# Divergence & Curl

Discussion in 'Homework Help' started by Student01, Mar 14, 2010.

1. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
Solved!

Hi all,

I'm trying to work out the following question:

I know to be solenoidal the divergence of the vector must be 0, however I am having a huge amount of trouble understanding how to differentiate this. I don't know when to treat things as a vector and when to treat them as a scalar.

Some further information: a is an arbitrary, constant vector. r is the position vector xi + yj + zk, and r is the magnitude of this vector.

So, to emphasise: my problem is when to treat things as a vector and when to treat them as a scalar. I know bold denotes a vector and the dot product of a.r turns that part into a scalar - but I'm lost after that.

I've been working on this question for about three hours so I'd really appreciate any help anyone can offer.

Last edited: Mar 15, 2010

Jul 7, 2009
1,585
142
You're most of the way there -- you just need to figure out what things are scalars. In the expression, if a.r means a dot product (this is somewhat non-standard notation), then it is a scalar. Thus, you can rewrite things as cr + da where c and d are constants (scalars). The divergence of this is

$c \nabla \cdot \bf{r}+ d \nabla \cdot \bf{a}$

Think in your mind "I need to dot del with the vector in each term" -- i.e., the del operator "blows past" the scalars to differentiate only the vectors. You should be able to figure the divergence of the position vector in your head. Solve for the divergence of a, then ask yourself how that can be zero.

As soon as you see what's going on, you'll be able to look at the original equation and find the answer by inspection.

3. ### steveb Senior Member

Jul 3, 2008
2,433
469
This statement can be misinterpreted. It's important to distinguish between scalar functions and scalars that are constants. For example the dot product of the "r" and "a" vectors will indeed produce a scalar, but since the "a" vector is a constant vector and the "r" vector is not, the dot product will produce a function which depends on position. If this function is then multiplied by a vector, the vector components are each scaled by the function. Hence, the divergence operator can not "blow past" the function. Only when the scalar is a constant can it be factored out of the divergence operator.

Jul 7, 2009
1,585
142
Steve, you're quite right and my original statement should be taken out and shot. It is simply wrong. The OP could rewrite things in terms of components and then apply the differential operator (assuming Cartesian coordinates):

$c \nabla \cdot [(xa_x + ya_y + za_z)(x \bf{i} + y\bf{j} + z\bf{k})] + d \nabla \cdot \bf{a}$

5. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
Ok, so this is what I can work out so far:

nr$^{n-2}$(a.r)r + r^{n}a
=
nr$^{n-2}$(a$_{1}$x + a$_{2}$y + a$_{3}$z)r+ r^{n}a
=
n(n-z)r$^{n-3}$(a.r)r + nr$^{n-2}$(2a$_{1}$x + 2a$_{2}$y + 2a$_{3}$z) + nr$^{n-1}$a

nb:
subscript doesn't seem to be working. "a1", "a2", "a3" are supposed to have the number in subscript (to represent components of a)

Now what?

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
I do think you were on a good track, and your resulting equation can be useful in this problem. It really gives a clue to the answer. The relation you showed will have a value of zero if d=0 and c is any constant. It turns out that n=0 will achieve this.

While this does not prove that there are no other values of n that work, it does get one started.

Last edited: Mar 14, 2010
7. ### steveb Senior Member

Jul 3, 2008
2,433
469
Oops! ... Caution again. You can not factor $r^n$ out of the divergence. Note that $r=\sqrt{x^2+y^2+z^2}$, hence it is a function of position also. The only case where this works is n=0, which is clearly a valid answer.

Jul 7, 2009
1,585
142
Bang!

Mrs. someonesdad: we took someonesdad out behind the building and put one in his ear. He was too lazy to work the problem out on paper and give the OP good advice. He should have known better. Thanks to steveb for politely notifying us to the need for this drastic but probably welcome action.

9. ### steveb Senior Member

Jul 3, 2008
2,433
469
Dear Mrs. someonesdad. I would like to politely express my regret for your drastic action. Please understand that engineers often think too quickly and try to skip a step to arrive at the answer that seems right intuitively. All of us here have done this many times. Hopefully you can get him to a hospital quickly. We hope to see him healed up and back here soon.

The answer given by someonesdad can be viewed as a prototype for finding one of the solutions. We can see that even this simple case requires special conditions (i.e. c=0) to be a solution. And the problem reduces to this prototype case when n=0, by inspection. This makes it clear that it will be difficult (but not impossible) for any other n to provide a solution, but we can check either by brute force, or by applying a vector relation.

I wrote it out by brute force, and it's a little messy. I didn't see another solution, but this is not a reliable method and I could be missing something. I'll try again using some vector identities. If this yields anything I'll post it.

Last edited: Mar 15, 2010
10. ### steveb Senior Member

Jul 3, 2008
2,433
469
This appears to be a bit of a tricky question. Working it out with some vector identities seems to show n=-10 as a solution as well as n=0. I've attached my attempt at the solution. Please double check the work carefully. I could have made an algebraic mistake, but I think the basic approach is valid.

EDIT: I wasn't happy with the answer n=-10 because it didn't seem intuitively correct. So I went back and double checked the algebra and found a mistake. I've updated the file. I now get n=-3 and n=0. This seems more likely to be correct, but I'd still recommend double checking my work.

• ###### vector problem.pdf
File size:
9.5 KB
Views:
21
Last edited: Mar 15, 2010
11. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
This is the answer I was able to finally arrive at also. Your working confirmed it. Thanks for your help.

12. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
I don't wish to be greedy, but I have another question that I cannot answer (this one I've dedicated approximately 4 or 5 hours to trying to solve and researching techniques to solve). If you could help me out, I'd be mighty appreciative:

Find the function f(r) for which f(r)(a^r) ^ r is curl free.

I know curl free means the curl must equal zero, and I also know you can apply the BAC - CAB identity for triple vector products. Then things get crazy and I end up with sums that are too long for the page and my head hurts.

13. ### steveb Senior Member

Jul 3, 2008
2,433
469
Can you be more clear about the notation here? Is the ^ symbol supposed to be a wedge operator denoting a cross product in 3 dimensional space?

That seems to be the only thing to that makes sense to me, but then the problem is too easy. The reason being that axrxr is always zero, and any function f(r) will work.

14. ### Student01 Thread Starter Active Member

Apr 15, 2009
35
0
My apologies, I presumed there was only one meaning for "^" as it is one of the more distinct symbols within the context of what I've been studying this semester. Yes, it's the cross product of two vectors in three dimensions.

I didn't know (a ^r) ^ r is always zero. I suspected this, but dismissed it as being too easy and looked for another way of approaching the problem. I guess I have my answer. Could you please state why it is always zero in a more general form. I know it can't be "the curl of a curl is always zero" because this is not true (according to the identities we've covered in lectures. It then must be something to do with the two instances of r.

15. ### steveb Senior Member

Jul 3, 2008
2,433
469
No apologies necessary. I also think that there is only one meaning of ^ in that context, but it's rare to see the wedge operator in 3d vector analysis nowadays. It's common in some old books and in differential geometry, but I just wanted to make sure I wasn't missing something.

OK, I think I understand the question now. However, I think I'm wrong about (axr)xr being zero. I was thinking that "a" and "r" would be colinear, hence axr=0, but now I realize that "r" should be any position vector and "a" is any arbitrary constant vector. So, in general this is not true. I'm assuming that "a" is still a constant vector, as in the previous problem.
By the way, note that ax(rxr) does equal zero.

Yes, that would have been too easy of a problem. So, I'll have to give this some thought tomorrow.

16. ### steveb Senior Member

Jul 3, 2008
2,433
469
Sorry I didn't get back to this sooner, but I was busy doing last minute tasks in order to leave for vacation. Today I departed and had some time on the plane to look at this problem. I determined a couple of methods to attack this, but most of them were tedious. I did find one method that is perhaps a little more elegant than the others. I don't want to give a full solution because the policy here is to help in steps and for you to post attempts at the solution first. So, I'll only be able to give a hint.

This is not necessarily the best approach, but it's the best I could come up with. I recommend that you start with the following vector identity.

$\nabla \times (\nabla g) = 0$

This identity tells you that if you have a curl of a vector equal to zero, then you can express that vector as the gradient of a scalar field. This then becomes a new equation to consider to find the allowed functions "f" that satisfy the relation. The new relation to consider is as follows.

$f (\vec a \times \vec r) \times \vec r = \nabla g$

It really doesn't matter what the function g looks like, as long as it is one function that produces all three of the correct vector components when you take the gradient of it.