aLIGO Surface Cleanliness Requirements

Absorption Requirements

COC DRD T000127-v2, Table 5, total cummulative HR surface absorption:
<1 ppm for ITM, ETM
<4 ppm for BS, RM
(where the HR coating absorption was assumed to be < 0.5 ppm)

TCS CO2P FDD T1100570-v3, section 3.2, TCS is designed for 0.5 ppm HR surface absorption of the ITM & ETM, with a safety factor of 2, so < 1 ppm HR surface absorption of TMs (consistent with the COC DRD)

Measurements of as-built coating absorption:
https://nebula.ligo.caltech.edu/optics/
0.3 ppm generally, max 0.4 ppm for ITM, ETM
≤ 0.7 ppm generally, max 1.0 ppm for PRM, PR2, PR3
≤ 0.6 ppm for BS

Consequently the allowable absorption due to contamination is:
≤ 0.7 ppm generally for ITM, ETM
≤ 3.3 ppm generally for BS, PRM, PR2, PR3
This allowable absorption must be allocated (budgeted) to both hydrocarbon deposition and particulates. Measurements to date (E1000193-x0, LIGO UHV Qualification Test Results) show no measureable absorption due to hydrocarbon film deposition, so assume:
≤ 0.1 ppm due to hydrocarbon film deposition
Consequently:
≤ 0.6 ppm for particulates on ITMs, ETMs
≤ 3.2 ppm for particulates on BS, PRM, PR2, PR3

N.B.: These allowable absorption levels are considerably higher than the value (≤0.1 ppm) assumed in section 2 of T080067-v1, “Protecting installed core optics from particulates”

Gaussian Laser beam at the test mass

w = beam waist (diameter) (m)
P = laser beam power (W)
r = radial coordinate (m)
i[r] = intensity as a function of r (W/m^2)

w=0.06 Sqrt[2];
i[r_]:=(2 P/(Pi w^2)) Exp[-2 r^2/w^2];
Plot[i[r]/P,{r,0,0.1}]

i[0]

Integrate[i[r] 2 Pi r,{r,0,Infinity}]

Particle Size Distribution

Complementary Cummulative Distribution Function (aka “tail distribution”) for particle number, areal density, N[x] (#/0.1 m^2 greater than size x for cleanliness level L):
N = 10^(0.926*[Log^2(L) - Log^2(x)])
from IEST-STD-CC1246D

n[L_,x_]:=10^(0.926*(Log10[L]^2-Log10[x]^2));

LogLogPlot[1-n[300,x]/n[300,1],{x,1,100},Filling->Axis,Frame->True,FrameLabel->{"Particle Size, X (microns)","CDF"},Axes->{False,False}]

LogLogPlot[1-n[50,x]/n[50,1],{x,1,100},Filling->Axis,Frame->True,FrameLabel->{"Particle Size, X (microns)","CDF"},Axes->{False,False}]

dc=ProbabilityDistribution[{"CDF",1-n[L,x]/n[L,1]},{x,1,∞}]

Plot[Evaluate[PDF[dc/.{L->300},x]],{x,1,10},PlotRange->{{1,Auto},All},Filling->Axis,Frame->True,FrameLabel->{"Particle Size, X (microns)","PDF"},Axes->{False,False}]

Plot[Evaluate[PDF[dc/.{L->50},x]],{x,1,10},PlotRange->{{1,Auto},All},Filling->Axis,Frame->True,FrameLabel->{"Particle Size, X (microns)","PDF"},Axes->{False,False}]

Evaluate[ n[L,1] (1-CDF[dc,x])/.{L->300,x->20}]

plt1=Plot[Evaluate[n[L, x] /. {L -> 300}],{x,1,10}, PlotRange -> All, PlotStyle -> {Black}];
plt2=Plot[Evaluate[n[L,1] (1-CDF[dc,x])/.{L->300}],{x,1,10},PlotRange->All,PlotStyle->{{Red,Thick,Dashing[0.05]}}];
Show[plt1, plt2]

Absorption

e = albedo
F = cross-sectional area
eF i[r]/P < 0.1 ppm

e = 1.0;

Assuming a uniform intensity distribution over diameter w

AreaBeam=Pi w^2/4;
AreaParticle=6 10^-7 AreaBeam;
diaParticle=10^6 2 Sqrt[AreaParticle/Pi]

Assuming a Guassian intensity distribution, particle at center

AreaParticle=6 10^-7 P/i[0];
diaParticle=10^6 2 Sqrt[AreaParticle/Pi]

Integral with the particle distribution

Note that the peak contribution to area blockage from particulates is

Plot[Evaluate[PDF[dc/.{L->50},x]] Pi x^2/4 ,{x,1,50},PlotRange->All]

Plot[Evaluate[CDF[dc/.{L->50},x]]  ,{x,1,200},PlotRange->All]

Evaluate[Integrate[Evaluate[PDF[dc/.{L->300},x]] Pi x^2/4,{x,1,300}]]

areaIntegral=Table[Evaluate[n[65,1]/(0.1 (10^6)^2) NIntegrate[Evaluate[PDF[dc/.{L->65},x]] Pi x^2/4,{x,1,i}]],{i,1,300}];

Length[areaIntegral]-Length[Select[areaIntegral,#>.9 areaIntegral[[Length[areaIntegral]]]&]]

ListLogLogPlot[areaIntegral,PlotRange->{{1,1000},{10^-9,10^-6}},Joined->True,PlotStyle->Thick,GridLines->Automatic]

areaIntegral0=Table[Evaluate[n[65,1]/(0.1 (10^6)^2) NIntegrate[Evaluate[PDF[dc/.{L->65},x]] Pi x^2/4,{x,i,200}]],{i,200,1,-1}];

areaIntegral1=Table[Evaluate[n[65,1]/(0.1 (10^6)^2) NIntegrate[Evaluate[PDF[dc/.{L->65},x]] Pi x^2/4,{x,i,300}]],{i,300,1,-1}];

areaIntegral2=Table[Evaluate[n[65,1]/(0.1 (10^6)^2) NIntegrate[Evaluate[PDF[dc/.{L->65},x]] Pi x^2/4,{x,i,500}]],{i,500,1,-1}];

ListLogPlot[{Reverse[areaIntegral0],Reverse[areaIntegral1],Reverse[areaIntegral2]},PlotRange->All,GridLines->Automatic,Joined->True,PlotStyle->{Thick,Thick,Thick},Frame->True,FrameLabel->{"Particle Size, X (microns)","Obscuration Ratio (L=65)"},Axes->{False,False}]

ListPlot[{Reverse[areaIntegral0],Reverse[areaIntegral1],Reverse[areaIntegral2]},PlotRange->{{0,20},All},GridLines->Automatic,Joined->True,PlotStyle->{Thick,Thick,Thick}]

Evaluate[n[65,1]/(0.1 (10^6)^2) NIntegrate[Evaluate[PDF[dc/.{L->65},x]] Pi x^2/4,{x,1,300}]]

check integral by discrete summation

discrete summation of areal fraction of dust matches integral of PDF, albeit with slow convergence:
0.0002538 for L=300, x increment of 0.001 vs 0.0002531 by PDF integral

L=65;
inc=0.001;
nt=Table[n[L,x],{x,1,5 L,inc}];
ntd=nt-RotateLeft[nt];
ntd[[Length[ntd]]]=0;
at=Table[ntd[[i]] Pi (1+i inc)^2/4,{i,1,Length[ntd]}];
Total[at]/(0.1 (10^6)^2)