative sign, as you should.

I chose two examples that are a bit involved to illustrate the use of the method. You can find more identities here and practice the method if you wish.

Edit
This post has been edited to correct a lapse of judgment. My thanks to @MatinSAR for pointing out the problem.

 

[MatinSAR]

Don't be so sure. A long time ago I came across "Feynman's vector calculus trick" (I think somewhere in one of the Feynman lectures books) where he says that you can as long as you follow a simple rule, "Don't leave any operator without something to operate on." An example of violation of this rule is the fourth equation in your problem that has a dangling ∇. Operator ∇ always must have something to operate on and only to its right.

I have his lectures and I wish I had enough time to read them.

When you use Feynman's trick, it helps to introduce subscripts indicating which vector ∇ operates on. In post #11 @Orodruin underlines the vectors to achieve the same goal but I prefer subscripts. Furthermore, keep in mind that if you have a cross product and you flip the order of A→ and B→, you may have to introduce a negative sign.

Thank you I really appreciate your help.

I chose two examples that are a bit involved to illustrate the use of the method. You can find more identities here and practice the method if you wish.

I will. Thanks a lot for your time.

 

[MatinSAR]

Then you swap the "vectors" between the parentheses, but when you do this, you must change the sign as demanded by the cross product in the original expression.

Sorry but I don't understand why the sign should change. I have changed the sign and I get:
$$\vec{\nabla}\times(\vec A \times \vec B) = -(\vec B \cdot \vec {\nabla})\vec A-\vec B(\vec{\nabla}\cdot \vec A)+\vec A(\vec{\nabla}\cdot \vec B)+(\vec A\cdot \vec{\nabla})\vec B$$

But then I did not change the sign and I get same result as arfken mathematical methods for physicists. Can't we say that according to $\vec A.\vec B = \vec B.\vec A$ the sign does not need to change?

 

[haruspex]

"Don't leave any operator without something to operate on."

Unless the result can reasonably be an operator.

 

[kuruman]

Unless the result can reasonably be an operator.

Sure. The quotations are not for the verbatim Feynman statement but to separate the rule (as I remember it) from the rest of the text in the post. My memory isn't what it used to be.

 

[MatinSAR]

You are mixing examples.

In Example 1 the swap of symbols $A$ and $B$ has even symmetry because $$~\vec{\nabla}(\vec A \cdot \vec B)=\vec{\nabla}(\vec B \cdot \vec A).$$ This means that when you have something like $(\vec{\nabla_A}\cdot \vec B)\vec A$ and you need to swap to get $\vec B$ out of the way, when you swap, you must preserve the even symmetry and write $$(\vec{\nabla_A}\cdot \vec B)\vec A\rightarrow+(\vec B\cdot \vec{\nabla_A})\vec A$$In Example 2 the swap of symbols $A$ and $B$ has odd symmetry because $$~\vec{\nabla}(\vec A \times \vec B)=-\vec{\nabla}(\vec B \times \vec A).$$ This means that when you have something like $(\vec{\nabla_A}\cdot \vec B)\vec A$ and you need to swap to get $\vec B$ out of the way, when you swap, you must preserve the odd symmetry and write $$(\vec{\nabla_A}\cdot \vec B)\vec A\rightarrow-(\vec B\cdot \vec{\nabla_A})\vec A.$$

Thanks. But Why your answer is different from the book?

Arfken mathematical methods for physicists:

 

[kuruman]

Thanks. But Why your answer is different from the book?

Because I led myself astray and did not follow the rules to treat the vectors properly. When you swap vectors, you change signs only when you have a "cross" and not when you have a "dot" between them. I edited my post with the two examples to fix the problem. I also deleted the post that you refer to in #21 because it was nonsense and I did not want to confuse readers in the future.

Thank you for pointing out the error of my ways.

 

[MatinSAR]

Because I led myself astray and did not follow the rules to treat the vectors properly. When you swap vectors, you change signs only when you have a "cross" and not when you have a "dot" between them. I edited my post with the two examples to fix the problem. I also deleted the post that you refer to in #21 because it was nonsense and I did not want to confuse readers in the future.

Thank you for pointing out the error of my ways.

Everything is completely clear now. Thanks a lot.