Divergence of vector field: Del operator/nabla

[whatisreality]

Homework Statement

Let ν(x,y,z) = (xi + yj + zk)r^k where v, i, j, k are vectors
The k in r^k∈ℝ and r=√(x²+y²+z²).
Show that ∇.v=λr^k 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)r^k? 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.

 

[BiGyElLoWhAt]

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 $\vec{\nabla} <=> \nabla$

 

[fourier jr]

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.

 

[BiGyElLoWhAt]

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.

 

[whatisreality]

So it's not 3r^k? 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)r^k?

 

[whatisreality]

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 r^k+xr^-k? Because it's ∂(xr^k)/∂x, which is x√x²+y²+z²

Edit: No, I take that back... There should be a k somewhere.

 

[BiGyElLoWhAt]

Yes you are, but what you actually have is this:
$\vec{v} = (x\sqrt{x^2+y^2+z^2}^k\hat{i}+y\sqrt{x^2+y^2+z^2}^k\hat{j}+z\sqrt{x^2+y^2+z^2}^k\hat{k})$

 

[fourier jr]

Oops I forgot that r depended on x, y & z so it's not as easy as I thought but yes $$\nabla \cdot v = \frac{\partial (xr^k)}{\partial x} + the rest$$

 

[BiGyElLoWhAt]

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.

 

[BiGyElLoWhAt]

What Fourier said ^

 

[whatisreality]

Right. I need to sort the partial derivatives first, then dot those with ν.
∂/∂x is r^k + kx²r^-k, I think.
∂/∂y is r^k + ky²r^-k
∂/∂z is r^k + kz²r^-k

Then the dot product of those with (xi + yj + zk) r^k

 

[whatisreality]

What Fourier said ^

I think I did what Fourier said... But then, I didn't get something of the form λr^k. Instead:
xr^2k+kx³+yr^2k+ky³+zr^k+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?

 

[BiGyElLoWhAt]

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.

 

[CAF123]

Hey whatisreality,

Right. I need to sort the partial derivatives first, then dot those with ν.
∂/∂x is r^k + kx²r^-k, I think.
∂/∂y is r^k + ky²r^-k
∂/∂z is r^k + 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 $\hat r$).

 

[whatisreality]

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 kx²r^k-1 instead?

 

[BiGyElLoWhAt]

Should the second term be kx²r^k-1 instead?

Yes it should. Good job W.I.R.!

 

[whatisreality]

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.

 

[BiGyElLoWhAt]

Yes you do, but you've already done the x one, so you need the second two terms.

 

[whatisreality]

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 = r^k+kyr^k-1, ∂/∂z = r^k + kzr^k-1.

 

[BiGyElLoWhAt]

They should be in the same form, but you're missing a squared.

 

[whatisreality]

They should be in the same form, but you're missing a squared.

Oops, silly mistake. And that is ∇.v?

 

[BiGyElLoWhAt]

That is indeed del v. And yea, it was just a careless mistake, of the order of forgetting a negative sign

 

[BiGyElLoWhAt]

So now how can you rearrange that to get it in terms of a constant (lambda) k and r?

 

[whatisreality]

So now how can you rearrange that to get it in terms of a constant (lambda) k and r?

If I factorise out r^k, then that leaves
∇.ν = r^k(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?

 

[BiGyElLoWhAt]

pretty dang close. try doing one more factorization. I'm talking about the kx^2 + ... bit. (leave the 3 there.

 

[whatisreality]

r^k(3+k/r (x²+y²+z²))
And x²+y²+z² is (r^k)^2/k
So I can probably do something with that... Yep.

 

[BiGyElLoWhAt]

Oh no -.-
Somewhere we lost a degree of r... Those careless mistakes. Let me finish a few things up here at work and I'll try to find the mistake.

 

[whatisreality]

r^k(3+kr)!

 

[BiGyElLoWhAt]

What you have is consistent, though. So job well done on that part.

 

[BiGyElLoWhAt]

Yes that is consistent with what we've said so far, but according to the problem statement and wolfram the answer is (k+3)r^k, not (kr + 3) ...

 

[whatisreality]

It's very close to it though! I'm pretty happy, I understand that so much better.
The derivatives are right according to wolfram. So that's one thing ruled out.

 

[BiGyElLoWhAt]

Aha! It is in the derivative (rather the substitution). See if you can find it.

Spoiler: No peeking until you've tried

So what I actually had was
$x\frac{k}{2}u^{\frac{k}{2}-1}*2x= x^2*u^{\frac{k}{2}-1}$
Subbing back in for r (r = u^1/2) that should be $x^2r^{k-2}$ rather than k-1

 

[whatisreality]

Aha! It is in the derivative (rather the substitution). See if you can find it.

Spoiler: No peeking until you've tried

So what I actually had was
$x\frac{k}{2}u^{\frac{k}{2}-1}*2x= x^2*u^{\frac{k}{2}-1}$
Subbing back in for r (r = u^1/2) that should be $x^2r^{k-2}$ rather than k-1

I used the chain rule on (x²+y²+z²)^0.5k though, and taking this approach I don't see where a k-2 comes from...

 

[BiGyElLoWhAt]

the chain rule is actually being applied twice. If you see my chain rule that I posted earlier, you ultimitely need to take the derivative with respect to u. that's where that expression comes from. So what you have there is fine, but that's u^k/2, not r. You need to substitute that back in. So you have $\frac{k}{2}u^{0.5k - 1}$ with me so far?
this is equal to $\frac{k}{2}\frac{u^{.5k}}{u}$ you need to use the fact that u = r^2 (because r has the square root in it)
$u^.5k = r^k$ and $1/u = r^{-2}$ so all together it's $\frac{k}{2}r^{k-2}$ I might have lost some constants along the way, I'm kinda in a hurry atm, walking out the door at work, but that's where the 2 comes in, and that's what gets rid of the r in the lambda.

 

[whatisreality]

Oh, got it! And that leads to the 3+k. Phew! What a question. I really, really appreciate all your time and help, thank you :)