Divergence and solenoidal vector fields

[MathematicalPhysics]

I want to find which values of n make the vector field

$$\underline{F} = {|\underline{r}|}^n\underline{r}$$ solenoidal.

So I have to evaluate the divergence of this vector field I think, then show for which values of n it is zero?

Im starting by substituting:

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

getting..

$$\underline{F} = {(x^2 + y^2 + z^2)}^{n/2}\sqrt{x^2 + y^2 + z^2}$$

How can I extract

$$\underline{F_x}, \underline{F_y}, \underline{F_z}$$?

It's probably really simple but I can't see it! Thanks in advance.

 

[MathematicalPhysics]

Yeah I was being daft, after sleeping on it I came up with:

$$\underline{F} = [{(x^2 + y^2 + z^2)}^{n/2}x, {(x^2 + y^2 + z^2)}^{n/2}y, {(x^2 + y^2 + z^2)}^{n/2}z]$$

That's better, yeah?

so
$$\underline{F}_x = {(x^2 + y^2 + z^2)}^{n/2}x$$
$$\underline{F}_y = {(x^2 + y^2 + z^2)}^{n/2}y$$
$$\underline{F}_z = {(x^2 + y^2 + z^2)}^{n/2}z$$

now
$$\frac{\partial\underline{F}_x}{\partial x} = nx^2{(x^2 + y^2 + z^2)}^{(n/2)-1} + {(x^2 + y^2 + z^2)}^{n/2}$$

Hmm if n = 0 the first term is zero as required but the second term would become 1. So how can I find the values of n for which this is zero?

 

[e(ho0n3]

$$\frac{\partial\underline{F}_x}{\partial x} = nx^2{(x^2 + y^2 + z^2)}^{(n/2)-1} + {(x^2 + y^2 + z^2)}^{n/2}$$

Hmm if n = 0 the first term is zero as required but the second term would become 1. So how can I find the values of n for which this is zero?

Calculating $\nabla \cdot \vec{F}$ should give you:

$$3(x^2+y^2+z^2)^{n/2} + n(x^2+y^2+z^2)^{n/2}$$

Set that equal to zero and solving for n whould give you n = -3 as long as $(x^2+y^2+z^2)^{n/2} \ne 0$.

I think...

 

[MathematicalPhysics]

Thanks, I was having trouble simplifying my expression for divF, knowing what I was aiming for gave me the confidence to proceed lol! Cheers,

Matt.

 

[MathematicalPhysics]

Following on I'm trying to find the value of $$\lambda$$ which makes

$$\frac{\lambda\underline{a}}{|\underline{r}|^3} - \frac{(\underline{a}.\underline{r})\underline{r}}{|\underline{r}|^5}$$

solenoidal. Where a is uniform.

I think I have to use div(PF) = PdivF + F.gradP (where P is a scalar field and F a vector field)

and grad(a.r) = a for fixed a.

So when calculating Div of the above, there should the a scalar field in there somewhere that I can separate out?!

I need some pointers please!

 

[e(ho0n3]

I don't understand what you mean when you say that 'a is uniform'. Fiddle around with the algebra and show me how far you get.

 

[MathematicalPhysics]

If this is wrong I can post my working (but its tedious to keep latexing my results!)

I get the expression down to $$\frac{(\lambda -1)a}{x^2+y^2+z^2}$$

so could I just say that if $$\lambda = 1$$ this would give "$$\underline{F}$$" say to be zero which implies dF/dx, dF/dy, dF/dz are all zero so divF = 0 which means it is solenoidal?

Thanks for being patient.

edit** is the (lambda -1)a the scalar field P to plug into that formula?

 

[e(ho0n3]

Where did you get that formula from? It makes no sense to me (i.e. you can't distribute an operator acting on different objects, scalars and vectors in this case). I know it's tedious to show your work but I can't really tell what your doing. For example, you obtained an expression which isn't even a vector!

 

[MathematicalPhysics]

sorry it was meant to be..

$$\frac{(\lambda -1)a}{(x^2 + y^2 + z^2)^{3/2}} = \frac{(\lambda -1)a}{|\underline{r}|^3}$$

I forgot the 3/2 power which didn't make it a vector as you pointed out!