Physics 301 11-Dec-2002 35-1The Dispersion RelationWhen considering the diffusion equation in the last lecture, we found that an expansionin cosine components was quite useful. One might also ask under what conditions is a planewave a solution of the diffusion equation? Supposeτ (r,t)=τ0ei(k · r −ωt).Plug into the diffusion equation and get−k2τ0ei(k · r − ωt)+iωDτ0ei(k · r − ωt)=0,orDk2= iω .A relation between k and ω is called a dispersion relation. This comes from the fact thatif k and ω are not proportional then a wave pulse disperses (or spreads out).In the example of the equilibrating bar, the spatial parts of the solutions are composedof pure oscillations. This means k is real and k2is positive. The dispersion relation tellsus that ω is pure imaginary,ω = −iDk2.When plugged back into the plane wave solution, the negative i in ω times the negative iin the plane wave exponent gives −1, so we get an exponential decay in time.We’ll consider some other consequences of the dispersion relation later.Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 11-Dec-2002 35-2Random Walks and DiffusionA simple one-dimensional random walk consists of N steps. Each step is of length but may be to the left or right. The probability for either direction is the same, 1/2. Thequestion is, after N steps, how far from the origin (where you started) have you randomlywalked? The probability of n steps to the right and (N − n) steps to the left is just thebinomial distribution (see lecture 3)P (n)=Nn12N.The distance from the origin is the number of positive steps minus the number of negativesteps times the length per stepd =[n − (N −n)] =(2n − N).The average distance is 0. What’s the root mean square distance? d2 = (4 (n − N/2)22= 4(Np(1 − p) = √N.In other words, the mean position is still the origin, but the spread around the meanposition grows as the square root of the number of steps. Now suppose each step isaccomplished at speed ¯c. Since each step is of length , the time per step is /¯c.Iftherandom walk lasts for time t, the number of steps is N = t/(/¯c)=t¯c/)and d2 =√t¯c .Notice that what’s multiplying t inside the square root is basically the diffusivity! So, oursimple random walk satisfies d2 = Dt ,The variance in position is proportional to the time and the proportionality constant is D.What does this have to do with transport and the diffusion equation, you ask? Recall,our picture for transport is that a molecule travels (on the average) at speed ¯c for aboutonemeanfreepath. At that point (on the average), it has a randomizing collision andgoes off in a random direction. This is three dimensional, but other than that it’s therandom walk we’ve been discussing.Let’s consider again the example of the equilibrating bar. This time, we’ll think of itas energy that must randomly walk until equilibrium is achieved. That is, until the energyis sufficiently spread out (sufficiently large variance) that it appears uniform on the scaleof interest. Consider a mode with wave number kn= nπ/L. The characteristic distancefor this mode is k−1n. Our random walk model says that the energy will random walk tothis rms distance in time tngiven bytn=1Dd2n=L2Dn2π2,Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 11-Dec-2002 35-3and sure enough, tnis the decay time for mode n. This is a characteristic propertyof diffusion processes and random walks: doubling the length scale quadruples the timescale.Sample Solution: A Temperature OscillationSuppose we consider a slab of material which has a boundary at x = 0. The materialextends to x>0. We suppose everything is uniform in the y and z directions. Suppose thatthe x = 0 boundary is forced (by some external agent) to undergo a sinusoidal temperatureoscillation,τ (0,t)=τ0Ree−iωt.Since the diffusion equation is linear, the real part of a complex solution is a solution, sowe’ll usually drop the indication that we’re considering the real part and take it to beimplied.We want to know the temperature distribution, τ (x, t) in the slab. We’ll have a planewave propagating in the x direction. The wave number is determined by the dispersionrelation,k2=iωD.We can take the square root,k = ±√iωD= ±ω2D(1 + i)=±k1(1 + i) ,where k1= ω/2D is a handy abbreviation. So the solution isτ (x, t)=τ0e±ik1x ∓ k1x − iωt= τ0e∓k1xe±ik1x −iωt.So the solution is a plane wave (the last factor above) times an exponential decay (orgrowth) in the x-direction (the middle factor above).The exponential growth of the wave into the material is unphysical, and we must takethe upper sign to be sure the wave decays as we go into the slab. We have a wave, butit is damped with a damping length comparable to its wavelength. This means that thewave “doesn’t get very far” (in terms of wavelengths). K&K give some numerical examplesshowing that not much dirt is needed to insulate underground pipes from the day-nightcycle or the summer-winter cycle!Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 11-Dec-2002 35-4The Diffusion of a One Dimensional BumpWhen we looked at a one dimensional random walk, we found that the variance inposition grows in proportion to the time. We’ve also argued that microscopically, diffusionis just a random walk. So, we might expect that if we start with an excess of energyconcentrated at x = 0, it will diffuse away from the origin in such a way that the meanposition of the energy remains at the origin, but the variance of the location of the energygrows with time. Furthermore, since the macroscopic energy distribution is determined bymany small events, we might expect the distribution to be Gaussian. So, we guess thatthe one-dimensional temperature distributionτ (x, t)=1√2πate−x22at,might be a solution of the diffusion equation. Here, a is a constant that must have thesame dimensions as a diffusion constant. This distribution is a Gaussian, centered at 0,with variance at. To find out if this is a solution, we must plug into the diffusion equationfor the temperature. We find∂2τ∂x2=1√2πate−x2/2at−1at+x2a2t2,∂τ∂t=1√2πate−x2/2at−12t+x22at2,The one-dimensional diffusion equation for the temperature is∂2τ∂x2−1D∂τ∂t=0,which