Vector Identity Using Index Notation

[Saladsamurai]

Homework Statement

I am supposed to verify that

[tex]\nabla\cdot(\mathbf{u}\times\mathbf{v}) = \mathbf{v}\cdot\nabla\times\mathbf{u} - \mathbf{u}\cdot\nabla\times\mathbf{v}\qquad(1)[/itex]

I want to use index notation (and I think I am supposed to, though it does not say to explicitly) to verify this.

I know that we can write $\mathbf{w}=\mathbf{u}\times\mathbf{v}$ as $w_i = \epsilon_{ijk}u_jv_k$ hence we would write:


$$\nabla\cdot(\mathbf{u}\times\mathbf{v}) = \partial{}_i\epsilon_{ijk}u_jv_k$$


Now I am trying to figure out how I would get a difference to show up in this to match the right hand side of (1). I know that the point of index notation is that it is essentially all just scalar algebra ... but I could use a hint on this one.

 

[fzero]

Instead of trying to guess how it works, just do the computation. When you convert back from index notation to dot and cross products, you will end up permuting indices on the Levi-Civita symbol, dictating overall signs.

 

[Saladsamurai]

Instead of trying to guess how it works, just do the computation. When you convert back from index notation to dot and cross products, you will end up permuting indices on the Levi-Civita symbol, dictating overall signs.

I am not trying to guess, I am trying to do the computation. I am asking what is left to do with $\partial{}_i\epsilon_{ijk}u_jv_k$.

 

[fzero]

I am not trying to guess, I am trying to do the computation. I am asking what is left to do with $\partial{}_i\epsilon_{ijk}u_jv_k$.

Write it as

$$\partial{}_i\epsilon_{ijk}u_jv_k = \epsilon_{ijk} \partial{}_i(u_jv_k)$$

and distribute.

 

[Saladsamurai]

Write it as

$$\partial{}_i\epsilon_{ijk}u_jv_k = \epsilon_{ijk} \partial{}_i(u_jv_k)$$

and distribute.

Ok, I think that I see where this is going. We have

$$\epsilon_{ijk} \partial{}_i(u_jv_k) = \epsilon_{ijk}[v_k\partial{}_iu_j+u_j\partial{}_iv_k]$$

$$= \epsilon_{ijk}v_k\partial{}_iu_j+\epsilon_{ijk}u_j\partial{}_iv_k$$

Now, my new question is this: Since we are dealing with scalars in index notation, we can use the rules of scalar algebra such as commutativity and so on. Does that apply to the derivative operators as well?

That is, is this valid: $v_k\partial{}_iu_j = u_j\partial{}_iv_k$ ?

 

[fzero]

Now, my new question is this: Since we are dealing with scalars in index notation, we can use the rules of scalar algebra such as commutativity and so on. Does that apply to the derivative operators as well?

That is, is this valid: $v_k\partial{}_iu_j = u_j\partial{}_iv_k$ ?

No, what function the derivative acts on still matters. However, commutativity means that

$$v_k\partial{}_iu_j =(\partial{}_iu_j) v_k.$$

 

[tiny-tim]

Hi Saladsamurai!

Just use the product rule on the RHS (the index version).

 

[Saladsamurai]

No, what function the derivative acts on still matters. However, commutativity means that

$$v_k\partial{}_iu_j =(\partial{}_iu_j) v_k.$$

Ok great. So I should treat $\partial{}_iu_j$ as a single entity.

Hi Saladsamurai!

Just use the product rule on the RHS (the index version).

Okie dokie TT Like I did in post #5? I am correct in treating $epsilon_{ijk}$ as a constant here, right? Seems pretty obvious. Stupid question. Nevermind.

 

[tiny-tim]

Okie dokie TT Like I did in post #5?

oops! I missed that post!

 

[Saladsamurai]

Ok, splendid! Thanks folks. Much obliged. I do have another question though. Rather not start a new thread, but we'll see what happens. I have figured out how to set these up now. For example, I know that if I want to find $\mathbf{t}\cdot(\mathbf{u}\times\mathbf{v})$ I know that the 1st subscript of $\epsilon$ needs to match the subscript on t. Now have done another vector identity and i think that I have the right answer, but I am unsure of how to convert back to symbolic notation from here.

I need to show that $\nabla\times(\mathbf{u}\times \mathbf{v}) = \mathbf{v}\cdot\nabla\mathbf{u} - \mathbf{u}\cdot\nabla\mathbf{v} + \mathbf{u}\nabla\cdot \mathbf{v} - \mathbf{v}\nabla\cdot \mathbf{u}$.

In index notation I have showed that

$$\nabla\times(\mathbf{u}\times \mathbf{v}) = \epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k = u_p\partial{}_kv_k + v_q\partial{}_qu_p - u_q\partial{}_qv_p - v_p\partial{}_qu_q \qquad(2)$$


Now I am just unsure of the 2 types of terms here:

I am quite sure that $u_p\partial{}_kv_k = \mathbf{u}\nabla\cdot\mathbf{v}$.

But I am not too sure of how to reason out that $v_q\partial{}_qu_p=\mathbf{v}\cdot/nabla\mathbf{u}$.


Ok ... I figured it out as I was typing! Nevermind. In order to add, they must all be vectors, so it must be.

Yay! I am starting to actually like index notation!

 

[Fredrik]

In index notation I have showed that

$$\nabla\times(\mathbf{u}\times \mathbf{v}) = \epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k = u_p\partial{}_kv_k + v_q\partial{}_qu_p - u_q\partial{}_qv_p - v_p\partial{}_qu_q \qquad(2)$$

The first equality is a bit messed up, because you're setting a vector equal to it's pth component. If we write the basis vectors as e_i, we have x=x_ie_i, but x isn't equal to x_i. So you should write

$$\nabla\times(u\times v)=e_p\epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k$$

or

$$(\nabla\times(u\times v))_p=\epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k$$

I usually choose the second form just so I don't have to include the basis vector in every step. I also prefer to drop the special notation for vectors (bold, or an arrow on top) when I use index notation, because I find it weird to e.g. write $\mathbf x$ for a vector, $x_i$ for it's ith component, and then have to deal with components of $\bold x \times\bold y$. It looks weird to write $(\mathbf x\times\mathbf y)_i$ when we don't write $\mathbf x_i$. At least that's how I feel, but this is a minor point anyway. You don't have to do everything the way I do.

I didn't check the second equality.

I am quite sure that $u_p\partial{}_kv_k = \mathbf{u}\nabla\cdot\mathbf{v}$.

To be more precise, it's $=u_p(\nabla\cdot v)$

But I am not too sure of how to reason out that $v_q\partial{}_qu_p=\mathbf{v}\cdot/nabla\mathbf{u}$.


Ok ... I figured it out as I was typing! Nevermind. In order to add, they must all be vectors, so it must be.

$\nabla$ doesn't act on vectors. ($\nabla\cdot$ and $\nabla\times$ do). Since $\nabla u_p=(\partial_1 u_p,\partial_2 u_p,\partial_3 u_p)$, we have $(\nabla u_p)_q=\partial_q u_p$ and therefore

[tex]v_q\partial{}_qu_p=v\cdot\nabla u_p[/itex].

Yay! I am starting to actually like index notation!

It's pretty cool. I think it's by far the easiest way to deal with these things (once you get comfortable with the summation convention, and have figured out the identities for products of epsilons). This is an example that really shows the power of this notation:

$$\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k=0$$

The first equality is just the definitions, and the second is immediately obvious when you have some experience with these things. The proof goes like this:

$$\partial_i\varepsilon_{ijk}\partial_j F_k=\partial_j\varepsilon_{jik}\partial_i F_k=-\partial_i\varepsilon_{ijk}\partial_j F_k\quad\Rightarrow\quad \partial_i\varepsilon_{ijk}\partial_j F_k=0$$

But you would only do this sort of calculation once. The next time you see a similar expression you should think something like "symmetric times antisymmetric is always zero" and just write "=0" on the right. (The second expression is the same as the first. I just chose different symbols for the summation variables. The second equality follows from the commutativity of partial derivatives and the definition of epsilon).

 

[Fredrik]

I didn't check the second equality.

I have now, because I became suspicious after writing that comment about how $\nabla$ doesn't act on vectors. I see that the identity you're trying to prove has two instances of $\nabla$ acting on vectors:

I need to show that $\nabla\times(\mathbf{u}\times \mathbf{v}) = \mathbf{v}\cdot\nabla\mathbf{u} - \mathbf{u}\cdot\nabla\mathbf{v} + \mathbf{u}\nabla\cdot \mathbf{v} - \mathbf{v}\nabla\cdot \mathbf{u}$.

What do those terms mean?

In index notation I have showed that

$$\nabla\times(\mathbf{u}\times \mathbf{v}) = \epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k = u_p\partial{}_kv_k + v_q\partial{}_qu_p - u_q\partial{}_qv_p - v_p\partial{}_qu_q \qquad(2)$$

How do you get four terms? Aren't you using the identity $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$? That should only give you two (the two that don't have $\nabla$ acting on a vector).

 

[Saladsamurai]

I have now, because I became suspicious after writing that comment about how $\nabla$ doesn't act on vectors. I see that the identity you're trying to prove has two instances of $\nabla$ acting on vectors:
What do those terms mean?


How do you get four terms? Aren't you using the identity $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$? That should only give you two (the two that don't have $\nabla$ acting on a vector).

Hi Frederik In $\partial{}_qu_jv_k$, I applied the product rule

 

[Saladsamurai]

The first equality is a bit messed up, because you're setting a vector equal to it's pth component. If we write the basis vectors as e_i, we have x=x_ie_i, but x isn't equal to x_i. So you should write

$$\nabla\times(u\times v)=e_p\epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k$$

or

$$(\nabla\times(u\times v))_p=\epsilon_{pqi}\partial{}_q\epsilon_{ijk}u_jv_k$$

I usually choose the second form just so I don't have to include the basis vector in every step. I also prefer to drop the special notation for vectors (bold, or an arrow on top) when I use index notation, because I find it weird to e.g. write $\mathbf x$ for a vector, $x_i$ for it's ith component, and then have to deal with components of $\bold x \times\bold y$. It looks weird to write $(\mathbf x\times\mathbf y)_i$ when we don't write $\mathbf x_i$. At least that's how I feel, but this is a minor point anyway. You don't have to do everything the way I do.

I didn't check the second equality.


To be more precise, it's $=u_p(\nabla\cdot v)$

Hi again!

Unfortunately, in Fluid mechanics, we are even lazier than that! In my text we use v = v_i and apparently the basis vectors are implicit as well. It makes i easier to write, but more difficult to grasp at first (at least for me).

$\nabla$ doesn't act on vectors. ($\nabla\cdot$ and $\nabla\times$ do). Since $\nabla u_p=(\partial_1 u_p,\partial_2 u_p,\partial_3 u_p)$, we have $(\nabla u_p)_q=\partial_q u_p$ and therefore

[tex]v_q\partial{}_qu_p=v\cdot\nabla u_p[/itex].

As for $\nabla$ acting on vectors, I believe it does, but I have never actually done so; nor does my text explain this formula or its utility. From the left hand side, we know that we are looking for the answer to be a vector. So if I had to guess, I would say that $\nabla\mathbf{u}$ yields a tensor and hence $\mathbf{u}\cdot\nabla\mathbf{v}$ must be a vector.

Not sure though

Edit: http://en.wikipedia.org/wiki/Vector_calculus_identities#Gradient_of_a_vector_field

 

[Fredrik]

Hi Frederik In $\partial{}_qu_jv_k$, I applied the product rule

Probably not the first time I've made that mistake.

As for $\nabla$ acting on vectors, I believe it does, but I have never actually done so; nor does my text explain this formula or its utility. From the left hand side, we know that we are looking for the answer to be a vector.

I decided to find out what those terms mean by doing the calculation. I get the desired result if I define

$$\mathbf v\cdot\nabla\mathbf u=\mathbf{e}_i(\mathbf v\cdot\nabla u_i)$$

and similarly for the other term. I guess it makes sense to define the notation on the left this way, if we have to deal with a bunch of expressions that contain expressions like the one on the right.