Stability for the Gödel and Jensen hierarchies Notations: Let $L_\alpha$ stand for the Gödel constructible hierarchy ($L_0=\varnothing$ and $L_{\alpha+1} = \mathrm{def}(L_\alpha)$ is the set of definable subsets of $L_\alpha$ and $L_\delta = \bigcup_{\beta<\delta} L_\beta$ for limit $\delta$), and $J_\alpha$ for the Jensen hierarchy ($J_0=\varnothing$ and $J_{\alpha+1} = \mathrm{rud}(J_\alpha)$ is the rud-closure of $J_\alpha \cup \{J_\alpha\}$ and $J_\delta = \bigcup_{\beta<\delta} J_\beta$ for limit $\delta$).  Let $M \mathrel{\preceq_1} N$ (if $M$ and $N$ are transitive sets mean “$(M,{\in})$ is a $\Sigma_1$-elementary submodel of $(N,{\in})$”.
Consider the following two relations between two ordinals $\sigma<\gamma$:


*

*say that “$\sigma$ is $\gamma$-stable” when $L_\sigma \mathrel{\preceq_1} L_\gamma$,

*say that “$\sigma$ is $\gamma$-J-stable” when $J_\sigma \mathrel{\preceq_1} J_\gamma$.

Question: Are the above relations equivalent?  If not, is there still a way to define “$\sigma$ is $\gamma$-stable” in terms of the Jensen hierarchy and/or  “$\sigma$ is $\gamma$-J-stable” in terms of the Gödel hierarchy?

Clearly, any one of the above relations implies that $L_\sigma = J_\sigma$ (because $\sigma$ is, at the very least, admissible $>\omega$), so the problem concerns the right-hand side, as $\gamma$ is not required to be admissible, or even a limit ordinal.  Certainly $J_\sigma \mathrel{\preceq_1} J_\gamma$ implies $L_\sigma \mathrel{\preceq_1} L_\gamma$, so the question is whether the converse holds, or, if not, whether we can still find a way to express stability for one hierarchy in terms of the other.
I know that $\sigma$ is $(\sigma+1)$-stable iff it is $\Pi_m$-reflecting for each $m$, for example, but I don't see whether this also applies to being $(\sigma+1)$-J-stable (or, if not, what the corresponding criterion would be).
Bonus question: What if we replace $1$ by $n$ (i.e., $\Sigma_1$ by $\Sigma_n$) throughout?
 A: As remarked the non-trivial direction is to show $L_\sigma\prec_{\Sigma_1}L_\gamma$ implies $J_\sigma\prec_{\sigma_1}J_\gamma$. Let's take the extreme case that $\gamma=\sigma+1$. Suppose $J_{\sigma+1}\models \exists u \varphi(u,x)$ where $\varphi\in\Sigma_0$ and $x\in J_\sigma = L_\sigma$. The short answer is to copy everything onto subsets of $\sigma$ and use $\mathcal{P}(L_\sigma)\cap J_{\sigma+1}= L_{\sigma+1}$. But in more detail: as $J_{\sigma+1}=\bigcup_{n\in\omega} S_{\omega\sigma+n}$ (see Jensen's original Fine Structure paper, or Devlin's "Constructibility" for the definition of the $S_\nu$-hierarchy,  and these and other assertions) we have for some $n$ that $S_{\omega\sigma+n}\models \exists u \varphi(u,x)$. (Here we have $\omega\sigma=\sigma$ as well as the fact that the Goedel pairing function $Q$ is $\Delta_1^{L_\sigma}$). By adding some extra `Basic Functions' to the usual list, we can assume each $S_{\omega\sigma+n}$ is transitive. Now use the fact that there is an onto function $F:\omega\sigma \rightarrow S_{\omega\sigma+n}$ in $J_{\sigma+1}$. Use this to get a subset $f\subseteq \omega\sigma$ with
$$\langle \sigma, Q  \mbox{''}f \rangle \cong \langle S_{\omega\sigma+n},\in \rangle $$
and $f\in J_{\sigma+1} $ and so also in $L_{\sigma+1} $. We now have an existential statement of the form $L_{\sigma+1} \models \mbox{''}\exists \sigma\exists f \subseteq \sigma \ldots $''. So reflect this to some $\tau<\sigma$ and obtain:
$$\langle \tau, Q  \mbox{''} \bar f \rangle \models
\exists u \varphi(u,Q(\xi)_1)
 $$ 
where we ensure for some $\xi \in \bar f$, $ Q(\xi)_1 $ represents $x$ in the structure say. Taking the transitive collapse of $\langle \tau, Q  \mbox{''}\bar f\rangle
 $ then yields a transitive model $U\in J_\sigma$ of $\exists u \varphi(u,x)$ as needed.
[Edit: added to address comment. Now use this argument repeatedly to prove by induction on $\delta$ for $\sigma+1\leq\delta<\gamma$ that $L_\sigma\prec_{\Sigma_1}L_\gamma$ implies $J_\sigma\prec_{\sigma_1}J_\gamma$. The successor case of a $\delta+1$ is just a variant of the one given and the  case of a limit $\delta$ is of course trivial.]
For the Bonus Q just note that for $n>1$, if $L_\sigma\prec_{\Sigma_n}L_\gamma$, then both $L_\sigma$ and $L_\gamma$ are limits of admissible ordinals, so then $L_\sigma=J_\sigma$ and $L_\gamma=J_\gamma$ and there is nothing to do!
