I don't understand simple Nabla operators

[Addez123]

Homework Statement

f = r ^3
A = (x^2, y^2, z^2)

Calculate div(fA)

Relevant Equations

Nabla operator rules such as
$$\nabla \cdot (\phi A) = (\nabla \phi) \cdot A + \phi \nabla \cdot A$$

Using the formula in 'relevant equations' I calculate
$$div(fA) = \nabla(fA) = (\nabla f) \cdot A + f \nabla \cdot A$$
$$3r^2 \cdot (x^2, y^2, z^2) + r^3 \cdot (2x + 2y + 2z)$$But the answer is
$$3r \cdot (x^3 + y^3 + z^3) + r^3 \cdot (2x + 2y + 2z)$$

I find no way of easily turning $3r^2 \cdot (x^2, y^2, z^2)$ into $3r \cdot (x^3 + y^3 + z^3)$ and even if I could I don't see why they would even bother obfuscating the answer like that. I assume f is a scalar, yet in the first exercise $grad(fg) = {g = 1/r^2} = e_r$

Overall I am just throughly confusing by something so seemingly obvious.

 

[anuttarasammyak]

You must be skeptical to your result the first term of which is a vector and the second term of which is a value. Try

\nabla f(r) =\frac{df}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z})

 

[Addez123]

$$\frac{df}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z}) \cdot (x^2, y^2, z^2) = $$
$$\frac{df}{dr}(\partial r 2x + \partial r 2y + \partial r 2z) = 2\partial f (x + y + z)$$
which is absolute garbage nonsense.

Seriously, I obviously lack fundamental understanding of this subject. Can you see what I am doing wrong here?

 

[anuttarasammyak]

What is the relation between r and x,y,z ? Why don't you write it down to calculate partial differentiations ?

 

[Addez123]

I don't see how I'm suppose to assume r means radius unless it's specifically written somewhere.
In this book they've denoted vector fields as capital letters and scalars as lower case letters.

What they expect me to do?
Substitute x with spherical coorindates? How would I know it's not written in cylindrical coorinates? That also utilize r as a variable.

 

[anuttarasammyak]

What is the relation between r and x,y,z ? Why don't you write it down to calculate partial differentiations ?

(Hint) The formula of sphere of radius r.

 

[Addez123]

$$\frac{df}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z}) =$$
$$3r^2 (cos(u)sin(v), sin(u)sin(v), cos(u))$$
That really doesn't simplify anything.

 

[Orodruin]

It is pretty standard to use $r$ to denote the magnitude of the position vector. In other words, what is the magnitude of $(x,y,z)$?

 

[Orodruin]

You should also note that using $\cdot$ should be reserved for inner products in order to avoid confusion. Do not use it to denote regular scalar multiplication (for which the typical standard is to not write any symbol at all, ie, denote scalar multiplication as $ab$ rather than $a\cdot b$).

 

[anuttarasammyak]

$$\frac{df}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z}) =$$
$$3r^2 (cos(u)sin(v), sin(u)sin(v), cos(u))$$
That really doesn't simplify anything.

Take it easy. It is not helpful to introduce new paratemers u, v when we think of partial differentiation of r wrt parameters x, y and z.

(Hint) The formula of sphere of radius r.

(Hint) r^2=...

 

[Addez123]

$$r^2 = x^2 + y ^ 2 + z ^2$$
but i don't see any use of that equation here..

 

[anuttarasammyak]

Take a differentiation of it
2rdr=..

 

[Addez123]

$$r^2 \frac{d}{dr} = 2r = 2(cos(u)sin(v), sin(u)sin(v), cos(u))$$
?

 

[anuttarasammyak]

Your RHS is a vector which cannot be. You do not seem to be accostomed to differentiation.
d(r^2)=2rdr
So d(x^2)=... ?
d(r^2)=d(x^2+y^2+z^2)=...?

 

[Addez123]

Your RHS is a vector which cannot be

isn't r a vector? Then it's derivative will also be vector, right?

I don't understand d(r^2) = 2r dr
you're taking the derivative of r^2 yet are left with a dr term?
d/dr of r^2 = 2r, not 2r dr

d(x^2) = 2x dx, but I don't see any logic behind the whole concept.

 

[anuttarasammyak]

> isn't r a vector?
No it is a value of distance as you showed $r^2=x^2+y^2+z^2$.

I assume you are familiar with differentiation
(x^2)'=\frac{d(x^2)}{dx}=2x
By regarding it as if it is a fraction which physics people do often, though timid criticism may come from mathematicians,
d(x^2)=2x dx
So in our relation
d(r^2)=d(x^2+y^2+z^2)=d(x^2)+d(y^2)+d(z^2)
2r dr =2xdx+ ... ?

 

[topsquark]

isn't r a vector? Then it's derivative will also be vector, right?

I don't understand d(r^2) = 2r dr
you're taking the derivative of r^2 yet are left with a dr term?
d/dr of r^2 = 2r, not 2r dr

d(x^2) = 2x dx, but I don't see any logic behind the whole concept.

$r$ is a vector. But $r^2 = r \cdot r$ is a scalar...

-Dan

 

[Steve4Physics]

Hi @Addez123. You seem to have hit a bit of a blind spot.See if this helps.

Here's a strategy. In order to use the identity "$div(fA) = (\nabla f) \cdot A + f \nabla \cdot A$" we need to evaluate $\nabla f$ and $\nabla \cdot A$.

Let me start you off.

$r= (x^2 +y^2+z^2)^{\frac 12}$
$r^2 = x^2 +y^2+z^2$
$f= r^3$
$f = (x^2 +y^2+z^2)^{\frac 32}$

$\nabla f = \nabla (x^2 +y^2+z^2)^{\frac 32}$
$= 3x(x^2 +y^2+z^2)^{1/2} \ \hat i$
$+ 3y(x^2 +y^2+z^2)^{1/2} \ \hat j$
$+ 3z(x^2 +y^2+z^2)^{1/2} \ \hat k$
$= 3(x^2 +y^2+z^2)^{1/2} \ (x\hat i + y\hat j + z \hat k)$
$= 3r<x,y,z>$ (A bit long-winded but hopefully easy to follow)

You now find $\nabla \cdot A$ and take it from there.

 

[Addez123]

2r dr =2xdx+ ... ?

$2r dr =2xdx+ 2ydy + 2zdz$

$$(\nabla \Phi) = 2xdx+ 2ydy + 2zdz$$
Now you must do the dot product on a number and a vector, because
$$\nabla \cdot (\Phi A) = (\nabla \Phi) \cdot A + \Phi \nabla \cdot A$$
which makes no sense still.

@Steve4Physics
$$\nabla \cdot A = (d/dx, d/dy, d/dz) \cdot (x^2, y^2, z^2) = 2x + 2y + 2z$$

$$3r(x, y, z) \cdot (x^2, y^2, z^2) + r^3 (2x + 2y + 2z) =$$
$$3r(x^3 + y^3 + z ^3) + 2r^3(x + y + z)$$
which is the right answer. But BOI I'm never going to be able to figure that out on my own.

Thanks for all the help guys!
Really needed it! This book really doesn't cover everything.

 

[Orodruin]

$2r dr =2xdx+ 2ydy + 2zdz$

That is correct.

$$(\nabla \Phi) = 2xdx+ 2ydy + 2zdz$$

No, this cannot be. On the left-hand side you have the gradient of a scalar field $\Phi$, on the right hand side you have a number. It is unclear what you are attempting to do.

 

[anuttarasammyak]

$2r dr =2xdx+ 2ydy + 2zdz$

$$(\nabla \Phi) = 2xdx+ 2ydy + 2zdz$$

Be careful. You should know that the second equation cannot hold because
LHS is a vector and ordinary quantity, but
RHS is not a vector and infinitely small quantity.

The first formula
$2r dr =2xdx+ 2ydy + 2zdz$ is right.
The reason that I suggest you this formula is that you can get partial differential
$\frac{\partial r}{\partial x}$ that I wrote in post #2 from it.
For partial difference wrt x, y and z are constant i.e. dy=dz=0 so the formula becomes
$2r dr =2xdx$
$\frac{dr}{dx}=\frac{x}{r}$
Noting that this relation holds in the condition of y and z are constant, we write it by "round d"
\frac{\partial r}{\partial x}=\frac{x}{r}
Similarly
\frac{\partial r}{\partial y}=\frac{y}{r},\ \frac{\partial r}{\partial z}=\frac{z}{r}
Try these substitutions in the formula in post #2 to know whether you get answer in the text.

 

[FactChecker]

I'm never going to be able to figure that out on my own.

Every subject has a handful of tricks. After you see them a few times, they become routine.

 

[vela]

But BOI I'm never going to be able to figure that out on my own.

I find It's helpful to go back over a solution to boil it down to the essentials.

In math, it's often quite helpful to know the definitions. It can help you avoid mistakes and at the very least give you a place to start. For example, I'm sure your book defines the gradient of $f$ as
$$\nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right).$$ Taking the gradient of a scalar function results in a vector quantity. In this problem, you needed to calculate
$$\nabla f = \left(\frac{\partial r^3}{\partial x},\frac{\partial r^3}{\partial y},\frac{\partial r^3}{\partial z}\right).$$

From what else you learned in this thread about the meaning of $r$, it shouldn't be a big mystery now how to calculate the three partial derivatives. In other words, do you understand what @Steve4Physics did above? Hopefully you do, but if it is confusing, that's an indication you have a little work to do.

So knowing those two things—the definition of gradient and the definition of $r$—do you really still think you couldn't have figured it out on your own?

 

[Amentia]

Hello,

I don't know if the author will see this answer but I would like to point out that I do not find their idea to use spherical coordinates to calculate $\nabla f$ nonsensical.

Indeed, from the expression of the gradient operator in spherical coordinates: https://mathworld.wolfram.com/SphericalCoordinates.html

i.e.:

$$\boldsymbol{\nabla}{f} = \mathbf{e}_r \frac{\partial{f}}{\partial{r}}
+
\mathbf{e}_\theta \frac{1}{r } \frac{\partial{f}}{\partial{\theta}}
+
\mathbf{e}_\phi \frac{1}{r\sin{\theta}} \frac{\partial{f}}{\partial{\phi}}$$

Since $f=r^{3}$ does not depend on $\theta$ and $\phi$, only the first term contributes to $\nabla f$. It is simply equal to:

$$\nabla f = \mathbf{e}_r \frac{\partial{f}}{\partial{r}} = \mathbf{e}_r 3r^{2} $$,

which is exactly the result @Addez123 found in their first post, but forgetting that it should be multiplied with the unit vector $\mathbf{e}_r$. It only remains to express $\mathbf{e}_r$ back in cartesian coordinates as:

$$\mathbf{e}_r = \frac{\boldsymbol{r}}{r} = \frac{1}{r} (x,y,z)$$

Then we are left with:
$$\nabla f = 3r(x,y,z)$$

and the inner product with the vector $\boldsymbol{A}$ provides the expected result.

 

[Amentia]

What I mean above is that the message quoted below is directly leading to the correct answer if explicited correctly. We simply do not need all the sine and cosine, we replace them with x/r, y/r and z/r.

$$\frac{df}{dr}(\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z}) =$$
$$3r^2 (cos(u)sin(v), sin(u)sin(v), cos(u))$$
That really doesn't simplify anything.

 

[vela]

I don't know if the author will see this answer but I would like to point out that I do not find their idea to use spherical coordinates to calculate $\nabla f$ nonsensical.

I think most here didn't think the OP's problem was the choice of coordinate system and simply forgetting a unit vector. It was that he or she didn't understand the basics of calculating a gradient (in any coordinate system), as the OP admitted to in post #3.

 

[Amentia]

I think most here didn't think the OP's problem was the choice of coordinate system and simply forgetting a unit vector. It was that he or she didn't understand the basics of calculating a gradient (in any coordinate system), as the OP admitted to in post #3.

I think they are right, it is even written in the title of the thread. What I am trying to do is to give a solution that could help the OP understand better how it works by providing a derivation that is closer to what they were trying to achieve.

Sometimes, it is helpful to see a correction of what you have already tried and which leads to the correct solution, rather than a good solution but which looks much less familiar because you did not think at all about it while working on the problem.

It is just the way I envision the learning process in most people brains when dealing with a new topic.