An ordinal $\alpha$ is "metadefinable" by some formula $\varphi$ without free variables if:
$$
\begin{cases}
L_\alpha \models\varphi \\
\forall\beta < \alpha \, L_\beta \not\models \varphi
\end{cases}
$$
Is there an usual terminology for such "metadefinable" ordinals?

1$\begingroup$ This is related to implicit definability. But I don't think there's a name for this specific property. $\endgroup$– WojowuApr 21 at 17:55

$\begingroup$ @Wojowu I don't quite see the connection with implicit definability—could you explain? Are you referring to the notion of implicit definability that I had introduced with Cole Leahy? Or do you refer to the classic modeltheoretic notion used in Beth's theorem? $\endgroup$– Joel David HamkinsApr 23 at 12:14
2 Answers
Let me say first that your concept is similar in spirit to the notion of sententially categorical cardinal appearing in my joint paper
 J. D. Hamkins and R. Solberg, Categorical large cardinals and the tension between categoricity and settheoretic reflection, arxiv:2009.07164, 2020.
We were interested in investigating the circumstances when Zermelo's quasicategoricity theorem rises to actual categoricity. Namely, a cardinal $\kappa$ is (firstorder) sententially categorical if there is some firstorder sentence $\sigma$ such that $V_\kappa\models\text{ZFC}_2+\sigma$, but this is not true in any other $V_\beta$. This is equivalent to saying that there is a sentence $\sigma$ for which $V_\kappa\models\text{ZFC}_2+\sigma$, but this is not true in any smaller $V_\beta$. These are equivalent since we can simply replace $\sigma$ with the assertion that $\sigma$ holds, but not earlier.
The difference between your notion and ours is that you are using the $L$ hierarchy and do not insist on $\text{ZFC}_2$, but the background categoricity idea seems fundamentally similar. In our paper, we consider also theory categoricity, secondorder sentential categoricity and secondorder theory categoricity.
Meanwhile, the same trick about categoricity works for your notion. That is, I claim an ordinal $\alpha$ is metadefinable on your definition if and only if there is some sentence $\varphi$ true in $L_\alpha$ and not in any other $L_\beta$ for $\beta\neq\alpha$. The reason is that if $\varphi$ is true for the first time at $L_\alpha$, then you can replace $\varphi$ with the assertion "$\varphi$ and this is not true earlier," and this will be true only in $L_\alpha$.
For this reason, what you have is a notion of sentential cateogoricity—you are identifying ordinals $\alpha$ for which $L_\alpha$ is characterized among all $L_\beta$ as the only one satisfying a certain sentence. Therefore, unlesss there is already an established terminology for your ordinals coming from finestructure theory (see below), I would suggest the terminology: $\alpha$ is sententially categorical with respect to the constructible hierarchy.
But second, your context is much lower down. All your ordinals, for example, are countable, since by condensation we can take a countable elementary substructure of $L_\alpha$, which will collapse below $\omega_1$, but have the same theory. And since there are only countably many metadefinable ordinals, and this is observable in $L$, they are bounded below $\omega_1^L$.
Indeed, every metadefinable ordinal is exhibiting a $\Sigma_1$property, since $L$ can see that $\alpha$ is metadefinable by observing that there is a sentence $\varphi$ true in $L_\alpha$ but not earlier. Thus, every metadefinable ordinal is below the first $1$stable ordinal, the smallest ordinal $\delta$ for which $L_\delta\prec_{\Sigma_1}L$.
Set theorists doing fine structure are often looking at the first stage $L_\alpha$ where a new $\Sigma_1$ fact becomes true (allowing parameters), and the supremum of these stages is the first $1$stable ordinal. But you do not allow parameters in your sentence, and so yours is a very special case. The finestructure analysis also has notions of $n$stable and so on. I am not sure if they isolate your notion exactly with terminology, but that is where I would look.

2$\begingroup$ Joel, it seems to me that the sup of these metadefinable ordinals is exactly the first 1stable. The key observation is that it's $\Sigma_1$ to say that a specific metadefinable ordinal is countable, and so there'll be an ordinal metadefined by the ability to see this collapse map. This implies that the supremum $\sigma$ will have $L_\sigma$ ($\Sigma_1$)pointwise definable. But then any $\Sigma_1$ formula true in $L$ with parameters from $L_\sigma$ is equivalent to a $\Sigma_1$ sentence. And so it must be true in some $L_\beta$ below $L_\sigma$ and hence in $L_\sigma$ by upward absoluteness. $\endgroup$ Apr 23 at 3:35

$\begingroup$ I looked around a bit and it happened to be also very close from the notion of characterizability mostly used in infinitary and second order logic. I gave some examples of uses in my answer. $\endgroup$– JohanMay 15 at 13:06
I have seen the concept of "characterizable" used for both cardinals and ordinals. It appears to stem from the Ph.D. dissertation of S.J. Garland (1967) part of which can be found in GARLAND, S. J. (1974), Secondorder cardinal characterizability and has been reused in e.g. George, B.R. (2006) SecondOrder Characterizable Cardinals and Ordinals and Kunen (1971), Indescribability and the continuum.
It boils down to the following definition, with some minor variations: A structure $M$ is (secondorder) characterizable if there is a formula $\varphi$ such that $M\models \varphi$ and all model of $\varphi$ are isomorphic to $M$. From there, an ordinal $\alpha$ is deemed characterizable if the structure $(\alpha, \in)$ is characterizable and it seems natural to extend it to either $L_\alpha$ or $V_\alpha$.
This is very close to what JD Hamkins suggested but the name maybe has the advantage of being more straightforward.
And another similar notion can be found in Barwise (p.257): An admissible set $A$ is selfdefinable if there is a theory $T$ such that some extension of $A$ into a structure for $T$ is a model for $T$ and any extension of $B$ being model of $T$ implies that $A \simeq B$.