The Pitch drop experiment has pitch, a fluid with extremely high viscosity, flowing through a funnel. In 85 years, only 9 drops have fallen. Photographs of the drops, however, show an appearance just like drops of fluids with much lower viscosity like jelly or syrup. Indeed, the pitch drops look like regular drops "frozen in time".

My question, then, is this: All else being equal, is the flow of a high-viscosity fluid the same as that of a low-viscosity fluid, just slower? That is, if a camera had been filming the pitch drop experiment for its entire duration and the footage was played back at a few hundred thousand times its recording speed, would it still be possible to distinguish pitch from honey?

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    $\begingroup$ I know that in a Stokes flow, where the viscosity is high and Reynolds number is low, the Navier-Stokes equations can be approximated as linear. For linear equations a simple analogy like this might be true, but I'm not sure. In the general case where non-linear terms are important, low viscosity and high viscosity fluids can create very different flows in the same geometry. $\endgroup$ – MonkeysUncle Mar 11 '15 at 12:33
  • $\begingroup$ It would be possible to distinguish it from honey because of how fast the separated drops would fall. Gravity still acts the same, so seeing the drop fall super fast would indicate it's been sped up artificially. $\endgroup$ – Jim Mar 11 '15 at 20:49

Not necessarily, it depends on how different the viscosities are.

@MonkeysUncle got it right. If the Reynolds number is < 2,000 the flow is laminar; if it's > 4,000 the flow is turbulent. Since the Reynolds number depends on viscosity, if the viscosity of the two fluids is different enough that it changes the flow from laminar to turbulent, then you wouldn't get matching flows simply by speeding up the video of the slow one. Water and lubricating oil have different enough viscosities, as one example.

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    $\begingroup$ Those critical Reynolds numbers you quote are only true for pipe flow. Drop formation is a free-boundary problem and any flow with Reynolds number $\gtrsim 1$ will have inertial effects in addition to viscous effects. $\endgroup$ – Dai Mar 17 '15 at 10:04
  • $\begingroup$ @Dai thanks for the comment. Could you provide an online source? I would like to look into this. $\endgroup$ – pentane Mar 17 '15 at 11:19
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    $\begingroup$ For the pipe flow bit the wikipedia reference you cite explains things quite nicely. The key is that inertial effects (what eventually cause turbulence) kick in when the Re is greater than 1. In pipe flow, viscosity is generally able to dampen out perturbations until about Re = 2000. Jets (free-shear) are generally said to be turbulent for Re > 1000 (Pope). See also the Oseen correction and subsequent improvements to the Stokes drag equation to account for fluid inertia. $\endgroup$ – Dai Mar 17 '15 at 13:41

Quite clearly time itself doesn't slow down, but the time scale over which the drops form and fall is strongly dependent on the viscosity of the fluid. The times cale for viscous flow through a constriction under the action of gravity is given through dimensional arguments as

$$ \tau \sim \nu / (dg) $$

where $\nu$ is the kinematic viscosity of the fluid, $d$ is the diameter of the constriction (i.e. the end of the funnel in the pitch drop experiment) and $g$ is gravity.1

So plugging in values for pitch, water and corn syrup for comparison (wiki viscosity, density pitch, density of water & corn syrup) we have time scales of

Corn syrup : 9.65E-3 s

Water : 9.11E-6 s

Pitch : 2.04E+6 s = 3.4 weeks

based on a diameter of 1 cm (my guess). This doesn't mean that a drop will form or fall in exactly this time, it's just the characteristic time for any motion in this kind of system. There may also be other effect on the flow such as capillary forces, visco-elastic forces etc. which would change the formulation of $\tau$ but this is the basic analysis.


Your question is ultimately about scaling. Scaling is best explored by non-dimensionalizing equations.

Let's look at a free-fall equation of a mass $m$ in the absence of air friction : $$ m \frac{d^2 z}{dt^2} = - m g $$ Obviously the time to fall of a given height will depend on gravity $g$. But your question is, will it be the same up to a scaling factor in time? (Of course in this case you'll easily guess that it's the case -- but we'll go to more complex cases later)

You can non-dimensionalize it by defining a characteristic height $H$ and time $T$. Say $H$ will be the height of your camera's viewing field and $T$ the lag between two frames of your movie. You get (getting rid also of $m$): $$ \frac{d^2 \tilde z}{d\tilde t^2} = - \frac{g T^2}{H} $$ Here, $\frac{g T^2}{H}$ is a non-dimensional group that characterizes the ratio between inertia and gravity force. You can easily see that, if you double $T$ when $g$ is divided by 4, you get exactly the same number: the movie with $g/4$ with frames every $2T$ will thus look exactly the same as the one with $g$ and frames every $T$.

That's the same for a purely viscous flow, if there is no other force present at all. Rather than looking at the flow itself, let's look at a heavy particle sedimenting in the fluid: its equation is $$ \frac{dz}{dt} = \frac{(\rho_p-\rho_f)gd^2}{18\eta} $$ to which you can apply the same non-dimensional process.

Now let's introduce a second force, e.g. friction in the free-fall (or inertia in the above sedimenting particle). You have 3 terms in the equation, such as: $$ m \frac{d^2 z}{dt^2} = - c \frac{d z}{d t} - m g $$ This means that you'll build two non-dimensional groups: one will be $\frac{c T}{m }$ and the other $\frac{g T^2}{H}$. This time, if $g$ is changed, you cannot tune $T$ so as to keep everything unchanged as this would modify the other non-dimensional group. Only, if $c$ is so small that its effect is negligible for either $g$ or $g/4$ conditions, then the effect will not be visible in your movie and you can continue tuning $T$ according to $g$ only.

For a viscous flow, many forces in addition to viscous ones can come at play. Inertia is one, but if you look at a reasonnably-sized flow of honey or pitch, inertia will be very low in both cases.

What may actually be more relevant is surface forces (interfacial tension), which will not vary in direct proportion to viscosity. But you could try to find a sort of honey that would have the same viscosity to surface tension as pitch, and preserve the scaling!


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