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7.3: Random walks - Viscosity and heat flow

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    The great mathematician George Pólya, later to become author of How to Solve It [38], was staying at a bed and breakfast in his adopted country of Switzerland and walking daily in the garden between its tall hedge rows. He kept running into a newlywed couple taking their walk. Were they following him, or was it a mathematical necessity? From this question was born the study of random walks.

    Random walks are everywhere. In the card game War, cards wander between two players, until one player gets the whole deck. How long does the game last, on average? A molecule of neurotransmitter is released from a synaptic vesicle. It wanders in the 20-nanometer gap, the synaptic cleft, until it binds to a muscle cell; then your leg muscle twitches. How long does the molecule’s journey take? On a winter day, you stand outside wearing only a thin layer of clothing, and your body heat wanders through the clothing. How much heat do you lose? And why do large organisms have circulatory systems? Answering these questions requires understanding random walks.

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    7.3.1 Behavior of random walks: Lumping and probabilistic reasoning

    For our first random walk, imagine a perfume molecule wandering in a room, moving in a straight line until collisions with air molecules deflect it in a random direction. This randomness reflects our incomplete knowledge: Knowing the complete state of the colliding molecules, we could calculate their paths after the collision (at least, in classical physics). However, we do not have that knowledge and do not want it!

    Even without that information, the random motion of one molecule is still complicated. The complexity arises from the generality—that the direction of travel and the distance between collisions can have any value. To simplify, we’ll lump in several ways.

    Distance. Let’s assume that the molecule travels a typical, fixed distance between collisions. This distance is the mean free path \(\lambda\).

    Direction. Let’s assume that the molecule travels only along coordinate axes. Let’s also study only one-dimensional motion; thus, the molecule moves either left or right (with equal probability).

    Time. Let’s assume that the molecule travels at a typical, fixed speed \(v\) and that every collision happens at regularly spaced clock ticks. These ticks are therefore separated by a characteristic time \(\tau = \lambda / v\). This time is called the mean free time.

    In this heavily lumped, one-dimensional model, a molecule starts at an origin (x = 0) and wanders along a line. At each tick it moves left or right with probability 1/2 for each direction.

    As time passes, the molecule spreads out. Actually, the molecule itself does not spread out! It has a particular position, but we just don’t know it. What spreads out is our belief about the position. In the notation of probability theory, this belief is a set of probabilities—a probability distribution—based upon our knowledge of the molecule’s starting position:

    As time passes, the molecule spreads out. Actually, the molecule itself does not spread out! It has a particular position, but we just don’t know it. What spreads out is our belief about the position. In the notation of probability theory, this belief is a set of probabilities—a probability distribution—based upon our knowledge of the molecule’s starting position:

    \[\textrm{Pr (molecule is at position } x \textrm{ at time } t \vert \textrm{ it was at } x = 0 \textrm{ at } t = 0).\]

    The changing beliefs are represented by a sequence of probability distributions, one for each time step. For example, at 2\(\tau\) (after two ticks), the molecule has probability 1/2 of being at the origin, by either going right then left or left then right. At 3\(\tau\), it has probability 0 of being at the origin (why?), but probability 3/8 of being at \(x = + \lambda\). To quantify the spread, which is shown in the following figure, we need an abstraction and a notation.

    clipboard_e3b5c87dc4579c00acc3e171743e9d9e6.png

    The position of the molecule will be x. Its expected position will be ⟨x⟩. The expected position is the weighted average of the possible positions, weighted by their probabilities. Both x and ⟨x⟩ are functions of time or, equivalently, of the number of ticks. However, because motion in each direction is equally likely, the expected position does not change (by symmetry). So ⟨x⟩, which starts at zero, remains zero.

    A useful measure is the squared position x2—more useful because it is never negative, making moot the symmetry argument that made ⟨x⟩ = 0. Analogous to ⟨x⟩, the expected or mean squared position ⟨x2⟩ is the average of the possible values of x2, weighted by their probabilities.

    Let’s see how ⟨x2⟩ changes with time. At t = 0, the only possibility is x = 0, so ⟨x2t=0 = 0. After one clock tick, at \(t = \tau\), the possibilities are also limited: \(x = + \lambda\) or \(-\lambda\). In either case, \(x^{2} = \lambda^{2}\). Therefore, \(\langle x^{2} \rangle_{t = \tau} = \lambda^{2}\).

    It may be the mark of a savage, in Pólya’s phrase [37], to generalize a pattern from only two data points. So let’s find \(t = 2 \tau\) and then guess a pattern. At \(t = 2\tau\), the position x could be \(-2 \lambda\), 0, or \(\lambda\), with probabilities 1/4, 1/2, and 1/4, respectively. The weighted average ⟨x2⟩ is \(2 \lambda^{2}\):

    \[\langle x^{2} \rangle = \frac{1}{4} \times (-2 \lambda )^{2} + \frac{1}{2} \times (0 \lambda)^{2} + \frac{1}{4} \times (+ 2 \lambda)^{2} = 2 \lambda^{2}.\]

    The pattern seems to be

    \[\langle x^{2} \rangle _{t= n \tau} = n \lambda ^{2}.\]

    This conjecture is correct. (You can test it at \(t = 3 \tau\) and \(4 \tau\) in Problem 7.13.) Each step contributes the squared step size \(\lambda^{2}\) to the squared spread ⟨x2⟩. We saw this pattern in Section 7.2.1, when we combined plausible ranges. The half width of the plausible range for an area \(A = h w\) was given by

    \[\sigma_{A}^{2} = \sigma_{h}^{2} + \sigma_{w}^{2},\]

    where \(\sigma_{x}\) is the half width of the plausible range for the quantity x. The half widths are step sizes in a random walk—random because the estimate is equally like to be an underestimate or an overestimate (representing stepping left or right, respectively). Therefore, the half widths, like the step sizes in a random walk, add via their squares (“adding in quadrature”).

    The number of ticks \(n = t/\tau\) so \(\langle x^{2} \rangle \), which is (n \lambda^{2}\), is also \(t \lambda^{2}/\tau\). Thus,

    \[\frac{\langle x^{2} \rangle}{t} = \frac{\lambda ^{2}}{\tau}.\]

    As time marches on, ⟨x2⟩/t remains \(\lambda^{2}/\tau\)! The invariant \(\lambda^{2}/\tau\) is all that we need to know about the details of a random walk.

    This abstraction is known as the diffusion constant. It is usually denoted D and has dimensions of L2T−1. The table gives useful approximate diffusion constants for a particle wandering in three dimensions. In d dimensions, the diffusion constant is defined with a dimensionless prefactor:

    \[D = \frac{1}{d} \frac{\lambda^{2}}{\tau}.\]

    clipboard_e153cdb8aa5ec99df7ea9737bbbb85bcd.png

    Because \(\lambda/ \tau\) is the speed \(v\) of the molecule between its randomizing collisions, a useful alternative form for estimating D is

    \[D = \frac{1}{d} \lambda v.\]

    Exercise \(\PageIndex{1}\): Testing the random-walk dispersion conjecture

    Make the probability distribution for the particle at \(t = 3 \tau\) and \(t = 4 \tau\), and compute each ⟨x2⟩. Do your results confirm that \(\langle x^{2} \rangle _{t= n \tau} = n \lambda ^{2}\)?

    Exercise \(\PageIndex{2}\): Higher dimensions

    For a molecule starting at the origin and wandering in two dimensions, \(\langle r^{2} \rangle = n \lambda^{2}\), where \(r^{2} = x^{2} + y^{2}\). Confirm this statement for \(t = 0...3\tau\).

    A random and a regular walk are analogous in having an invariant. For a regular walk, it is ⟨x⟩/t: the speed. For a random walk, it is ⟨x2⟩/t: the diffusion constant. (The diffusion constant is to a random walk as the speed is to a regular walk.) However, the two walks differ in a scaling exponent. For a regular walk, time is 1. For a random walk, we’ll use the rms (root-mean-square) position ⟨x⟩ ∝ t: The scaling exponent connecting position and

    \[x_{rms} \equiv \sqrt{\langle x^{2} \rangle}\]

    as a measure analogous to ⟨x⟩ for a regular walk. Because ⟨x2⟩ ∝ t, the rms position xrms is proportional to t1/2: In a random walk, the scaling exponent connecting position and time is only 1/2. This scaling exponent has profound effects on heat, drag, and diffusion.

    As an example of the effect, let’s apply our knowledge of diffusion and random walks to a familiar situation. Across a room someone opens a bottle of perfume or, if your taste in problems is pessimistic, a lunch of leftover fish.

    clipboard_e886f5601e28e8843534e74eea41f997c.png

    How long until odor molecules reach your nose?

    As time passes, the molecule wanders farther afield, with its rms position growing proportional to t1/2. As a lumping approximation, imagine that the molecule is equally likely to be anywhere within a distance xrms of the source (the perfume bottle or leftover fish). For the molecule to have a significant probability to be at your nose, xrms should be comparable to the room size L. Because \(\langle x^{2} \rangle = Dt\), the condition is \(L^{2} \sim Dt\), and the required diffusion time is \(t \sim L^{2}/D\). (For another derivation, see Problem 7.16.)

    For perfume molecules diffusing in air, D is roughly 10−6 square meters per second. For a 3-meter room, the diffusion time is roughly 4 months:

    \[t \sim \frac{L^{2}}{D} \sim \frac{(3m)^{2}}{10^{-6}m^{2}s^{-1}} \approx 10^{7} s \approx 4 \: months.\]

    This estimate does not agree with experiment! After perhaps a minute, you’ll notice the aroma, whether perfume or fish. Diffusion is too slow to explain why the odor molecules arrive so quickly. In reality, they make most of the journey using a regular walk: The small, unavoidable air currents in the room transport the molecules much farther and faster than diffusion can. The speedup is a consequence of the change in scaling exponent, from 1/2 (for random walks) to 1 (for regular walks).

    Exercise \(\PageIndex{3}\): Diffusion constant of air

    Estimate the diffusion constant for air molecules diffusing in air by using

    \[D \sim \frac{1}{3} \times \textrm{mean free path} \times \textrm{travel speed}\]

    and the mean free path of air molecules (Section 6.4.5). This value is also its thermal diffusivity \(\kappa_{air}\) and its kinematic viscosity \(v_{air}\)!

    Exercise \(\PageIndex{4}\): Dimensional analysis for the diffusion time

    Use dimensional analysis to estimate the diffusion time t based on L (the relevant characteristic of the room) and D (the characteristic of the random walk).

    A similar estimate explains the existence of circulatory systems. Imagine an oxygen molecule diffusing through our body to a muscle cell, where its services are needed to burn glucose and produce energy. The diffusion distance (our body size) is L ∼ 1 meter. The diffusion constant for an oxygen molecule in water(a small molecule in water) is roughly 10−9 square meters per second. The diffusion time is roughly 109 seconds or 30 years:

    \[t \sim \frac{L^{2}}{D} \sim \frac{(1m)^{2}}{10^{-9}m^{2}s^{-1}} = 10^{9} \approx 30 \: years.\]

    Over long distances—long compared to the mean free path \(\lambda\)—diffusion is a slow method of transport! Large organisms, especially warm-blooded organisms with high metabolic rates, need another solution: a circulatory system. It transports oxygen much more efficiently than diffusion can, just as air currents do for perfume. The circulatory system, a branching network of ever-smaller capillaries, ends once the typical distance between the smallest capillaries and a cell is small enough for diffusion to be efficient.

    Another biological example of short-distance diffusion is the gap between two neighboring neurons, called the synaptic cleft. Its width is only L ∼ 20 nanometers. Signals between neighboring nerve cells, or between a neuron and a muscle cell, travel chemically, as neurotransmitter molecules.

    Let’s estimate the diffusion time for a neurotransmitter molecule. A neurotransmitter is a large molecule, and it is diffusing in water, so D ∼ 10−10 square meters per second. Its diffusion time is 4 microseconds:

    \[t \sim \frac{L^{2}}{D} \sim \frac{(2 \times 10^{-8}m)^{2}}{10^{-10} m^{2} s^{-1}} = 4 \times 10^{-6} s.\]

    Without a comparison, this time means little: We cannot say right away whether the time is long or short. However, because it is much smaller than the spike-timing accuracy of neurons (about 100 microseconds), the time to cross the synaptic cleft is small enough not to affect nerve-signal propagation. For transmitting signals between neighboring neurons, diffusion is an efficient and simple solution.

    Exercise \(\PageIndex{5}\): Probability of being at the origin

    Pólya’s analysis of his encounters with the newlywed couple required first finding the probability \(p_{n}\) that a random walker is at the origin after n clock ticks (p0 = 1). For a d-dimensional random walk, find the scaling exponent \(\beta (d)\) in \(p_{n} \propto n^{\beta (d)}\).

    Exercise \(\PageIndex{6}\): Expected number of visits to the origin

    Using your result from Problem 7.17, estimate the expected number of visits a random walker in one and two dimensions makes to the origin (summed over all ticks n ≥ 0). Thereby explain Pólya’s theorem [36], that a random walker in one or two dimensions always returns to the origin. What makes a three-dimensional random walk different from one or two dimensions?

    7.3.2 Types of diffusion constants

    Because random walks are everywhere, there are several kinds of diffusion constants. They are named differently depending on what is diffusing, but they share the mathematics of the random walk. Therefore, they all have dimensions of length squared per time (L2T−1).

    clipboard_e3a7bd0ef847520d92953db8f8edd284b.png

    A handful of useful diffusion constants (for particles) were tabulated on page 251. To complement that table, here are a few useful, approximate thermal diffusivities and kinematic viscosities.

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    In air, all three diffusion constants—D for molecules, \(\kappa\) for energy, and \(\nu\) for momentum—are roughly 1.5 × 10−5 square meters per second. Their similarity is no coincidence. The same mechanism (diffusion of air molecules) transports molecules, energy, and momentum.

    In water, however, the molecular diffusion constant D is several orders of magnitude smaller than the heat- and momentum-diffusion constants (\(\kappa\) and \(\nu\), respectively). Even the momentum- and heat-diffusion constants differ roughly by a factor of 7. This dimensionless ratio, \(\nu/\kappa\), is the Prandtl number Pr. For water, I usually remember \(\nu\), because it is just a power of ten in SI units, and remember the Prandtl number—the lucky number 7—and use these values to reconstruct \(\kappa\).

    7.3.3 Thermal diffusivities of liquids and solids

    The large discrepancy between the molecular and thermal diffusion constants in water indicates that our model for diffusion in water is not complete. The problem is not limited to water. If we had made a similar comparison for any solid, comparing the molecular and thermal diffusion constants (D and \(\kappa\)), the discrepancy would have been even larger.

    Indeed, in liquids and solids, in contrast to gases, heat is not transported by molecular motion. In a solid, the molecules sit at their sites in the lattice. They vibrate but scarcely wander. In a liquid, molecules wander but only slowly. Their tight packing keeps the mean free path short and the diffusion constant small. Yet, as everyday experience and the large \(\kappa/D\) ratio suggest, heat can travel quickly in liquids and solids. The reason is that heat is transported by miniature sound waves rather than by molecular motion. The sound waves are called phonons.

    By analogy to photons, which represent the vibrations of the electromagnetic field, phonons represent the vibrations of the lattice entities (the atoms or molecules of the liquid or solid). One molecule vibrates, shaking the next molecule, which shakes the next molecule. This chain is the motion of a phonon. Phonons act like particles: They travel through the lattice, bouncing off impurities and other phonons. Like ordinary particles, they have a mean free path and a propagation speed. These properties of their random walk determine the thermal diffusivity:

    \[\kappa \approx \frac{1}{3} \times \textrm{phonon mean free path} \times \textrm{propagation speed}.\]

    The propagation speed is the speed of sound cs, because phonons are tiny sound waves (familiar sound waves contain many phonons, just as light beams contain many photons). Sound speeds in liquids and solids are much higher than the thermal speeds, so we already see one reason why \(\kappa\), the diffusion constant for heat, is larger than D, the diffusion constant for particles.

    The mean free path \(\lambda\) measures how far a phonon travels before bouncing (or scattering) and heading off in a random direction. Let’s write \(\lambda = \beta a\), where a is the typical lattice spacing (3 ångströms) and \(\beta\) is the number of lattice spacings that the phonon survives

    Then the thermal diffusivity becomes

    \[\kappa = \frac{1}{3} c_{s} \beta a.\]

    For water, our favorite substance,cs∼ 1.5 kilometers per second (as you estimated in Problem 5.58). Then the predicted thermal diffusivity becomes

    \[\kappa_{water} \sim \beta \times 1.5 \times 10^{-7} m^{2} s^{-1}.\]

    Because the actual thermal diffusivity of water is 1.5 × 10−7 square meters per second, our estimate is exact if we use \(\beta = 1\). This choice is easy to interpret and remember: In water, the phonons travel roughly one lattice spacing before scattering in a random direction. This distance is so short because water molecules do not sit in an ordered lattice. Their disorder provides irregularities that scatter the phonons. (At the same time, this mean free path is much larger than the mean free path of tightly packed atoms or molecules, which move a fraction of an ångström before getting significantly deflected. Therefore, even in a liquid, \(\kappa\) is much larger than the molecular diffusion constant D.)

    To estimate \(\kappa\) for a solid, let’s use \(\kappa_{water}\) along with the scaling relation

    \[\kappa \propto \lambda c_{s}.\]

    In a typical solid, the sound speed cs is 5 kilometers per second—roughly a factor of 3 faster than in water. The mean free path \(\lambda\) is also longer than in the totally disordered lattice of a liquid. In a solid without too many lattice defects, and at room temperature, a phonon travels a few lattice spacings before scattering—compared to just one lattice spacing in water.

    These two differences, each contributing a factor of 3, make the typical thermal diffusivity of a solid a factor of 10 larger than that of water:

    \[\kappa_{solid} \sim \kappa_{water} \times 10 \approx 1.5 \times 10^{-6} m^{2} s^{-1}.\]

    Rounded to 10-6 square meters per second, this value is our canonical thermal diffusivity of a solid--for example, sandstone or brick. The table also shows a new phenomenon: For metals, \(\kappa\) is much larger than our canonical value. Although a small discrepancy could be explained by a few missing numerical factors, this significant discrepancy indicates a missing piece in our model.

    clipboard_ed7154cb59214a659681818cd144541c7.png

    Indeed, in metals, heat can also be carried by electron waves, not just phonons (lattice waves). Electron waves travel much faster and farther than phonons. Their speed, known as the Fermi velocity, is comparable to the orbital speed of an atom's outer-shell electron. As you found in Problem 5.36, for hydrogen this speed is \(\alpha c\) where \(\alpha\) is the fine-structure constant (~10-2) and c is the speed of light. This speed, roughly 1000 kilometers per second, is much faster than any sound speed! As a result, the thermal diffusivity in metals is large. As you can see in the table, for a good conductor, such as copper or gold, \(\kappa\) ∼ 10−4 square meters per second.

    7.3.3.1 Heating a skillet

    To feel a thermal diffusivity, place a thin cast-iron skillet on a hot stove.

    How long does it take for the top surface to feel hot?

    The hot stove supplies blobs of heat (of energy) that wander back and forth: The heat blobs perform a random walk. In a random walk, a particle with diffusion constant D wandering for a time t reaches a distance \(z \sim \sqrt{Dt}\); we used this lumping model in Section 7.3.1 to estimate the diffusion time across a sunaptic cleft. Here, the particle is a blob of heat, so the diffusion constant is \(\kappa\). Thus, the hot front reaches a distance \(z \sim \sqrt{\kappa t}\) into the skillet.

    In this lumping picture, the temperature profile is a rectangle with a moving right edge—representing the heat wave moving upward and into the skillet. For my 1-centimeter-thick cast-iron skillet, the hot front should reach the top of the skillet in roughly 4 seconds:

    \[t \sim \frac{L^{2}}{\kappa_{Fe}} \approx \frac{(10^{-2}m)^{2}}{2.3 \times 10^{-5} m^{2} s^{-1}} \approx 4s.\]

    clipboard_eac67c75ae1101e6bfa77e134ff696a55.png

    Don’t try the following experiment at home. But for the sake of lumping and probabilistic reasoning, I set our flattest cast-iron skillet on a hot electric stove while touching the top surface. After 2 seconds, my finger involuntarily jumped off the skillet.

    The discrepancy of a factor of 2 between the predicted and actual times is not bad considering the simplicity, or crudity, of the lumping approximations that it incorporates. However, there is a bigger discrepancy. The model predicts that, forthe first 4 seconds, the top of the skillet remains at room temperature. One feels nothing until, wham, the full temperature of the stove hits at 4 seconds.

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    However, experience suggests that the skillet’s temperature starts rising before the skillet becomes too hot to touch, and then it monotonically approaches the stove temperature—as sketched in the figure.

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    One reason for this discrepancy could be the skillet’s top surface. In our model, the skillet has only a bottom surface and is infinitely thick. The top surface might alter the heat flow. However, the infinite-slab assumption isn’t the fundamental problem (correcting the assumption turns out to speed up the heating process by a factor of 2). Even if we fix it, the model would still make the bogus prediction of a sudden jump to the stove temperature. It’s hard to believe after the exhortations on the power of lumping, but we have lumped too much.

    To improve the model, let’s incorporate a more realistic temperature profile. Beyond the canonical lumping shape of a rectangle, the next simplest shape is a triangle (the integral of a rectangle). We will therefore replace the rectangular temperature profile with a triangle having the same area as the rectangle. Because of the factor of 1/2 in the area of a triangle, the hot zone now extends to 2\(\sqrt{\kappa t}\) instead of to \(\sqrt{\kappa t}\).

    The area of the triangle is proportional to the heat transferred to the skillet. By matching the triangle’s area to the rectangle’s area, we preserve this integral quantity. When making lumping models, preserving integral quantities is usually more robust than is preserving differential quantities (such as slope).

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    In the triangular-lumping model, you feel nothing until the tip of the triangle reaches the top of the skillet. Because the triangular hot front extends a factor of 2 farther than the rectangular hot front (2 \(\sqrt{\kappa t}\) versus \(\sqrt{\kappa t}\)), and because diffusion times are proportional to distance squared, the required time falls by a factor of 4. Thus, it falls from 4 seconds to 1 second. At 1 second, the hot front arrives at the top of the skillet, which starts to feel warm. Then the temperature slowly increased toward the stove temperature. This next simplest model makes quite realistic predictions!

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    Exercise \(\PageIndex{7}\): Cooling the Moon

    How long does it take for the Moon, with a radius of 1.7 × 106 meters to cool significantly by heat diffusion through rock? Given that the Moon is now cold, what do you conclude about the mechanism of cooling?

    Exercise \(\PageIndex{8}\): Diffusion with a bit of drift: Breaking the bank at Monte Carlo

    You can play the card game blackjack such that your probability of winning a hand is p = 0.51 and of losing is 1 − p = 0.49. You start with N betting units, the stake; and you bet 1 unit on each hand. The goal of this problem is to estimate the threshold N such that you are more likely to break the bank than to lose the stake.

    Let xn be your balance after the nth hand. Thus, \(x_{0}=N\). Losing your stake (the N units) corresponds to x = 0. To estimate N, extend the random-walk model to account for drift: that the probabilities of moving left and right are not equal.

    a. What is the symbolic expression for breaking the bank?

    b. Sketch your expected balance ⟨x⟩ versus the number of hands n (on linear axes).

    c. Sketch, on the same axes, the dispersion xrms versus n.

    d. Explain graphically “a significant probability of breaking the bank.”

    e. Thus, estimate the required stake ��.

    7.3.3.2 Baking

    A skillet on a stove gets heated from one side. An equally important kind of cooking, and one that helps us practice and extend our lumping model of random walks, is heating from two sides: baking. As an example, imagine baking a slab of fish that is L = 1 inch (2.5 centimeters) thick.

    How long should it bake in the oven?

    A first, and quick, analysis predicts \(L^{2}/\kappa\), the characteristic time for heat diffusion, where \(\kappa\) is the thermal diffusivity of water (organisms are mostly water). However, this simple model predicts an absurd time:

    \[t \sim \frac{L^{2}}{\kappa_{water}} \approx \frac{(2.5 cm)^{2}}{1.5 \times 10^{-3} cm^{2} s^{-1}} \sim 70 \textrm{ minutes.}\]

    After more than an hour in the oven, the fish will be so dry that it might catch fire, never mind not being edible. The model also ignores an important quantity: the oven temperature. Fixing this hole in the model will also improve the time estimate.

    Why does the oven temperature matter?

    The inside of the fish must get cooked, which means that the proteins denature (lose their folded shape) and the fats and carbohydrates change their chemistry enough to become digestible. This process happens only if the food gets hot enough. Thus, a cold oven could not cook the fish, even after the fish reached oven temperature. What temperature is hot enough? From experience, a thin piece of meat on a hot skillet (at, say, 200C) cooks in less than a minute. At the other extreme, if the skillet is at 50C, unpleasantly hot to the touch but not much hotter than body temperature, the meat never cooks. A round, intermediate temperature of 100 ∘C, enough to boil water, should be enough to cook meat thoroughly.

    If we set the oven to, say, 180C, the fish will then be cooked once its center reaches the midpoint of the room (20∘C) and oven temperatures, which is 100∘C. In this improved model of cooking, the interior of the fish starts at Troom ≈ 20C, and the oven holds the top and bottom surfaces of the fish at Toven = 180C (360F). With this model, and using triangular temperature profiles, we will estimate the time required for the center of the fish to reach the midpoint temperature of 100C.

    clipboard_eaa5f5ae71c9104265a556c6086011af2.png

    Two triangular hot zones, one from each surface, move toward the center of the fish. Before the zones meet, the center of the fish, at \(z = L/2\), is cold (room temperature). The fish, unless it is very fresh, is not ready to eat. But soon the triangles will meet.

    clipboard_ec4c014feae2f1f14e23414d90fa3ed41.png

    Each triangle extends a distance \(2 \sqrt{\kappa t}\). They first meet (at \(x = L/2\)) when \(2 \sqrt{\kappa t} = L/2\). Thus, \(t_{meet} \sim L^{2}/16 \kappa\). From that moment, the center warms up as the triangle fronts overlap ever more. The fish is cooked when the center reaches the mean of the room and oven temperatures (which is 100°C).

    clipboard_e61c0936c200af4d18166b678e98966d6.png

    In dimensionless temperature units, where Troom corresponds to \(\overline{T} = 0\) and Toven to \(\overline{T} = 1\), the cooking criterion is \(\overline{T} = 1/2\). Each triangle wave therefore contributes \(\overline{T} = 1/4}\). The left triangle, representing the hot front invading from the bottom surgace, then passes through the points (z=0, \: \overline{T} =1\)) and (\(z = L/2, \: \overline{T} = 1/4\)). It reaches the z axis (\(\overline{T} = 0\)) when \(z = 2L/3\). The corresponding triangle-zone diffusion time is given by \(2 \sqrt{\kappa t} = 2L/3\), whose solution is

    \[t_{cooked} \sim \frac{L^{2}}{9 \kappa}.\]

    clipboard_e12edda94d36d5e123f79d91d6b963c02.png

    For our 2.5-centimeter-thick fish, the time is roughly 7 minutes:

    \[t \sim \frac{1}{9} \times \frac{(2.5 cm)^{2}}{1.5 \times 10^{-3} cm^{2} s^{-1}} \sim 7 \textrm{ minutes.}\]

    This estimate is reasonable. My experience is that a fish fillet of this thickness requires about 10 minutes in a hot oven to cook all the way through (and further baking only dries it out).

    Exercise \(\PageIndex{9}\): Baking too long

    In the model of two approaching triangle hot fronts, the central temperature starts to rise at \(t \sim L^{2}/16 \kappa\), and it reaches the halfway temperature at \(t_{cooked} \sim L^{2}/9 \kappa\). When does it reach the oven temperature? Sketch the central temperature versus time, labeling interesting values.

    Exercise \(\PageIndex{10}\): Cooking an egg

    Baking a 6-kilogram turkey requires, from experience, 3 to 4 hours (you get to predict this time in Problem 7.34). Use proportional reasoning to estimate the time required to boil an egg.

    7.3.4 Boundary layers

    In cooking, the hot zone diffuses inward from the hot surface. For the most tangible example of this random walk, rub your finger on the blade of a (stationary!) window fan. Your finger comes away dusty and leaves a dust-free streak on the blade. But why is any dust on the blade at all? When the fan was turning, why didn’t the air streaming by the blade blow off the dust?

    The answer lies in the concept of the boundary layer. For the cooking examples (Sections 7.3.3.1 and 7.3.3.2), this layer is the expanding hot zone. It arises from the boundary constraint (the temperature), which diffuses into the skillet or the fish. For the fan blade, the analogous constraint is that, next to the blade, the fluid has zero velocity with respect to the blade. The condition, called the no-slip boundary condition, has the justification that the fluid molecules next the surface get caught by the inevitable roughness at the surface. (For a historical and philosophical discussion of the subtleties of this boundary condition, see Michael Day’s article on “The no-slip condition of fluid dynamics” [8].)

    Starting at the blade surgace, a zero-speed or, equivalently, zero-momentum zone diffuses into the fluid--just as the stove temperature diffuses into the skillet or the oven temperature into the fish fillet. After a growth time t, the zero-momentum front has diffused a distance \(\delta \sim \sqrt{\nu t}\), where \(\nu\) is the diffusion constant for momentum (the kinematic viscosity). The distance \(\delta\) is the boundary-layer thickness. Within the boundary layer, the fluid moves more slowly than the fluid in the stream. Using a rectangular lumping picture, the fluid speed is zero within the layer and full speed outside it. Therefore, dust particles entirely in the layer remain on the blade.

    To estimate the boundary-layerthickness, imagine a window fan with a blade width l and rotation speed \(v\) at the widest part of the blade. The growth time is \(t \sim l/v\). For a generic window fan sweeping out a diameter of 0.5 meters, the blade width l may be roughly 0.15 meters. If the fan rotates at roughly 15 revolutions per second or \(\omega \sim 100\) radians per second, the blade speed \(v\) is 10 meters per second:

    \[v \sim \underbrace{0.1m}_{\textrm{arc radius}} \times \underbrace{100 \textrm{ rad s}^{-1}}_{\omega} = 10 ms^{-1}.\]

    clipboard_edf82e7b8412d1085722cb56a74dd16d1.png

    At this speed, the growth time t is 0.015 seconds. In that time, the zero-momentum constraint diffuses roughly 0.5 millimeters from the fan blade:

    \[\delta \sim (\underbrace{0.5 cm^{2} s^{-1}}_{\nu_{air}} \times \underbrace{0.015s}_{t})^{1/2} \approx 0.05 cm.\]

    Dust particles that are significantly smaller than 0.5 millimeters feel no air flow (in this simplest lumping picture) and thus remain on the fan blade until your finger rubs them off. For the same reason (boundary layers), simply rinsing dirty dishes in water will not remove the thin layer of food next to the surface. Proper cleaning requires scrubbing (a sponge) or soap (which gets underneath the food oils).

    Exercise \(\PageIndex{11}\): Reynolds number in the boundary layer

    Based on the boundary-layer thickness \(\delta \sim \sqrt{\nu t}\), estimate the Reynolds number in a boundary layer on an object of size L (a length) moving in a fluid at speed \(v\).

    Exercise \(\PageIndex{12}\): Turbulence in the boundary layer

    At Reynolds numbers comparable to 1000, flows usually become turbulent. Use Problem 7.23 to estimate the main-flow Reynolds number at which the boundary layer becomes turbulent. For a smooth ball similar in size to a golf ball, what flow speed is required? Suggest an explanation for the dimples on golf balls.


    This page titled 7.3: Random walks - Viscosity and heat flow is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .

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