Solving Operator Nabla Example Problem

[prehisto]

Homework Statement

So I have this rather komplex example and I am looking for help.
∇(3(r*a)r)/R⁵ -a/R⁵)
r=xe_x+ye_y+ze_z
a-constant vector
R=r^1/2

Homework Equations

The Attempt at a Solution

So the nabla " works" on every member individualy,and i have to careful here:(r*∇a),because of analogue with derivative rule,am I correct?

So 9(r*a)/R^5+9r/R^5+9r/R^5-15(r*a)r*r/R^5-∇a/R^5

1. Homework Statement [/

Homework Equations

The Attempt at a Solution

 

[Goddar]

Homework Statement

So I have this rather komplex example and I am looking for help.
∇(3(r*a)r)/R⁵ -a/R⁵)

Hi again.
Your notation here is not clear: are r and a vectors? In which case (\vec{r}\cdot\vec{a})\vec{r} is a vector and you
cannot take its gradient (unless you're defining a matrix form, which i doubt you are)...
Now you can act on it with ∇\cdot (divergence) or ∇\times (curl), or you can have [∇(\vec{r}\cdot\vec{a}/R⁵)]\vec{r}. So could you clarify your notation?

 

[prehisto]

Yes,r and a are vectors.
\vec{∇}(3(\vec{r}*\vec{a})\vec{r}/R⁵-\vec{a}/R⁵)

Hmm,why can't I act on (\vec{r}*\vec{a})\vec{r} because its a vector..I can act on \vec{r},which is a vector,so why not ?

 

[Goddar]

No you can't (again, unless you're defining a matrix)!
The gradient turns a scalar function into a vector, the divergence does the opposite and the curl turns a vector into another vector. So you have to make sure exactly where you put your parenthesis and adopt a precise notation, here...

 

[ChrisVer]

it would be better to use the nabla operator in spherical coordinates... I think it will simplify your problem :)

(plus I don't see any gradient or divergence in the OP's post)

 

[Goddar]

it would be better to use the nabla operator in spherical coordinates... I think it will simplify your problem :)

(plus I don't see any gradient or divergence in the OP's post)

Yes, using r is indeed an incentive to use spherical coordinates, for which you need the corresponding form for ∇. But the first issue here is to determine exactly what operation is asked, and on what type of object because there's something wrong in the problem as stated...

 

[ChrisVer]

yes, because most of times you will not see div acting on scalars, as you won't see grad acting on vectors... So in that case intuitively you choose the correct action XD.
If someone has a vector, he'll use the div for the vector, and when they have a scalar they'll use the grad...

 

[prehisto]

Ok, now I am starting to see that there is problem in essence of example.
But I have to solve it, in given cordinates.

I assume that
∇R=dR/dx*e_x+dR/dy*e_y+dR/dz*e_z
where R is modul of vector r
and
∇r=dx/dx*e_x+dy/dy*e_y+dz/dz*e_z

 

[Goddar]

Ok, now I am starting to see that there is problem in essence of example.
But I have to solve it, in given cordinates.

I assume that
∇R=dR/dx*e_x+dR/dy*e_y+dR/dz*e_z
where R is modul of vector r

Yes

∇r=dx/dx*e_x+dy/dy*e_y+dz/dz*e_z

No, if r is the position vector then:
∇\cdot\vec{r}=(dx/dx)(e_x*e_x)+(dy/dy)(e_y*e_y)+(dz/dz)(e_z*e_z)= dx/dx + dy/dy + dz/dz = 3
And ∇\vec{r} is a matrix.

 

[Chestermiller]

You still haven't told us whether you are taking the divergence of a vector, or the gradient of the vector. The gradient of the vector is a second order tensor, while the divergence of the vector is a scalar. So, which is it? (Irrespective of what coordinate system you are using)

Chet