Solving Gradient Problem (c): LHS ≠ RHS

[NewtonApple]

Homework Statement

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

2. Homework Equations

Let $\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

and $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$

The Attempt at a Solution

Consider left side of the inequality.

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

L.H.S. $=\overrightarrow{r}.(\nabla.\overrightarrow{r})$

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


Now consider right side of the inequality.

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

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

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

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

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

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

Hence L.H.S. $\neq$ R.H.S.

 

[vela]

Homework Statement

View attachment 77205 [/B]
Solving part (c) which should be
$\overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r$

That still can't be right. See below.

Consider left side of the inequality.

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

L.H.S. $=\overrightarrow{r}.(\nabla.\overrightarrow{r})$

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

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.

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

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

You can't change the order of $r$ and $\nabla$. The expression on the RHS is equal to $\nabla r$, not $r \nabla$.

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

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

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

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. $=r\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}=\overrightarrow{r}$

Hence L.H.S. $\neq$ R.H.S.

The RHS you started with was $r\nabla r$. At the end, you incorrectly say it's $r(r\nabla)$.

 

[NewtonApple]

Thx Vela! I'll give it another try.

There is a misprint in the problem statement.

I think (c) part should be $\overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r$
Are you OK with it?

 

[vela]

No.

 

[haruspex]

$r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$

In the photocopied text, all occurrences of 'r' are in bold, implying vectors.

 

[BvU]

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 :)