Divergence of vector field: Del operator/nabla

  • #1

Homework Statement


Let ν(x,y,z) = (xi + yj + zk)rk where v, i, j, k are vectors
The k in rk∈ℝ and r=√(x2+y2+z2).
Show that ∇.v=λrk except at r=0 and find λ in terms of k.

Homework Equations


As far as I understand it, ∇.v=∂/∂x i + ∂/∂y j + ∂/∂z k, but this may very well be wrong.

The Attempt at a Solution


That would mean that ∇.v = (i+j+k)rk? So λ= i+j+k. And the as to finding λ in terms of k, no idea! I don't know how to manipulate an equation with a ∇.v in it.
 

Answers and Replies

  • #2
BiGyElLoWhAt
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Hey, trippy name XD
It seems to me as though there're actually a few issues we need to get through here. One is the definition of the del operator and the other is the definition of the dot product.
Del is defined (in cartesian coords.) as ##\vec{\nabla} = <\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z},>##
The dot product is defined (in all vector spaces) as ##\vec{A}\cdot \vec{B} = |A||B|cos(\theta_{A \to B})##
I wrote del as a vector, which it is, but it's commonly denoted without the vector notation, as it is always a vector (at least to my knowledge[?]).
so ##\vec{\nabla} <=> \nabla##
 
  • #3
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In this case I got ##\nabla \cdot v = r^k + r^k + r^k = 3r^k## so it looks like ##\lambda = 3##. Remember divergence is a "dot product" which is a scalar. The x derivative of the first component plus the y derivative of the 2nd component, etc. What to do from there I'm not sure if it's as simple as ##\frac{\nabla \cdot v}{r^k} = \lambda## or what. The letter k is used in two different ways & it took me a while to sort out what was what. :oldtongue:
 
  • #4
BiGyElLoWhAt
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Way to hand that out on a silver platter... -.-
Also, r is a function of x, y, and z, so you must apply the product rule during differentiation. That is not correct for an r dependant on x, y, & z.
 
  • #5
So it's not 3rk? Very annoying. I was just about to say that I see where that came from, so clearly I still don't understand.
So I am doing the partial differential dotted with (xi + yj + zk)rk?
 
  • #6
Way to hand that out on a silver platter... -.-
Also, r is a function of x, y, and z, so you must apply the product rule during differentiation. That is not correct for an r dependant on x, y, & z.
So is the partial derivative ∂/∂x rk+xr-k? Because it's ∂(xrk)/∂x, which is x√x2+y2+z2

Edit: No, I take that back... There should be a k somewhere.
 
  • #7
BiGyElLoWhAt
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Yes you are, but what you actually have is this:
##\vec{v} = (x\sqrt{x^2+y^2+z^2}^k\hat{i}+y\sqrt{x^2+y^2+z^2}^k\hat{j}+z\sqrt{x^2+y^2+z^2}^k\hat{k})##
 
  • #8
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Oops I forgot that r depended on x, y & z so it's not as easy as I thought but yes $$\nabla \cdot v = \frac{\partial (xr^k)}{\partial x} + the rest$$
 
  • #9
BiGyElLoWhAt
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Here, let me give you an example. Fourier had the right answer assuming r^k was independant of x,y,z , but it's not.
 
  • #10
BiGyElLoWhAt
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What fourier said ^
 
  • #11
Right. I need to sort the partial derivatives first, then dot those with ν.
∂/∂x is rk + kx2r-k, I think.
∂/∂y is rk + ky2r-k
∂/∂z is rk + kz2r-k

Then the dot product of those with (xi + yj + zk) rk
 
  • #12
What fourier said ^
I think I did what Fourier said... But then, I didn't get something of the form λrk. Instead:
xr2k+kx3+yr2k+ky3+zrk+kz3. So my partial derivatives or my dot product is wrong! But only if I actually dot product the derivatives with the vector.
Wikipedia makes it sound like ∇.ν is just the partial derivatives not dotted with anything, unless I misunderstood it?
 
Last edited:
  • #13
BiGyElLoWhAt
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not quite. the first term is right, but the secont term is not so much.
You need to apply the chain rule.
## \frac{\partial}{\partial x} = \frac{\partial}{\partial r}\frac{\partial r}{\partial u}\frac{\partial u}{\partial x}## where ##u=x^2 + y^2 + z^2## for this case only.
 
  • #14
CAF123
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Hey whatisreality,
Right. I need to sort the partial derivatives first, then dot those with ν.
∂/∂x is rk + kx2r-k, I think.
∂/∂y is rk + ky2r-k
∂/∂z is rk + kz2r-k
The computation of ##\frac{\partial}{\partial x} \sqrt{(x^2+y^2+z^2)}^k = (x^2+y^2+z^2)^{\frac{k}{2}}## and similarly for the other two derivatives has been done incorrectly. But you are on right track now.

Alternatively, without converting to cartesian coordinates, just write ##\mathbf v = r^{k+1}\hat r## and compute ##\nabla \cdot \mathbf v## using the product rules for the divergence of a product of a scalar field (here ##r^{k+1}##) and a vector field (here ##\hat r##).
 
  • #15
not quite. the first term is right, but the secont term is not so much.
You need to apply the chain rule.
## \frac{\partial}{\partial x} = \frac{\partial}{\partial r}\frac{\partial r}{\partial u}\frac{\partial u}{\partial x}## where ##u=x^2 + y^2 + z^2## for this case only.
Should the second term be kx2rk-1 instead?
 
  • #17
Yes it should. Good job W.I.R.!
Got there! You're being so patient, thank you :)
Now, I multiply ∂/∂x by xr^k, and ∂/∂y by yr^k etc.
 
  • #18
BiGyElLoWhAt
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Yes you do, but you've already done the x one, so you need the second two terms.
 
  • #19
Yes you do, but you've already done the x one, so you need the second two terms.
Shouldn't they be of the same form? As in ∂/∂y = rk+kyrk-1, ∂/∂z = rk + kzrk-1.
 
  • #20
BiGyElLoWhAt
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They should be in the same form, but you're missing a squared.
 
  • #21
They should be in the same form, but you're missing a squared.
Oops, silly mistake. And that is ∇.v?
 
  • #22
BiGyElLoWhAt
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That is indeed del v. And yea, it was just a careless mistake, of the order of forgetting a negative sign :wink:
 
  • #23
BiGyElLoWhAt
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So now how can you rearrange that to get it in terms of a constant (lambda) k and r?
 
  • #24
So now how can you rearrange that to get it in terms of a constant (lambda) k and r?
If I factorise out rk, then that leaves
∇.ν = rk(3 + kx2/r + ky2/r + kz2/r)
Which annoyingly has an r inside the brackets. I don't think r is constant, is it? It's the magnitude of ν. Oh, so maybe that is constant. Then λ would be the bit inside the brackets?
 
  • #25
BiGyElLoWhAt
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pretty dang close. try doing one more factorization. I'm talking about the kx^2 + ... bit. (leave the 3 there.
 

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