Proof of a vector identity in electromagnetism

[Ishika_96_sparkles]

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
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?

 

[PeroK]

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

This is wrong. The LHS is a vector and the RHS doesn't make much sense. In any case:
$$\vec{\nabla}(\vec a \cdot \vec b) \ne (\vec \nabla \cdot \vec{a})\vec b + (\vec \nabla \cdot \vec{b})\vec a$$

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}##

Method 2 is correct.

 

[PeroK]

To tidy this up, we have:
$$\vec \nabla (\frac 1 {r}) = -\frac{\vec r}{r^3}$$$$\vec{\nabla}(\vec M \cdot \vec r) = \vec M$$$$\vec \nabla (\frac 1 {r^3}) = -\frac{3\vec r}{r^5}$$Hence:
$$ \vec{\nabla} \big[\vec{M}\cdot\vec{\nabla} \big(\frac{1}{r}\big)\big] = - \vec{\nabla} \big[\frac{\vec{M}\cdot\vec r}{r^3}\big]$$$$-\big[\vec{\nabla}(\vec{M}\cdot \vec{r}) \frac{1}{r^3} + (\vec{M}\cdot \vec{r}) \big(\vec{\nabla} \frac{1}{r^3}\big) \big] = -\big[\frac{\vec M}{r^3} - (\vec{M}\cdot \vec{r})\frac{3\vec r}{r^5}\big ]$$$$= \frac{3(\vec{M}\cdot \vec{r})\vec r}{r^5} - \frac{\vec M}{r^3}$$

 

[Ishika_96_sparkles]

Thank you so much!!!

 

[Orodruin]

This is wrong. The LHS is a vector and the RHS doesn't make much sense.

Not so. The RHS will make sense if it is interpreted properly, which is by using the tensor formalism and realising that ##\vec \nabla \vec M## (ie, the gradient of ##\vec M##) should be interpreted as a rank 2 tensor. The application of this tensor on the position vector with the dot product is correct.

In index form
$$
\nabla(\vec M \cdot \vec r) = \vec e_i \partial_i (M_j x_j) = \vec e_i (\partial_i M_j) x_j + \vec e_i M_i
$$
where ##\partial_i M_j## would be the components of ##\nabla \vec M##.

Granted, this will often not be covered in introductory vector analysis and randomly moving the dot product around is certainly not the solution.

It should also be noted that the first term will always disappear as ##\vec M## is considered constant.

 

[PeroK]

Not so. The RHS will make sense if it is interpreted properly, which is by using the tensor formalism and realising that ##\vec \nabla \vec M## (ie, the gradient of ##\vec M##) should be interpreted as a rank 2 tensor. The application of this tensor on the position vector with the dot product is correct.

In index form
$$
\nabla(\vec M \cdot \vec r) = \vec e_i \partial_i (M_j x_j) = \vec e_i (\partial_i M_j) x_j + \vec e_i M_i
$$
where ##\partial_i M_j## would be the components of ##\nabla \vec M##.

Granted, this will often not be covered in introductory vector analysis and randomly moving the dot product around is certainly not the solution.

It should also be noted that the first term will always disappear as ##\vec M## is considered constant.

Back with a bang!

 

[Ishika_96_sparkles]

Thank you @PeroK and @Orodruin for your interest in my query, your valuable insights and for your time.

@PeroK I came across the following identity (No idea who derived it and how it was constructed) but it solves my question straightaway and also gives birth to another query. Let us see the following identity
$$ \vec{\nabla}(\vec{A}\cdot\vec{B}) = \vec{A}\times (\vec{\nabla} \times \vec{B} ) + \vec{B} \times (\vec{\nabla} \times \vec{A} ) +(\vec{A}\cdot\vec{\nabla})\vec{B}+(\vec{B}\cdot\vec{\nabla}) \vec{A}$$
Where, if we use ##\vec{A}=\vec{M}## and ##\vec{B}=\vec{r}##, the cross-product terms vanish, and we are left with the two dot-product terms. Out of these terms only one is non-zero i.e., ##(\vec{M}\cdot\vec{\nabla}) \vec{r}## which gives ##\vec{M}## as the output.

Now, here is my question for both of you, in the light of the following statement

Granted, this will often not be covered in introductory vector analysis and randomly moving the dot product around is certainly not the solution.

why the gradient operator changed to the curl and dot product in this identity, while my assumption in the OP about the dot-product in the terms ##\vec{\nabla}\vec{M}## and ##\vec{\nabla}\vec{r}## gave wrong result?

@Orodruin Are these terms i.e., ##\vec{\nabla}\vec{M}## and ##\vec{\nabla}\vec{r}## called the dyads in ancient scientific literature?

 

[PeroK]

Thank you @PeroK and @Orodruin for your interest in my query, your valuable insights and for your time.

@PeroK I came across the following identity (No idea who derived it and how it was constructed) but it solves my question straightaway and also gives birth to another query. Let us see the following identity
$$ \vec{\nabla}(\vec{A}\cdot\vec{B}) = \vec{A}\times (\vec{\nabla} \times \vec{B} ) + \vec{B} \times (\vec{\nabla} \times \vec{A} ) +(\vec{A}\cdot\vec{\nabla})\vec{B}+(\vec{B}\cdot\vec{\nabla}) \vec{A}$$

It's not too hard to prove. Just a bit messy. It's a standard identity in vector calculus.

 

[Orodruin]

why the gradient operator changed to the curl and dot product in this identity, while my assumption in the OP about the dot-product in the terms ∇→M→ and ∇→r→ gave wrong result?

Are you familiar with the BAC-CAB rule? This identity has the same vector structure with the addition that the derivative acts on a product so you get four terms due to the Leibniz rule rather than just two terms.

@Orodruin Are these terms i.e., ∇→M→ and ∇→r→ called the dyads in ancient scientific literature?

Yes. It is not a notation I am particularly fond of, I prefer index notation - it is usually much clearer.

 

[Ishika_96_sparkles]

Are you familiar with the BAC-CAB rule? This identity has the same vector structure with the addition that the derivative acts on a product so you get four terms due to the Leibniz rule rather than just two terms.

yes i am aware of the rule and proved it as a part of exercise. So, this one uses two pieces of knowledge
1) BAC-CAB rule and 2) LEIBNITZ RULE. I get it now that its just a mater of notation or dressing up but the structure is the same...just a nicer way of writing an identity.

Thank you very much for the reply.