# Vector Calculus: Index Notation

## Homework Statement

[PLAIN]http://img585.imageshack.us/img585/526/indexnotation.jpg [Broken]

## The Attempt at a Solution

How do I proceed?

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Hurkyl
Staff Emeritus
Gold Member
Using index notation sounds like a good place to start....

Using index notation sounds like a good place to start....
Which notation? I know the Kronecker Delta and Levi-Civita symbols...

This is the first time I've done anything with index notation and I don't really 'get' it at the moment.

For the first one this seems to be what I've been told so where do I go from here? Because there are repeated 'j' indices does this imply a summation?

$\nabla \cdot \mathbf{r} = \partial _j r_j$

So does this equal $\sum_{j=1}^3\partial _j r_j$

and does $\partial _1 r_1 = \partial _2 r_2 = \partial _3 r_3 = 1$ ?

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dextercioby
Homework Helper
I would say that r has the components $x,y,z$ or $x_i$. So the divergence of r is the <scalar product> between the del operator and the r.

So $$\mbox{Div} {}\mathbf{r} = (\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j)$$

So complete the calculation.

Along the same lines you'll solve the 2nd point as well.

I would say that r has the components $x,y,z$ or $x_i$. So the divergence of r is the <scalar product> between the del operator and the r.

So $$\mbox{Div} {}\mathbf{r} = (\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j)$$

So complete the calculation.

Along the same lines you'll solve the 2nd point as well.
So how do I evaluate $(\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j)$ ?

Is it just $\mathbf{e}_1 \partial_1 x_1\mathbf{e}_1 + \mathbf{e}_2 \partial_2 x_2\mathbf{e}_2 + \mathbf{e}_3 \partial_3 x_3\mathbf{e}_3$ (this doesn't look right at all)

hunt_mat
Homework Helper
To start:
$$\mathbf{r}=x^{i}\mathbf{e}_{i}$$
and div is:
$$\nabla\cdot =\partial_{i}(e_{i}\cdot )$$
so:
$$\nabla\cdot\mathbf{r}=\sum_{i=1}^{3}\partial_{i}(\mathbf{e}_{i}\cdot (x^{i}\mathbf{e}_{i}))=...$$

To start:
$$\mathbf{r}=x^{i}\mathbf{e}_{i}$$
and div is:
$$\nabla\cdot =\partial_{i}(e_{i}\cdot )$$
so:
$$\nabla\cdot\mathbf{r}=\sum_{i=1}^{3}\partial_{i}(\mathbf{e}_{i}\cdot (x^{i}\mathbf{e}_{i}))=...$$
Where are these $\mathbf{e}$ vectors coming from? In all my solutions to these questions on index notation I never see an $\mathbf{e}$ appearing.

A similar question to the first one is Show $\nabla (\mathbf{a} \cdot \mathbf{r} ) = \mathbf{a}\;,\;\mathbf{a}\in\mathbb{R}^3$

and the solution is $\partial_j a_k x_k = a_k \partial_j x_k = a_k \delta_{jk} = a_j$

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dextercioby
Homework Helper
So how do I evaluate $(\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j)$ ?

Is it just $\mathbf{e}_1 \partial_1 x_1\mathbf{e}_1 + \mathbf{e}_2 \partial_2 x_2\mathbf{e}_2 + \mathbf{e}_3 \partial_3 x_3\mathbf{e}_3$ (this doesn't look right at all)
It's correct, because $\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}$, so you get what you wrote. But of course $\partial_1 x_1 =1$ and as well for the other 2 components. So it's not difficult to reach the desired conclusion.

EDIT: The bolded e's are the unit vectors along the coordinate axes. in R^3 they are usually denoted i,j,k and have modulus = to 1.

hunt_mat
Homework Helper
Where are these $\mathbf{e}$ vectors coming from? In all my solutions to these questions on index notation I never see an $\mathbf{e}$ appearing.

A similar question to the first one is Show $\nabla (\mathbf{a} \cdot \mathbf{r} ) = \mathbf{a}\;,\;\mathbf{a}\in\mathbb{R}^3$

and the solution is $\partial_j a_k x_k = a_k \partial_j x_k = a_k \delta_{jk} = a_j$
The $$\mathbf{e}_{i}$$ vectors are the basis vectors of three space, so $$\mathbf{e}_{1}=\mathbf{i},\mathbf{e}_{2}=\mathbf{j},\mathbf{e}_{3}=\mathbf{k}$$

hunt_mat
Homework Helper
Oh and also: $$x^{1}=x,x^{2}=y,x^{3}=z$$

OK so how do I write the 2nd one in terms of the Levi-Civita symbol?

Is it $\varepsilon_{jmn} \partial_m \varepsilon_{nkl} a_k x_l$ ?

If this is right it goes to $a_k ( \delta_{jk} \delta_{ml} - \delta_{jl} \delta_{mk}) \delta_{ml}$

fzero
Homework Helper
Gold Member
OK so how do I write the 2nd one in terms of the Levi-Civita symbol?

Is it $\varepsilon_{jmn} \partial_m \varepsilon_{nkl} a_k x_l$ ?
It's less confusing if you use parentheses to keep track of what the derivative acts on:

$\varepsilon_{jmn} \partial_m (\varepsilon_{nkl} a_k x_l)$

If this is right it goes to $a_k ( \delta_{jk} \delta_{ml} - \delta_{jl} \delta_{mk}) \delta_{ml}$
This is true as long as $$a_k$$ are constants, which is probably intended. Now you should try to compute $$\delta_{ml} \delta_{ml}$$.

It's less confusing if you use parentheses to keep track of what the derivative acts on:

$\varepsilon_{jmn} \partial_m (\varepsilon_{nkl} a_k x_l)$

This is true as long as $$a_k$$ are constants, which is probably intended. Now you should try to compute $$\delta_{ml} \delta_{ml}$$.
Is this all OK?:

$[\nabla \times (\mathbf{a} \times \mathbf{r})]_j = \varepsilon_{jkl} \partial_k (\mathbf{a} \times \mathbf{r})_l = \varepsilon_{jkl} \partial_k (\varepsilon_{lmn} a_m x_n) = a_m \varepsilon_{jkl} \varepsilon_{lmn} \partial_k x_n$ (since the $a_m$ are constant and so the derivative doesn't act on them)

$= a_m \varepsilon_{jkl} \varepsilon_{lmn} \delta_{kn} = a_m \varepsilon_{jkl} \varepsilon_{lmk} = a_m (\delta_{jm} \delta_{kk} - \delta_{jk} \delta_{km} ) = a_m (3\delta_{jm} - \delta_{jk} \delta_{km} ) = a_m (3\delta_{jm} - \delta_{jm} ) = 2a_m \delta_{jm} = 2a_j = 2\mathbf{a}$

dextercioby