Simple question about antisymmetrization

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  • #1
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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.
 

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  • #2
PeterDonis
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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.
 
  • #3
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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
Ibix
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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
Ibix
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Swaps of pairs of indices, that is.

Apologies for double post - the edit function is not working on my phone for some reason.
 
  • #6
ChrisVer
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In general it depends on what you want to antisymmetrize. If you want to antisymmetrize it to all the indices, you have to apply:
[itex] \frac{1}{N!} \epsilon^{abcd} T_{abcd} [/itex].
Where N for 4 indices is 4.

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

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 [itex]S_N[/itex] ).

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

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 - ...)
 
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  • #7
Matterwave
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In general it depends on what you want to antisymmetrize. If you want to antisymmetrize it to all the indices, you have to apply:
[itex] \frac{1}{N!} \epsilon^{abcd} T_{abcd} [/itex].
Where N for 4 indices is 4.
Wouldn't this be a full contraction, and thereby not result in a tensor, but a scalar?
 
  • #8
ChrisVer
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Wouldn't this be a full contraction, and thereby not result in a tensor, but a scalar?
Well the object [itex]T_{12...n} [/itex] is a number, not a tensor.
If you put tensor in the game, like writing : [itex]T_{[ab...m]}[/itex] then on the righthand side you have to put the appropriate Levi-Civita: [itex]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} [/itex]
 

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