Vector Calculus: Index Notation

In summary: So you should end up with something like 2a_{j}.In summary, the conversation is about using index notation to solve a problem involving the divergence operator and the cross product. The conversation includes discussions on how to proceed, the use of Kronecker Delta and Levi-Civita symbols, and the steps to solve the problem using index notation. The solution involves computing derivatives and simplifying the expression using the defined symbols.
  • #1
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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
Using index notation sounds like a good place to start...
 
  • #3
Hurkyl said:
Using index notation sounds like a good place to start...

Which notation? I know the Kronecker Delta and Levi-Civita symbols...
 
  • #4
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
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
bigubau said:
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
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
hunt_mat said:
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
Ted123 said:
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
Ted123 said:
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
Oh and also: [tex]x^{1}=x,x^{2}=y,x^{3}=z[/tex]
 
  • #12
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
Ted123 said:
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
fzero said:
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
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>.
 

1. What is index notation in vector calculus?

Index notation is a method of writing vector and tensor equations using indices to represent the components of the vectors or tensors. It is also known as Einstein notation or tensor notation.

2. How is index notation used in vector calculus?

In index notation, each index represents a specific dimension of a vector or tensor. Repeated indices in an equation indicate summation over those dimensions. This notation allows for a more concise and elegant representation of vector and tensor equations.

3. What are the benefits of using index notation in vector calculus?

Index notation allows for a clearer and more compact representation of vector and tensor equations, making them easier to manipulate and understand. It also helps to avoid errors in calculations and allows for a more elegant and efficient way of solving problems.

4. What are some common operations in vector calculus that use index notation?

Some common operations in vector calculus that use index notation include vector addition, dot product, cross product, and gradient, divergence, and curl operations. These operations can be expressed more succinctly and precisely using index notation.

5. Are there any drawbacks to using index notation in vector calculus?

One of the drawbacks of index notation is that it can be confusing for beginners to understand and use. It also requires a solid understanding of vectors and tensors and their properties. Additionally, some vector calculus operations, such as the determinant, cannot be expressed using index notation.

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