How Do You Expand the Antisymmetrized Tensor \( T_{1234} \)?

[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.

 

[PeterDonis]

If

##T_{12} = \frac{1}{2} {(T_1 T_2 - T_2 T_1)}​

That mean antisymmetrization.

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.

How would I expand then

$T_{1234}$​

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.

 

[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?

 

[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.

 

[Ibix]

Swaps of pairs of indices, that is.

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

 

[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 - ...)

 

[Matterwave]

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.

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

 

[ChrisVer]

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

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: $T_{[a_1 a_2 ... a_i]} = \frac{1}{(n-i)!} \epsilon_{a_1 a_2 ... a_i b_1 b_2 ... b_{n-i}} \frac{1}{i!} \epsilon^{i_1 i_2 ... i_i b_1 b_2 ... b_{n-i}} T_{i_1 i_2 ... i_i}$