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## 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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- Thread starter Ted123
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[PLAIN]http://img585.imageshack.us/img585/526/indexnotation.jpg [Broken]

How do I proceed?

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- #2

Hurkyl

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Using index notation sounds like a good place to start....

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Using index notation sounds like a good place to start....

Which notation? I know the Kronecker Delta and Levi-Civita symbols...

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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?

[itex]\nabla \cdot \mathbf{r} = \partial _j r_j[/itex]

So does this equal [itex]\sum_{j=1}^3\partial _j r_j[/itex]

and does [itex]\partial _1 r_1 = \partial _2 r_2 = \partial _3 r_3 = 1[/itex] ?

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?

[itex]\nabla \cdot \mathbf{r} = \partial _j r_j[/itex]

So does this equal [itex]\sum_{j=1}^3\partial _j r_j[/itex]

and does [itex]\partial _1 r_1 = \partial _2 r_2 = \partial _3 r_3 = 1[/itex] ?

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- #5

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So [tex] \mbox{Div} {}\mathbf{r} = (\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j) [/tex]

So complete the calculation.

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

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rhas the components [itex] x,y,z [/itex] or [itex] x_i [/itex]. So the divergence ofris the <scalar product> between the del operator and ther.

So [tex] \mbox{Div} {}\mathbf{r} = (\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j) [/tex]

So complete the calculation.

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

So how do I evaluate [itex](\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j) [/itex] ?

Is it just [itex]\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[/itex] (this doesn't look right at all)

- #7

hunt_mat

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[tex]

\mathbf{r}=x^{i}\mathbf{e}_{i}

[/tex]

and div is:

[tex]

\nabla\cdot =\partial_{i}(e_{i}\cdot )

[/tex]

so:

[tex]

\nabla\cdot\mathbf{r}=\sum_{i=1}^{3}\partial_{i}(\mathbf{e}_{i}\cdot (x^{i}\mathbf{e}_{i}))=...

[/tex]

- #8

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[tex]

\mathbf{r}=x^{i}\mathbf{e}_{i}

[/tex]

and div is:

[tex]

\nabla\cdot =\partial_{i}(e_{i}\cdot )

[/tex]

so:

[tex]

\nabla\cdot\mathbf{r}=\sum_{i=1}^{3}\partial_{i}(\mathbf{e}_{i}\cdot (x^{i}\mathbf{e}_{i}))=...

[/tex]

Where are these [itex]\mathbf{e}[/itex] vectors coming from? In all my solutions to these questions on index notation I never see an [itex]\mathbf{e}[/itex] appearing.

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

and the solution is [itex]\partial_j a_k x_k = a_k \partial_j x_k = a_k \delta_{jk} = a_j[/itex]

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- #9

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So how do I evaluate [itex](\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j) [/itex] ?

Is it just [itex]\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[/itex] (this doesn't look right at all)

It's correct, because [itex] \mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij} [/itex], so you get what you wrote. But of course [itex] \partial_1 x_1 =1 [/itex] 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.

- #10

hunt_mat

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Where are these [itex]\mathbf{e}[/itex] vectors coming from? In all my solutions to these questions on index notation I never see an [itex]\mathbf{e}[/itex] appearing.

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

and the solution is [itex]\partial_j a_k x_k = a_k \partial_j x_k = a_k \delta_{jk} = a_j[/itex]

The [tex]\mathbf{e}_{i}[/tex] vectors are the basis vectors of three space, so [tex]\mathbf{e}_{1}=\mathbf{i},\mathbf{e}_{2}=\mathbf{j},\mathbf{e}_{3}=\mathbf{k}[/tex]

- #11

hunt_mat

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Oh and also: [tex]x^{1}=x,x^{2}=y,x^{3}=z[/tex]

- #12

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Is it [itex]\varepsilon_{jmn} \partial_m \varepsilon_{nkl} a_k x_l[/itex] ?

If this is right it goes to [itex]a_k ( \delta_{jk} \delta_{ml} - \delta_{jl} \delta_{mk}) \delta_{ml}[/itex]

- #13

fzero

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

Is it [itex]\varepsilon_{jmn} \partial_m \varepsilon_{nkl} a_k x_l[/itex] ?

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

[itex]\varepsilon_{jmn} \partial_m (\varepsilon_{nkl} a_k x_l)[/itex]

If this is right it goes to [itex]a_k ( \delta_{jk} \delta_{ml} - \delta_{jl} \delta_{mk}) \delta_{ml}[/itex]

This is true as long as [tex]a_k[/tex] are constants, which is probably intended. Now you should try to compute [tex]\delta_{ml} \delta_{ml} [/tex].

- #14

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

[itex]\varepsilon_{jmn} \partial_m (\varepsilon_{nkl} a_k x_l)[/itex]

This is true as long as [tex]a_k[/tex] are constants, which is probably intended. Now you should try to compute [tex]\delta_{ml} \delta_{ml} [/tex].

Is this all OK?:

[itex][\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 [/itex] (since the [itex]a_m[/itex] are constant and so the derivative doesn't act on them)

[itex]= 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}[/itex]

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