Disclaimer: I am not familiar with the formalities of Turing machines, and am quite possibly talking out of my ass, and probably not thinking along the same lines as Eliezer here. But it might be possible to salvage the ideas into something more formal/correct.
Consider a model containing exactly the natural numbers and the starred chain. Then we might have a Turing machine which starts at 0 and 0 , halts if it is fed 0, and continues to the successor otherwise. Then it never halts on the natural chain, but halts immediately on the starred chain. Here, a Turing machine presumably operates on every chain in a model meeting the first-order Peano axioms.
So in general, it might be meaningful to talk of a Turing machine acting within a model containing chains, which is closed on every given chain (e.g. it can’t jump from 0 to 0 ), and which could therefore be said to be associated by a ‘halt time’ function, h, which maps each chain (or each chain’s zero, if you like) to a nonnegative number in that chain which is the halting time on that chain. So in my above example, we might leave h(0) undefined, because the machine never halts on the naturals, and h(0 )=0*, because it halts immediately on that chain. This would then completely define the halting time over chains. (In fact, we could probably drop closure if we wanted to.)
(Edited:) I think you’re conflating the natural numbers and the tape that the Turing machine runs on. Interpreting “nonstandard halting time,” the way I think Eliezer is using the term, doesn’t require changing our notion of what a tape is; it just requires translating the statement “this Turing machine is in state s at time t” into a statement in Peano arithmetic (where t is a natural number) and then interpreting it in a nonstandard model.
Disclaimer: I am not familiar with the formalities of Turing machines, and am quite possibly talking out of my ass, and probably not thinking along the same lines as Eliezer here. But it might be possible to salvage the ideas into something more formal/correct.
Consider a model containing exactly the natural numbers and the starred chain. Then we might have a Turing machine which starts at 0 and 0 , halts if it is fed 0, and continues to the successor otherwise. Then it never halts on the natural chain, but halts immediately on the starred chain. Here, a Turing machine presumably operates on every chain in a model meeting the first-order Peano axioms.
So in general, it might be meaningful to talk of a Turing machine acting within a model containing chains, which is closed on every given chain (e.g. it can’t jump from 0 to 0 ), and which could therefore be said to be associated by a ‘halt time’ function, h, which maps each chain (or each chain’s zero, if you like) to a nonnegative number in that chain which is the halting time on that chain. So in my above example, we might leave h(0) undefined, because the machine never halts on the naturals, and h(0 )=0*, because it halts immediately on that chain. This would then completely define the halting time over chains. (In fact, we could probably drop closure if we wanted to.)
(Edited:) I think you’re conflating the natural numbers and the tape that the Turing machine runs on. Interpreting “nonstandard halting time,” the way I think Eliezer is using the term, doesn’t require changing our notion of what a tape is; it just requires translating the statement “this Turing machine is in state s at time t” into a statement in Peano arithmetic (where t is a natural number) and then interpreting it in a nonstandard model.