If lightning is caused by ionisation of air, why does it only last briefly?

Question

I'm comparing lightning and fire - both are related to ionisation of air but lightning happens so fast in a blink of an eye while fire goes on until it runs out of fuel. My question is: despite being plasma just like a fire, why does lightning only occur for such a brief time?

Answer 1

Lightning is an electrostatic discharge. After that phenomenon, the atmosphere and ground temporarily neutralize themselves. Hence, it is over in a blink of an eye.

Fire is a byproduct of some chemical reaction and hence can go on for a longer time.

https://www.quora.com/Scientifically-speaking-is-there-any-connection-between-fire-and-lightning

Answer 2

Fire is a chemical reaction fueled by some product that undergoes combustion, generating energy by breaking (usually) carbon chain molecules down into $CO_{2}$ and $H_{2}O$. So long as fuel plus oxygen is available and no extinguishing forces are present, the fire will continue to burn. Lightning, in contrast, requires an electrostatic discharge to breakdown the neutral particles into an ionized gas. This results from a very large electric field. Once discharged, the electric field energy is gone so there is nothing to continue to sustain the lightning pulse (i.e., flow of current).

Ionization, dissociation, and recombination all occur on extremely fast time scales in an atmosphere with number densities pushing ~$10^{19} \ cm^{-3}$. Fire is, at best, a VERY weakly ionized, dusty plasma (e.g., see discussion at https://physics.stackexchange.com/a/340276/59023).

As noted at https://physics.stackexchange.com/a/288810/59023, the recombination time is proportional to $\propto \sqrt{ \tfrac{ T_{e} }{ n_{e}^{2} } }$, where $T_{e}$ is the electron temperature and $n_{e}$ is the electron number density. The value for $n_{e}$ in a lightning discharge is on the order of ~$10^{17}--10^{18} \ cm^{-3}$ and the temperature is ~0.7-4.3 eV, which means the recombination time scales are going to be tiny (i.e., slowest I would expect would be microseconds or less).

Answer 3

For the fire the source of energy is the "chemical" energy stored in the reacting compounds whereas for lightning the stored of energy is electrostatic in nature due to the separation of charges.

Comparing power, gas fire - $4\times 10^3$ watts and lightning strike - $3\times 10^8$ volts and $3\times 10^4$ amps $\Rightarrow \approx 10^{13}$ watts, you will note that the rate at which energy is dissipated in a lightning strike is much, much greater than that in a gas fire.
Thus one would expect the lightning strike to last for a much shorter period of time than that for a chemical fire.

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@Peter-ReinstateMonica has made a valid comment that I only compared powers so here is an analysis resulting in a time for the event answer.

A gas stove has a power rating of about $4\,\rm kW$ and suppose that it is run from a cylinder of propane which contains $10\,\rm kg$ of gas with a calorific value of $50\,\rm MJ\,kg^{-1}$.
So if run continuously the cylinder will last about one day.

The paper Measurement of the electrical properties of a thundercloud through muon imaging by the GRAPES-3 experiment has some estimates on page 4 regarding thunderclouds.
There is an estimate of $\ge 720 \,\rm GJ$ stored at a potential of approximately $1\,\rm GV$.
Taking the figure for the current during a lightning strike to be $30\,\rm kA$ gives an estimated time for a complete discharge (which is unlikely) of the cloud to be $\mathbf 20\, \rm ms$.

Answer 4

Imagine looking at a fire, let’s say a candle flame. There is a region with partially ionised gas at the edge of the visible flame in the lower part of the flame. Think about the gas in that spot. It is a mixture of air from the surroundings and evaporated wax that came from the wick, undergoing chemical reactions that heat it up. The crucial issue is that the gas is not at rest, but moving upwards through the flame at significant speed.

Where were the gas molecules that are reacting with each other in the flame right this instant a fraction of a second before? Some of them were in the surrounding air, being drawn toward the flame, and some were inside the wick in the form of liquid wax which then evaporated. And a fraction of a second later, they will be rising up above the flame and no longer emitting noticable amounts of light.

The atoms and molecules which are ionised at any given time are so only for a short interval of time. The gas inside the part of the flame where there is significant ionisation is constantly moving and gas which moves out of this zone is being replaced by other gas moving inside it. This means that the duration of the ionisation is also quite short in the case of the flame – when you look at it from the point of view of the gas molecules – similar to the lightning case.

The difference is that in the case of lightning, the whole gas volume is ionised almost at once, then it recombines and the process is over, whereas in the case of a flame it’s a steady-state process with fresh reactants moving into the flame and reaction products moving out continuously, but the duration of ionisation of individual molecules is also short, as explained in the answer by honeste_vivere.

Answer 5

Like many (if not the majority) of question posters on this site you proceed from a false premise. Lightning is not caused by the ionization of air. It is caused by polarization of static charges at the bottom of the cloud and the ground. Lightning does not even need air to propagate through, or any medium for that matter. Although the discharge is a result of dielectric breakdown of the air within the potential of the charge difference, a vacuum also has a dielectric constant which can be easily overcome by the large voltages generated by the static polarization of these charges.