When we derive the Navier-Stokes Equation, we come across a a common assumption made by Stokes that makes the two quantities namely Mechanical pressure and Thermodynamic pressure equal to each other.

It is claimed that Thermodynamic pressure represents the energy over all the degrees of freedom of the molecules while the Mechanical pressure is related to only translational degrees of freedom. When a system is subjected to abrupt change in the properties, the molecules don't get enough time to adjust to that changes that is they don't have enough time to convert all its energy from all the degrees of freedom into translational ones.

J. Chem. Phys. 39, 654 (1963); https://doi.org/10.1063/1.1734304 39, 654 © 1963 American Institute of Physics.
Formal Kinetic Theory of Transport Phenomena in Polyatomic Gas Mixtures Cite as: J. Chem. Phys. 39, 654 (1963); https://doi.org/10.1063/1.1734304 Submitted: 05 April 1963 .
Published Online: 29 June 2004 L. Monchick, K. S. Yun, and E. A. Mason

It has long been known, from a phenomenological point of view, that the volume viscosity is proportional to a relaxation time, T, which is a measure of the rate of transfer of molecular energy between internal and translational degrees of freedom.

In this citation the author speaks about the relation of bulk viscosity with Relaxation time for the molecules. This clearly hints that mechanical pressure is a result of only translational degrees of freedom. As when the bulk viscosity cancels only then the thermodynamic pressure equals mechanical pressure like in the case of monoatomic ideal gas that exhibits only translational degrees of freedom.

My question here is how mechanical pressure is related to translational degrees of freedom? Any insights from Kinetic Theory of gases or statistical mechanics would be appreciated.


2 Answers 2


Thermodynamic pressure is defined on the basis of thermodynamic relations. E.g., from the basic thermodynamic relation $dU = TdS - PdV$ we obtain $$ P=-\left(\frac{\partial U}{\partial V}\right)_S.$$ As not all the changes in energy are related to the kinetic energy - they can involve other degrees of freedom or interactions between the gas/liquid particles - the pressure defined in this way may not be equal to the mechanical pressure.

Mechanical pressure is just the force acting per surface element (e.g., in Newton laws, when deriving the hydrodynamic equations). Most statistical physics texts define pressure in this sense, and calculate it as a result of particle collisions against a surface. For ideal gas, where all the energy is the kinetic energy of its atoms/molecules, it can then be shown that thus defined pressure is equivalent to the thermodynamic pressure, resulting in the ideal gas law, $PV=nRT$.

Hydrodynamic pressure The above definition of mechanical pressure however becomes meaningless in a liquid or gas, where one cannot meaningfully speak about a surface against which the molecules are scattered (see this thread). This is where the hypothesis mentioned in the OP comes into play: the pressure is defined thermodynamically, but when deriving the equations-of-motion for the liquid we treat it as mechanical. Same is true (although perhaps a bit less obvious) when deriving the equations in elasticity theory.

The solution lies in obtaining hydrodynamic equations from more fundamental Boltzmann equation (or, more precisely, the BBGKY hierarchy of equations, see the links in this answer). In essence, hydrodynamic equations assume local equilibrium, i.e., thermodynamic parameters and mechanical values slowly changing in space in time, as compared to the diffusion length and the speed of sound. That is, we can apply these equations only to volumes and times on which every molecule would experience multiple collisions. One can then think of pressure in a gas/liquid as the momentum transfer between the layers of thickness greater than the diffusion length: all the molecules that entered a layer from adjacent layers collide with the layer molecules and are reflected back, transferring their momentum and thus creating pressure - just like in deriving the ideal gas law.

Spatial or temporal changes abrupt in comparison to the scales imposed by the diffusion length and the speed of sound violate the assumption of local equilibrium and the equality of mechanical and thermodynamic pressures.

  • $\begingroup$ Okay I understand the Thermodynamic pressure, why and how it involves all the translational as well as internal degrees f freedom but after making the system go through abrupt change why is it so that all the internal degrees of freedom converts into translational ones after some relaxation time thereby giving rise to mechanical pressure and how in the first place mechanical pressure is a result of only translational degrees of freedom of the molecules. $\endgroup$ Jan 20, 2022 at 5:53

Bulk viscosity effects arises when there is a change in the volume of fluid, i.e, either compression or expansion. Whenever any expansion or compression happens, the system either gains or losses energy by means of external work. For a dilute (i.e., low-density) gas, this energy loss or gain of the system only affects the translational motion. That means, if the work is being done on the system in a compression process, then the gained energy first goes to the translational motion of the system and then by means of inter-molecular collisions redistributes among other internal (i.e, translational and rotational) modes of energy, e.g., rotational and vibrational. This redistribution of energy between translational and other modes is a slow process and requires several intermolecular collisions and hence some finite time. Mechanical pressure is the instantaneous pressure of the system that, say, a pressure probe would measure. And thermodynamic pressure can be understood as mechanical pressure if the system is hypothetically brought to equilibrium through an isentropic process. During the expansion or compression, since energy exchange between translational and internal mode is not instantaneous and energy change in the system due to external work only affects the translational motion, the translational kinetic energy, hence mechanical pressure, at equilibrium would be different at equilibrium than that during the expansion or compression. The same is represented as bulk viscosity from a macroscopic standpoint.

For a better understanding, you can refer to the introduction section of the following articles.





Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.