1. Feb 7, 2015

### PhyAmateur

If $$T_{12} = \frac{1}{2} {(T_1 T_2 - T_2 T_1)}$$

That mean antisymmetrization.

How would I expand then $$T_{1234}$$ I find it complicated, it is written on wikipedia for the general case but I can't still deal with these general notations http://en.wikipedia.org/wiki/Antisymmetric_tensor, I mean I can't yet expand it. You would help me if you could expand it or guide me through expanding it using {1,2,3,4} rather than {i,j,k,l} and kronocker delta and those stuff.

2. Feb 7, 2015

### Staff: Mentor

No; antisymmetrization is this:

$$T_{[12]} = \frac{1}{2} \left( T_{12} - T_{21} \right)$$

Perhaps that is what you meant to write down; but care with notation is important, particularly if you are trying to expand out more complicated cases.

The general rule is that you add together all the components with the same index, with a plus sign in front of components whose indexes are an even permutation, and a minus sign in front of components whose indexes are an odd permutation. Then you put a factor of 1 / n! in front, where n is the number of indexes.

It should be obvious that what I wrote down for $T_{[12]}$ above (note that I put brackets around the indexes; that is the standard notation for antisymmetrization) satisfies the general rule just given.

For three indexes, we would have

$$T_{[123]} = \frac{1}{6} \left( T_{123} - T_{132} + T_{231} - T_{213} + T_{312} - T_{321} \right)$$

You should be able to expand out $T_{[1234]}$ using the general rule as above. Note that there will be 4! = 24 terms. The number of terms is the reason more compact notation for this was invented.

3. Feb 8, 2015

### PhyAmateur

I linked this answer in the wikipedia link, I was hoping you could help me with the 4 termed one because I can't understand what an odd swap is or an even one?

4. Feb 8, 2015

### Ibix

Count the number of swaps needed to get there from the start position. Is it odd or even?

Start position is 1234
1243 is odd
2143 is even
etc.

5. Feb 8, 2015

### Ibix

Swaps of pairs of indices, that is.

Apologies for double post - the edit function is not working on my phone for some reason.

6. Feb 8, 2015

### ChrisVer

In general it depends on what you want to antisymmetrize. If you want to antisymmetrize it to all the indices, you have to apply:
$\frac{1}{N!} \epsilon^{abcd} T_{abcd}$.
Where N for 4 indices is 4.

You can check out that this is true for the 2 indices too...
$\frac{1}{2!} \epsilon^{ab} T_{ab} = \frac{1}{2} [ T_{12} - T_{21} ] = T_{[12]}$

In general antisymmetrization can be seen as using Permutations, and that's the origin of the factor N! ... Because for N indices, you can have N! number of permutations (the number of the elements of the Symmetric Group $S_N$ ).

If you have then to write:
$T_{[12...n]} = \frac{1}{N!} \epsilon^{i_1, i_2, ... , i_n } T_{i_1, i_2 , ... , i_n}$

In an almost similar way you can work out the antisymmetrization of less than all the indices.

For the 4 then you have a lot, because the symmetric group has 24 elements (so you have 24 terms to put with + or - ...)

Last edited: Feb 8, 2015
7. Feb 9, 2015

### Matterwave

Wouldn't this be a full contraction, and thereby not result in a tensor, but a scalar?

8. Feb 9, 2015

### ChrisVer

Well the object $T_{12...n}$ is a number, not a tensor.
If you put tensor in the game, like writing : $T_{[ab...m]}$ then on the righthand side you have to put the appropriate Levi-Civita: $T_{[a_1 a_2 ... a_i]} = \frac{1}{(n-i)!} \epsilon_{a_1 a_2 ... a_i b_1 b_2 ... b_{n-i}} \frac{1}{i!} \epsilon^{i_1 i_2 ... i_i b_1 b_2 ... b_{n-i}} T_{i_1 i_2 ... i_i}$