User:Brian P. Josey/Notebook/2010/07/09: Difference between revisions
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== | ==Ch 4 Random Walks, Friction, and Diffusion== | ||
Continuing from my notes yesterday. | |||
===Random Walks and Diffusion=== | |||
'''Random Walk''' is a process where something moves in steps, where each step is determined by random chance. For example, if you have a pawn on a chess board, and you have two coins that you flip. Depending on how one the coins lands, you move it either to the left or right one square, and similarly up and down based on the second coin. It then becomes impossible to determine exactly where the pawn will move before flipping each coin, and impossible to determine exactly where the pawn will be after n-number of turns. Essentially, you cannot say anything exact about its motion, and looking at its history will not help you determine where it will end up. However, you can say probabilistically where it could end up. | |||
'''Diffusion Law''' is a mathematical law that predicts the displacement of a particle after a number of steps in a random walk process. Using the example of a marker being moved back and forth depending on how a coin lands after flipping, the expectation value of the location of the marker is given by: | |||
<math> | |||
\langle (x_N)^2 \rangle = \langle (x_{N-1} + k_N L)^2 \rangle = \langle (x_{N-1})^2 \rangle +2L \langle x_{N-1} k_N \rangle + L^2 \langle (k_N)^2 \rangle | |||
</math> | |||
Where: | |||
* '''x<sub>i</sub>''' is the displacement of the marker after '''i''' number of steps | |||
* '''L''' is the length of each step, assumed to be constant for both forward and backwards and each iteration | |||
* '''N''' is the total number of steps | |||
* '''k<sub>N</sub>''' is either +1 or -1, representing a movement in the positive or negative direction depending on how the coin lands, notice the last term becomes just L<sup>2</sup> in the above equation | |||
This can then be generalized to: | |||
<math> | |||
\langle (x_N)^2 \rangle = NL^2 | |||
</math> | |||
Then if we wait a time '''t''', the marker will move N=t/Δt steps, where Δt is the time between steps. Then by introducing a constant, '''D''', called the diffusion constant, we get the diffusion law for one-dimension, which states: | |||
* The mean-square displacement increases linearly in time for a one-dimensional random walk given by: | |||
<math> | |||
\langle (x_N)^2 \rangle = 2Dt | |||
</math> | |||
* where the constant '''D''' equals L<sup>2</sup>/(2Δt) | |||
But it is very important to remember that the brackets indicate an average, and any particular example will not conform exactly to the diffusion law, instead the average over a large number of examples will conform to the law. This law then can be generalized for higher dimensions. To do this, x is swapped out with the position vector, and using the fact that the step size in a 2 dimensional diagonal is actually <math> L \sqrt {2} </math>, the diffusion law in two-dimensions is: | |||
<math> | |||
\langle (\vec r_N)^2 \rangle = 4Dt | |||
</math> | |||
and similarly, in three-dimensions it is: | |||
<math> | |||
\langle (\vec r_N)^2 \rangle = 6Dt | |||
</math> | |||
This diffusion law, however, is fairly simplistic, and in the real world random walks don't take discrete steps of uniform size, there is a continuum of step sizes. Fortunately, the ''diffusion law is universal,'' as long as there is some distribution of random, independent steps. Working in one-dimension, given a set of numbers P<sub>k</sub>, the probabilities of taking steps of length kL, where k is an integer. The length k<sub>j</sub> of the ''j''th step can be negative or possitive, we can calculate ''u'' which is the mean value of ''k<sub>j</sub>'': | |||
<math> | |||
u= \langle k_j \rangle = \sum {k} kP_k | |||
</math> | |||
This value, ''u'', then describes the average drift motion that is superimposed on the random walk. This can then be put into the above equation used to describe the distance traveled by the marker (top equation on the page) to find the mean position of the marker: | |||
<math> | |||
\langle x_N \rangle = \langle x_{N-1} \rangle + L \langle k_N \rangle = \langle x_{N-1} \rangle + uL= NuL | |||
</math> | |||
Then by measuring the variance in the motion, omitted to save space and time, the variance in the mean displacement is found by: | |||
<math> | |||
variance (x_N) = 2Dt | |||
</math> | |||
where | |||
<math> | |||
D= \frac {L^2} {2 \delta t} \times variance(k) | |||
</math> | |||
If the drift, ''u'', is zero, then the equation just simplifies to the simpler diffusion law given above. From that, it is clear that the diffusion law is model independent. | |||
===Friction=== | |||
Diffusion is essentially the result of random fluctuations, and it can be argued that friction is also the result of the same interactions. As a simple example, imagine a body, mass ''m'', moving in one dimension, denoted by ''x'' under a constant force, ''f''. As this body moves, it collides with other particles once every ''Δt'', and given the starting value of the velocity as ''v<sub>0,x</sub>''. The uniformally accelerated motion of the particle can then be described by: | |||
<math> | |||
\Delta x= v_{0,x} \Delta t + \frac {f} {2m} (\Delta t)^2 | |||
</math> | |||
Since each collision is independent of all other collisions, then each step, ''v<sub>0,x</sub>'', is either positive or negative and it's average value <math> \langle v_{0,x} \rangle </math> is zero. Then the average displacement from each collision is then <math> \langle \Delta x \rangle = (f/2m)(\Delta t)^2 </math>. Even though the particle is buffered by random collisions, it still neverless has a net drift velocity equal to the average change in position over the time between collisions or: | |||
<math> | |||
v_{drift}=\frac {f} {\zeta} | |||
</math> | |||
where | |||
<math> | |||
\zeta= \frac {2m} {\Delta t} | |||
</math> | |||
This ''ζ'' is the '''viscous friction coefficient''' and the above example illustrates that a particle under a constant force does come to a terminal velocity that is proportional to the force, keeping consistent with observations of larger bodies. And with this it is possible to measure the viscous friction coefficient of a fluid experimentally by just measuring how long it takes for a particle to settle under gravity. However, the viscous friction coefficient is also related to the size of the particle. For a circular particle with radius ''R'', it gives us Stokes equation: | |||
<math> | |||
\zeta = 6 \pi \eta R | |||
</math> | |||
and plugging in the relationship between the drift velocity, force and zeta I get: | |||
<math> | |||
v_{drift} f = 6 \pi \eta R | |||
</math> | |||
which has made more than its fair share of appearances in my notebook. Of course, ''μ'' represents the viscosity. From these it is possible to measure the mass of a particle, as long as you understand how to calculate the time between collisions, and the length of the collisions. Einstein did this by observing that ''(L/Δt)<sup>2</sup> = (v<sub>0,x</sub>)<sup>2</sup> '', and given that the ideal gas laws states that: | |||
<math> | |||
\langle (v_{0,x})^2 \rangle = \frac {k_B T} {m} | |||
</math> | |||
it then comes down to Einstein's relation: | |||
<math> | |||
\zeta D =k_B T | |||
</math> | |||
Which all assembled together allows for the mass of molecules to be estimated. | |||
'''Some Notes on What I've Written So Far''' | |||
* When I assumed that the collisions wiped away all memory of all older collisions, I was assuming the initial momentum of the particle was comparable to the momentum of the molecules striking it. A bullet fired from a gun will not come to a sudden stop if fired into water simply because one fast moving water molecule hit it in the sweet spot. It will however slow down as many particles hit it. | |||
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Ch 4 Random Walks, Friction, and DiffusionContinuing from my notes yesterday. Random Walks and DiffusionRandom Walk is a process where something moves in steps, where each step is determined by random chance. For example, if you have a pawn on a chess board, and you have two coins that you flip. Depending on how one the coins lands, you move it either to the left or right one square, and similarly up and down based on the second coin. It then becomes impossible to determine exactly where the pawn will move before flipping each coin, and impossible to determine exactly where the pawn will be after n-number of turns. Essentially, you cannot say anything exact about its motion, and looking at its history will not help you determine where it will end up. However, you can say probabilistically where it could end up. Diffusion Law is a mathematical law that predicts the displacement of a particle after a number of steps in a random walk process. Using the example of a marker being moved back and forth depending on how a coin lands after flipping, the expectation value of the location of the marker is given by: [math]\displaystyle{ \langle (x_N)^2 \rangle = \langle (x_{N-1} + k_N L)^2 \rangle = \langle (x_{N-1})^2 \rangle +2L \langle x_{N-1} k_N \rangle + L^2 \langle (k_N)^2 \rangle }[/math] Where:
This can then be generalized to: [math]\displaystyle{ \langle (x_N)^2 \rangle = NL^2 }[/math] Then if we wait a time t, the marker will move N=t/Δt steps, where Δt is the time between steps. Then by introducing a constant, D, called the diffusion constant, we get the diffusion law for one-dimension, which states:
[math]\displaystyle{ \langle (x_N)^2 \rangle = 2Dt }[/math]
But it is very important to remember that the brackets indicate an average, and any particular example will not conform exactly to the diffusion law, instead the average over a large number of examples will conform to the law. This law then can be generalized for higher dimensions. To do this, x is swapped out with the position vector, and using the fact that the step size in a 2 dimensional diagonal is actually [math]\displaystyle{ L \sqrt {2} }[/math], the diffusion law in two-dimensions is: [math]\displaystyle{ \langle (\vec r_N)^2 \rangle = 4Dt }[/math] and similarly, in three-dimensions it is: [math]\displaystyle{ \langle (\vec r_N)^2 \rangle = 6Dt }[/math] This diffusion law, however, is fairly simplistic, and in the real world random walks don't take discrete steps of uniform size, there is a continuum of step sizes. Fortunately, the diffusion law is universal, as long as there is some distribution of random, independent steps. Working in one-dimension, given a set of numbers Pk, the probabilities of taking steps of length kL, where k is an integer. The length kj of the jth step can be negative or possitive, we can calculate u which is the mean value of kj: [math]\displaystyle{ u= \langle k_j \rangle = \sum {k} kP_k }[/math] This value, u, then describes the average drift motion that is superimposed on the random walk. This can then be put into the above equation used to describe the distance traveled by the marker (top equation on the page) to find the mean position of the marker: [math]\displaystyle{ \langle x_N \rangle = \langle x_{N-1} \rangle + L \langle k_N \rangle = \langle x_{N-1} \rangle + uL= NuL }[/math] Then by measuring the variance in the motion, omitted to save space and time, the variance in the mean displacement is found by: [math]\displaystyle{ variance (x_N) = 2Dt }[/math] where [math]\displaystyle{ D= \frac {L^2} {2 \delta t} \times variance(k) }[/math] If the drift, u, is zero, then the equation just simplifies to the simpler diffusion law given above. From that, it is clear that the diffusion law is model independent. FrictionDiffusion is essentially the result of random fluctuations, and it can be argued that friction is also the result of the same interactions. As a simple example, imagine a body, mass m, moving in one dimension, denoted by x under a constant force, f. As this body moves, it collides with other particles once every Δt, and given the starting value of the velocity as v0,x. The uniformally accelerated motion of the particle can then be described by: [math]\displaystyle{ \Delta x= v_{0,x} \Delta t + \frac {f} {2m} (\Delta t)^2 }[/math] Since each collision is independent of all other collisions, then each step, v0,x, is either positive or negative and it's average value [math]\displaystyle{ \langle v_{0,x} \rangle }[/math] is zero. Then the average displacement from each collision is then [math]\displaystyle{ \langle \Delta x \rangle = (f/2m)(\Delta t)^2 }[/math]. Even though the particle is buffered by random collisions, it still neverless has a net drift velocity equal to the average change in position over the time between collisions or: [math]\displaystyle{ v_{drift}=\frac {f} {\zeta} }[/math] where [math]\displaystyle{ \zeta= \frac {2m} {\Delta t} }[/math] This ζ is the viscous friction coefficient and the above example illustrates that a particle under a constant force does come to a terminal velocity that is proportional to the force, keeping consistent with observations of larger bodies. And with this it is possible to measure the viscous friction coefficient of a fluid experimentally by just measuring how long it takes for a particle to settle under gravity. However, the viscous friction coefficient is also related to the size of the particle. For a circular particle with radius R, it gives us Stokes equation: [math]\displaystyle{ \zeta = 6 \pi \eta R }[/math] and plugging in the relationship between the drift velocity, force and zeta I get: [math]\displaystyle{ v_{drift} f = 6 \pi \eta R }[/math] which has made more than its fair share of appearances in my notebook. Of course, μ represents the viscosity. From these it is possible to measure the mass of a particle, as long as you understand how to calculate the time between collisions, and the length of the collisions. Einstein did this by observing that (L/Δt)2 = (v0,x)2 , and given that the ideal gas laws states that: [math]\displaystyle{ \langle (v_{0,x})^2 \rangle = \frac {k_B T} {m} }[/math] it then comes down to Einstein's relation: [math]\displaystyle{ \zeta D =k_B T }[/math] Which all assembled together allows for the mass of molecules to be estimated. Some Notes on What I've Written So Far
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