Proving Symmetric Tensor Equation: S=[0.5(T+TT)]

[manicwhite]

Homework Statement

S : T = S:[0.5(T+T^T)]
S is a symetric tensor
show for any tensor T the above is valid

Homework Equations

The Attempt at a Solution

what i think i know
S^T=S
S:T=tr[S^TT] =tr[ST]
einstein notation
tr[S_ijT_jk]
[S_ijT_jk]_ii
but i can't really see this leading to the LHS

so i changed tact

LHS=RHS
T=0.5(T+T^T)
now i can see this being true if T was symetric but that isn't the question

thanks

 

[hunt_mat]

This is quite easy:
<br /> S:\left(\frac{1}{2}(T+T^{T}\right) =\frac{1}{2}S^{ab}(T_{ab}+T_{ba})=\frac{1}{2}S^{ab}T_{ab}+\frac{1}{2}S^{ab}T_{ba}=\frac{1}{2}S^{ab}T_{ab}+\frac{1}{2}S^{ba}T_{ba}<br />
The last follows as S is symmetric

 

[Fredrik]

S:T=tr[S^TT] =tr[ST]
einstein notation
tr[S_ijT_jk]
[S_ijT_jk]_ii

This notation doesn't make sense. (That doesn't mean that I would be surprised if your book or your professor uses it or something very similar). This makes sense:

S:T=\operatorname{tr}(S^T T)=(S^T T)_{ii}=(S^T)_{ij}T_{ji}=S_{ji}T_{ji}

Note that I'm just using your definition of S:T, the definition of trace, the definition of matrix multiplication, and the definition of the transpose. (And obviously the summation convention too).

hunt_mat's suggestion is an easy way to solve this problem. Another approach, which may be more useful in the long run, is to prove that

a) if S is symmetric and A is antisymmetric (i.e. A_ij=-A_ji), then SA=0. (The terms "skew symmetric", "skew" or "alternating" are often used instead of "antisymmetric").

b) Every tensor (with two indices) can be expressed as a sum of a symmetric and an antisymmetric tensor.

Part a) is an easy exercise that I suggest that you do. Part b) is obvious once you've seen the trick, but it can take some time to see it:

T=\frac{T+T^T}{2}+\frac{T-T^T}{2}

As an example of how this can simplify things, consider the following proof of the identity \vec x\cdot(\vec x\times \vec y):

\vec x\cdot(\vec x\times \vec y)=x_i\varepsilon_{ijk}x_jy_k=0

The conclusion is immediate, since \varepsilon is antisymmetric in i and j (i.e. changes sign under an exchange of i and j), and x_ix_j is symmetric in i and j.

 

[manicwhite]

thanks very, you have both been a great help.

it easy when you know how. hopefully i do know now and this will stick in my head.

the notation i used was my attempt to derive this myself. i actually found the final line"SijTij"quoted in my notes in the end

once again thanks very much
merry xamss