# Homework Help: Curl and vectorial laplacian

1. Oct 13, 2009

### Vicfred

1. The problem statement, all variables and given/known data
I want to calculate $$\nabla\times[\vec{F}(r)]$$ and $$\nabla^2[\vec{F}(r)]$$ where F if a function that depends of r, and $$r = \sqrt{x^2+y^2+z^2}$$

2. Relevant equations
$$1)\nabla \times \vec A = \left|\begin{matrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \\ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \\ \\ A_x & A_y & A_z \end{matrix}\right|$$
$$2)\nabla^2\vec A = \nabla(\nabla\cdot\vec A)-\nabla\times(\nabla\times\vec A)$$

3. The attempt at a solution
$$\left( \frac{\partial}{\partial y}\vec F_{z}(r)-\frac{\partial}{\partial z}\vec F_{y}(r)\right)\hat{i} + \left( \frac{\partial}{\partial x}\vec F_{z}(r)-\frac{\partial}{\partial z}\vec F_{x}(r)\right)\hat{j} + \left( \frac{\partial}{\partial x}\vec F_{y}(r)-\frac{\partial}{\partial y}\vec F_{x}(r)\right)\hat{k}$$

I need help calculating the partials I tried this:
$$\frac{1}{r}\left(y - z \right) + \frac{1}{r}\left(x - z \right) + \frac{1}{r}\left(x - y \right) \Rightarrow \frac{2x - 2z}{r}$$ What happened to the i, j, k? I don't know =P! I suppose I'm wrong for vanishing them...

When I'm trying to calculate the vectorial laplacian I get stuck when I have to calculate $$\nabla\times\vec {F}(r)$$ so...

2. Oct 13, 2009

### lanedance

can you look at the spherical coordinate representations?

3. Oct 13, 2009

### Vicfred

Well, today I started to learn about spherical and cylindrical coordinates and I still don't know how to transform nabla, dot or vector product to another coordinate system... is that necessary to solve this problem?

4. Oct 13, 2009

### lanedance

its not necessary, but if F is only a function r, then the spherical symmetry of the problem may make things a easier if you can just use the spherical representation of the lapacian or curl & F, though maybe not if F is in caretsian form

is
$$\vec{F}(r)$$
just an arbitrary vector function in cartesian form?

maybe starting by caculating the first partials may help? what are:
$$\frac{\partial F_{x_i}}{\partial x_j}$$

5. Oct 13, 2009

### Vicfred

yes, $$\vec{F}(r)$$ is an arbitrary vector function in cartesian form that depends of x, y and z. Maybe I was wrong to say that $$r = \sqrt{x^2+y^2+z^2}$$ maybe $$\left( \frac{\partial}{\partial y}\vec F_{z}(r)-\frac{\partial}{\partial z}\vec F_{y}(r)\right)\hat{i} + \left( \frac{\partial}{\partial x}\vec F_{z}(r)-\frac{\partial}{\partial z}\vec F_{x}(r)\right)\hat{j} + \left( \frac{\partial}{\partial x}\vec F_{y}(r)-\frac{\partial}{\partial y}\vec F_{x}(r)\right)\hat{k}$$ could be the general asnwer to calculate the curl of any function that depends of r, for example: $$\left( \frac{r^2\sqrt{csc^2(r)}}{ ln (3r)}\right)\hat i + \left( 3\sqrt[5]{r^2}\cdot e^{3r^2}\right)\hat j + \left( \frac{6r^3}{cot^3(r)}\right) \hat k$$ (or any other)

Last edited: Oct 13, 2009
6. Oct 13, 2009

### lanedance

that could be it, the only way i think you could maybe simplify is to look at

$$\frac{\partial F_{x_i}(r)}{\partial x_j} = \frac{d F_{x_i}(r)}{d r} \frac{\partial r}{\partial x_j}$$

Last edited: Oct 13, 2009
7. Oct 13, 2009

### Vicfred

Ok, thanks I'll try it, it's too late here (3:55am), for the vector laplacian I think I'll have to do a BIG matrix... I guess I'll have to learn something about linear algebra... Thanks for your help.

8. Oct 13, 2009

### lanedance

vector laplacian in cartesian coordinates becomes
$$\nabla^2\vec{F}(r) = (\nabla^2 \vec{F_x}(r),\nabla^2 \vec{F_y}(r),\nabla^2 \vec{F_z}(r)$$)
and you can show this with the equation you have

9. Oct 17, 2009

### Vicfred

ok then$$\frac{\partial}{\partial y} F_{z}(r) = \frac{d }{d r} F_{z}(r) \frac{\partial r}{\partial z}$$ ?

Last edited: Oct 17, 2009
10. Oct 17, 2009

### Vicfred

if someone have the equations for that rotational post it that way at least I'd know what I have to get...

11. Oct 18, 2009

### lanedance

might be a little notational confusion, but in this case I thought the sunscript was implying a vector componet, whilst r is the function of the varibales x,y,z, so if:
$$\vec{F}(r) = (F_x(r), F_y(r), F_z(r))$$
with
$$r = r(x,y,z)$$

then the derivative, you want, from teh chain rule is
$$\frac{\partial}{\partial y} F_{z}(r) = \frac{d }{d r} F_{z}(r) \frac{\partial r}{\partial y}$$

also one more thought - due to the depndence on r, you may be able to use symmetry & intergal relations (divergence & curl theorems), or other properties to simplify the expressions

what exactly is the quetsion you are tyring to answer?

12. Oct 18, 2009

### Vicfred

I'm trying to find an equation like this:

$$\nabla\left[F(r) \right] = \frac{d}{dr}F(r) \frac{\vec r}{r}$$
$$\nabla\cdot\left[ \vec F(r) \right] = \frac{d}{dr}F(r) \cdot \frac{\vec r}{r}$$
$${\nabla}^2 \left[\vec F(r) \right] = 2 \frac{d}{dr}F(r) + \frac{d^2}{dr^2}F(r)$$

hmm sometimes I say rotational, sometimes curl... I meant the curl, they are the same thing right?

Last edited: Oct 18, 2009
13. Oct 18, 2009

### lanedance

ok now i understand what we're going for, so take the first one, the gradient of a scalar function F(r)
$$\nabla\left[F(r) \right] = (\frac{\partial F(r)}{\partial x}, \frac{\partial F(r)}{\partial y},\frac{\partial F(r)}{\partial z})$$

using the chain rule on F(r(x,y,z)) gives
$$\nabla\left[F(r) \right] = (\frac{d F(r)}{dr}\frac{\partial r}{\partial x}, \frac{d F(r)}{dr}\frac{\partial r}{\partial y},\frac{d F(r)}{dr}\frac{\partial r}{\partial z}) = \frac{d F(r)}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y},\frac{\partial r}{\partial z})$$

with
$$r = \sqrt{x^2 + y^2 + z^2}$$

then once again using the chain rule
$$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \frac{x}{|r|}$$

and you should be able to do similar for the others

now substitute back in for the derivatives of r
$$\nabla\left[F(r) \right] = \frac{d F(r)}{dr} (\frac{x}{|r|},\frac{y}{|r|},\frac{z}{|r|}) = \frac{d F(r)}{dr}\frac{1}{|r|} (x,y,z) = \frac{d F(r)}{dr}\frac{\vec{r}}{|r|}$$
as required,
hopefully you can use a similar method for the curl & laplacian problem

not sure, maybe? i've only really been taught & used the curl

also the last equation you gave i assume should be the vector F

Last edited: Oct 18, 2009
14. Oct 18, 2009

### Vicfred

I already had the gradient, the divergence and the scalar laplacian, what I can't find still is the curl and the vector laplacian for the curl I have this, I'm not sure if it's correct.

$$\left(\frac{d}{dr} F_{z}(r)\frac{\partial y}{\partial r} -\frac{d}{dr} F_{y}(r)\frac{\partial z}{\partial r}\right)\hat{i} + \left(\frac{d}{dr} F_{z}(r)\frac{\partial x}{\partial r}-\frac{d}{dr} F_{x}(r)\frac{\partial z}{\partial r}\right)\hat{j} + \left(\frac{d}{dr} F_{y}(r)\frac{\partial x}{\partial r}-\frac{d}{dr} F_{x}(r)\frac{\partial y}{\partial r}\right)\hat{k} = \frac{\partial}{\partial r} \vec F(x) \times \frac{d}{dr} \vec F(r)$$

15. Oct 19, 2009

### lanedance

noooo... the components look right however the partials should be inverted (use the chain rule and refer to the previous post)

so to make the notation a little easier, let
$$\vec{F}(r) = u(r)i + v(r)j + w(r)k$$
where i,j,k denote the unit vectors

remembering the r dependence, we get to
$$\nabla \times \vec{F}(r) = (\frac{d w}{dr} \frac{\partial r}{\partial y} -\frac{dv}{dr} \frac{\partial z}{\partial r})i + (\frac{d u}{dr} \frac{\partial r}{\partial z} -\frac{dw}{dr} \frac{\partial r}{\partial x})j + (\frac{d v}{dr} \frac{\partial r}{\partial x} -\frac{du}{dr} \frac{\partial r}{\partial y})k$$

now i would calculate & substitute in for the partial derivatives of r w.r.t. x,y&z and see where that gets you,

hint: i'm guessing you'll get close to something like (dF/dr) x (r/|r|)

hopefully this doesn't confuse things, but if you know how to write the cross product in component form, this one will be much easier to evaluate

sorry for the notation change, but its required, so let
$$\vec{F}(r) = (F_1, F_2, F_3) = F_j$$
$$\vec(r) = (x,y,z) = (x_1, x_2, x_3) = x_i$$
where i,j are dummy subscripts that wll be use to express the cross prouct as a sum

then
$$\nabla \times \vec{F}(r) = \frac{\partial}{\partial x_i} F_j \epsilon_{ijk} = \frac{\partial F_j}{\partial x_i} \epsilon_{ijk} = \frac{dF_j}{dr}\frac{\partial r}{\partial x_i} \epsilon_{ijk}$$
where k represents the component of the cross product

once again subtitute in for the partial derivatives and carry on, this should be easily identifiable as the cross product

Last edited: Oct 19, 2009
16. Oct 19, 2009

### Vicfred

I got
$$\frac{\vec r}{r} \times \frac{d}{dr} \vec F(r)$$

17. Oct 19, 2009

### lanedance

yeah i think thats right,
so from before
$$\frac{\partial r}{\partial x_i} = \frac{x_i}{r}$$

and carrying on (with the notation for the kth component of the vector), i get the same
$$\nabla \times \vec{F}(r) = \frac{\partial}{\partial x_i} F_j \epsilon_{ijk} = \frac{\partial F_j}{\partial x_i} \epsilon_{ijk} = \frac{d F_j}{dr}\frac{\partial r}{\partial x_i} \epsilon_{ijk} = \frac{d F_j}{dr}\frac{x_i}{r}\epsilon_{ijk} = \frac{x_i}{r} \frac{d F_j}{dr} \epsilon_{ijk} = \frac{\vec{r}}{r} \times\frac{ d\vec{F}}{dr}$$

sorry, almost missed a minus before as i hadn't done the whole thing

Last edited: Oct 19, 2009
18. Oct 19, 2009

### lanedance

19. Oct 20, 2009

### Vicfred

ok, from that proof I've got
$$\nabla^2\vec{F}(r) = \nabla^2 {u} \hat e_x + \nabla^2 {v} \hat e_y + \nabla^2 {w} \hat e_z$$
$$\nabla^2 {u} = \nabla \cdot \left[ \nabla u \right]$$
$$\nabla u = \frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z} = \frac{x}{r} + \frac{y}{r} + \frac{z}{r} = \frac{\vec r}{r}$$
$$\nabla \cdot \left( \frac{\vec r}{r}\right) = \frac{\partial}{\partial x} \frac{\vec r}{r} + \frac{\partial}{\partial y} \frac{\vec r}{r} + \frac{\partial}{\partial z} \frac{\vec r}{r}$$

hmm... I'm wrong right? U doesn't depend of (r,r,r)... maybe $$u_x,u_y,u_z$$?

20. Oct 20, 2009

### lanedance

i think you've complicated the laplacian and messed the differentiation - what happened to the chain rule?

using the notation you've chosen (i still like the subscripts better, bit easier)
$$\vec{F} = (u(r), v(r), w(r))$$

and from the proof you know
$$\nabla^2 \vec{F} = (\nabla^2 u(r), \nabla^2 v(r), \nabla^2 w(r))$$

considering only the x component
$$\nabla^2 u(r) = (\frac{\partial^2}{\partial x^2 } + \frac{\partial^2}{\partial y^2 } +\frac{\partial^2}{\partial z^2 })u(r)$$

so we only really need to consider one of the 2nd partial derviatives to evaluate the form, the first derivative is, as before:
$$\frac{\partial}{\partial x} u(r) = \frac{du(r)}{dr}\frac{\partial r}{\partial x} = \frac{du(r)}{dr}\frac{x}{r}$$
now eveaulating the 2nd partial
$$\frac{\partial^2}{\partial x^2} u(r) = \frac{\partial }{\partial x}( \frac{\partial u(r)}{\partial x}) = \frac{\partial }{\partial x}(\frac{du(r)}{dr}\frac{x}{r})$$
so, using the product rule & chain rule, what does that give you?

after evaluating subtitute back in for your whole vector

21. Oct 20, 2009

### Vicfred

$$\frac{du(r)}{dr}\frac{\partial}{\partial x}\frac{x}{r} + \frac{x}{r}\frac{\partial}{\partial x}\frac{du(r)}{dr}$$? hmmm...

Last edited: Oct 20, 2009
22. Oct 20, 2009

### lanedance

ok so I would use the chain & product rules to further evaluate those derivatives

23. Oct 20, 2009

### lanedance

i think it might help if you think of the chain rule when acting on an scalar function depenedent only on r(x,y,z) given by
$$f(r(x,y,z))$$

then the partial derivative w.r.t x (which assumes y&z stay constant) is:
$$\frac{\partial f(r(x,y,z))}{\partial x} = \frac{\partial r(x,y,z)}{\partial x}\frac{df(r)}{dr}$$

you can see this because the total change in f, for a change in r, by dr, is given by
$$df(r) = \frac{df(r)}{dr}dr$$

and the total change in r, for a change in x,yz, by dx,dy,dz is given by
$$dr(x,y,z) = \frac{\partial r(x,y,z)}{\partial x}dx + \frac{\partial r(x,y,z)}{\partial y}dy + \frac{\partial r(x,y,z)}{\partial z}dz$$

you can see this because the total change in r, for a change in x,yz, by dx,dy,dz is given by
$$dr(x,y,z) = \frac{\partial r(x,y,z)}{\partial x}dx + \frac{\partial r(x,y,z)}{\partial y}dy + \frac{\partial r(x,y,z)}{\partial z}dz$$

so the total change in f, for a change in x,yz, by dx,dy,dz is given by
$$df(r((x,y,z))) = \frac{df(r)}{dr}(\frac{\partial r(x,y,z)}{\partial x}dx + \frac{\partial r(x,y,z)}{\partial y}dy + \frac{\partial r(x,y,z)}{\partial z}dz)$$

so when dz=dy=0, ie in the case of a partial derivative w.r.t. x, the partial becomes (the y,z meaning they are kept constant:
$$df(r((x,y,z)))_{y,z} = \frac{df(r)}{dr}\frac{\partial r(x,y,z)}{\partial x}dx$$

so in the case where you have a function only of r(x,y,z) case you can think of the partial w.r.t. x as an operator given by
$$\frac{\partial}{\partial x} = \frac{\partial r(x,y,z)}{\partial x}\frac{d}{dr}$$

wow this is a long one... ;)

Last edited: Oct 20, 2009
24. Oct 20, 2009

### Vicfred

$$\frac{\partial }{\partial x}(\frac{du(r)}{dr}\frac{x}{r}) = u\frac{dr}{dr} \frac{x}{r} = \frac{\partial}{\partial x}\frac{ux}{r} = \frac {r(u{\frac{\partial x}{\partial x}} + x\frac{\partial u}{\partial x}) + (ux)\frac{\partial r}{\partial x}}{r^2}$$

25. Oct 20, 2009

### lanedance

updated last post