What is the 'molecular' origin of the viscosity? The molecular origin of elasticity is almost clear for me: at the very bottom the 'elasticity' comes from the attraction and repulsion between atoms when they are reasonably far from each other or very close (like Lennard-Jones potential). Is there a similar description for the viscosity as well? Where does it come from?

  • $\begingroup$ Elasticity in a gas is not a matter of attraction and repulsion, it is a matter of continual thermodynamic collisions. For viscosity, all the answers are good. $\endgroup$ – Mike Dunlavey Aug 6 '14 at 19:27

Imagine two two trains side by side - one going faster than the other. Frictionless rails. Start shoveling coal from the slow train to the fast one, and from the fast to the slow one. Every shovel of coal results in a transfer of momentum - until the two trains move at the same speed.

In the same way, when layers of liquid move past one another at different velocities (in other words, there is shear in the liquid), then thermal motion of molecules from one layer to the other tends to equalize the velocities - just like viscous drag does. The larger the shovel, the faster you equalize.

Now instead of shoveling coal, imagine that you had trains with trees growing out of the side of them. As the fast train tries to overtake the slow train, the branches intertwine and the train velocities tend to equalize again - more efficiently than the coal.

In these two analogies you see some of the essential features of molecules that cause viscosity. Small molecules can diffuse, and in this way facilitate the exchange of momentum between fluid layers moving at different velocities. Larger molecules with branches on them have much higher viscosity.

Imagine the trains further having magnets on the side - as they get close together, these really stick the trains together and provide even more efficient momentum transfer.

Looking at the paper Prediction of liquid viscosity for organic compounds by a quantitative structure±property relationship you get a thorough description of what you are asking about - but in essence it comes down to the principles I described above. The authors find that five factors relating to the structure of organic molecules can go a long way to predicting viscosity.

These factors include

  1. Hydrogen-bonding donor charged surface area - roughly the capacity to form hydrogen bonds (the "magnets" on the trains)
  2. Gravitational index - a measure of "density" of the molecule (mass of atoms divided by bond length) - related to the size of the shovel of coal
  3. Number of rings - related to the complexity of the shape of the molecule ("branches" in the train analogy)

I recommend that you read the entire paper (and some of its references) to get a more complete picture of this.

  • $\begingroup$ Also a good answer! You're post got me thinking. It might be easier to motivate the dynamic viscosity, rather than the kinematic as I did. If there is a velocity gradient $\nabla v$, molecules will on average transfer momentum $m\lambda \nabla v$ from one side to the other, so the extra momentum transferred across a unit area per unit time (the stress) is $(nv) (m\lambda \nabla v) = (\rho \lambda v) \nabla v$, from which we read off the dynamic viscosity $\mu = \rho \lambda v $. $\endgroup$ – alemi Aug 6 '14 at 4:16
  • $\begingroup$ @alemi did you have a chance to look at that reference? $\endgroup$ – Floris Aug 6 '14 at 4:23
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    $\begingroup$ just for fun, the number of people who tripped over their own two feet and died correlates with the total revenue generated in the US at golf courses to 0.84: link $\endgroup$ – alemi Aug 6 '14 at 4:49
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    $\begingroup$ @alemi I really enjoy spurious correlations. They come up with the damndest things. And every time I see one I starting trying to build a change of inference that offers a possible explanation. In this case I thought "Hmmm ... old people trip and old people play golf..."*. $\endgroup$ – dmckee Aug 6 '14 at 17:57
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    $\begingroup$ I like your train analogies. $\endgroup$ – Mike Dunlavey Aug 6 '14 at 19:28

Yes there is. Let's focus on the kinematic viscosity ($\nu$), which is defined as the diffusion constant for momentum in the fluid. That is, it tells us how quickly a momentum disturbance would diffuse through the rest of the fluid. Or, in particular, it gives us the linear dependence on the mean square propagation of the momentum as a function of time, giving the mean square of the distance $L$ a momentum disturbance travels as: $$ \langle L^2 \rangle \sim \nu t $$

Ignoring dense fluids like water for a second, let's talk about gases. In a gas, the only way to transport momentum is by letting the gas molecules themselves move from place to place. So, let's try to estimate how quickly they can diffuse.

For the most part, we can model a dilute gas as a bunch of particles all bouncing around and hitting each other, undergoing some kind of random walk. How far do they travel between collisions? Well, let's call this the mean free path $\lambda$. Assuming we have a certain density of gas molecules $n$, each moving with some velocity $v$, and each having a certain cross sectional area $\sigma$, in a small amount of time $\tau$, a single particle will sweep out a volume of size $$ V = \sigma v \tau = \sigma \lambda $$ the number of molecules contained in that volume will be $nV$, and if we set that to 1, we should get a decent estimate for the time between collisions ($\tau$) and the distance traveled between collisions $\lambda$ $$ 1 = nV = n \sigma v \tau = n \sigma \lambda $$ so notice, in particular $\lambda \sim \frac 1 {n \sigma}$ and $\tau \sim \frac{1}{n\sigma v} \sim \frac{\lambda}{v}$. So, assuming our gas molecules are undergoing a random walk, between each collision they travel a distance $\lambda$, after $N$ collisions they will go a characteristic distance $\lambda \sqrt N$ (as is the nature of random walks). So we have $$ L^2 \sim \left( \lambda^2 \right) \left( \frac{t}{\tau} \right) \sim \lambda v t $$ That is, we go a $\lambda^2$ distance on average for each collision, and we have $t/\tau$ collisions in a time $t$.

From this we can read off the kinematic viscosity. If we had done all of the particular calculations just right in statisical mechanics, a magic factor of $1/3$ pops in, giving us

$$ \boxed{ \nu = \frac 1 3 \lambda v = \frac 1 3 \frac{v}{n\sigma} }$$ If we try to validate this for air, using the thermal speed $ v = \sqrt{ kT / m }$ and ideal gas density $n=P/kT$ and assuming $\sigma = 10 \text{ angstroms}^2 $ we get $\mu \sim 0.4 \text{ cm}^2/s$, compared to the measured value of $\nu = 0.15 \text{ cm}^2/s$, not bad for a rough calculation.

Now, this was for gases, though the molecular reason for dense fluids is the same, fluids are more complicated. They are nearer together, so the details of the calculation are going to get a little screwy. Though if we use the speed of sound in water for $v$, and $n = (1 \text{ g/cm}^3)/(18 m_p)$, we are only a factor of 6 off from the observed value for the kinematic viscosity. The discrepancy will be due to the details of the intermolecular forces due to the close contact of the water molecules.

  • $\begingroup$ A very thorough answer as usual. Always enjoy reading your answers. $\endgroup$ – Floris Aug 6 '14 at 4:01

In a fluid the molecules collide and interact randomly. As a consequence, local physical features typically dampen out over timescales of a few collisions.

However, quantities that are conserved in collisions form an exception to this rule. Quantities such as mass and momentum can not be 'destroyed' and therefore do not dampen out over collision timescales. Instead, as a result of collisions, mass and momentum both get randomly redistributed. The result is that mass and momentum disturbances diffuse.

Mass diffusion is referred to simply as 'diffusion', momentum diffusion is referred to as 'viscosity'. Momentum diffusion is the effect that causes fluids to come to rest.


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