Divergence and solenoidal vector fields

In summary, the conversation is about finding the values of n that make a given vector field, F, solenoidal. The process involves substituting \underline{r} = \sqrt{x^2 + y^2 + z^2} into F and evaluating the divergence. It is noted that when n = 0, the first term of the divergence is zero, but the second term becomes 1, so finding the values of n for which this is zero is the goal. The conversation also touches on using the formula div(PF) = PdivF + F.gradP, where P is a scalar field and F is a vector field, to simplify the expression and find a solution.
  • #1
MathematicalPhysics
40
0
I want to find which values of n make the vector field

[tex]\underline{F} = {|\underline{r}|}^n\underline{r}[/tex] 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:

[tex]\underline{r} = \sqrt{x^2 + y^2 + z^2}[/tex]

getting..

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

How can I extract

[tex]\underline{F_x}, \underline{F_y}, \underline{F_z}[/tex]?

It's probably really simple but I can't see it! Thanks in advance.
 
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  • #2
Yeah I was being daft, after sleeping on it I came up with:

[tex] \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] [/tex]

That's better, yeah?

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

now
[tex]\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} [/tex]

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?
 
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  • #3
MathematicalPhysics said:
[tex]\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} [/tex]

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 [itex]\nabla \cdot \vec{F}[/itex] should give you:

[tex]3(x^2+y^2+z^2)^{n/2} + n(x^2+y^2+z^2)^{n/2}[/tex]

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

I think...
 
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  • #4
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.
 
  • #5
Following on I'm trying to find the value of [tex]\lambda[/tex] which makes

[tex] \frac{\lambda\underline{a}}{|\underline{r}|^3} - \frac{(\underline{a}.\underline{r})\underline{r}}{|\underline{r}|^5} [/tex]

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!
 
  • #6
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.
 
  • #7
If this is wrong I can post my working (but its tedious to keep latexing my results!)

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

so could I just say that if [tex] \lambda = 1 [/tex] this would give "[tex] \underline{F}[/tex]" 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?
 
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  • #8
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!
 
  • #9
sorry it was meant to be..

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

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

1. What is the difference between divergence and solenoidal vector fields?

Divergence and solenoidal vector fields are two types of vector fields used in vector calculus. The main difference between them is that divergence measures the net flow of a vector field through a given point, while solenoidal vector fields have a net flow of zero through any point.

2. How can I determine if a vector field is solenoidal or not?

To determine if a vector field is solenoidal, you can use the divergence theorem, which states that the integral of the divergence of a vector field over a closed surface is equal to the net flow of the vector field through that surface. If the net flow is equal to zero, the vector field is solenoidal.

3. What is the physical significance of divergence and solenoidal vector fields?

Divergence and solenoidal vector fields have important physical significance in fields such as fluid mechanics, electromagnetism, and heat transfer. Divergence represents the creation or destruction of a quantity at a point, while solenoidal vector fields represent the conservation of a quantity.

4. Can a vector field have both divergence and solenoidal components?

Yes, a vector field can have both divergence and solenoidal components. In fact, most vector fields have both components. This is because the divergence and curl of a vector field are independent of each other, meaning one can exist without the other.

5. How are divergence and curl related to divergence and solenoidal vector fields?

The divergence and curl of a vector field are related to divergence and solenoidal vector fields through the Helmholtz decomposition theorem. This theorem states that any vector field can be decomposed into a sum of a solenoidal vector field and a divergence vector field. This means that any vector field can be broken down into its solenoidal and irrotational components.

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