What is the difference between a system being in a steady state and thermodynamic equilibrium? Can a system be in steady state but not in thermodynamic equilibrium and vice-versa? Do they have different definitions on the basis of the master equation?
$\begingroup$
$\endgroup$
4
-
5$\begingroup$ Thermodynamic equilibrium is a stronger condition than steady state. A system coupled to two baths at different temperatures (or different chemical potentials) can never be in equilibrium but will usually develop a steady state. $\endgroup$– Sebastian RieseCommented Jun 23, 2015 at 13:55
-
$\begingroup$ @ Sebastian Riese- Is there a way to detect whether a system is in a steady state or equilibrium? What is the characteristic difference between a steady state and an equilibrium state? Can we define the steady state to be one in which the thermodynamic variables do not change with time? But this also happens in thermodynamic equilibrium. Then how to define thermodynamic equilibrium so as to distinguish it from steady state? $\endgroup$– SRSCommented Jun 23, 2015 at 14:02
-
2$\begingroup$ @SRS A stead state is a state which does not change in time. Thermodynamic equilibrium is a stead state in which all parts of the system are in some form of thermal contact but no net transfer of heat take place. $\endgroup$– By SymmetryCommented Jun 23, 2015 at 14:05
-
$\begingroup$ @SebastianRiese Can statistical mechanics distinguish between equilibrium and steady state? If yes, what is the criterion? $\endgroup$– SRSCommented Mar 23, 2017 at 7:00
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
To illustrate the difference take two blocks in contact with each other, with block A at a higher temperature than block B initially. As soon as we bring A in contact with B, energy will flow from A to B, the temperature of A will decrease, and the temperature of B will increase. Given enough time, the two blocks will reach the same temperature, and the energy flux will cease. They are said to be in equilibrium.
Now take our system and instead pin one end of block A at temperature $T_1$, connect the other end to block B, and pin the other end of B at $T_2 < T_1$. Note that at the interface there is initially a temperature difference $T_1 - T_2$. Once again energy will flow from A to B through that interface. However we have fixed the temperature of either end of the two block system. You will see that over time, given that the two blocks have the same temperature-independent thermal conductivity, a linear temperature profile will appear with slope $\frac{T_1-T_2}{L}$ where $L$ is the end to end distance. Since there's still a nonzero temperature gradient everywhere in the system, energy is still flowing from the $T_1$ end to the $T_2$ end. However, the bulk properties (temperature) are not changing with time at any point in the system. This system is said to be in a steady state condition, but it is not in thermal equilibrium since there is a spatial temperature variation.
You can apply this to problems in fluid mechanics. Water can be flowing through a pipe at steady state, but not at (chemical) equilibrium.
To sum up, for a given driving "force" (not a true force of course) - temperature for thermal, pressure for mechanical, chemical potential or concentration for chemical - if that quantity doesn't change with time at any point in the system, then your system is at steady state. If additionally, it doesn't vary spatially either (it is uniform throughout), then it is in equilibrium.
This ignores microscopic fluctuations, of course.
answered Jun 23, 2015 at 14:39
-
$\begingroup$ Can the thermodynamics variables be defined in steady state like in equilibrium? $\endgroup$– AntonCommented Mar 23, 2021 at 13:02
-
$\begingroup$ A very good answer. Just to add: "steady-state equilibrium" and "statistical equilibrium" are the same thing. They both mean that the (statistical) distribution of states is constant in time (for a given spatial region). $\endgroup$– NoldorinCommented Nov 17, 2021 at 23:45
$\begingroup$
$\endgroup$
2
Using entropy changes where the variation of entropy is the sum of production and flows
$$\frac{dS}{dt} = \frac{d_iS}{dt} + \frac{d_eS}{dt}$$
a steady or stationary state is one for which entropy is constant ${dS}/{dt} = 0$, and an equilibrium state is one for which ${dS}/{dt} = {d_iS}/{dt} = 0$.
All equilibrium states are stationary states, but not all stationary states are equilibrium states.
-
$\begingroup$ And how is entropy defined for those stationary (steady?) states in the case that there is no thermodynamics equilibrium? $\endgroup$ Commented Mar 30, 2018 at 19:36
-
1$\begingroup$ @no_choice99 If the state is not very far from equilibrium and processes aren't too fast one uses Thermodynamics of Irreversible Processes (TIP) to get the expression for the entropy; e.g. the production term is given by the well-known bilinear product of forces and fluxes $d_iS/dt = \sum X_aJ_a$. Otherwise one uses Extended Irreversible Thermodynamics (EIT). $\endgroup$– juanrgaCommented Apr 4, 2018 at 10:27