Bitometry said:
I'm hoping someone can clarify for me, I have seen the following used:
\frac{\partial}{\partial g^{ab}}\left( g^{cd} \right) = \frac{1}{2} \left( \delta_a^c \delta_b^d + \delta_b^c \delta_a^d\right)
I understand the two half terms are used to account for the symmetry of the metric tensor.
Yes, when you differentiate any tensor \mathbf{T} with respect to itself, you get the identity tensor of the tensor product space: \frac{\delta \mathbf{T}}{\delta \mathbf{T}} = \mathbf{I} \otimes \mathbf{I} . So, for symmetric tensor you have \frac{\delta \mathbf{g}}{\delta \mathbf{g}} = \frac{1}{2} \left( \mathbf{I} \otimes \mathbf{I} \oplus \mathbf{I} \otimes \mathbf{I} \right) . In terms of components, the above equations become \frac{\delta T^{a b}}{\delta T^{c d}} = \delta^{a}_{c} \delta^{b}_{d} , \ \ \ \ \ \ (1) \frac{\delta g^{ab}}{\delta g^{cd}} = \frac{1}{2} ( \delta^{a}_{c} \delta^{b}_{d} + \delta^{a}_{d} \delta^{b}_{c} ) . \ \ \ (2) Okay, now you try to write the following equations in components: \frac{\delta \mathbf{T}}{\delta \mathbf{T}^{-1}} = - \mathbf{T} \otimes \mathbf{T} , \frac{\delta \mathbf{g}}{\delta \mathbf{g}^{-1}} = - \frac{1}{2} \left( \mathbf{g} \otimes \mathbf{g} \oplus \mathbf{g} \otimes \mathbf{g} \right) .
However, it appears to only half account for terms where a \ne b.
To demonstrate what I mean by example, consider a 2D metric where (a,b,c,d = 1,2)
\frac{\partial}{\partial g^{11}}\left( g^{11} \right) = \frac{1}{2} \left( \delta_1^1 \delta_1^1 + \delta_1^1 \delta_1^1\right) = 1
but
\frac{\partial}{\partial g^{12}}\left( g^{12} \right) = \frac{1}{2} \left( \delta_1^1 \delta_2^2 + \delta_2^1 \delta_1^2\right) = \frac{1}{2}
I expected (possibly incorrectly) the latter would also be 1.
Is the contribution of g^{cd} evenly distributed between \frac{\partial}{\partial g^{ab}} and \frac{\partial}{\partial g^{ba}}, and thus for a=b the value is double the others?
No, here you need to be careful with the functional dependence. For any fixed value of c and d, you have the following function g^{c d} = \bar{g}^{c d} ( g^{1 1} , g^{(1 2)} , g^{22} ) \equiv g^{c d} \left( g^{11} , g^{12} ( g^{(12)} ) , g^{21}( g^{(12)} ) , g^{22} \right) , where g^{(12)} = \frac{1}{2} ( g^{12} + g^{21} ) = g^{12} = g^{21} . So, \frac{\partial \bar{g}^{cd}}{\partial g^{(12)}} = \frac{\partial g^{cd}}{\partial g^{12}} \frac{\partial g^{12}}{\partial g^{(12)}} + \frac{\partial g^{cd}}{\partial g^{21}} \frac{\partial g^{21}}{\partial g^{(12)}} , or \frac{\partial \bar{g}^{cd}}{\partial g^{(12)}} = \frac{\partial g^{cd}}{\partial g^{12}} + \frac{\partial g^{cd}}{\partial g^{21}} = \delta^{c}_{1} \delta^{d}_{2} + \delta^{c}_{2} \delta^{d}_{1} .
In practise the metric tensor comes always contracted with other tensors, so you never need to use Eq(2), because the (symmetric) metric kills the anti-symmetric part of the contraction. So, it is always safe to use Eq(1) for the metric. To see that, let’s consider two examples and try to calculate the functional derivative with respect to g^{n m}
The first is the simple trace equation S = A^{a}_{a} = A_{ab}g^{ab} . Since g^{ab} = g^{ba}, the anti-symmetric part of A_{ab} will get killed by the metric. So, we may take A_{ab} to be symmetric, i.e. A_{ab} = A_{(ab)}. Now \frac{\delta S}{\delta g^{nm}} = A_{(ab)} \frac{\delta g^{ab}}{\delta g^{nm}} = A_{(ab)} \delta^{a}_{n} \delta^{b}_{m} = A_{(nm)} . Notice that Eq(1) and Eq(2) give you the same answer. The second example which is a bit more complicated (but important), is the square of rank-2 tensor S = F^{2} = F_{ab} F^{ab} = F_{ac} F_{bd} g^{ab} g^{cd} . I will use Eq(1) to calculate the derivative
\frac{\delta S}{\delta g^{nm}} = F_{ac} F_{bd} \left( g^{ab} \frac{\delta g^{cd}}{\delta g^{nm}} + \frac{\delta g^{ab}}{\delta g^{nm}} g^{cd} \right) . Thus, by Eq(1), \frac{\delta S}{\delta g^{nm}} = F_{ac} F_{bd} ( g^{ab} \delta^{c}_{n} \delta^{d}_{m} + g^{cd} \delta^{a}_{n} \delta^{b}_{m} ) . Finally, by changing the dummy indices in the second term, we get \frac{\delta S}{\delta g^{nm}} = g^{ab} ( F_{an} F_{bm} + F_{na} F_{mb} ) . This is the final result for arbitrary rank-2 tensor. Furthermore, for symmetric or anti-symmetric F_{ab}, the above expression simplifies to \frac{\delta S}{\delta g^{nm}} = 2 g^{ab} F_{an} F_{bm} . Now you repeat the calculation using Eq(2) instead.
Sam