(Everything is assuming $V=L$.)


Fix an uncountable regular cardinal $\kappa$, and let $$E\_\kappa=\{\mu<\kappa: \mbox{there is an elementary substructure of $L\_\kappa$ isomorphic to $L\_\mu$}\},$$ and let $$E\_\kappa^+=\{\mu<\kappa: \mbox{$\exists$ an elementary substructure of $(L\_{\kappa^+}, L\_\kappa)$ which is isomorphic to $(L\_\kappa, L\_\mu)$}\}.$$ (In each definition $\mu$ ranges over ordinals.) Here "$(L\_\alpha, L\_\beta)$" denotes the structure $(L\_\alpha; \in)$ augmented by a predicate for $L\_\beta$.


Clearly $E\_\kappa^+\subseteq E\_\kappa$. In the course of a problem I'm working on, I assumed that $E\_\kappa$ and $E\_\kappa^+$ are quite different, but thinking about it in more detail this seems unjustified. My question is:



> 
> Is $E\_\kappa^+$ necessarily a proper subset of $E\_\kappa$?
> 
> 
> 


I would also be interested in specific cases, e.g. $\kappa=\omega\_1$.




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EDIT: Oh dear, this is what I get for doing math late at night: the question above has a terrible typo - $E\_\kappa^+$ is supposed to be the set of $\mu$ such that *for some $\alpha<\kappa$* we have $(L\_\alpha,L\_\mu)$ is isomorphic to an elementary substructure of $(L\_{\kappa^+}, L\_\kappa)$. This completely changes the meaning of the question, of course. 


I'm leaving the question as is and accepting Joel's answer, since it is a good answer; but I've asked a corrected version of the question at [Levels of L resembling each other, take 2](https://mathoverflow.net/questions/235286/levels-of-l-resembling-each-other-take-2).



Fine structure question: when do levels of $L$ look "a lot" like each other?
Under your assumptions, the set $E\_\kappa^+$ is empty. Indeed, we don't even need the predicates that you mention. (In any case, since $\kappa$ is definable in $L\_{\kappa^+}$, the predicate for $L\_\kappa$ in $\langle L\_{\kappa^+},\in,L\_\kappa\rangle$ would be definable and therefore offer no additional expressive power.) 


What I claim is that if $\kappa$ is a cardinal, then there is no elementary substructure of $\langle L\_{\kappa^+},\in\rangle$ in $L$ that is isomorphic to $\langle L\_\kappa,\in\rangle$. 


To see this, suppose that $\kappa$ is a cardinal and there is an elementary substructure $X\prec L\_{\kappa^+}$ in $L$ that is isomorphic to $L\_\kappa$. This amounts to an elementary embedding $j:L\_\kappa\to L\_{\kappa^+}$. Since $\kappa$ is definable in the target structure, we must have $j(\mu)=\kappa$ for some $\mu<\kappa$, and so $j$ has a critical point $\delta\leq\mu$. Since all the subsets of $\delta$ in $L$ appear in $L\_\kappa$, we may define in $L$ a measure on $\delta$ via $X\in U\iff \delta\in j(X)$. So $\delta$ is a measurable cardinal in $L$, contradicting Scott's theorem that there are no measurable cardinals in $L$. 


