# Beers Law with Spherical Symmetry

1. Aug 20, 2009

### John Creighto

In essence, Beers law describes the behavior of a quantity where the probability of a unit of that quantity over a distance changing follows an exponential distribution. This can be for instance, the probability of a particle being absorbed or scattered over a distance x.

Problems of spherical symmetry are interesting, because they could describe many things, like neutrons escaping from some particle mixture, photons escaping from the sun or radiative cooling of an object. I've created an excel spread sheet to try and numerically calculate the probability of particles emitted from one shell of a sphere being scattered by particles in another shell. I conjecture, that the fraction of the particles scattered by shell B which are emitted by shell A should follow the equation:

$$\frac{V_B}{V_A}\frac{w_B}{L_M}exp(-abs(r_B-r_A)/L_M)$$

where:

$$V_A$$ is the volume of shell A
$$V_B$$ is the volume of shell B
$$r_A$$ is the radius of shell A
$$r_B$$ is the radius of shell B
$$w_B$$ is the width of shell B
$$L_m$$ is the mean free path

My conjecture puts me off by about a factor of 2 in comparison to my numeric calculations. My numeric calculations tell me that twice this much is absorbed. My numeric calculations seem correct because, if I sum up the total fraction absorbed I get 1. What am I missing in my logic? I would think for a thin shell, half the particles should go in each direction and the amount absorbed should be roughly, the width of the shell divided by the mean free path. Any Suggestions? Bellow is my code:

Code (Text):
Function pAbsorb(r1, r2, dr1, dr2, Lm, Optional No) As Double
'r1 the radius of the sphere where the molecule is emitted from
'r2 the radius of the sphere where the molecule is absorbed
'dr1 the width of the spherical shell where the molecule is emitted from
'dr2 the width of the spherical shell where the molecule is absorbed
'Lm The mean free path
'N The number of Steps per Lm
NTau = 4  ' Number of decay constants to integrate over
If IsMissing(No) Then
No = 20

End If
Pi = WorksheetFunction.Pi()
'(x1,y1,z1) The point where the molecule is emitted
'(x2,y2,z2) The Point where the molecules are absorbed
x2 = r2 * Cos(0) * Sin(0)
y2 = r2 * Sin(0) * Cos(0)
z2 = r2 * Cos(0)
V1 = 4 * Pi * r1 ^ 2 * dr1
V2 = 4 * Pi * r2 ^ 2 * dr2
pAbsorb = 0

If Abs(r1 - r2) < NTau * Lm Then

If r1 + r2 < NTau * Lm Then
maxLat = 180
Else
maxLat = (180 / Pi) * WorksheetFunction.Acos((r1 ^ 2 + r2 ^ 2 - (NTau * Lm) ^ 2) / (2 * r1 * r2))
End If
Dim NLat, NLong As Integer
NLat = WorksheetFunction.Max(10 * r1 / (Lm / No) * (maxLat) / 180, 40)
'  NLong = NLat
For i = 1 To NLat

'     For j = 0 To NLong
'       longitude = j * 360 / NLong - 180 'phi

lattitude = i * maxLat / NLat ' theat (Note, This give north pole as zero lattitude and south pole as 180)
'dV=r^2 sin(theata) dr dtheata dphi
dv1 = Abs(r1 ^ 2 * Sin(lattitude * Pi / 180) * dr1 * (maxLat / NLat * Pi / 180) * (2 * Pi))

'dv2 = r2 * Sin(lattidude) * dr2 * (180 / N) * (360 / N)
'The point where the molecules are emitted
x1 = r1 * Cos(0) * Sin(lattitude * Pi / 180)
y1 = r1 * Sin(0) * Cos(lattitude * Pi / 180)
z1 = r1 * Cos(lattitude * Pi / 180)
dx = Sqr((x1 - x2) ^ 2 + (y1 - y2) ^ 2 + (z1 - z2) ^ 2)
pAbsorb = pAbsorb + Exp(-dx / Lm) * (V2 / Lm) / (4 * Pi * dx ^ (2)) * dv1 / V1

'    Next j
Next i
Else
pAbsorb = 0
End If
End Function

As a side note, I tackled this problem via integration, alternatively I think it would also possible to express the problem in terms of partial differential equations. See my post:

Continuity Like Equation

#### Attached Files:

• ###### Beer-Law.xls
File size:
87.5 KB
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Last edited: Aug 20, 2009