Diffusion equation question in 1D?

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SUMMARY

The discussion revolves around solving the 1D diffusion equation represented by the formula n'(x,t) = N/sqrt(4piDt) * exp(-x^2/4DT), where n'(x,t) denotes particle concentration, N is the total particle count, and D is the diffusion coefficient. Participants address various problems related to the diffusion equation, including calculating particle distribution in a slab, determining probabilities, and deriving the mean square displacement as 2Dt. The conversation also touches on Gaussian distributions and normalization of probability density functions (PDFs) in the context of diffusion.

PREREQUISITES
  • Understanding of the diffusion equation in one dimension
  • Familiarity with Gaussian distributions and their properties
  • Knowledge of integration techniques in calculus
  • Basic concepts of probability theory and random walks
NEXT STEPS
  • Study the derivation of the diffusion equation and its applications in physics
  • Learn about Gaussian distribution normalization and its significance in probability
  • Explore the concept of mean square displacement in random walks
  • Investigate the combinatorial aspects of random walks and their implications in statistical mechanics
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Students in physics, particularly those studying statistical mechanics, mathematicians interested in probability theory, and researchers working on diffusion processes and random walks.

magicuniverse
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Homework Statement



The solution to the diffusion equation in 1D may be written as follows:

n'(x,t) = N/sqrt(4piDt) * exp(-x^2/4DT)

where n'(x,t) is the concentration of the particles at position x at time t, N is the total number of particles and D is the diffusion coefficient.


a) Write down an expression for the number of particles in a slab of thickness dx located at position x.

b) What is the probability that a particle is in the slab?

c) If the particle is in the slab, what distance has it traveled from the origin?

d) Show that the mean square displacement of a particle is 2Dt.

e) Sketch the form of the solution to the diffusion equation for two times t1 and t2 where t2 = 4t1. How will the width of the curves be related at these two times?

f) Consider a 1D random walk where the particle had equal possibilities of moving from left to right. For a journey of 3 steps, a particular sequence of steps might be, for example RRR. Write down all possible sequences for a yourney of 3 steps.

g) What is the total number of journeys possible for a walk of N steps?

h) Write down an expression for the number of journeys in which exactly NL steps are taken to the left if the total number of steps is N. If the random walker starts at the origin, after 4 steps what is the probability that the particle has returned to the origin?


Homework Equations



At the start N.A. as I go through the question I know that I need standard integrals and the combinations formula but ill raise these when needed.


The Attempt at a Solution



I am struggling with much of this but can do some of it. I think the most sensible thing to so it to go through it in order.

So I don't know how to integrate the first function between x+dx and x! Any ideas? thanks
 
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magicuniverse said:
So I don't know how to integrate the first function between x+dx and x! Any ideas? thanks

The integral of any function f(x) between x and x+dx is nothing but f(x)dx.

Hope this starts you off.
 
Note that the integrale of any function f(x) between x and x +dx is simply f(x) dx!

\int_x^{x+dx} f(x') dx' \approx f(x) dx
 
So = N/sqrt(4piDt) * exp(-x^2/4DT)dx ?

If that's right I don't know how to do the sencond part. I know that since expx^2 term the distirbution is gaussian and that it can be normalised to 1. However what do i get for an answer?
 
someone please help!
 
magicuniverse said:
I know that since expx^2 term the distirbution is gaussian and that it can be normalised to 1. However what do i get for an answer?

If after normalization the PDF is P(x), then the reqd probability is again nothing but P(x)dx.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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