Understanding Index Notation in Tensor Calculus?

[Peregrine]

I am playing around with learning index notation for tensors, and I came across the following where C is a 0th order tensor:

E_{ijk} \partial_j \partial_k C = 0

I believe this equates to \nabla \times \nabla C. I don't understand why this comes out to 0. Any ideas?

Also, I am trying to understand in index notation how to represent the grad of a vector. The reason I am confused is that it seems that, taking C as a 0th order tensor, V as a 1st order tensor and T as a 2nd order tensor:

div T = \nabla \cdot T = \partial_iT_{ij}
div V = \nabla \cdot V = \partial_iV_i
And of course, div C does not make sense as it would be a -1st order tensor.

But, since:
grad C = \nabla C = \partial_iC
I don't follow how to represent grad V = \nabla V or grad T = \nabla T in index notation; from what I have it seems there would be no difference in notation between grad and div! Any help would be greatly appreciated. Thanks!

 

[coalquay404]

I am playing around with learning index notation for tensors, and I came across the following where C is a 0th order tensor:

E_{ijk} \partial_j \partial_k C = 0

I believe this equates to \nabla \times \nabla C. I don't understand why this comes out to 0. Any ideas?

The first thing to note is that your notation here is a bit non-standard. If you assume that E_{ijk} is just some arbitrary (0,3) tensor then \sum_{j,k}E_{ijk}\partial_j\partial_kC\ne 0 in general. However, if you take E_{ijk}=\epsilon_{ijk}, the totally antisymmetric or permutation tensor, then \sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC=0 is trivially satisfied. To see why this is so, note that since partial derivatives commute we can write

\begin{equation*}<br /> \begin{split}<br /> \sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC<br /> &amp;= \sum_{j,k}\frac{1}{2}\epsilon_{ijk}(\partial_j\partial_kC + \partial_k\partial_jC) \\<br /> &amp;= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk}\partial_j\partial_kC + \epsilon_{ijk}\partial_k\partial_jC) \\<br /> &amp;= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk}\partial_j\partial_kC + \epsilon_{ikj}\partial_j\partial_kC) \\<br /> &amp;= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk} + \epsilon_{ikj})\partial_j\partial_kC.<br /> \end{split}<br /> \end{equation*}

However, since one has \epsilon_{ijk}=-\epsilon_{ikj} by definition, one can then write

\sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC = \sum_{j,k}\frac{1}{2}(\epsilon_{ijk} - \epsilon_{ijk})\partial_j\partial_kC = 0.<br />

You are then correct to say that \nabla\times\nabla C=\sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC=0.

Also, I am trying to understand in index notation how to represent the grad of a vector. The reason I am confused is that it seems that, taking C as a 0th order tensor, V as a 1st order tensor and T as a 2nd order tensor:

div T = \nabla \cdot T = \partial_iT_{ij}
div V = \nabla \cdot V = \partial_iV_i
And of course, div C does not make sense as it would be a -1st order tensor.

Correct.

But, since:
grad C = \nabla C = \partial_iC
I don't follow how to represent grad V = \nabla V or grad T = \nabla T in index notation; from what I have it seems there would be no difference in notation between grad and div! Any help would be greatly appreciated. Thanks!

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[Jheriko]

I am playing around with learning index notation for tensors, and I came across the following where C is a 0th order tensor:

E_{ijk} \partial_j \partial_k C = 0

I believe this equates to \nabla \times \nabla C. I don't understand why this comes out to 0. Any ideas?
...

Not sure if this is the source of your confusion, but notice that the 0 is a component of a tensor with rank(0,1) since only the j and k indices are repeated. i.e.

A_{i} = E_{ijk} \partial_j \partial_k C = 0

 

[Peregrine]

coalquay,

You are correct that I intended \epsilon_{ijk}. Sorry for the bad notation. But I greatly appreciate the help, I did not think of that approach. Thanks!

 

[mathwonk]

sillyme, i hoped this was about the index of an elliptic operator, but its the same old same old.

 

[coalquay404]

A question about index theorems would be nice, wouldn't it? I suspect, however, we'll be waiting a while before we see one...