I Application of the first law of thermodynamics

[abdossamad2003]

In Halliday's physics book, there is an example of the first law of thermodynamics that shows its application. The figure below explains this example:

Here is a question, if the element alone is chosen as the system, doubts arise in the first law, because in this system, Q<0 (because heat is transferred from the element to the water) and W=0 (because there is no external work on the system is not done) as a result the internal energy of the system (the element alone) should decrease while it does not because the temperature of the element is constant.!!!Part of the text:

 

[Dale]

W=0 (because there is no external work on the system is not done)

There is external work done on the system (the resistor). The electrical work done is $$W=\int I(t)\ V(t) \ dt$$ Where $I$ is the current through the resistor and $V$ is the voltage across it.

 

[Chestermiller]

For the combination of weight and coil, external work is done by gravity on the weight (by the earth), as the weight lowers. An approximately equal amount of heat is delivered from the coil to the water. The book author is a little wrong, because the coil gets a little hotter as the water gets a little hotter. But this increase in internal energy of the coil is negligible compared to the increase in internal energy of the water. The assumption is that the velocity of the weight does not increase significantly during the test, so that its kinetic energy doesn't need to be accounted for.

 

[abdossamad2003]

There is external work done on the system (the resistor). The electrical work done is $$W=\int I(t)\ V(t) \ dt$$ Where $I$ is the current through the resistor and $V$ is the voltage across it.

I thought that in the first law of thermodynamics, W external mechanical work is not necessarily mechanical work and can be electrical work and so on.

 

[Dale]

I thought that in the first law of thermodynamics, W external mechanical work is not necessarily mechanical work and can be electrical work and so on.

Yes, the expression I wrote is the expression for electrical work. It is not zero.

 

[abdossamad2003]

My second question is, if the initial and final state are not in equilibrium, then the first law of thermodynamics is not correct. Can you give an example?

 

[Chestermiller]

My second question is, if the initial and final state are not in equilibrium, then the first law of thermodynamics is not correct. Can you give an example?

View attachment 334300

An example is a gas within an insulated cylinder, featuring a massless frictionless piston above, where a weight is sitting on top of the piston, and part of the weight is suddenly removed; the gas expands rapidly and irreversibly. All intermediate states are not thermodynamic equilibrium states, and the first law of thermodynamics can not be used to determine the internal energy for these states. It can only be used for the final state in which the gas is again at thermodynamic equilibrium.

 

[Squizzie]

An example is a gas within an insulated cylinder, featuring a massless frictionless piston above, where a weight is sitting on top of the piston, and part of the weight is suddenly removed; the gas expands rapidly and irreversibly. All intermediate states are not thermodynamic equilibrium states, and the first law of thermodynamics can not be used to determine the internal energy for these states. It can only be used for the final state in which the gas is again at thermodynamic equilibrium.

But in an ideal case, wouldn't the piston oscillate indefinitely, as PE and KE are exchanged cyclically as the piston moves up and down under the influence of the pressure in the cylinder?

 

[Chestermiller]

But in an ideal case, wouldn't the piston oscillate indefinitely, as PE and KE are exchanged cyclically as the piston moves up and down under the influence of the pressure in the cylinder?

No. Even in the ideal gas limit, gases exhibit viscous behavior which would act to damp the oscillation. See Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 1, Section 1.3 Temperature and Pressure Dependence of Viscosity and Fig. 1.3-1. Viscous behavior will prevail as long as the mean free path of the molecules is small compared to the physical dimensions of the system (e.g., the piston diameter).

Viscous stresses are also the reason why the ideal gas law cannot be used for rapidly deforming irreversible deformations. In such cases, even though the gas may exhibit ideal gas behavior at thermodynamic equilibrium (and in reversible deformations, which are always comprised of a continuous sequence of thermodynamic equilibrium states), in irreversible deformations, the ideal gas does not exhibit ideal gas behavior (except in the initial- and final thermodynamic equilibrium states).

 

[Squizzie]

No. Even in the ideal gas limit, gases exhibit viscous behavior

But doesn't "ideal" require an ideal gas: one that obeys PV=nRT over all ranges of temperature and pressure, with no viscosity?

 

[Chestermiller]

But doesn't "ideal" require an ideal gas: one that obeys PV=nRT over all ranges of temperature and pressure, with no viscosity?

No. An ideal gas is the limit of real gas behavior at low pressures (but not so low that the mean free path of the molecules is on the same order or larger than the physical dimensions of the system). Bird, et al. state "The chart (Fig. 1.3-1) shows that the viscosity of a gas approaches a limit (the low density limit) as the pressure becomes smaller; for most gases, this limit is nearly attained at 1 atm. pressure."

Because of viscosity, an ideal gas obeys PV=nRT only at thermodynamic equilibrium.

 

[Squizzie]

No. An ideal gas is the limit of real gas behavior at low pressures (but not so low that the mean free path of the molecules is on the same order or larger than the physical dimensions of the system). Bird, et al. state "The chart (Fig. 1.3-1) shows that the viscosity of a gas approaches a limit (the low density limit) as the pressure becomes smaller; for most gases, this limit is nearly attained at 1 atm. pressure."

Because of viscosity, an ideal gas obeys PV=nRT only at thermodynamic equilibrium.

I think you'll find that Bird, et al. were referring to the viscosity of a real gas, not an ideal gas.
Tuckerman refers to an ideal gas as "an ideal gas, which is defined (thermodynamically) as a system whose equation of state is PV − nRT = 0,"
and "An ideal gas is defined to be a system of particles that do not interact."

 

[Chestermiller]

I think you'll find that Bird, et al. were referring to the viscosity of a real gas, not an ideal gas.
Tuckerman refers to an ideal gas as "an ideal gas, which is defined (thermodynamically) as a system whose equation of state is PV − nRT = 0,"
and "An ideal gas is defined to be a system of particles that do not interact."

Well, with all due respect to Tuckerman (whoever that is), you are not going to be able to explain mechanistically what happens in Joule expansion in a rigid container, or Joule Thomson change in flow through a porous plug or valve, or Joule heating by an impeller without taking into account the viscous behavior of the gas.

I guess we are just going to have to agree to disagree.

 

[Squizzie]

Well, with all due respect to Tuckerman (whoever that is),

Tuckerman, Mark E. (2010). Statistical Mechanics: Theory and Molecular Simulation (1st ed.). ISBN 978-0-19-852526-4.

 

[Chestermiller]

Tuckerman, Mark E. (2010). Statistical Mechanics: Theory and Molecular Simulation (1st ed.). p. 87. ISBN 978-0-19-852526-4.

Statistical Mechanics?? That applies only to thermodynamic equilibrium conditions, and we have been considering here irreversible processes in which the system passes through non-equilibrium states (not described by statistical mechanics). How does statistical mechanics treat irreversible processes. And when molecular simulation is done, this has to take into consideration molecular interactions. But, as you said (which I disagree with), An ideal gas is defined to be a system of particles that do not interact." So how can statistical mechanics deal with particles that do not interact (and yet describe an irreversible process for such a system)?

 

[Squizzie]

Statistical Mechanics??

Young and Friedman (2018) University Physics with Modern Physics, 15th Edition Sears & Zemansky

 

[Chestermiller]

View attachment 335199
Young and Friedman (2018) University Physics with Modern Physics, 15th Edition Sears & Zemansky

Only for thermodynamic equilibrium!

 

[Squizzie]

Only for thermodynamic equilibrium!

If that is the case, then if along with assuming the "ideal" states of an insulated cylinder and a massless, frictionless piston, we assume that the gas is an ideal gas, how should we model your example?

An example is a gas within an insulated cylinder, featuring a massless frictionless piston above, where a weight is sitting on top of the piston, and part of the weight is suddenly removed; the gas expands rapidly and irreversibly.

And why is the expansion of the gas "irreversible"?

 

[Chestermiller]

If that is the case, then if along with assuming the "ideal" states of an insulated cylinder and a massless, frictionless piston, we assume that the gas is an ideal gas, how should we model your example?

Which example is that?

And why is the expansion of the gas "irreversible"?

Do you understand how an irreversible process is defined. Please articulate your understanding.

 

[Squizzie]

Which example is that?

Your example at #7:

An example is a gas within an insulated cylinder, featuring a massless frictionless piston above, where a weight is sitting on top of the piston, and part of the weight is suddenly removed; the gas expands rapidly and irreversibly.

Do you understand how an irreversible process is defined. Please articulate your understanding.

You stated in the example that the process was irreversible. It seemed to me that, by specifying an insulated cylinder and frictionless piston, you had specified purely adiabatic process. My question was requesting an explanation of why (or more accurately, how?) the described process would be irreversible.

 

[Squizzie]

Just for giggles, I have simulated the example in python and have run it for an initial mass of 10 tonnes, removing 1 tonne :

Interesting things happen when most of the weight is removed. Here's removing 9,800 kg (all but 100 kg) and running the simulation for 30,000 iterations:

I think the process is reversible.

""" "An example is a gas within an insulated cylinder, featuring a massless frictionless piston above, where a weight is sitting on top of the piston, and part of the weight is suddenly removed." Modelling: The cylinder has a cross-section of 1 metre The initial weight is 1 tonne. There is sufficient quantity of gas to support the pistom 1 metre from the base, giving an initial gas volume of 1 cubic metre. A mass (coded as deltaM below) is suddenly removed. A numerical solution """ import numpy as np import matplotlib.pyplot as plt g = 9.8 # m/sec^2 Gravity. gamma = 1.4 # the adiabatic index of air M = 10000 # kg Initial mass deltaM = 1000 # kg, mass removed newMass = M - deltaM gf = newMass * g # kilo newtons gravitational force of new mass step = 0.0010 # seconds. The time interval for the iteration time_steps = 10000 # the number of iterations time = np.linspace(0, step * time_steps, time_steps) xPosition = np.zeros(time_steps) #the array of x-values at each time interval def oscillatePiston(): for i in range(time_steps): if i == 0: # initial conditions initialForce = M * g # initial pressure force from 1 cu metre compressed with the initial mass. netForce = initialForce - gf x=1 # meters. start position v=0 # initial velocity m/sec else: v1 = v + a * step # m/sec x = x + (v1 + v) / 2 * step # use the average of previous and current velocity to calculate the next position pf = initialForce * (x**(-gamma)) # The volume is proportional to the value of x pressure force netForce = pf - gf # net force on piston is weight of the mass minus the pressure force of the gas. v = v1 # set v for next iterations a = netForce / newMass # acceleration (force = mass * acceleration) xPosition[i] = x oscillatePiston() plt.plot(time, xPosition) plt.xlabel('Time (seconds)') plt.ylabel('Position (meters)') plt.title('Piston Oscillation') plt.show()

 

[Chestermiller]

Your example at #7:

I'll present a. solution to that after you answer my question about your understanding of the definition of an irreversible process.

You stated in the example that the process was irreversible. It seemed to me that, by specifying an insulated cylinder and frictionless piston, you had specified purely adiabatic process. My question was requesting an explanation of why (or more accurately, how?) the described process would be irreversible.

Again, what is your definition of an irreversible process?

 

[Squizzie]

Again, what is your definition of an irreversible process?

I assumed that you were referring to the standard thermodynamics interpretation of an irreversible process. Here's one from Cengel Y. A. and Boles M. A.(2015) Thermodynamics: An Engineering Approach McGraw Hill p. 292:
"A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings (Fig. 6–29). That is, both the system and the surroundings are returned to their initial states at the end of the reverse process. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process. Processes that are not reversible are called irreversible processes."

 

[Chestermiller]

I have n moles of gas inside an insulated cylinder of area A sitting in a vacuum chamber, with its axis oriented vertically. The piston is frictionless and massless, but there is initially a mass M sitting on top of the piston, so that the initial pressure of the gas is $P_1=\frac{Mg}{A}$; the initial. temperature is $T_1$. I suddenly remove mass m<M from the top of the piston, and let the gas expand and re-equilibrate so that its final pressure is $P_2=\frac{(M-m)g}{A}$. Since viscous forces will damp the motion of the mass during the process, and it will eventually come to rest. Use the 1st law of thermodynamics to determine the final temperature.

In this process, the total amount of work done by the gas in lifting the weight is $$W=(M-m)g\Delta h=\frac{(M-m)g}{A}(V_2-V_1)=P_2(V_2-V_1)$$Combining this with the ideal gas law gives:$$W=nRT_2-P_2\frac{nRT_1}{P_1}=nR\left(T_2-\frac{P_2}{P_1}T_1\right)$$For an. ideal gas, the change in internal energy is $$\Delta U=nC_v(T_2-T_1)$$ so that $$nC_v(T_2-T_1)=-nR\left(T_2-\frac{P_2}{P_1}T_1\right)$$Solving for the final equilibrium temperature $T_2$ gives $$T_2=\frac{C_v+R(P_2/P_1)}{C_p}T_1$$
Questions?

If this were a reversible process, since the system is adiabatic, the change in entropy would be zero.

 

[Squizzie]

Questions?

Is your gas an ideal gas?

 

[Chestermiller]

Is your gas an ideal gas?

Did I use the ideal gas law and the equation for the internal energy of an ideal gas?

 

[Squizzie]

Did I use the ideal gas law and the equation for the internal energy of an ideal gas?

Yes, you did use the ideal gas law in

For an. ideal gas, the change in internal energy is $$\Delta U=nC_v(T_2-T_1)$$

So I think it's a fair question: Is your gas an ideal gas?

 

[Chestermiller]

What do you think?

 

[Squizzie]

What do you think?

I think it's a pretty straightforward question. Is your answer modelled on an ideal gas?

 

[Chestermiller]

I think it's a pretty straightforward question. Is your answer modelled on an ideal gas?

Do you think I did?

 

[Squizzie]

Do you think I did?

I'm just asking a simple question about a detail about your example which is quite significant to the understanding of your explanation.

 

[Chestermiller]

I'm just asking a simple question about a detail about your example which is quite significant to the understanding of your explanation.

It is an ideal gas according to the definition of an ideal gas used by chemical engineers.

 

[vanhees71]

Statistical Mechanics?? That applies only to thermodynamic equilibrium conditions, and we have been considering here irreversible processes in which the system passes through non-equilibrium states (not described by statistical mechanics). How does statistical mechanics treat irreversible processes. And when molecular simulation is done, this has to take into consideration molecular interactions. But, as you said (which I disagree with), An ideal gas is defined to be a system of particles that do not interact." So how can statistical mechanics deal with particles that do not interact (and yet describe an irreversible process for such a system)?

I think, Boltzmann would be pretty sad, hearing this. Of course statistical mechanics is by no means restricted to thermostatics.

 

[Chestermiller]

I think, Boltzmann would be pretty sad, hearing this. Of course statistical mechanics is by no means restricted to thermostatics.

So how would Boltzmann have handled a specific problem involving an irreversible process between an initial and final thermodynamic equilibrium state. Please provide an example with details.

 

[vanhees71]

You mean the problem described in #1? I guess, he'd solve the equation of motion for $m$ together with the AC circuit problem. Then the total energy consumed in the resistor is transferred to the water as heat. I don't see, where in this entire problem an ideal gas occurs in the first place.

 

[Chestermiller]

You mean the problem described in #1? I guess, he'd solve the equation of motion for $m$ together with the AC circuit problem. Then the total energy consumed in the resistor is transferred to the water as heat. I don't see, where in this entire problem an ideal gas occurs in the first place.

I think squizzle and I were referring to the irreversible gas expansion problem in posts #16 and forward.

 

[Chestermiller]

@Squizzle You indicate that you are skeptical about the Chemical Engineers' definition of an ideal gas being taken as the limiting behavior of a real gas at low pressures. This includes approaching a non-zero viscosity that is a function only on temperature. I submit that this is a more realistic definition of an ideal gas than the more restrictive definition used by physicists where the ideal gas viscosity is taken as zero. I also submit that the inviscid definition of an ideal gas is not capable of properly analyzing the Joule Thomson effect of an ideal gas flowing through a porous plug with a higher pressure on the upstream side of the plug and an lower pressure on the downstream side of the plug.

Have you ever had a course in fluid mechanics, including viscous Newtonian gases and liquids? I suggest that it is time for you to start learning about this.

 

[Squizzie]

@Squizzle You indicate that you are skeptical about the Chemical Engineers' definition of an ideal gas

My scepticism arises from my inability, despite researching a number of texts_[1][2][3], to identify a Chemical Engineers' definition of an ideal gas that differs from the classical physics definition quoted above.

[1] Smith J. M. (1970) Chemical Engineering Kinetics, Mcgraw Hill
[2] Denn M. M. (2012) Chemical Engineering an Introduction, Cambridge University Press
[3] Backhurst J.R and Harker J. H (2001) Chemical Engineering, Butterworth-Heinemann

 

[Chestermiller]

My scepticism arises from my inability, despite researching a number of texts_[1][2][3], to identify a Chemical Engineers' definition of an ideal gas that differs from the classical physics definition quoted above.

[1] Smith J. M. (1970) Chemical Engineering Kinetics, Mcgraw Hill
[2] Denn M. M. (2012) Chemical Engineering an Introduction, Cambridge University Press
[3] Backhurst J.R and Harker J. H (2001) Chemical Engineering, Butterworth-Heinemann

Transport Phenomena is a book that has stood the test of time, written by the department head and two prominant professors from the chemical engineering department at the university of Wisconsin.

 

[Chestermiller]

Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics: Chapter 2, page 40: "The oscillations of the piston assembly are damped out because the viscous nature of the gas gradually converts gross direct motion of the molecules into chaotic molecular motion. This dissipative process transforms for of the World initially done by the gas in accelerating the piston back into internal energy of the gas. Once the process is initiated, no infinitesimal change in external conditions can reverse its direction; the process is irreversible."

Chapter 2, page 64: "In engineering calculations, gases at pressures up a few bars may often be considered ideal."

 

[Squizzie]

Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics: Chapter 2, page 40: "The oscillations of the piston assembly are damped out because the viscous nature of the gas gradually converts gross direct motion of the molecules into chaotic molecular motion. This dissipative process transforms for of the World initially done by the gas in accelerating the piston back into internal energy of the gas. Once the process is initiated, no infinitesimal change in external conditions can reverse its direction; the process is irreversible."

The example quoted is using "a gas". Not an "ideal gas" as is made abundantly clear in the paragraph following:
"All processes carried out in finite time with real substances are accompanied in some degree by dissipative effects of one kind or another, and all are therefore irreversible." (my emphasis)

 

[Chestermiller]

The example quoted is using "a gas". Not an "ideal gas" as is made abundantly clear in the paragraph following:
"All processes carried out in finite time with real substances are accompanied in some degree by dissipative effects of one kind or another, and all are therefore irreversible." (my emphasis)

Well, what good does it do to use a mathematical model of a substance that does not capture simple the first order picture of how the substance behaves in actual physical situations. Saying that the piston oscillates forever when we know that it would not is just silly, especially when we can easily calculate what the final steady state would be when the oscillation is damped out . Saying that, in the expansion of a gas into half a chamber initially under vacuum, the mechanism for the temperature remaining constant is not related to viscous forces is likewise silly. And in Joule Thomson process, flowing a gas through a porous plug, an inviscid gas model would result in no pressure change while we know that the pressure change in the porous plug is the result of viscous forces; so neglecting viscous stresses would prevent us from modeling the Joule Thomson effect; there would be no Joule Thomson effect without gas viscosity. Saying that the Joule Thomson coefficient for an ideal gas is zero could not be done because the pressure drop being zero would make the Joule Thomson coefficient 0/0.

It all boils down to what you want to use as the definition of an ideal gas. I contend that considering an ideal gas to have viscosity (as real gases have in the limit of low pressures) allows simple calculations and explanations of gas behavior at low pressures which cannot be done with an inviscid ideal gas model.

 

[Squizzie]

Well, what good does it do to use a mathematical model of a substance that does not capture simple the first order picture of how the substance behaves in actual physical situations. Saying that the piston oscillates forever when we know that it would not is just silly, especially when we can easily calculate what the final steady state would be when the oscillation is damped out .

Yes, it would be silly to say that. But that is not what is being said. What I am suggesting is that in the idealised world of frictionless, massless, perfectly insulated cylinders, ideal gas etc., the piston would oscillate forever.
The analysis of such a system can provide an insight into the fundamental properties of pressure, temperature, mass, momentum, energy and, incidentally, the fundamental differences between the various states of matter.
Its practical application has contributed immeasurably to the solution of uncountable practical engineering challenges of the modern industrial and technological world.
I'm sure Boyle, Charles, Gay Lussac, Kelvin, Joule and all the teachers of thermodynamics, and indeed of science generally, would be disappointed to hear you say that this methodology was silly.

Incidentally, you would also have to allow the cylinder's insulation to be less than ideal to allow the "real" viscosity to be able to damp the oscillations.

Thank you for clarifying the real nature of the gas in the experiment.
[EDIT] I know it's a subjective view, but I would suggest the issues of friction, sealing and insulation would rate higher than the minute influence of viscosity as first order omissions from reality.

 

[Chestermiller]

Yes, it would be silly to say that. But that is not what is being said. What I am suggesting is that in the idealised world of frictionless, massless, perfectly insulated cylinders, ideal gas etc., the piston would oscillate forever.
The analysis of such a system can provide an insight into the fundamental properties of pressure, temperature, mass, momentum, energy and, incidentally, the fundamental differences between the various states of matter.
Its practical application has contributed immeasurably to the solution of uncountable practical engineering challenges of the modern industrial and technological world.
I'm sure Boyle, Charles, Gay Lussac, Kelvin, Joule and all the teachers of thermodynamics, and indeed of science generally, would be disappointed to hear you say that this methodology was silly.

Incidentally, you would also have to allow the cylinder's insulation to be less than ideal to allow the "real" viscosity to be able to damp the oscillations.

Thank you for clarifying the real nature of the gas in the experiment.
[EDIT] I know it's a subjective view, but I would suggest the issues of friction, sealing and insulation would rate higher than the minute influence of viscosity as first order omissions from reality.

Would you also suggest that, if indeed your gas is approximated as having zero viscosity, your assumption that the temperature, pressure, and density of the ideal gas in your cylinder are spatially uniform (as the piston oscillates forever) is valid? When you suddenly release your massless frictionless piston from rest, do you think that pressure-, temperature-, and density waves within the gas will develop that propagate in the axial direction along the cylinder?

Let me guess...you're not an engineer, right.

 

[Squizzie]

Let me guess...you're not an engineer, right.

An astute observation. Neither am I a biologist, architect, surgeon, or lawyer. My first love and academic qualification is physics, which is what drew me to this forum.

 

[Frabjous]

My first love and academic qualification is physics, which is what drew me to this forum.

As a physicist, what do you think that you have contributed to the 2 thermodynamic threads that you have been replying to?
Most physicists would be thrilled to learn of ways to generalize the ideal gas law to increase is applicability. You seem more interested in textbook definitions than the phenomena itself.

 

[vanhees71]

I highly recommend Landau+Lifshitz vol. 5 (Stat. Phys. 1) and vol. 6 (hydrodynamics). There you get the physicists' point of view. An ideal gas is, as the name suggests, an idealization. It describes situations of fluid, where the changes are so slow that you can assume that it's always instantaneously in local (!) thermal equilibrium, i.e., there's no heat transfer between fluid cells. This means there's no friction and no dissipation. The result is the perfect-fluid (Euler) equation for hydro, and that's for sure an idealization. In reality you have dissipation. In non-relativistic physics you can describe this in the next step of the approximation by viscous hydrodynamics (Navier-Stokes equation).

This can be studied from the next more fundamental level by employing transport equations like the Boltzmann equation. Hydrodynamics can be derived as an effective description for the gas staying close to local thermal equibrium. At 0th order you get perfect-fluid hydro. There you neglect the collision term completely, because you assume that the fluid cells are instantaneously in thermal equilibrium. It turns out that this neglects all kinds of irreversibility, i.e., you have an adiabatic equation of state. The next step is to take into account deviations from local thermal equilibrium at the linear order. This leads to the occurance of transport coefficients, i.e., the bulk and shear viscosity and the Navier-Stokes equation.

That's usually sufficient for the non-relativsitic case. In the relativistic case you run into trouble, when doing this naively. For a long time one thus has believed that one needs at least a 2nd-order hydro. Using the relaxation-time approximation for the collision term in the Boltzmann equation you are lead to Israel-Steward 2nd-order hydrodynamics, which is relativistically consistent. In very recent years, it has been figured out, that also a viable first-order Navier-Stokes-like relativistic hydrodynamics is possible by exploiting all kinds of "matching conditions" consistently.

For a detailed introduction to the relativistic case, see

G. S. Denicol and D. H. Rischke, Microscopic Foundations of
Relativistic Fluid Dynamics, Springer, Cham (2021),
https://doi.org/10.1007/978-3-030-82077-0