I saw people write $\frac{\partial( F^{ab} F_{ab})}{\partial g^{ef}}$ as $\frac {\partial (g^{ca}g^{db}F_{cd}F_{ab})}{\partial g^{ef}}$ in a way that exposes the dependence on the metric. but exactly what it means?
So we have $$\frac {\partial (g^{ca}g^{db}F_{cd}F_{ab})}{\partial g^{ef}}=\frac{\partial g^{ca}}{\partial g^{ef}}g^{db}F_{cd}F_{ab}+\frac{\partial g^{db}}{\partial g^{ef}}g^{ca}F_{cd}F_{ab}+\frac{\partial F_{cd}}{\partial g^{ef}}g^{ca}g^{db}F_{ab}+\frac{\partial F_{ab}}{\partial g^{ef}}g^{ca}g^{db}F_{cd}.$$ Is it correct?
$$\frac{\partial g^{ca}}{\partial g^{ef}}=\frac{1}{2}(\delta^c_{e} \delta^a_{f}+\delta^c_{f} \delta^a_{e})$$ Is it correct?
What's happening for $\frac{\partial F_{cd}}{\partial g^{ef}}$?
Also I have no idea about varying a vector field as $N_\mu(x^\nu)$ with respect to metric or$\nabla_\mu N_\nu$ with respect to the metric.
Asked
Active
Viewed 477 times
1
-
1
- I don't understand what you want to know. 2. & 3. Check-my-work questions are off-topic 4. & 5. seems to be off-topic as homework-like.
– ACuriousMind Apr 21 '15 at 10:57 -
1
- Is answered here: http://physics.stackexchange.com/q/149066/
– innisfree Apr 21 '15 at 15:06 -
what is the variation of $N_\mu$ and $\nabla_\nu N_\mu$ with respect to inverse metric? $N_\mu$ is unit time like vector field. i'm confused!!! – the_doors Apr 21 '15 at 11:20
1 Answers
2
When we vary $F^{ab}F_{ab}$ with respect to the metric, we must also specify what we are holding fixed. Assuming that the context is that of electromagnetism, we consider the four-potential $A_b$ as an independent variable, and therefore under variations of other variables (such as the metric), it is held fixed, as is $F_{ab} = \partial_a A_b - \partial_b A_a$.
But $F^{ab} = g^{ac}g^{bd}F_{cd}$, and it is therefore dependent on the metric. Thus, when varying $$F^{ab}F_{ab} = g^{ac}g^{bd}F_{cd}F_{ab}$$ with respect to the metric while holding $F_{ab}$ fixed, it is convenient to explicitly express the dependence on the metric.
AV23
- 3,193
-
-
It's zero: there is no dependence of $F_{ab}$ on the metric, for these variations. – AV23 Apr 21 '15 at 11:15
-
my problem is exactly here , why there is no dependence on metric ?. for example , if $F_{12}=r^4$ and $g^{33}=r^2$ then we will have 2 r. – the_doors Apr 21 '15 at 11:38
-
1
-
so is this correct : $\frac{\partial (N^\mu N_\nu)}{\partial g^{\rho\zeta}} = \frac{\partial g^{\mu\sigma}}{\partial g^{\rho\zeta}} (N_\sigma N_\nu) $ – the_doors Apr 21 '15 at 13:02
-
Once again, that depends on the kind of partial derivative i.e. whether you are holding $N^{\mu}$ or $N_{\mu}$ fixed. So it's correct only if $N$ is 'naturally' a co-vector i.e. if you consider $N_{\mu}$ to be independent of the metric. – AV23 Apr 21 '15 at 13:20