• Support PF! Buy your school textbooks, materials and every day products Here!

Gradient problem

  • #1

Homework Statement


image024.gif
[/B]
Solving part (c) which should be
[itex]\overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r[/itex]

2. Homework Equations


Let [itex]\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}[/itex]

and [itex] \overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}[/itex]
[itex] r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}[/itex]

The Attempt at a Solution



Consider left side of the inequality.

Now [itex]\nabla.\overrightarrow{r}=
(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}
).\left(x\hat{i}+y\hat{j}+z\hat{k}\right)=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3[/itex]

L.H.S. [itex]=\overrightarrow{r}.(\nabla.\overrightarrow{r})[/itex]

L.H.S. [itex]= \left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).3=3\overrightarrow{r}[/itex]


Now consider right side of the inequality.

[itex]\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)[/itex]

[itex]\left(r\nabla\right) =\hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}[/itex]

[itex]\left(r\nabla\right)=\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

[itex]\left(r\nabla\right)=
\hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

[itex]\left(r\nabla\right)=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}[/itex]

R.H.S. [itex]=r\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}=\overrightarrow{r}[/itex]

Hence L.H.S. [itex]\neq[/itex] R.H.S.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,196

Homework Statement


View attachment 77205 [/B]
Solving part (c) which should be
[itex]\overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r[/itex]
That still can't be right. See below.

Consider left side of the inequality.

Now [itex]\nabla.\overrightarrow{r}=
(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}
).\left(x\hat{i}+y\hat{j}+z\hat{k}\right)=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3[/itex]

L.H.S. [itex]=\overrightarrow{r}.(\nabla.\overrightarrow{r})[/itex]

L.H.S. [itex]= \left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).3=3\overrightarrow{r}[/itex]
Your last line is incorrect. You found ##\nabla\cdot\vec{r} = 3##, so you should have ##\vec{r}\cdot(\nabla\cdot\vec{r}) = \vec{r}\cdot 3##, which doesn't make sense because you can't dot a vector with a scalar.


Now consider right side of the inequality.

[itex]\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)[/itex]

[itex]\left(r\nabla\right) =\hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}[/itex]
You can't change the order of ##r## and ##\nabla##. The expression on the RHS is equal to ##\nabla r##, not ##r \nabla##.

[itex]\left(r\nabla\right)=\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

[itex]\left(r\nabla\right)=
\hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

[itex]\left(r\nabla\right)=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}[/itex]
What you actually calculated is ##\nabla r = \hat{r}##. It's not for nothing, however, as you need this to correctly evaluate the righthand side.

R.H.S. [itex]=r\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}=\overrightarrow{r}[/itex]

Hence L.H.S. [itex]\neq[/itex] R.H.S.
The RHS you started with was ##r\nabla r##. At the end, you incorrectly say it's ##r(r\nabla)##.
 
  • #3
Thx Vela! I'll give it another try.

There is a misprint in the problem statement.
image024.gif

I think (c) part should be [itex] \overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r [/itex]
Are you OK with it?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,196
No.
 
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,722
5,029
##r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}##
In the photocopied text, all occurrences of 'r' are in bold, implying vectors.
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,019
3,009
In your chapter 1 in the Chow book, under "Formulas involving ##\nabla## " , you find ##\nabla\cdot {\bf r} = 3## and ##({\bf A}\cdot \nabla){\bf r} = {\bf A}## for A any differentiable vector field function. Substituting ##{\bf A} = {\bf r}## gives ##({\bf r}\cdot \nabla){\bf r} = {\bf r}##.

##{\bf r} \cdot (\nabla\cdot {\bf r}) ## would be a vector dotted into a scalar. Doesn't make sense.

And in ##({\bf r} \nabla){\bf r} = {\bf r}## the ##({\bf r} \nabla){\bf r}## (without the ##\cdot## ) could be a matrix for all we know.

As in 1.18(b), mr Chow seems to be a bit sloppy, or at least inconsistent with his notation. Best I can make of it is that he wants you to prove that
##{\bf r} \; (\nabla\cdot {\bf r}) \ne ({\bf r} \cdot \nabla)\;{\bf r} ##, which by now should be a piece of cake for you :)
 

Related Threads on Gradient problem

  • Last Post
Replies
1
Views
966
  • Last Post
Replies
1
Views
657
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
581
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
6
Views
4K
Top