Engrave a metal surface with a controlled low-energy arc because, like a laser, it can deliver very high power levels, but it’s a lot easier to build than a laser is. You can vaporize a little bit of metal, or you can melt it and then blow it away with air.
There are a couple of things that make it difficult to melt metals. One is that most metals don’t melt until they’re pretty hot, so you can’t melt them at all with things that only get moderately hot, like most flames. (Jet fuel, as they say, can soften steel beams until they can support almost nothing, but not melt them.) The other is that it’s a good heat conductor, so if you deliver the heat slowly, it gets conducted away and the metal never gets very hot. Aluminum especially is a good heat conductor.
The easiest approach is to bring an electrode, charged with a capacitor, toward the surface of the oppositely-charged metal until the air breaks down and the energy discharges in a spark.
Striking an arc in air needs on the order of 300 volts at a minimum at
a distance of about 13 microns. If you have a 1-microfarad low-ESR
capacitor supporting that arc, it has an energy of about 45
millijoules. Vaporizing aluminum takes about 14 kilojoules per gram
(((2470 - 20) K 24.20 J/mol/K + 10.71 kJ/mol + 284 kJ/mol) * 27.0
g/mol), the majority of which is from its heat of vaporization. That
means that this spark can vaporize about three micrograms of aluminum,
which sounds insignificant, but if it’s hemispherical, it’s actually a
crater about 160 microns across, which you will note is more than ten
times the distance from the electrode to the workpiece. A 160-micron
crater is clearly visible and palpable; it’s comparable to the kerf
you get from a laser cutter.
That’s kind of a best case, though, because some of the heat will go into heating the electrode and the air, some of it will be conducted away, and some of it will go into heating the already boiled metal in its gaseous form. If the workpiece is connected to the negative side of the circuit (the cathode), most of the heat of the arc will be deposited at the surface of the workpiece, as it is bombarded by ionized air, rather than on the marking electrode, which is receiving only electrons.
Thermal runaway concentrates the electrical current on the hottest part of the cathode, as that’s the part that can emit the largest number of electrons, so the spot that the arc heats can be very small indeed.
That still leaves the question of how fast the whole discharge happens, which depends crucially on the E/I curve of the arc, where most of the resistance in the circuit is found. The RC time constant of just the electrode and wires can easily be around a microsecond, which would imply a power of around a kilowatt and a power density of 50 gigawatts per square meter or 50 kilowatts per square millimeter, which is in the neighborhood of what metal-cutting lasers put out, so it should probably work okay. (WP says 1500 watts in a 25-micron laser spot is common, and you can cut 1-mm aluminum at 14.82 cm/s at 1000 W; if we figure the kerf is 200 μm, that works out to 12.5 mJ/μg, very close to the value of 14 I calculated above; and at that rate the laser is vaporizing a volume of material comparable to our crater every 36 microseconds, which should be an easily achievable speed for the spark.)