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I'm trying to understand the following conversion from vector form into Einstein summation notation, found on P2 of http://www.stanford.edu/~vkl/research/notes/index_not.pdf which states:

Show $\mathbf{v} \cdot \nabla\mathbf{v} = \nabla\left(\frac{|\mathbf{v}|^2}{2}\right)+(\color{brown}{\nabla \times \mathbf{v})} \times \mathbf{v}$

Proof: $v_a \partial_a v_b = \partial_b\left(\frac{v_av_a}{2} \right) + \epsilon_{bac}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c \tag{*}$ (Rest of proof omitted here)

$\Large{\text{Question #1.}}$How did they get the LHS of (*)? I don't think my course covers the gradient of a vector, so I don't know how to convert it into Einstein notation. Or is it supposed to be $\nabla \cdot \mathbf{v}$?

$\Large{\text{Question #2.}} $ The solution seems to be working with the $b$th component of $ \mathbf{v} \cdot \nabla\mathbf{v}$, but it doesn't say so. Because $\mathbf{u} \times \mathbf{v} = \epsilon_{acb}u_av_j\color{red}{\mathbf{\hat{e_b}}}$, is the solution missing $\color{red}{\mathbf{\hat{e_b}}}$ on the RHS: $$\partial_b\left(\frac{v_av_a}{2} \right) + \underbrace{\epsilon_{acb}}_{\Large{= \epsilon_{bac}}}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c\color{red}{\mathbf{\hat{e_b}} }?$$

Here are two supplementary questions in response to tom's answer:

$\Large{\text{Question #1.1.}} $ How did you realise that $\mathbf{v} \cdot \nabla\mathbf{v}$ should've been written as $(\mathbf{v} \cdot \nabla)\mathbf{v}$?

$\Large{\text{Question #2.1.}}$ I don't understand your answer. $((\color{brown}{\nabla \times \mathbf{v})} \times \mathbf{v})$is a vector so why does the given solution not convert this to $\epsilon_{acb}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c\color{red}{\mathbf{\hat{e_b}} }$?

Here are two supplementary questions in response to Muphrid's answer:

$\Large{\text{Question #1.2.}} $ What would be the "order of operations" which you mentioned in your answer?

$\Large{\text{Question #2.2.}}$ Sorry, I don't understand what you mean by "...it would otherwise appear as the same basis vector in each term." Could you please clarify?

Here are two supplementary questions in response to Muphrid's 2nd answer:

$\Large{\text{Question #2.3.}}$ Could you please explain how a free index ($\color{red}{b}$ here) means that we are looking at the component of this free index ($\color{red}{b}$th component here)?

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2 Answers2

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Q1:

Well $v\cdot \nabla v$ should be writen in form $(v \cdot \nabla) v$.

Nabla is $$\nabla = (\partial_x,\partial_y,\partial_z).$$ So

$$v\cdot \nabla = v_x\partial_x+v_y\partial_y+v_z\partial_z$$

Of course you can define gradient of vector valued function see Jacobi matrix

Q2:

Well no it is not. If by $\hat{e}_b$ you mean b-th basis vector.

$\partial_b\left(\frac{v_av_a}{2} \right)$ is scalar so $\epsilon_{bac}{(\epsilon_{adf}\partial_dv_f)}v_c $ has to be scalar too

Q1.1: Don't know what to say, but may I ask what is your definition of $\textbf{v} \cdot \nabla \textbf{v}$ ?

For me it is $\textbf{v} \cdot \nabla \textbf{v} = (v_x\partial_x+v_y\partial_y+v_z\partial_z)\textbf{v}$ so there is not much to talk about.

Q2.1: Yes you are right that $\mathbf{u} \times \mathbf{v} = \epsilon_{abc}u_av_b\mathbf{\hat{e}_c}$ the same is with $\mathbf{v} = v_a \mathbf{\hat{e}_a}$ and $\nabla = \mathbf{\hat{e}_a} \partial_a $.

So $\nabla\left(\frac{|\mathbf{v}|^2}{2}\right) = \frac{1}{2}\mathbf{\hat{e}_a}\partial_a (v_bv_b)$ and $(\nabla \times \mathbf{v}) \times \mathbf{v} = \epsilon_{abc} (\epsilon_{bef} \partial_e v_f) v_c \mathbf{\hat{e}_c}$.

For more info on term $v\cdot \nabla v$ see Convective acceleration in Navier-Stokes equations

tom
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  • Thank you for your response. Could you please look at two supplementary questions in response to your answer, with which I've updated my original post? –  May 06 '13 at 20:29
  • Edited. I had mistake in original answer which should be fixed now. – tom May 06 '13 at 20:53
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(1) It's $(v \cdot \nabla) v$, the directional derivative. This is a common notation for it (dropping the brackets) because the gradient of a vector field doesn't fall under the scope of vector calculus--there is nothing else it could be. You could also consider it an order of operations sort of thing, if you like.

(2) No, $\hat e_b$ is not missing, or rather, if it is missing, it is "missing" from every term. The basis vector has been dropped from both sides for convenience. You can verify that $b$ is the only free index on both sides, and it would otherwise appear as the same basis vector in each term.

Muphrid
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  • Thank you for your answer again. Could you please look at two supplementary questions (#2.1 and #2.2) in response to your answer in my original post? –  May 07 '13 at 21:36
  • 1.2) Merely that the dot product has higher precedence than the derivative acting on the object to its right. I don't think there's a formal set of conventions here in vector calculus--generally, brackets are preferred. The only reason those brackets are dropped here is, again, because the gradient of a vector field is not defined in vector calculus. (2.2) I mean every term has $\hat e_b$, and every term has had it canceled. – Muphrid May 07 '13 at 21:47
  • Thank you very much again. Could you please look at another supplementary question (#2.3) added to my original post? –  May 09 '13 at 11:01
  • I don't feel like I have much of an answer to that question other than "that simply is what it means." Write out a vector by itself: $v = v_b \hat e_b$. Because the "free index" is always matched up with a basis vector with the same label, it's redundant information to have the basis vector there too, so it's almost always dropped (this also neatly avoids the question of what two free indices mean, and how two basis vectors can be present in an expression like that--that question is also interesting!). – Muphrid May 09 '13 at 14:06