As is familiar to anyone who has inadvertently spilt a few drops of water onto a highly heated pan, rather than boiling away in a flash the water gathers itself up into globules and begins to "dance" over the surface as they evaporate away slowly. On momentary contact with the highly heated surface a vapour layer is quickly formed between the droplets and the surface that lay beneath. The vapour layer, being a poor conductor of heat, is able to insulate the drops (to a degree) and instead of being vaporised leads to their slow evaporation instead. Drops in this state I call Leidenfrost drops.
Leidenfrost drops are highly mobile due to the reduction in friction (reduction in friction between the liquid drops and the solid surface, there will of course still be a very small amount of friction between the liquid drop and its vapour).
If water is thrown onto a surface in a random fashion, given the almost frictionless nature of drops in the Leidenfrost state, I would imagine that a component of the velocity initially horizontal to the surface is enough to get the drops dancing.
Now imagine drops of water are ever so carefully placed onto a highly heated surface. Assume the solid surface is perfectly horizontal and its surface very smooth. Are the drops able to start dancing, that is, be self-propelled, in such a situation? And if so, what would generate the propelling force that acts on the drops causing them to move? Does the size of the drop play a role?
I ask the question since it is known liquid drops carefully placed onto a liquid surface experience self-propulsion. Liquid drops placed onto a so-called ratchet, a solid horizontal surface consisting of a series of saw-teeth, are also able to be self-propelled. But what about in the case of a perfectly smooth solid surface?