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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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Which notation? I know the Kronecker Delta and Levi-Civita symbols...Using index notation sounds like a good place to start....
So how do I evaluate [itex](\mathbf{e}_i \partial_i)\cdot (x_j \mathbf{e}_j) [/itex] ?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.
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.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]
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.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)
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]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]
It's less confusing if you use parentheses to keep track of what the derivative acts on: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] ?
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].If this is right it goes to [itex]a_k ( \delta_{jk} \delta_{ml} - \delta_{jl} \delta_{mk}) \delta_{ml}[/itex]
Is this all OK?: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].