Vector Calculus: Index Notation

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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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  • #2
Hurkyl
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Using index notation sounds like a good place to start....
 
  • #3
Ted123
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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...
 
  • #4
Ted123
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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] ?
 
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  • #5
dextercioby
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I would say that r has the components [itex] x,y,z [/itex] or [itex] x_i [/itex]. So the divergence of r is the <scalar product> between the del operator and the r.

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.
 
  • #6
Ted123
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I would say that r has the components [itex] x,y,z [/itex] or [itex] x_i [/itex]. So the divergence of r is the <scalar product> between the del operator and the r.

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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To start:
[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
Ted123
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To start:
[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
dextercioby
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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
Ted123
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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] ?

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
Ted123
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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]
 
  • #15
dextercioby
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Looks ok to me up to the last equal sign. The bolding part at the end you may want to drop though, because you were making the computation only for the arbitrary component <j>.
 

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