Talk:Tensor derivative (continuum mechanics)
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References present but no citation
[edit]There are references here and I'm sure that they check out, the proofs look good but could someone cite the proof (i.e. from which chapter of which books can we find these proofs?).
Inconsistency in the gradient definition
[edit]The text says:
- The gradient, , of a tensor field in the direction of an arbitrary constant vector is defined as:
This expression assumes that a dot product exists between the gradient and the constant vector c. Therefore the gradient of a generic tensor should always be a vector. But then the text says:
- The gradient of a tensor field of order is a tensor field of order .
I assume therefore that the given definition is not the general one. Am I missing something?
--Juansempere (talk) 18:54, 10 October 2012 (UTC)
The gradient will be a tensor field of rank n+1, and then when you dot with a vector you reduce the rank again by one, leaving a tensor field of rank n. So this checks out.
If you're concerned about the dot product being well-defined, the definition of the dot product between a tensor and a vector is given in coordinates:
129.32.11.206 (talk) 16:59, 11 October 2012 (UTC)
- I didn't know the dot product extension. Thanks. --Juansempere (talk) 22:13, 11 October 2012 (UTC)
Inconsistency with directional derivative page
[edit]Under Derivatives of scalar valued functions of vectors, the text states "[...] is the vector defined as [...]". Similarly, under Derivatives of vector valued functions of vectors we have "[...] is the second order tensor defined as [...]". In the directional derivative page, the same two definitions are given, but the type of the first isn't given, while the second is defined to be a vector.
It looks to me like these two definitions ought to yield a scalar and vector respectively, but I am not confident enough to go ahead and update them; nor do I have a suitable reference to verify.
94.174.139.227 (talk) 23:51, 18 December 2013 (UTC)
- I've fixed the directional derivative page. Note that vector.vector = scalar and 2-tensor.vector = vector (without going into details of one-forms and two-forms.) Bbanerje (talk) 02:53, 19 December 2013 (UTC)
- Yes, I think I see my error now—I was reading it as the derivative being the product, rather than being defined by it. Thank you. 94.174.139.227 (talk) 18:20, 19 December 2013 (UTC)