Prove a vector identity for $\boldsymbol{\nabla} \left( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime \right)$

Question

Context

I am studying radiation fields using [1]. In Eq. 20.98 in [1] I came across an identity (see Eq 1). I have attempted to use the the chain rule, and product rule and other identities found [2], [3], and [4] to demonstrate this identity and have not succeeded.

Question

Given two vector fields $\mathbf{r}$ and $\mathbf{r}^\prime$, where $\mathbf{r} \in \mathbb{R}^n$ and $\mathbf{r}^\prime \in \mathbb{R}^n$, the unit vector $\hat{\mathbf{r}}$, and that the differential operator $\boldsymbol{\nabla}$ operates on $\mathbf{r}$, can you demonstrate that $$\boldsymbol{\nabla} \left( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime \right) = \hat{\mathbf{r}} \times\left( \hat{\mathbf{r}} \times \frac{\mathbf{r}^\prime}{r} \right) ?\tag{1}$$

My Attempt

I know from [2] that $$ \nabla( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime ) =\ ( \hat{\mathbf{r}} \cdot \nabla)\mathbf{r}^\prime \,+\, (\mathbf{r}^\prime \cdot \nabla) \hat{\mathbf{r}} \,+\, \hat{\mathbf{r}} {\times} (\nabla {\times}\mathbf{r}^\prime) \,+\, \mathbf{r}^\prime {\times} (\nabla {\times} \hat{\mathbf{r}} ) . $$ The third term on the right is zero, because the del operator acts on $\mathbf{r}$ and does not act on $\mathbf{r}^\prime$. The fourth term is zero according to the definition of the curl in spherical coordinates. Now, I know also from [2] that in spherical coordinates \begin{align} (\mathbf{A} \cdot \boldsymbol{\nabla})\mathbf{B} = \left( A_r \frac{\partial B_r}{\partial r} + \frac{A_\theta}{r} \frac{\partial B_r}{\partial \theta} + \frac{A_\varphi}{r\sin\theta} \frac{\partial B_r}{\partial \varphi} - \frac{A_\theta B_\theta + A_\varphi B_\varphi}{r} \right) &\hat{\mathbf r} \\ + \left( A_r \frac{\partial B_\theta}{\partial r} + \frac{A_\theta}{r} \frac{\partial B_\theta}{\partial \theta} + \frac{A_\varphi}{r\sin\theta} \frac{\partial B_\theta}{\partial \varphi} + \frac{A_\theta B_r}{r} - \frac{A_\varphi B_\varphi\cot\theta}{r} \right) &\hat{\boldsymbol \theta} \\ + \left( A_r \frac{\partial B_\varphi}{\partial r} + \frac{A_\theta}{r} \frac{\partial B_\varphi}{\partial \theta} + \frac{A_\varphi}{r\sin\theta} \frac{\partial B_\varphi}{\partial \varphi} + \frac{A_\varphi B_r}{r} + \frac{A_\varphi B_\theta \cot\theta}{r} \right) &\hat{\boldsymbol \varphi}. \end{align} Thus, \begin{align} (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla})\mathbf{r}^\prime = \mathbf{0} \end{align} I arrive at $$ \boxed{ \boldsymbol{\nabla}( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime ) = (\mathbf{r}^\prime \cdot \boldsymbol{\nabla}) \hat{\mathbf{r}} .} $$ This is not that which I was supposed to prove.

Bibliography

[1] Zangwill, Modern Electrodynamics, 2013, p. 733. [2] https://en.wikipedia.org/wiki/Vector_calculus_identities

[3] https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

[4] https://en.wikipedia.org/wiki/Triple_product

Answer 1

Let me write $\mathbf{a} = \mathbf{r}'$ to better distinguish $\mathbf{r}$ and $\mathbf{r}'$. Then by BAC-CAB rule,

$$ \hat{\mathbf{r}} \times \left( \frac{\hat{\mathbf{r}}}{r} \times \mathbf{a} \right) = \frac{\hat{\mathbf{r}} (\hat{\mathbf{r}} \cdot \mathbf{a}) - \mathbf{a} (\hat{\mathbf{r}} \cdot \hat{\mathbf{r}})}{r}. \tag{1} $$

On the other hand, writing $\mathbf{r} = (x_1, x_2, x_3)$ and noting that $r = \sqrt{x_1^2 + x_2^2 + x_3^2}$,

\begin{align*} \nabla (\hat{\mathbf{r}} \cdot \mathbf{a}) &= \sum_{i} \mathbf{e}_i \frac{\partial}{\partial x_i} \left( \frac{\mathbf{r}}{r} \cdot \mathbf{a} \right) \\ &= \sum_{i} \mathbf{e}_i \left[ \left( \frac{\mathbf{e}_i}{r} - \frac{\mathbf{r}x_i}{r^3} \right) \cdot \mathbf{a} \right] \\ &= \frac{\sum_{i} \mathbf{e}_i (\mathbf{e}_i \cdot \mathbf{a})}{r} - \frac{\left( \sum_{i} \mathbf{e}_i x_i \right)(\mathbf{r} \cdot \mathbf{a})}{r^3} \\ &= \frac{\mathbf{a}}{r} - \frac{\mathbf{r} (\mathbf{r}\cdot\mathbf{a})}{r^3} \tag{2} \end{align*}

From this, it actually turns out that

$$ \nabla (\hat{\mathbf{r}} \cdot \mathbf{a}) = -\hat{\mathbf{r}} \times \left( \frac{\hat{\mathbf{r}}}{r} \times \mathbf{a} \right). $$

Mathematica 13 also verifies this:

Answer 2

$ \renewcommand\vec\mathbf \newcommand\uvec[1]{\hat{\vec#1}} $First $$ \nabla(\uvec r\cdot\vec　r') = \nabla(\frac{\vec r}r\cdot\vec r') = (\nabla\frac1r)(\vec r\cdot\vec r') + \frac1r\nabla(\vec r\cdot\vec r'). $$ The second term we get from the fundamental identity $$ \nabla(\vec r\cdot\vec r') = \vec r'. \tag{$*$} $$ For the first term we use the chain rule to get $$ \nabla\frac1r = \frac{-\nabla r}{r^2} = \frac{-\uvec r}{r^2}. $$ Thus all together $$ \nabla(\uvec r\cdot\vec　r') = \frac{-\uvec r}r(\uvec r\cdot\vec r') + \frac{\vec r'}r. $$ Because $\uvec r\cdot\uvec r = 1$ we can rearrange and write $$ \nabla(\uvec r\cdot\vec　r') = \frac1r(\uvec r\cdot\uvec r)\vec r' - \frac1r(\uvec r\cdot\vec r')\uvec r $$ so finally by the BAC-CAB identity $$ \nabla(\uvec r\cdot\vec　r') = \frac1r\uvec r\times(\vec r'\times\uvec r) $$ which actually differs from the desired identity by a minus sign.

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This is indeed the exact same thing as $$ (\vec r'\cdot\nabla)\uvec r $$ because we have another fundamental identity closely related to ($*$): $$ (\vec r'\cdot\nabla)\vec r = \vec r'. $$ If you expand $(\vec r'\cdot\nabla)\uvec r$ by writing $\uvec r = \vec r/r$ like I did above, then you will quickly see that we arrive at the same result.

Answer 3

I was really confused by the notations and might be completely lost without reading other answers. It actually ended up that the $\mathbf{r}^\prime$ is a constant vector if I am wrong otherwise. $$ \nabla(\hat{\mathbf{r}}\cdot\mathbf{r}^\prime )=\require{cancel}\cancel{(\hat{\mathbf{r}}\cdot \nabla)\mathbf{r}^\prime}+\require{cancel} \cancel{(\mathbf{r}^\prime \cdot \nabla) \hat{\mathbf{r}}}+\hat{\mathbf{r}} {\times} (\nabla {\times}\mathbf{r}^\prime)+\require{cancel} \cancel{\mathbf{r}^\prime {\times} (\nabla {\times} \hat{\mathbf{r}})}. $$ As mentioned several times by Nicholas Todoroff. This involves indeed some "fundamental identities" which we are not aware of. For instance for an $m-$dimensional functional $f(\mathbf{r})$ and $\mathbf{r}=[x_1,\ldots,x_m]$, the following is a list I have derived one by one by hand and hope experts can help to check and make it useful. \begin{align} \nabla\cdot(r^n\boldsymbol{\hat{\textbf{r}}})&=\nabla\cdot(r^{n-1}\mathbf{r})=r^{n-1}\nabla\cdot\mathbf{r}+\mathbf{r}\cdot\nabla r^{n-1}=(m+n-1)r^{n-1}\\ \nabla\cdot\frac{\boldsymbol{\hat{\textbf{r}}}}{r^n}&=\nabla\cdot\frac{\mathbf{r}}{r^{n+1}}=\frac{1}{r^{n+1}}\nabla\cdot\mathbf{r}+\mathbf{r}\cdot\nabla\frac{1}{r^{n+1}}=\frac{m-1-n}{r^{n+1}}\label{eq:Diverger^-n}\\ \nabla{r^n}&=n{r^{n-1}}\nabla r=n{r^{n-1}}\boldsymbol{\hat{\textbf{r}}}=n{r^{n-2}}\mathbf{r}\\ \nabla\frac{1}{r^n}&=\frac{-n\nabla r}{r^{n+1}}=\frac{-n\mathbf{r}}{r^{n+2}}\\ \nabla^2\frac{1}{r^n}&=\nabla\cdot\nabla\frac{1}{r^n}=\frac{n(n+2-m)}{r^{n+2}}\\ \nabla\times\mathbf{r}&=0\\ \nabla\times\boldsymbol{\hat{\textbf{r}}}&=\frac{1}{r}\nabla\times\mathbf{r}+\nabla\frac{1}{r}\times\mathbf{r}=0-\frac{\mathbf{r}\times\mathbf{r}}{r^3}=0 \end{align}

\begin{align} \nabla(\mathbf{A} \cdot\mathbf{B})&=\underbrace{(\mathbf{A} \cdot\nabla)\mathbf{B}+\mathbf{A} {\times}(\nabla{\times}\mathbf{B})}_{=\mathbf{A} \cdot\nabla\!\!\mathbf{B}}+\underbrace{(\mathbf{B}\cdot\nabla)\mathbf{A} +\mathbf{B}{\times}(\nabla{\times}\mathbf{A} )}_{=\mathbf{B}\cdot\nabla\!\!\mathbf{A} }\\ &=\mathbf{A} \cdot\mathbf{J}^{\!\!\top}_{\!\!\mathbf{B}}+\mathbf{B}\cdot\mathbf{J}^{\!\!\top}_{\!\!\mathbf{A} }=\mathbf{A} \cdot\nabla\!\!\mathbf{B}+\mathbf{B}\cdot\nabla\!\!\mathbf{A} \end{align} where $\mathbf{J}^{\!\!\top}_{\!\!\mathbf{A} }=\nabla\!\!\mathbf{A} = (\partial A_j/\partial x_i)_{ij}=(\partial_iA_j)$ denotes the transposed Jacobian matrix of the vector field $\mathbf{A} =(A_1,\ldots,A_n)$, and in the last expression the $\cdot$ operations are understood not to act on the $\nabla$ directions (different definition in Wiki:Vector_calculus_identities, but to inner-product of the vector immediately after $\nabla$. Alternatively, using Feynman subscript notation, \begin{align} \nabla(\mathbf{A} \cdot\mathbf{B}) = \nabla_{\mathbf{A} }(\mathbf{A} \cdot \mathbf{B}) + \nabla_{\mathbf{B}}(\mathbf{A} \cdot\mathbf{B}). \end{align}

Note the difference: \begin{align} \mathbf{A} \cdot\nabla\mathbf{B}&=\nabla_{\!\mathbf{B}}(\mathbf{A} \cdot\mathbf{B})=\mathbf{A} \cdot\mathbf{J}^{\!\!\top}_{\!\!\mathbf{B}}=A_j\left({\frac{\partial B_j}{\partial x_i}}\right)=\sum_jA_j\partial_iB_j\\ (\mathbf{A} \cdot\nabla)\mathbf{B}&=\left(A_j{\frac{\partial}{\partial x_j}}\right)B_i=\mathbf{A} \cdot\mathbf{J}_{\!\!\mathbf{B}}=A_j\left({\frac{\partial B_i}{\partial x_j}}\right)=\sum_jA_j\partial_jB_i \end{align} Note: the squeezed notation is to emphasize the connection and please remove it if annoying.