Levi-Civita & Jacobi: Meaning & Question

[zn52]

hey Folks,
please have a look at the attached Ex from MTW. does somebody know what is the meaning of the parallel bars in the first levi civita symbol ? Is there a typo in this EX perhaps? I would have expected that on the right hand side one would see the product which is shown in the first formula and not the Jacobian...
Thank you,

 

[henry_m]

The lines mean antisymmetrisation. A more common notation is to use square brackets, so
$$T_{[\mu\nu]}=\frac{1}{2}\left(T_{\mu\nu}-T_{\nu\mu}\right)$$
and more generally, if there are n indices in the brackets, it means a sum over all permutations of the indices, with a minus sign for odd permutations, all divided by n!, which is the total number of permutations. You might also see ordinary round brackets () which mean symmetrisation, in which case the signs are all +.

 

[Matterwave]

Isn't the Levi-Civita symbol a totally anti-symmetric (pseudo)-tensor, so anti-symmetrizing it...shouldn't change it though...right...

 

[zn52]

thank you very much.
I have tried to expand the Levi-Civita symbol as you mentioned above, but how would the anti-symmetrisation yield the half on the RHS of the formula ?

 

[zn52]

I agree at least for 2 D...

 

[Matterwave]

The 1/2 is to make sure that if you have a tensor which is already anti-symmetric, then you don't have to do anything and get the same tensor back. Similarly, that's why you have the 1/n! for the symmetrization.

 

[zn52]

The issue in this very example is that on the LHS, if you would expand the Levi-Civita symbol using the antisymmetrisation you mentioned above, you will end up with the same symbol which is on he RHS. This would mean that the 1/2 has no place ! I'm may be confused or I'm I missing something ? I hope you see what I mean.
Thanks.

 

[henry_m]

Apologies, I've reread the question and it looks like the convention the question uses for antisymmetrisation doesn't include the factor of n!. The equality between the first line and the final answer is unambiguously true though, so just make sure you can derive that.

The point of having the factor is that if you have an antisymmetric tensor $A_{\mu\nu}$, you can write it as $A_{\mu\nu}=A_{\left[\mu\nu\right]}=\frac{1}{2}\left(A_{\mu\nu}-A_{\nu\mu}\right)$. The half appears because there are two terms, all of them equal. The same would hold for n indices, in which case there would be n! equal terms.

 

[zn52]

I thank you so much for your clarification and guidance. Now I have moved on after having worked intensively on the Levi-civita and determinants. I had to work through simple cases in 3D and 2D in order to understand these mathematical identities...