Contraction of Tensors

  • #1
Hi all! I've got a short question concerning a minor notational issue about tensor contraction I've run across recently.

Let A be an antisymmetric (0,2)-tensor and S a symmetric (2,0)-tensor.
Then their total contraction is zero: [itex]C_1^1C_2^2\,A \otimes S=0[/itex].
As a proof one simply computes: [itex]A_{ij}S^{ij}=-A_{ji}S^{ji}=-A_{ij}S^{ij}[/itex]
When I first saw this, I was a bit confused about the second equality. Of course, a scalar is a symmetric tensor…but is it not an abuse of notation? I mean this seems to run into conflict with the way one handles components of antisymmetric tensors…as for me, for someone who's just got accustomed to the components manipulation machinery, I was disturbed when I saw this. Am I alone?

This is not a big deal…but are there alternatives to expressing stuff like that? Any comments?
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,490
1,227
No, there is no problem. Remember, i and j are dummy summed-over indices, and thus can be relabeled. Is there a problem with

[tex]-A_{ji}S^{ji}=-A_{\alpha \beta}S^{\alpha \beta}?[/tex]
 
  • #3
WannabeNewton
Science Advisor
5,815
543
George answered your question but I just wanted to add something else. In the abstract index notation, ##S_{ab}## and ##A_{ab}## are actual tensors; the indices are called "abstract" indices because they simply label the rank 2-tensor in index form-they don't refer to components. In that sense, all the usual algebraic operations you know regarding tensors in the usual index-free formalism can be carried over the abstract index formalism in a natural way (so it has nothing to do with components) e.g. the tensor product of ##S## and ##A## can be written as ##A_{ab}S_{cd}## and the contraction as ##A_{ab}S^{ab}##. So if it helps, you can think of things in terms of the abstract index notation instead of coordinate component indices. See here for more: http://en.wikipedia.org/wiki/Abstract_index_notation

I hope that helps, cheers!
 
  • #4
To George:
Aha, ok, if you put it this way then I agree that it becomes a bit less controversial…but still I would then feel inclined to relabel the indices on initial step and then it would actually look all the same, but with relabeled indices. Alright, perhaps I just need a bit more practice in order not to see oddity in things like these. Thanks

To WannabeNewton:
OK, thank you for the info, too. I'll keep it mind.
 
  • #5
WannabeNewton
Science Advisor
5,815
543
Alright, perhaps I just need a bit more practice in order not to see oddity in things like these.

You should be able to get ample practice with index gymnastics via say proper general relativity texts. Ask away if you want some recommendations or anything. Have fun and good luck!
 
  • #6
21,628
4,911
Hi all! I've got a short question concerning a minor notational issue about tensor contraction I've run across recently.

Let A be an antisymmetric (0,2)-tensor and S a symmetric (2,0)-tensor.
Then their total contraction is zero: [itex]C_1^1C_2^2\,A \otimes S=0[/itex].
As a proof one simply computes: [itex]A_{ij}S^{ij}=-A_{ji}S^{ji}=-A_{ij}S^{ij}[/itex]
When I first saw this, I was a bit confused about the second equality. Of course, a scalar is a symmetric tensor…but is it not an abuse of notation? I mean this seems to run into conflict with the way one handles components of antisymmetric tensors…as for me, for someone who's just got accustomed to the components manipulation machinery, I was disturbed when I saw this. Am I alone?

This is not a big deal…but are there alternatives to expressing stuff like that? Any comments?
Your first equality is incorrect. There should be a plus sign rather than a minus sign. The proof should read [itex]A_{ij}S^{ij}=A_{ji}S^{ji}=-A_{ij}S^{ij}[/itex]

Chet
 
  • #7
WannabeNewton
Science Advisor
5,815
543
Your first equality is incorrect. There should be a plus sign rather than a minus sign. The proof should read [itex]A_{ij}S^{ij}=A_{ji}S^{ji}=-A_{ij}S^{ij}[/itex]

Chet
What he had was fine and what you have is fine as well. You relabeled the indices first and then used the antisymmetry and symmetry of the respective tensors. He used the antisymmetry and symmetry of the respective tensors first and then relabeled. Relabeling and antisymmetry/symmetry are commutative operations so there's no issue.
 
  • #8
21,628
4,911
What he had was fine and what you have is fine as well. You relabeled the indices first and then used the antisymmetry and symmetry of the respective tensors. He used the antisymmetry and symmetry of the respective tensors first and then relabeled. Relabeling and antisymmetry/symmetry are commutative operations so there's no issue.

Thanks WBN. That's interesting. Another question: Shouldn't the (double) contraction of an arbitrary 2nd order tensor A with another arbitrary 2nd order tensor S, in terms of the covariant components of A and the contravariant components of S, be written as AijSji rather than as AijSij?
 
  • #9
WannabeNewton
Science Advisor
5,815
543
So what we have before contracting is the tensor ##A_{ab}S^{cd}##. Your first expression says to contract ##b## with ##c## and ##a## with ##d## whereas your second expression says to contract ##a## with ##c## and ##b## with ##d## i.e. ##A_{ab}S^{ba}## vs. ##A_{ab}S^{ab}##. So there is no single (i.e. unique) "choice" of contraction; you have to pick which indices you want to contract over (this is for the arbitrary case).

EDIT: By the way, the first chapter of the following notes on the foundations of general relativity by David Malament has a very thorough coverage of index based (but entirely coordinate free) tensor algebra and tensor calculus: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf so check them out if you're interested. They are very much in the spirit of Geroch's notes on general relativity. Have fun :)!
 
Last edited:
  • #10
21,628
4,911
So what we have before contracting is the tensor ##A_{ab}S^{cd}##. Your first expression says to contract ##b## with ##c## and ##a## with ##d## whereas your second expression says to contract ##a## with ##c## and ##b## with ##d## i.e. ##A_{ab}S^{ba}## vs. ##A_{ab}S^{ab}##. So there is no single (i.e. unique) "choice" of contraction; you have to pick which indices you want to contract over (this is for the arbitrary case).

EDIT: By the way, the first chapter of the following notes on the foundations of general relativity by David Malament has a very thorough coverage of index based (but entirely coordinate free) tensor algebra and tensor calculus: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf so check them out if you're interested. They are very much in the spirit of Geroch's notes on general relativity. Have fun :)!
Thanks WBN. I guess I'm biased by my fluid mechanics and rheology experience.
 
  • #11
WannabeNewton
Science Advisor
5,815
543
Sounds like fun stuff! I have no experience in that area whatsoever so I can't say I know what you're alluding to but as long as ya get the right numbers in the end eh :)
 

Related Threads on Contraction of Tensors

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
14K
Replies
4
Views
927
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
6K
Replies
5
Views
2K
  • Last Post
Replies
0
Views
3K
Top