Calculating Divergence of a Gradient in Cartesian Coordinates

[NewtonApple]

Homework Statement

Homework Equations

The Attempt at a Solution

(a)[/B] Divergence of a gradient is a Laplacian.

(b) I suppose to do it in Cartesian coordinates.

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

\hat{r}=\hat{i}+\hat{j}+\hat{k}

First calculate \nabla\left(\frac{1}{\overrightarrow{r}}\right)=\nabla\left(\frac{1}{\mid r\mid\hat{r}}\right)=\nabla\left(\frac{1}{\mid r\mid}\hat{r}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\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}}

Am I doing it the right way?

PS. cross posted at
https://www.physicsforums.com/threads/laplacian-of-a-vector.789514/#post-4959007

 

[Ray Vickson]

Homework Statement

View attachment 77030

Homework Equations

The Attempt at a Solution

(a)[/B] Divergence of a gradient is a Laplacian.

(b) I suppose to do it in Cartesian coordinates.

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

\hat{r}=\hat{i}+\hat{j}+\hat{k}

First calculate \nabla\left(\frac{1}{\overrightarrow{r}}\right)=\nabla\left(\frac{1}{\mid r\mid\hat{r}}\right)=\nabla\left(\frac{1}{\mid r\mid}\hat{r}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\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}}

Am I doing it the right way?

PS. cross posted at
https://www.physicsforums.com/threads/laplacian-of-a-vector.789514/#post-4959007

No. $\hat{r}$ is a vector, so $\nabla(\hat{r}/r)$ does not make sense: you need to take the gradient of a scalar, not of a vector. Anyway, why do you define $\hat{r}=\hat{i}+\hat{j}+\hat{k}$? I cannot see its relation to anything in the problem. If by $\hat{r}$ you mean the unit vector in the direction of $\vec{r}$, then that is most certainly not equal to what you wrote.

Also, never, never write something like $\frac{1}{|r|\hat{r}}$ because that is meaningless: it is a fraction of the form 1/vector, and those things do not exist in any usual form.

 

[BvU]

Dear Isaac,

If you yourself write $\vec r = \hat {\bf r} | {\bf r}|$ you really should go back to $\vec {r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$ and correct the hideous $\hat{r}=\hat{\imath}+\hat{\jmath}+\hat{k}$ to $\hat{r}={x\over \sqrt{x^{2}+y^{2}+z^{2}}} \hat{\imath} + {y\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{\jmath} + {z\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{k}$

[edit] sorry, pressed wrong button.

But, as Ray (And TSny) indicate, you don't need $\hat r$

In your defence: The original problem formulation is confusing because it uses a boldface r in the 1/r. Many of us are conditioned to see that as a vector.

 

[NewtonApple]

Dear Isaac,

If you yourself write $\vec r = \hat {\bf r} | {\bf r}|$ you really should go back to $\vec {r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$ and correct the hideous $\hat{r}=\hat{\imath}+\hat{\jmath}+\hat{k}$ to $\hat{r}={x\over \sqrt{x^{2}+y^{2}+z^{2}}} \hat{\imath} + {y\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{\jmath} + {z\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{k}$

[edit] sorry, pressed wrong button.

But, as Ray (And TSny) indicate, you don't need $\hat r$

In your defence: The original problem formulation is confusing because it uses a boldface r in the 1/r. Many of us are conditioned to see that as a vector.

Dear BvU, I think Author referred it as a vector. In the book (Mathematical Methods for Physicists by Tai L. Chow) scalars are mentioned as non bold, such as in same exercise page other problems are

 

[BvU]

I count it as a misprint. What could possibly be the interpretation of $1/\;\vec {\bf r}$ ?

 

[NewtonApple]

Ok, thanks for the input.

Homework Statement

Show that \nabla^{2}\left(\frac{1}{r}\right)=0

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

The Attempt at a Solution

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

\nabla\left(\frac{1}{r}\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}}

\nabla\left(\frac{1}{r}\right)=-\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}

\nabla.\nabla\left(\frac{1}{r}\right)=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).\left(-\hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)

\nabla^{2}\left(\frac{1}{r}\right)=\frac{\partial}{\partial x}\left(-x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)+\frac{\partial}{\partial y}\left(-y\left(x^{2}+y^{2}+z^{2}\right)\right)+\frac{\partial}{\partial z}-\left(z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)

\nabla^{2}\left(\frac{1}{r}\right)=-x\left(2x\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-y\left(2y\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}
- \left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-z\left(2z\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}

\nabla^{2}\left(\frac{1}{r}\right)=3x^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}+3y^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}
+3z^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}

\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}

\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}+1}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}

\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\nabla^{2}\left(\frac{1}{r}\right)=0

Hence Showed

 

[BvU]

You allowed to use spherical coordinates ? Makes this a piece of cake...

 

[NewtonApple]

Yes, but we've to it in both - Cartesian and Spherical coordinates.

 

[NewtonApple]

Yes, but we've to it in both - Cartesian and Spherical coordinates.

^ solve it

 

[NewtonApple]

Solving part (c). As suggested above it's also a misprint. It should be
\overrightarrow{r}.\left(\nabla.\overrightarrow{r}\right)\neq\left(r\nabla\right)r