phys 369 09 hw 9 solutions

phys 369 09 hw 9 solutions - Phys. 369, Fall 2009 Homework...

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1 Phys. 369, Fall 2009 Homework #9 Solutions 1. (14 points) a) Equation (7) of Einstein’s paper gives the diffusion coefficient for spheres of radius P diffusing through a fluid of viscosity k. As written by Einstein, it reads 1 6 RT D Nk P π = . ( 1 ) Reading back through the paper, we see that R is the gas constant, T the temperature, N Avogadro’s number, k the viscosity of the fluid, and P the radius of the particle. Translating into the notation of Tipler, this equation reads 1 6 A RT D Na πη = ( viscosity η = and sphere radius. a = ) (2) b) The probability distribution function is given by eq. (10) of Einstein’s paper 2 4 1 (,) 4 x Dt e fx t Dt = ( 3 ) Here, I have set the number of particles to 1, n = so that the distribution function is normalized to 1 rather than to the number of particles n as in Einstein’s paper. _____________________________________________________________________________ You will receive full credit for your solution to this part if you just copy down this result from Einstein’s paper as shown above. However, you should study and understand the following solution to the diffusion equation. The problem we are trying to solve is the following. We suppose that the position of a particle has been accurately measured at time 0, t = and we set the origin of the spatial coordinate, 0 x = , to be equal to this measured position. Therefore the distribution function must be (,0 ) () f xx δ = at time 0. t = For later times the particle distribution function (,) f xt must obey the diffusion equation 2 2 f f D tx ∂∂ = ( 4 ) As time evolves, the particle is buffeted by many collisions and therefore undergoes a random walk process. It follows that the distribution function at later times t must be a normalized Gaussian, and that the standard deviation σ of the distribution should grow as the square root of the time. (See the solution to problem 3.) So, we can say that f must be of the form 2 2 2 2 2( ) 2 1/2 11 ) 2 x x t A t t e e tA t πσ == ( 5 )
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2 where 1/2 () tA t σ = is the standard deviation, and A is a constant. x δ is any function that (i) has unit area, 1 x dx −∞ = , and (ii) has value zero everywhere except within an infinitesimal displacement from 0. x = We see that 2 2 2 1/2 0 1 (,0 ) l im 2 x At t f xe x At π == ( 6 ) since it has properties (i) and (ii). So it matches the required initial condition. To verify that eq. (5) solves the diffusion equation, we’ll need the following derivatives: 2 22 2 2 2 2 2 1/2 3/2 1/2 2 2 2 2 2 11 1 1 1 1 1 1 2 2 x xx x fe x ee t t At At At A t ex At A t ππ −− ⎛⎞ ⎜⎟ ∂∂ + ⎝⎠ =− ( 7 ) and 2 2 2 2 2 2 2 2 2 21 / 2 1 / 2 2 2 2 2 1/2 2 2 3 3/2 2 1 2 2 1 1 1 x x x x e x A t A t x A t x A t A t A t =+ = ( 8 ) Substituting from eq. (8) and (9) into eq. (4) gives 2 23 / 2 2 3 3 / 2 2 2 ff e x e x DD t x A t =⇒ = 2 2 2 DA D A ⇒= ( 9 ) So that 2 2 2 24 1/2 1/2 (,) x x A tD t fx t e e At D t ( 1 0 ) is the solution to the diffusion eq. (4) with initial condition (,0 ) . f = This agrees with the result given by Einstein.
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3 c) For your experiment, the radius of the particles is 4 diameter/2 0.25 microns 0.25 10 cm.
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phys 369 09 hw 9 solutions - Phys. 369, Fall 2009 Homework...

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