Curl of the transpose of a gradient of a vector: demonstration of an identity

[traianus]

I would like to demonstrate an identity with the INDICIAL NOTATION. I have attached my attempt. Please let me know where I made mistakes. Any suggestion? I am trying to understand tensors all by myself because they are the keys in continuum mechanics
Thanks

 

[Peeter]

What sort of vector product are you using here \hat{e}_i \hat{e}_j?

 

[traianus]

tensor product

 

[traianus]

Is this question so difficult? Please help me: I am trying to learn tensors and I would like to know what my mistake is. Thanks!

 

[traianus]

Any suggestion?

 

[traianus]

Can anyone suggest a forum to post my question? Thanks

 

[traianus]

anything? please help!

 

[traianus]

?

 

[traianus]

Is my question too difficult? Please advise.

 

[Landau]

I don't really understand what is meant by
\nabla(\nabla\times\mathbf{u})
and
\nabla \mathbf{u}.

For example, if \mathbf{u}=u_j\hat{e}_j, then \nabla \mathbf{u}=(\partial_i\hat{e}_i)(u_j\hat{e}_j)=\partial_iu_j\hat{e}_i\hat{e}_j.

But what is \hat{e}_i\hat{e}_j; the inner product between the unit basis vectors? Then the result would be a scalar instead of a vector.

 

[traianus]

Another demonstration is reported here http://en.wikiversity.org/wiki/Continuum_mechanics/Curl_of_a_gradient_of_a_vector . This may answer your question Landau.

I would like to understand where I make the mistake. Thanks

 

[traianus]

Any other input?

 

[traianus]

?

 

[Fredrik]

You should at least explain how you define \nabla u when u is a vector.

 

[traianus]

The problem is at the very bottom line in the definition of a curl of a tensor. I found 2 definitions which contradict to each other. Mine is one of them. I will email the authors.

 

[traianus]

I asked an expert. The question was not trivial. After a while I found out that there are different definitions of curl of a tensor.