Whenever we twist two wires together to make a circuit, there's no soldering or fusing-together of wires involved. So, in such a case do electrons jump across the microscopic air gaps which are present in between the two wires? Or is it just that they influence the electrons of the next wire to move?

If it is the latter case, what exactly constitutes an electric current? flow of electrons literally or just the transfer of energy etc. ?

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    $\begingroup$ Possible duplicate of tranfser of electrons between two metals and a wire $\endgroup$ – Bosoneando Aug 5 '16 at 9:35
  • $\begingroup$ The surfaces have to touch (otherwise the wires could freely slip) and those microscopic points will conduct most of the charge. For good conductors like copper there is probably some microscopic cold welding going on, anyway. Having said that, forming good electric contacts requires special surface preparation, which is why we usually do it with connectors rather than poorly defined mechanical twisting. $\endgroup$ – CuriousOne Aug 5 '16 at 9:36

Yes, they jump from one wire to the other. However there isn't a microscopic air gap to use your phrase. When you twist the wires together they will make contact at the high spots on their surface and those high spots will deform under the contact pressure and spread out to form a contact patch. The true area of contact will be a lot less than the apparent area of contact due to this surface roughness, but there will be a non-zero contact area.

Depending on the force you used to twist the layer there may be a very thin layer of oxide or some other surface contaminent between the two wires at the contact patch. However this will typically be of a thickness comparable to atomic dimensions and the electrons will easily tunnel through it to get from one wire to the other.

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