I asked Turing degree of a turing machine with access to an (arbitrary) nonstandard integer, not thinking about the possiblity that this could depend on the model used. The question was not formulated very well, but I did not want to delete it since I got some good answers. Now I will ask the question I originally had in mind.

Let $U$ be a ultrafilter on $\mathbb N$ and $f$ a partial computable function of two arguments, which can either accept or reject its input (or not halt) (I'm using a finite range so that the output is guaranteed to be standard, which should make things slightly simpler). We say that $f$ accepts $n$ with help from $U$ if $\{i \in \mathbb N : f(n,i)\text{ accepts}\} \in U$, and rejects or does not halt likewise (due to the definition of the ultrafilter, exactly one of these possibilities will be true). This can be seen as defining running $f$ in a model defined by a ultrapower of $\mathbb N$ with input $n$ (which is standard) and $H$=$(0,1,2,\dots)$. My question is, what is the turing degree associated with such functions, for a given ultrafilter $U$.

If the ultrafitler $U$ is principal, then its turing degree will just be $0$, so we will mostly consider nonprincipal ultrafilters.

For a nonprincipal ultrafilter $U$ (we will assume all ultrafilters are nonprincipal from this point on), we can construct a function $f(n,i)$ which interprets $n$ as a standard turing machine, and runs this machine for $i$ steps. If the machine halts, it will after a standard number of steps, and so $f$ will accept. If the machine has not halted after $i$ steps, it never will, and so $f$ will reject. $f$ has solved the halting problem for standard turing machines, and so the Turing degree of $U$ is at least $0'$. (Note that if the machine associated with $n$ is one that looks for a contradiction in, say, $ZFC$, $f$ is still guaranteed to give the right answer (according to whether or not $ZFC$ is consistient), even if the length of shortest contradiction is nonstandard in whatever metatheory we are using. That's because the ultraproduct model layers another layer of "nonstandardness" over what the metatheory already has.) We can also note that $U$'s turing degree must be at least $0'$, since it can be used to model an eventually correct machine.

On the other hand, for any set $S \subseteq \mathbb N$, we can construct an ultrafilter $U$ such that, with its help, we can compute $S$. Choose a computable encoding of finite sets of natural numbers.. Now, let $a$ be the set of encodings of subsets of $S$ of the form $S_{<n}$ for some $n$.

If $S$ is finite, then we can compute $S$ easily. Otherwise, $a$ will be infinite. Therefore, there will be some ultrafilter $U$ (assuming that at least one nonstandard ultrafilter exists) such that $a \in U$. Therefore, the function $f(n,i)$ defined as $n \in i$ (according to the encoding), will compute $S$. (To see this, note that for all $n \in S$, $f$ will accept for cofinitely of $i \in a$, and so $f$ will accept. Otherwise, $f$ will reject for all $i \in a$, and so will reject.)

This shows that ultrafilters can have arbitrarily high turing degrees, and in particular not all ultrafilters have the same turing degree.

So, my question is, for a given ultrafilter $U$, how can we determine the turing degree associated with it?