# Show the identity ##\vec{\nabla}(\vec{r} \cdot \vec{u})##

Homework Statement:
Show the identity ##\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})##, ##\vec{L} = \frac{\vec{r}}{i} \times \vec{\nabla}##
Relevant Equations:
##\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})##, ##\vec{L} = \frac{\vec{r}}{i} \times \vec{\nabla}##
First of all, sorry for the title I don't know the name of this formula and that's part of the problem, I can't find anything on google.
I have to show the identity above. Here's what I did. I don't know if this is correct so far.

##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})##

= ##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\frac{\vec{r}}{i} \times \vec{\nabla} \times \vec{u})##

= ##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{\nabla}(\frac{\vec{r}}{i} \cdot \vec{u}) - \vec{u}(\frac{\vec{r}}{i} \cdot \vec{\nabla}))##

= ##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + \vec{\nabla}(\vec{r} \cdot \vec{u}) - \vec{u}(\vec{r} \cdot \vec{\nabla})##

Is there a specific name for this identity?

Last edited:

kuruman
Homework Helper
Gold Member
I would start with the vector identity for ##\vec \nabla (\vec A\cdot \vec B)=\dots## and rewrite each of the 4 terms on the right-hand side.

ergospherical
Gold Member
Use suffix notation
\begin{align*}
L_j &= \dfrac{1}{i} \epsilon_{jlm} x_l \partial_m \\
\implies i (\mathbf{L} \times \mathbf{u})_i &= \epsilon_{ijk} \epsilon_{jlm} x_l \partial_m u_k \\
&= (\delta_{kl} \delta_{im} - \delta_{il} \delta_{km})x_l \partial_m u_k \\
&= x_k \partial_i u_k - x_i \partial_k u_k \\
&= x_k \partial_i u_k - (\nabla \cdot \mathbf{u}) x_i
\end{align*}also\begin{align*}
(\nabla(\mathbf{r} \cdot \mathbf{u}))_i = \partial_i (x_k u_k) &= u_k \partial_i x_k + x_k \partial_i u_k \\
&= u_i + x_k \partial_i u_k
\end{align*}
now can you see?

My bad, I forgot to mention that ##i = \sqrt{-1}##

That's why I replaced ##\vec{L}## to get rid of i's

ergospherical
Gold Member
there is no ambiguity in using the letter ##i## to denote both the imaginary unit and a vector index, since it is obvious from context

see post #3

Use suffix notation
\begin{align*}
L_j &= \dfrac{1}{i} \epsilon_{jlm} x_l \partial_m \\
\implies i (\mathbf{L} \times \mathbf{u})_i &= \epsilon_{ijk} \epsilon_{jlm} x_l \partial_m u_k \\
&= (\delta_{kl} \delta_{im} - \delta_{il} \delta_{km})x_l \partial_m u_k \\
&= x_k \partial_i u_k - x_i \partial_k u_k \\
&= x_k \partial_i u_k - (\nabla \cdot \mathbf{u}) x_i
\end{align*}also\begin{align*}
(\nabla(\mathbf{r} \cdot \mathbf{u}))_i = \partial_i (x_k u_k) &= u_k \partial_i x_k + x_k \partial_i u_k \\
&= u_i + x_k \partial_i u_k
\end{align*}
now can you see?
Is it the only way? I never used the suffix notation yet.

kuruman
Homework Helper
Gold Member
Is it the only way? I never used the suffix notation yet.
There is. See post #2.

There is. See post #2.
I was working on it, but I wasn't sure since I can't get it work.

On the left hand side I get ##\vec{\nabla}(\vec{r}\cdot \vec{u}) = \vec{\nabla}(\vec{r}\cdot \vec{u}) + \vec{\nabla}(\vec{u} + \vec{r}))##
It doesn't work, but I can't figure out where are my errors.

Using https://en.wikipedia.org/wiki/Vector_calculus_identities#Dot_product_rule

and "bac cab" for the triple cross product.

Last edited:
kuruman
Homework Helper
Gold Member
Use
##\vec \nabla (\vec A\cdot \vec B)=(\vec A \cdot \vec \nabla)\vec B+\vec A\times(\vec \nabla \times \vec B) + (\vec B \cdot \vec \nabla)\vec A+\vec B\times(\vec \nabla \times \vec A).##

Let ##\vec A = \vec r## and ##\vec B=\vec u##. This is the brute-force method. Write it all out and unless you really, really know what you're doing don't use bac-cab for the triple cross product.

Use
##\vec \nabla (\vec A\cdot \vec B)=(\vec A \cdot \vec \nabla)\vec B+\vec A\times(\vec \nabla \times \vec B) + (\vec B \cdot \vec \nabla)\vec A+\vec B\times(\vec \nabla \times \vec A).##

Let ##\vec A = \vec r## and ##\vec B=\vec u##. This is the brute-force method. Write it all out and unless you really, really know what you're doing don't use bac-cab for the triple cross product.
If I want to get rid of the cross products to have all the dot products to match the right hand side, I don't see how otherwise.
Maybe the answer I found for the right hand side isn't correct neither since I used bac-cab to get rid of the i's

I find the use of brute-force method and suffix notations both tedious and cumbersome. Why don't you try using first useful vector identities to simplify your equation. The BAC minus CAB rule is one such useful identity.
But I noticed in your OP where you wrote:
##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})## = ##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\frac{\vec{r}}{i} \times \vec{\nabla} \times \vec{u})##
that you didn't properly group together the factors in the third term on the right hand side. You should have written
##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})## = ##\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i[ (\frac{\vec{r}}{i} \times \vec{\nabla}) \times \vec{u} ]##

The OP posted this problem-to-prove question:
Homework Statement:: Show the identity ##\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})## , ##\vec{L} = \frac{\vec{r}}{i} \times \vec{\nabla}##

... don't know if ... correct ... ##\vec{\nabla}(\vec{r} \cdot \vec{u}) =\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\frac{\vec{r}}{i} \times \vec{\nabla} \times \vec{u})## ...
As I already said in post #11, the correct simplification should lead to$$\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + [~(\vec r \times \vec{\nabla}) \times \vec{u}~]$$But I was wondering if the identity to be proven shouldn't have been written$$\vec{\nabla}(\vec{r} \cdot \vec{u}) = 3\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + i(\vec{L} \times \vec{u})$$with an additional factor of 3 in the first term on the right hand side, so that it eventually simplifies to$$\vec{\nabla}(\vec{r} \cdot \vec{u}) = 3\vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + [~(\vec r \times \vec{\nabla}) \times \vec{u}~]$$It would be easy to get that additional factor if the term ##\vec{u}(\nabla \cdot\vec{r}~)## could somehow be incorporated in the equations while doing the vector algebra.

robphy