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In Riemannian geometry for some reasons we consider a variation of metric then we compute its time-depending derivative. I want to know why we do this? Is it similar to finding the critical points? If so what is the definition and intuition of critical points here?

Update(29-03-2021)

For example sometimes in the easiest case, we consider first-order deformation $$g_t=g+th,$$ for some symmetric tensor $h$. Is in this case $\dfrac{dg_t}{dt}|_{t=0}=h$? If so what is the point of considering $g_t$, instead we can work with $h$? more generally if we consider $$g_t=g+th+t^2k+\cdots ,$$ then I think the result is same as above. i.e. $\dfrac{dg_t}{dt}|_{t=0}=h$. Am I right?

Also I forgot from calculus that why we evaluate it at $t=0$ after taking derivative?

C.F.G
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    It's a tensor of the same type as the metric, whose components are given by the time derivatives of the correspondent components of the metric. – Matheus Andrade Jul 02 '20 at 08:16
  • Answer to last question: Whether you want to set $t=0$ or not depends on the context. Sometimes, the equation you want is only at $t=0$. This occurs, for example, when you want to study a variation of a given metric $g_0$. – Deane Mar 29 '21 at 19:50

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$g_t$ is a path in the space of all Riemannian metrics. With the appropriate topology and some differential calculus, it should have some kind of extremum (we should know that what we are measuring to know what the extremum means) and for this reason we should take derivative to see where they are.

For second part of question, Yes it is correct $\dfrac{dg_t}{dt}|_{t=0}=h$ in both cases. for last question see Deane's comment.

C.F.G
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