# Solving Gradient Problem (c): LHS ≠ RHS

• NewtonApple
In summary, the problem is to prove that ##{\bf r} \; (\nabla\cdot {\bf r}) \ne ({\bf r} \cdot \nabla)\;{\bf r}##. This can be done by using the formulas for ##\nabla \cdot {\bf r}## and ##{\bf A} \cdot \nabla{\bf r}## and substituting ##{\bf r}## for ##{\bf A}##.

## 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.

NewtonApple said:

## 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?

No.

NewtonApple said:
##r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}##
In the photocopied text, all occurrences of 'r' are in bold, implying vectors.

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 :)

NewtonApple

## 1. What is the "Solving Gradient Problem (c)"?

The "Solving Gradient Problem (c)" refers to a mathematical concept where the left-hand side (LHS) of an equation does not equal the right-hand side (RHS). This can occur in gradient problems, which involve finding the slope or rate of change of a function at a specific point.

## 2. Why is it important to solve the gradient problem?

Solving the gradient problem is important because it allows us to understand the behavior of a function at a specific point. This information is useful in various fields, such as physics, economics, and engineering, as it helps us make predictions and solve real-world problems.

## 3. What are some common reasons for LHS ≠ RHS in gradient problems?

There can be several reasons why the LHS and RHS do not match in a gradient problem. These can include errors in calculations, incorrect input values, or inconsistencies in the given function or equation.

## 4. How can I verify my solution to the gradient problem?

The best way to verify your solution to the gradient problem is to double-check your calculations and make sure you have used the correct input values. You can also try solving the problem using a different method or asking for feedback from a colleague or teacher.

## 5. Are there any tips for solving gradient problems?

Yes, there are some helpful tips for solving gradient problems. First, make sure you understand the basic concepts of gradients and slope. Then, carefully read and analyze the given problem to identify the unknown variables and what is being asked. Finally, use the appropriate mathematical formulas and techniques to solve the problem step-by-step.