1
$\begingroup$

I'm working on proving

For a second order tensor $\mathbf{A}$,$\mathbf{u}\cdot\mathbf{A}\cdot\mathbf{u}=0$ for all vectors $\mathbf{u}$ if and only if $\mathbf{A}$ is skew symmetric.

Now, I understand the proof for it fully. But I came up with (what I think) is a clever way to do it. The notation of what I did is what I'm having trouble with.

For the direction that $\mathbf{u}\cdot\mathbf{A}\cdot\mathbf{u}=0 \implies \mathbf{A}$ is skew symmetric, I defined vectors $\mathbf{u}^{(r,s)}$. Where the vectors defined in indicial notation are $u_i^{(r,s)}=\delta_{ri}+\delta_{si}$.

Some examples of my vector in three dimensions are $$\mathbf{u}^{(1,1)}=[2,0,0]$$ $$\mathbf{u}^{(2,3)}=[0,1,1]$$

Now, the superscript parenthesis is my own notation. $r$ and $s$ are supposed to be fixed as a free index. And I'm interested, first, in the case where $r=s$. So basically,

$$\mathbf{u}^{(r,r)}\cdot\mathbf{A}\cdot\mathbf{u}^{(r,r)}=A_{ij}u_i^{(r,r)}u_j^{(r,r)}=A_{ij}(2\delta_{ri})(2\delta_{rj})=4A_{rr}=0$$

I want this to show that the element in column $r$ and row $r$ is zero if we used the matrix representation.

But this makes it seem like it's now a dummy variable and that we are summing over all possible values of $r$.

What did I do wrong or is there a better way to notate what I wanted to?

Also, we are using the book "The Linearized Theory of Elasticty" by William Slaughter. The indicial notation is all subscripted (unlike conventional Einstein indicial notation which has both subscripts and superscripts)

After discussion with Avitus, i believe my question boils down more to:

"How do I represent the particular elements $A_{11},A_{22},\cdots$ in the indicial notation without making $i$ in $A_{ii}$ a dummy variable? And how do I pick my notation for $\mathbf{u}$ accordingly?"

$\endgroup$
0
$\begingroup$

Your strategy is correct: you probably need to change notation a bit. I suppose you want to prove

$$I:=\sum_{k,l=1}^n u_k A_{kl} u_l = 0~~\forall u\in\mathbb R^n \Rightarrow A_{kk}=0,~A_{kl}=-A_{lk},~~ k\neq l. $$

To prove the first skewness relation for $A$ we choose $u=(0,\cdots,1,\cdots,0)$, where the $1$ corresponds to the $i$-th coordinate, i.e. $u_k = \delta_{ik}$, $k=1,\dots,n$. Then

$$I=A_{ii}=0.$$

Repeating the same argument for all $i=1,\dots, n$ one arrives at the first skewness relation for $A$.

We are left to prove that $A_{ij}=-A_{ji},~~ i\neq j$. In this case we choose $u=(0,\cdots,1,\cdots,1,\cdots,0)$ where the $1$'s correspond to the $i$-th and $j$-th coordinate, with $i\neq j$. Then then sum $I$ consists of 2 terms: in the first one $i=k$ and $l=j$; in the second $k=j$ and $l=i$, leading to

$$I= A_{ij}+ A_{ji}=0.$$

Choosing all possible distinct pairs $i,j=1,\dots,n$ leads to the second skewness relation for $A$.

$\endgroup$
  • $\begingroup$ Thank you for showing the proof in a different notation. But, unfortunately, I didn't have a question about how to prove it. I'm just having notation issues in the indicial notation. Because in the notation, $A_{ii}=0$ is identical to $\sum_{i=1}^nA_{ii}=0$ which is not that same as $A_{11}=A_{22}=A_{33}=\cdots=0$. I suppose a more concise question would be: "How do I represent the particular elements $A_{11},A_{22},\cdots$ in the indicial notation without making $i$ in $A_{ii}$ a dummy variable." $\endgroup$ – BeaumontTaz Sep 8 '14 at 21:17
  • $\begingroup$ I would avoid the notation $A_{ii}=\sum_{i=1}^n A_{ii}$: it can generate confusion. In this case one can simply try to be coherent with the usual notation for matrix elements: if $M$ is a matrix, then $M_{ii}$ denotes the $i$-th diagonal element and not the trace of the matrix itself. $\endgroup$ – Avitus Sep 8 '14 at 21:20
  • $\begingroup$ I cannot avoid the notation $A_{ii}=\sum_{i=1}^nA_{ii}$ because that is the tensors index notation convention in this book. $\endgroup$ – BeaumontTaz Sep 8 '14 at 21:21
  • $\begingroup$ This is fine: once it is clear what $A_{ii}$ means, then all follows. As said, for coherence, for example, with matrix notation, I would avoid to denote summation in cases like $A_{ii}$. Is the book clear about this passage? Is there something like "a notation" section? $\endgroup$ – Avitus Sep 8 '14 at 21:23
  • $\begingroup$ Yes, there is a section: Most notably "If an index appears twice in a term of an indicial notation equation, summation over the range of the index is implied. Such an index is called a dummy index." It's essentially Einstein notation. $\endgroup$ – BeaumontTaz Sep 8 '14 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.