How does viscosity cause dissipation? [closed]

I am not a physicist so bear with me. I'm trying to understand the mechanism through which energy is dissipated in a fluid or solid. Often the explanation is that it happens due to viscosity or friction and that velocity dependent forces have to be added that reduce the total energy. However at molecular and thermodynamic level energy should be conserved. I need to understand better the nature of the dissipated heat and where it goes.

This question How is viscosity described on the molecular level? gave me some good insights but I'm still puzzled by a few things. Let's take a scenario of molecules interacting via Lennard-Jones potentials and let them fall under gravity (or even in zero g). In the real world this bunch of particles will always reach equilibrium, i.e. zero velocity and kinetic energy relative to its inertial frame. Highly viscous materials will just do it quicker. As far as I understand this is because energy gets transformed to heat, although by energy we mean here actually "useful" energy which I'm not sure what it means.

My question is: what happens to the energy of the flow in a viscous fluid (at molecular level)? Does it get transformed to increased temperature, increased entropy, kinetic energy of hidden degrees of freedom, stored potential energy, heat dissipation into the environment or something else?

closed as too broad by valerio, Gert, sammy gerbil, knzhou, ACuriousMind♦Jul 16 '16 at 9:38

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Dissipative effects introduce irreversibility and increase the order of the derivatives (in the equations of motion) by an odd number (from first order derivatives), thus why viscosity is associated with a second-order derivative. I have some more details at http://physics.stackexchange.com/a/139436/59023. – honeste_vivere Jul 15 '16 at 16:49
• I am voting to close this question because it is asking too many questions covering a broad topic. This site is not a substitute for studying a textbook or equivalent, or engaging a tutor or teacher. There are many other resources for learning about viscosity besides the linked question. – sammy gerbil Jul 16 '16 at 1:52
• I think this was just mean. I thought this was exactly the purpose of this site. You're doing a great job of blocking access to knowledge. Instead you could have pointed me to a textbook. I thank the others who answered. Don't you think that if I had a tutor I wouldn't be asking here? Don't worry you and the others: I won't reformulate and I won't ask anything on this site again. – user123634 Jul 16 '16 at 15:30
• @honeste_vivere: I can see now why entropy increases and thus irreversibility, but I will need to look more closely at your math to understand properly the part with the derivative order. Thanks! – user123634 Jul 16 '16 at 15:46
• @zetzar - The easiest solution to the issue that caused your question to be closed is to ask a shorter, more concise single question for each post. If you ask multiple, complicated questions in one post any possible answer would be so long as to preclude it being read, let alone being useful for a wider audience. Does that make sense? – honeste_vivere Jul 16 '16 at 19:17

Total energy of a closed system is always conserved (time translation, Noether theorem). However, this energy might go from macroscopic motion (fluid flow) into random vibrations of the molecules.

I am not 100% sure of the thermodynamics, but I will try.

The total energy $U$ is still there but it has changed to heat. The velocity $v$ of the molecules also has not changed, however the directions got randomized!

What has also changed is the entropy $S$. There are only a few possible realizations of perfect laminar flow with a certain velocity. For random vibrations, you have way more possible configurations. So although the internal energy $U$ has not changed, the “useful” macroscopic energy $F$ (Helmholtz free energy) has changed: $$F = U - TS \,.$$ In a closed system we know that $U$ is fixed. So let us rewrite it as $$U = F + TS \,.$$ So we start with a configuration with low entropy $S$. This means that the usable energy is high. Then due to various relaxation processes the entropy $S$ will increase. Assuming that the temperature $T$ does not change a lot, we will then have a decrease in useful energy $F$.

So heat is still a form of energy, albeit not directly usable as kinetic energy of a macroscopic object (say a pendulum swinging).

• Thanks for the answer. I see your point and that's what I meant by useful energy. So if entropy increases, at constant T, free energy decreases, and this is the effect of viscosity - it dissipates F into heat Q=TS? But why do we need to add a special viscous drag term when writing the equations of motion for a collection of particles or Navier-Stokes to account for this phenomenon? Is the total energy/Hamiltonian equivalent in this case to F and the reason why it doesn't get conserved is that we cannot account for all the DOFs? – user123634 Jul 15 '16 at 13:17
• Let's go back to the water and honey example: if we simulate it at molecular level we don't need such terms and viscosity should emerge by itself. Clearly honey should flow more slowly. What is then a measure of damping/dissipation of energy? Is it the entropy or the kinetic energy? Cause what the eye perceives is slower velocities - does this imply honey flows with less kinetic energy than water? But if temperature correspond to the average kinetic energy, shouldn't viscosity be reflected in T somehow? – user123634 Jul 15 '16 at 13:17
• If you perform a simulation on the microscopic level, there is no friction involved and it should all emerge from the microscopic DOF.The total energy is $U$, so $\langle H \rangle = U$. There are just too many of them to do a macroscopic simulation. Navier-Stokes equation is an effective/phenomenological thing that describes the relevant DOF of the system. A measure for dissipation with relevant DOFs should be the increase of $S$. – Martin Ueding Jul 15 '16 at 13:57

Part of the answer is, at low Reynolds number, in streamline flow, there's little if any drag. At higher speeds, though, the wake of a moving object creates vortices (and does so in random fashion, a spontaneous symmetry breaking occurs). Those random vortices cause a velocity-squared retarding force, which is energy-losing whether your motion is northerly or southerly... it is a definitely non conservative force field. Drag can be decreased by some shapes that shed vortices (so the vortex doesn't impinge on the moving object).

In any case, vortices decay to thermal energy in a fluid. The water at the base of Niagara falls is slightly warmer than the water at the top, so the energy isn't only lost to solid walls.

• This is interesting: vortices are causing drag. I will dig more into it. Also, you are saying that the energy decay is found in the thermal energy, i.e. temperature increase, while Martin said it's just in the entropy. I will think more about it. Thanks anyway! – user123634 Jul 16 '16 at 15:41
• There is entropy in the spontaneous symmetry breaking; that's a part of thermalization. The entropic blending of cream into coffee similarly starts with vortex formation around the moving spoon... – Whit3rd Jul 20 '16 at 21:00