Using the Daniell Cell as an example, where in the steady state we expect zinc to be oxidised and copper to be reduced spontaneously due to the overall reduction in Gibbs free energy, how does the system "know" to start moving electrons from the anode to the cathode? What is the underlying physical mechanism?
When the two half cells are first connected via the external circuit, it seems that they would not yet "know" anything about each other; each half cell would not already "know" it is anode or cathode (e.g. zinc could be the cathode in a different galvanic cell) and so how would the system immediately start driving the electrons in the right direction?
In other answers on this site (e.g. https://chemistry.stackexchange.com/a/28638/98580) I've seen statements like "the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell".
However, as a physicist this is somewhat unsatisfying, as the phrase "induce reduction" seems to bundle up a whole lot of physical processes in the half cells and the external circuit into a very high level concept. Fundamentally, it must come down to electrical forces and thermodynamics to explain the overall behaviour of the system.
My understanding is that prior to connecting the external circuit, each electrode is in equilibrium with its electrolyte, undergoing both oxidation and reduction at the same time; so each electrode could both push and pull electrons to/from a connected wire. The moment the external circuit is connected, do electrons start flowing from the zinc electrode to the copper electrode because zinc has a greater "tendency" (than copper) to push electrons onto a connected wire, and copper has a greater "tendency" (than zinc) to pull electrons from a connected wire? That would then sound like the galvanic cell gets started via a statistical process and is sustained by electrical charging of the electrodes thereafter.