# Bracketed subscripts

I. What does

A_{[\alpha} B_{\beta]}

mean? How do you expand this?

II. How do you put TeX in board posts?

robphy
Homework Helper
Gold Member
bigplanet401 said:
I. What does

A_{[\alpha} B_{\beta]}

mean? How do you expand this?

II. How do you put TeX in board posts?
[ tex ] A_{[\alpha} B_{\beta]} [ /tex ] (remove the spaces in the tags)
$$A_{[\alpha} B_{\beta]}$$ (click the on the equation to see)

$$A_{[\alpha} B_{\beta]} =\frac{1}{2!}\left( A_{\alpha} B_{\beta} - A_{\beta} B_{\alpha} \right)$$ the "antisymmetric part of $$A_{\alpha} B_{\beta}$$"

dextercioby
Homework Helper
It's not necessary to use the numerical factor (we call it "weght"). See for example the em. tensor in vacuum. U can use the "no number convention" (i use it)

$$F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:\partial_{[\mu}A_{\nu]}$$

or the one Rob exemplified, when an ugly 2 comes up

$$F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:2 \partial_{[\mu}A_{\nu]}$$

Daniel.

dextercioby
Homework Helper
And for symmetrizing, we use round brackets. For example, the linearized graviton field has the irreducible Lagrangian infinitesimal gauge transformations

$$\delta_{\epsilon}h_{\left(\mu\nu\right)}(x) =\partial_{(\mu}\epsilon_{\nu)}(x)$$

or with the "2", if you use an analogue convention Rob used.

Daniel.

robphy
Homework Helper
Gold Member
While the combinatorial factor I used may be conventional, I believe it is the preferred convention. See, for example, http://mathworld.wolfram.com/AntisymmetricTensor.html

Note that the "symmetric part" of a matrix is $$\frac{1}{2}(A+A^T)$$ and the "antisymmetric part" of a matrix is $$\frac{1}{2}(A-A^T)$$. (Similarly, the "real part of a complex number z" is $$\frac{1}{2}(z+\bar z)$$ and "imaginary part of a complex number z" is $$\frac{1}{2i}(z-\bar z)$$.)

So, one can write the matrix equation
$$A= A_{SYM} + A_{ANTISYM}$$
and an analogous tensorial equation
\begin{align*} A_{ab} &= \frac{1}{2}( A_{ab} + A_{ba} )+ \frac{1}{2}( A_{ab} - A_{ba} ) \\ &= A_{(ab)} + A_{[ab]} \\ \end{align*}

Note, if A is antisymmetric, then we can write
$$A_{ab} = A_{[ab]}$$.

dextercioby said:
It's not necessary to use the numerical factor (we call it "weght"). See for example the em. tensor in vacuum. U can use the "no number convention" (i use it)

$$F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:\partial_{[\mu}A_{\nu]}$$

or the one Rob exemplified, when an ugly 2 comes up

$$F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:2 \partial_{[\mu}A_{\nu]}$$

Daniel.
I think there is an inconsistency in your use of the brackets in the "no number convention" $$F_{\mu\nu}\equiv F_{\left[\mu\nu\right]}=:\partial_{[\mu}A_{\nu]}$$

If "bracket" means "sum the alternating permutations without dividing by the combinatorial factor", then you must write for an antisymmetric F:
$$F_{\mu\nu}=\frac{1}{2}F_{\left[\mu\nu\right]}=\frac{1}{2}(F_{\mu\nu}-F_{\nu\mu})$$ or $$2F_{\mu\nu}=F_{\left[\mu\nu\right]}$$

dextercioby
Homework Helper
I didn't in any place claim that

$$F_{[\mu\nu]}=F_{\mu\nu}-F_{\nu\mu}$$

So no inconsistency. Whatsoever.

Daniel.

robphy
Homework Helper
Gold Member
dextercioby said:
I didn't in any place claim that

$$F_{[\mu\nu]}=F_{\mu\nu}-F_{\nu\mu}$$

So no inconsistency. Whatsoever.

Daniel.
So, maybe I am misunderstanding what your "no number convention" is.
What does $$\partial_{[\mu}A_{\nu]}$$ mean in your convention?
$$\partial_{[\mu}A_{\nu]}\stackrel{?}{=}\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$ or
$$\partial_{[\mu}A_{\nu]}\stackrel{?}{=}\frac{1}{2!}\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)$$?

dextercioby