How to differentiate a term with respect to metric

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
the_doors
Messages
17
Reaction score
0

Homework Statement


for example: ##\frac{\partial(F^{ab}F_{ab})}{ \partial g^{ab}} ## where F_{ab} is electromagnetic tensor.
or ##\frac{\partial N_{a}}{\partial g^{ab}}## where ##N_{a}(x^{b}) ## is a vector field.

Homework Equations

The Attempt at a Solution


i saw people write ##F^{ab}F_{ab}## as ## g^{ca}g^{db}F_{cd} F_{ab}## and say we write the term in a way that exposes the dependence on the metric. but exactly what it means ?
 
Physics news on Phys.org
You need to write your function in terms of the variables you have, in your case ##g^{ab}## and ##F_{ab}##. The contravariant ##F^{ab}## is not independent of those and so will have non zero partial derivatives wrt both. Writing it in terms of your variables let's you differentiate easier.

As a similar example: Take ##f(x,y) = xy##. The partial derivative ##\partial_yf## is simply ##\partial_y (xy) = x##. This is essentially the same. Do not be confused by the fact that you use the same letter to denote the covariant and contravariant tensors.
 
for example if ##N_{\mu}(x^\nu)## be unit time like dynamical four velocity, what's happening ? i don't know explicit form of ##N_{\mu}##.
 
ok we can write for first example:##\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}}=\delta^c_{e} \delta^a_{f}## is it correct ?

what's happening for ## \frac{\partial F_{cd}}{\partial g^{ef}}## ?