What is the formal definition of quasi static process?

I am accustomed with it a bit intuitively, i want to know the formal definition of this. At some source I found the definition of somewhat reversible adiabatic process which somehow also defines quasi static process

"def. 8: Adiabatic process is a process by which the system parameters change so slowly that the characteristic time of changing is much longer than the period of the slowest mode of natural oscillations; also, there should be no dissipative processes (where mechanical energy is converted to heat), e.g. friction. In the case of gases, this means that the speed of the container walls needs to be much smaller than the speed of sound, and also there should be no external heat supply. "

Basically the emphasized text defines quasi static process, Now my questions:

What is characteristic time of a process?

What is mode of natural oscillation of a system?

So what is formal definition of quasi static process?

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    $\begingroup$ Please give a link to the pdf so that people be able to answer your problems more accurately. $\endgroup$ Jan 4, 2019 at 15:41
  • $\begingroup$ ioc.ee/~kalda/ipho/Thermodyn.pdf $\endgroup$ Jan 4, 2019 at 15:42
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    $\begingroup$ Good. Now the definition you mention(Adiabatic) and what you ask in the title(Quasi static) are different things. What do you want to know about? $\endgroup$ Jan 4, 2019 at 15:46
  • $\begingroup$ go through the definition carefully though it states the author was defining adiabatic it was actually defined reversible adiabatic which includes quasi static in itself (that is why i have emphasized the part of definition dealing with quasi static case) $\endgroup$ Jan 4, 2019 at 15:49
  • $\begingroup$ So adiabatic processes can happen if you perform the process very fast. So any change of state happening is much slower than the time elapsed. By the slowest mode of oscillation, they mean to say the most minimal change of state that can occur in the system. $\endgroup$ Jan 4, 2019 at 15:58

4 Answers 4


The definition is actually mixing in a unique definition quasi-static and adiabatic transformation to provide the definition of a reversible adiabatic transformation.

Notice that the two definition can define separate processes: one may have a quasistatic isothermal, isocoric, isoenthalpic,... process, which are not adiabatic, of course, or it is possible to have a non-quasi-static adiabatic transformation.

Characteristic time of a process is basically the time interval between the start and the end of the process.

Mode of natural oscillation of a system is a non-completely-correct way of referring to the characteristic times of the system. Such characteristic times may look like oscillations (for instance a local fluctuation of density would relax through density waves, i.e. sound-like motion) but not necessarily: particle diffusion is not a wave-like phenomenon.

The definition of quasi-static process on which there is general agreement is any transformation slow with respect to the characteristic times of all the process which drive the system toward thermodynamic equilibrium.

For such transformations one is sure that the systems, even if its state is changing, at every time remains as closest as possible to an equilibrium state.

  • $\begingroup$ is the mode of natural oscillation of a system in here refers to characteristic time of the system or something else,@GiorgioP can you please have a look at the pdf file and read few paragraphs before the definition 8? $\endgroup$ Jan 4, 2019 at 16:03
  • $\begingroup$ ioc.ee/~kalda/ipho/Thermodyn.pdf here is the pdf $\endgroup$ Jan 4, 2019 at 16:04
  • $\begingroup$ in your definition: "The definition of quasi-static process on which there is general agreement is any transformation slow with respect to the characteristic times of all the process which drive the system toward thermodynamic equilibrium." what is meant formally my slow ? do not we have any formal definition of idealized quasi static process ? $\endgroup$ Jan 4, 2019 at 16:08
  • $\begingroup$ @BijayanRay Please, what do you mean by formal definition? $\endgroup$ Jan 4, 2019 at 16:15
  • $\begingroup$ @GiorgioP Is my comment an incorrect answer to the question? $\endgroup$ Jan 4, 2019 at 16:17

You can probably Google up a lot of definitions- though I’m not sure there are any “official” ones. I’ll describe what works for me.

Thermodynamic processes are driven by disequilibrium. Heat transfer requires thermal disequilibrium (temperature differences). Work requires unbalanced forces/pressures (mechanical equilibrium). And so forth. A quasi-static process is one that minimizes disequilibrium. It’s called quasi-static because if it were truly static it would mean there would be no temperature, pressure, etc., differences. If that were the case, the relevant process would stop.

When temperature and pressure differences get smaller and smaller, the rate at which the involved processes get slower and slower. For example, heat transfer rate is proportional to temperature difference (all other factors being the same). For this reason quasi-static processes generally proceed very slowly.

In order for a process to be reversible, a necessary (but not sufficient) condition it must be quasi-static. The definition you quoted is incorrect when it says a quasi-static process can involve no friction. You can have a quasi-static process with friction. It is, however, not a reversible process (which is why quasi-static is a necessary, but not sufficient, condition).

I’m not sure why the definition refers to natural oscillations. Perhaps it is something related to the rapid adiabatic expansion of a gas in a piston cylinder arrangement that proceeds so fast that the expansion has an overshoot followed by damped oscillation of the piston before equilibrium is reach. The damping is indicative of dissipative losses in the gas due to viscous friction. But that’s just a guess.

Anyway, I’m not sure if this answers your questions, but perhaps may provide a perspective that hopefully helps.


Quasistatic means that the process is in equilbrium at every point during the process. Let's think about what this means.

At every point in the process, the system is in mechanical equilibrium. This means there are no pressure gradients in the system. In other words, a packet of gas at one location has the same pressure as all other packets of gas. We can therefore say the work done is $PdV$, where we've chosen a single value $P$ to represent the pressure of the entire gas.

If the process was not quasistatic (i.e., close to the speed of sound), we might have pressure gradients since the gas will compress unevenly at different locations.

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    $\begingroup$ I like your answer, but I suppose you can look at quasi-static in two essentially equivalent was (in terms of net result). You can look at it as a series of discrete steps where the process "stops", reaches equilibrium, then "starts" again when a slight disequilibrium is allowed so the process can get to the next "stop". Or you can say that at each point along the way the amount of disequilibrium is infinitely small, which is how I characterized it. The dictionary (Oxford) of quasi is "apparently but not really" $\endgroup$
    – Bob D
    Jan 6, 2019 at 10:11

When a process proceeds in such a way that the system remains little close to an equilibrium state at all times , then it is called a quasi-static process.


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