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

1. Dec 19, 2009

### 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

#### Attached Files:

• ###### TENSOR_IDENTITY.pdf
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Last edited: Dec 19, 2009
2. Dec 19, 2009

### Peeter

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

3. Dec 19, 2009

### traianus

tensor product

4. Dec 21, 2009

### 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!

5. Jan 6, 2010

### traianus

Any suggestion?

6. Jan 10, 2010

### traianus

Can anyone suggest a forum to post my question? Thanks

7. Jan 13, 2010

### traianus

8. Jan 24, 2010

### traianus

????

9. Jan 26, 2010

### traianus

10. Jan 26, 2010

### 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.

Last edited: Jan 26, 2010
11. Jan 29, 2010

### traianus

12. Feb 6, 2010

### traianus

Any other input?

13. Feb 11, 2010

### traianus

?????

14. Feb 11, 2010

### Fredrik

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

15. Feb 22, 2010

### 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.

16. Apr 3, 2010

### 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.