# Homework Help: Mathematical physics, density of probability

1. May 20, 2012

### Jamalll

1. The problem statement, all variables and given/known data
So I have to calculate the transmission of light, which goes through random black and "transparent" blocks (p=0.5), but the thickness (x) of each block is distributed exponentially
dP/dx=1/t*exp(-x/t). When light goes through black block, it diminishes (I stands for intensity):
I(x)=I(0)exp(-cx), c is some constant, when it goes through transparent block I stays unchanged.

How can we get, how intensity is dependent upon the whole thickness, D, because there are many blocks for a certain D? I(D)=?

2. Relevant equations
dP/dx=1/t*exp(-x/t)
I(x)=I(0)exp(-cx)

2. May 21, 2012

### clamtrox

Re: from mathematical physics,

So if we denote the intensity after 1 block as I1 = I0 exp(-c1 x1), 2 blocks I2 = I1 exp(-c2 x2), etc... Then what's the intensity after n blocks?

3. May 21, 2012

### Jamalll

Re: from mathematical physics,

well not exactly:
when it goes through one block (length x1), it diminishes
I=I(o)exp(-c*x1)
when it goes through another (x2)
I=I(o)exp(-c*x2)...etc

but in half cases there are transparent blocks, and
the lengths (x1, x2, ....) are distributed exponentially

I know this one is not easy, I hope someone can figure it out...

4. May 21, 2012

### clamtrox

Re: from mathematical physics,

If the initial intensity is I(o), then after 1 block it's I(1)=I(0)exp(-c*x1), and after 2 blocks I(2)=I(1)exp(-c*x1). What is this in terms of I(0)? How about I(n)?

If you hope that someone is going to do the problem for you, then you will be disappointed! You will need to work it out yourself, we will only nudge you to the correct direction...

5. May 21, 2012

### Jamalll

Re: from mathematical physics,

@clamtrox

really appreciate your replies, but it is not laziness that got me here. I have been working on this for months (really!), and I came with one solutionof convolution of densities of probability
in the sense
l= thickness

if dP/dl= g(l)=k*exp(-k*l)......this is desnity of probability for thickness l
I=I(0)exp(-c*l)......transmission

Because there is 50/50 chance of transparent/black block, there are many ways
the overall "black" path is of length lc, this is writen as

w(lc)= integral(g(x1)*g(x2)*....*g(lc-x1-x2))dx1dx2... this is convolution which goes towards gauss if there are many ....

w(lc)=g(lc, D/2, sqrt(D/(2k)))......this is convolution theorem

so now

I=I(0) integral(from 0 to thickness D) (exp(-lc*c)*gauss (lc, D/2, sqrt(D/(2k))))

but this is somehow wrong, because graph comes out wrong

HOW WOULD ONE CALCULATE THIS IN TERMS OF n AS YOU SAID? Could this be done diffrently? thanks

6. May 21, 2012

### Jamalll

Re: from mathematical physics,

I actually got a tip (quite reliable), that the easiest way is to start with

x....is the thickness of the "whole thing"
a....somewhere between 0 and x, it could be either black block or transparent one

I(x)=Ʃ(w(a)*exp(-c*a)*I(x-a))+W(x)*exp(-c*x)

the sum goes through all black and transparent blocks
the W(x) is probability that it "goes out" or something

I really don't understand this, but it is supposed to somehow go in this way...

7. May 21, 2012

### clamtrox

Re: from mathematical physics,

That's kind of where I was going :) So let's suppose you have n different blocks. Then
I(n) = I(n-1)exp(-cn*xn) = I(n-2)exp (-cn-1*xn-1 -cn*xn) = ... = I(0) exp(- Ʃ ci xi), where ci is either c or 0... Does this make sense?

Now, if for a given random realization of the block structure, we have Ʃ ci xi = c*a, ie. the sum taken over black blocks is a, then I = I(0) exp(-ca). This is the first important thing to note. Now, our block structure is random, so we need to take some sort of expectation value over this quantity:

<I> = 1/N Ʃ a I(a) ρ(a),
where ρ(a) is the probability of a given length a, and N is the normalization, N = Ʃ ρ(a)

8. May 22, 2012

### Jamalll

Re: from mathematical physics,

well this is great, sounds it is the right path. However, the real problem lies in ρ(a).
If we have just distribution for each individual block thickness (x), say

w(x)= k*exp(-k*x),

how can we get probability, ρ(a), that the SUM over black blocks equals a?

9. May 22, 2012

### clamtrox

Re: from mathematical physics,

The number of black blocks follows binomial distribution, and the size of the blocks follows exponential distribution. Are you familiar with Poisson processes? If not, you might want to google that.

10. May 22, 2012

### Jamalll

Re: from mathematical physics,

exactly! number of black blocks follows binomial distribution which goes into Poisson distribution if the number of blocks is large (which we can take for granted), stil there is p=0.5 chance, but for Poissonian dist it has to be small enough so <N>=Z*p stays constant...so I am not sure about that...

But you are definitly on the spot about black blocks follow binomial distribution, and the size of the blocks follows exponential distribution. I just can't find how you can put those two together...

11. Jun 15, 2012

### Jamalll

1. The problem statement, all variables and given/known data

My problem is a problem of light going through
1: two different media, each with it's own absorption coefficient (let us say "black" ones have
μ and "transparent" have 0. )

2: when light passes through transparent media, it does not absorb, when it goes through "black", intensity diminishes like I=I(0)exp(-μd), where d is a thickness of "grain".

3: the question is, how is transmission I(x) dependent upon thickness of sample,x, if probability for thickness of each grain is exponential: ρ(d)=1/λexp(-λd)

2. Relevant equations

I=I(0)exp(-μd)
ρ(d)=1/λexp(-λd)

3. The attempt at a solution
We take a n-th grain, when light passes through n-th grain it diminishes by
I=I(n-1)exp(-μd), second grain: I=I(n-2)exp(-μd1)exp(-μd2)... and so on

let me remind you the thickness is exponentially distributed, and each grain has 50/50 chance of being transparent or black. I am stuck with this homework for a long time, and it is
1. The problem statement, all variables and given/known data

12. Jun 15, 2012

### Simon Bridge

The problem is under-specified: what is meant by "grain"?

Perhaps you have a solid block made out of granules of the two materials?
Perhaps you are shining light through a series of layers which may be one of two materials? (The second one sounds like it is closer to you description.)
Perhaps it is something else?

From what you have said - then for a very large number of grains, about half of them will be absorbing and the other half transparent. The net absorption will depend on the total thickness of the absorbing material present. If all the grains were equal size you'd expect (off the top of my head) about x/2 to absorb so I=I(0)exp(-μx/2) right?

But they are not equal thickness - how does the thickness distribution affect the calculation?

13. Jun 15, 2012

### Jamalll

Yes, you are right, let us say there are "blocks", layers of them.
But each layer is either "transparent" either "black" (this is binomially distributed)
AND the thickness of each layer (d) is distributed lie this: ρ(d)=1/λexp(-λd)

The question is, how is intensity that goes through dependent upon the whole thickness I(D).

14. Jun 15, 2012

### Jamalll

sstarting point should be of the form:

ths sum goes for black and transperent

I(D)=Ʃρ(d)exp(-μd)I(D-d)

can anyone solve this? This is from mathematical physics, we had poisson, binomial distributions etc...

15. Jun 15, 2012

### Ray Vickson

If I understand your problem correctly, it can be stated like this: we have n blocks, where each block is either transparent or absorbing, with probability 1/2 each, independent of other blocks. A single absorbing block has thickness that is exponentially distributed. The total output intensity is $I \equiv I_\text{out} = I_0 e^{-\mu T},$ where T is the total thickness of all absorbing blocks. (That is true because a transparent block does not alter the intensity, so we might as well lump all the absorbing blocks together in a single block.) Here, $T = \sum_{i=0}^K T_i,$ where K = number of absorbing blocks and the $T_i$ are independent, identically exponential random variables (the individual thicknesses). The random variable K = {0,1,2,...,n} is the number of absorbing blocks, and $$P(K = k) = {n \choose k} 1/2^n .$$ Note that K has a binomial distribution with parameters n and 1/2.

Given K = k > 0, T is the sum of k iid exponential random variables, and has a density $f_k(t).$ This density function is standard and can be found in any applied probability book; or you can search on-line. However, we need the probability density f(t) of T, which will be a mixture of the densities f_k, weighted by the binomial probabilities of the values of k. The result is doable, but only in terms of non-elementary functions (in this case, in terms of Whittaker M-functions, or hypergeometric functions). So, T has a mixed distribution, with point-mass at t = 0 having probability $2^{-n}$ (corresponding to all blocks transparent) and for t > 0 a density f(t) that is expressible in terms of hypergeometric functions. For large n we can neglect the point-mass at 0 and just take T to be continuous with density f(t). If we are given n and λ then we can determine values of f(t) numerically.

Anyway, if we somehow determine the density f(t) of T, the density of the output intensity is the density of the random variable $$I_0 e^{-\mu T},$$
so can be obtained from f by a standard change of variable formula for probability densities.

RGV

Last edited: Jun 15, 2012
16. Jun 15, 2012

### vela

Staff Emeritus