- #1

Ishika_96_sparkles

- 57

- 22

- Homework Statement
- Prove that ##\vec{\nabla} \big[\vec{M}\cdot\vec{\nabla} \big(\frac{1}{r}\big)\big]=\frac{3(\vec{M}\cdot\vec{r})\vec{r}}{r^5}-\frac{\vec{M}}{r^3}##

- Relevant Equations
- ##\vec{\nabla} \big(\frac{1}{r}\big) =-\frac{\vec{r}}{r^3}##

During the calculations, I tried to solve the following

$$ \vec{\nabla} \big[\vec{M}\cdot\vec{\nabla} \big(\frac{1}{r}\big)\big] = -\big[\vec{\nabla}(\vec{M}\cdot \vec{r}) \frac{1}{r^3} + (\vec{M}\cdot \vec{r}) \big(\vec{\nabla} \frac{1}{r^3}\big) \big]$$

by solving the first term i.e., ##\frac{1}{r^3} \vec{\nabla}(\vec{M}\cdot \vec{r}) ## by using the following ways

We keep in mind that ##\vec{A} ## here is a constant vector.

1) $$ \vec{\nabla}(\vec{M}\cdot \vec{r})= (\vec{\nabla}\vec{M}) \cdot \vec{r} +\vec{M}\cdot ( \vec{\nabla} \vec{r})$$

now, since ##\vec{\nabla}\vec{A}## only makes sense if we have a

$$(\vec{\nabla} \cdot \vec{M}) \cdot \vec{r} +\vec{M} \cdot ( \vec{\nabla} \cdot \vec{r})= 0 +\vec{M}\cdot ( \vec{\nabla} \cdot \vec{r})$$

since, ##\vec{M}## is a

However, by doing the method

$$\vec{\nabla}(\vec{M}\cdot \vec{r})= \vec{\nabla}(M_1 x+M_2 y+M_3 z)$$

This is a gradient of a scalar and thus could be written as ##(M_1\vec{\nabla} x+M_2 \vec{\nabla}y+M_3 \vec{\nabla}z) =(M_1 \hat{i}+M_2 \hat{j}+M_3 \hat{k})=\vec{M}##

What am I doing wrong here?

$$ \vec{\nabla} \big[\vec{M}\cdot\vec{\nabla} \big(\frac{1}{r}\big)\big] = -\big[\vec{\nabla}(\vec{M}\cdot \vec{r}) \frac{1}{r^3} + (\vec{M}\cdot \vec{r}) \big(\vec{\nabla} \frac{1}{r^3}\big) \big]$$

by solving the first term i.e., ##\frac{1}{r^3} \vec{\nabla}(\vec{M}\cdot \vec{r}) ## by using the following ways

**1)**##\frac{d (\vec{A}\cdot\vec{B})}{dt}=\frac{d \vec{A}}{dt} \cdot\vec{B} +\vec{A}\cdot \frac{d \vec{B}}{dt}## and by**2)**##\frac{d (\vec{A}\cdot\vec{B})}{dt}=\frac{d (A_xB_x+A_yB_y+A_zB_z)}{dt}##.We keep in mind that ##\vec{A} ## here is a constant vector.

1) $$ \vec{\nabla}(\vec{M}\cdot \vec{r})= (\vec{\nabla}\vec{M}) \cdot \vec{r} +\vec{M}\cdot ( \vec{\nabla} \vec{r})$$

now, since ##\vec{\nabla}\vec{A}## only makes sense if we have a

**dot product**between them i.e., the**divergence**##\vec{\nabla}\cdot\vec{A}##. Therefore, we proceed, by__assuming a dot product__here to get$$(\vec{\nabla} \cdot \vec{M}) \cdot \vec{r} +\vec{M} \cdot ( \vec{\nabla} \cdot \vec{r})= 0 +\vec{M}\cdot ( \vec{\nabla} \cdot \vec{r})$$

since, ##\vec{M}## is a

**constant vector**and ##\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}##. Now, ##\vec{\nabla} \cdot \vec{r}=3## and thus, we get the answer ##3\vec{M}## and hence ##-\frac{\vec{3M}}{r^3}##.However, by doing the method

**2)**we get$$\vec{\nabla}(\vec{M}\cdot \vec{r})= \vec{\nabla}(M_1 x+M_2 y+M_3 z)$$

This is a gradient of a scalar and thus could be written as ##(M_1\vec{\nabla} x+M_2 \vec{\nabla}y+M_3 \vec{\nabla}z) =(M_1 \hat{i}+M_2 \hat{j}+M_3 \hat{k})=\vec{M}##

What am I doing wrong here?

Last edited: