Runtime of Grover's algorithm What is the time complexity (not query complexity) of Grover's algorithm?  It seems clear to me that it is $\Omega(\log(N) \sqrt{N})$ since there are $\Omega(\sqrt{N})$ iterations and each iteration requires use of the reflection operation which in turn takes time $\Omega(\log(N))$ using any standard set of universal gates.

The problem is, I can't find even a single reference which says the time complexity of Grover's algorithm is $\Omega(\log(N) \sqrt{N})$.  Wikipedia, and several other web pages, say $O(\sqrt{N})$ time complexity.  Grover's paper claims $O(\sqrt{N})$ "steps".

Am I missing something?  Perhaps people define the reflection operation to take unit time.  But that doesn't make sense to me because if we can play the game of allowing arbitrary unitaries to take unit time then there would be no difference between query complexity and time complexity.
The question is usually taken to be moot, for the following reason.  Grover's algorithm is a combinatorial search algorithm to find a solution to an arbitrary predicate.   While, yes, $\Theta(\log N)$ is the quantum gate complexity in each stage of the black-box algorithm, the predicate needs to be computed too.  The quantum gate complexity of that is $\Omega(\log N)$, because otherwise it wouldn't read the whole input and you could discard some of the input bits from the search.  On the other hand, an interesting predicate could take a lot more time than that.  Hence, the number of calls to the predicate is taken to be the standard coin, just as it is for the classical analogue of Grover's algorithm, namely random guessing.