8.4 Finite mirrors

The treatment of a mirror as an infinite medium can be accepted when the readout beam is a sharp one, of width much smaller than the mirror’s dimensions. But it becomes highly questionable when the spatial extension of the mode is comparable to those dimensions, which is precisely the case when high-order LG modes or mesa beams are considered. This is why an analytic method for computing U in the case of finite size mirrors has been developed in BHV, then amended in [28Jump To The Next Citation Point] for one boundary condition. We change slightly the coordinate system with respect to the preceding thermal studies. The mirror is assumed to fill the cylindrical volume defined by r ∈ [0,a ], z ∈ [0, h]. The reflecting face (intracavity) is assumed to be at z = 0. As in the thermal studies, we take a displacement vector under the form of an FB series
( ∑ { ur(r,z) = sAs (z)J1(ksr) ( uϕ(r,z) = 0∑ (8.66 ) uz(r,z) = sBs (r,z)J0(ksr)
where As and Bs are unknown functions of z, and ks are constants to be determined.

8.4.1 Equilibrium equations

The Navier–Cauchy equations read

{ 2 2 μ(∂z − ks)As − (λ + μ )ks(∂zBs + ksAs) , (8.67 ) μ(∂2z − k2s)Bs + (λ + μ )∂z(∂zBs + ksAs)
which yields
(∂2 − k2)(∂ A + k B ) = 0. (8.68 ) z s z s s s
The general solution of which is
−ksz ksz ∂zAs + ksBs = Cse + Dse . (8.69 )
This allows one to compute Bs in terms of As and to substitute it in Equation (8.67View Equation), so that one gets
(∂2 − k2 )A = − λ-+--μ-k2(C e−ksz − D eksz) . (8.70 ) z s s λ + 2μ s s s
The solution of which is
−ksz ksz λ + μ ( − ksz ksz) As(z) = Mse + Nse + ---------ksz Cse + Dse , (8.71 ) 2(λ + 2μ)
where Ms and Ns are two more arbitrary constants. Now Bs is determined by
( ) ( ) B (z) = --λ +-3μ--C + M e−ksz + --λ-+-3μ--D − N eksz + s 2(λ + 2μ ) s s 2 (λ + 2 μ) s s λ + μ ( ) ----------ksz Cse−ksz − Dseksz (8.72 ) 2 (λ + 2 μ)

8.4.2 Boundary conditions

The boundary conditions we assume are

The pressure distribution can, as usual, be expanded on the complete orthogonal family J0(ksz):

1 ∑ p(r) = I(r) = ---2 psJ0(ksr). (8.79 ) πa s
Owing to the norm of the functions J (k r) 0 s, the FB coefficients p s are now defined by
∫ --2π-- a ps = J 2(ζ ) I(r)J0(ksr)rdr. (8.80 ) 0 s 0
We have already encountered this kind of integral. The general result for an LGn,m mode is
1 p(nm),s = -2----exp (− ys)Lm (y)Lm+n (y ) (ys ≡ k2sw2∕8) (8.81 ) J0(ζs)
and for a flat mode
2aJ1(ζsb∕a) pflat,s = -----2-----. (8.82 ) bJ0(ζs)
For a mesa mode, numerical integration is necessary. The Θrz and Θzz components of the stress tensor are obtained as FB series:
Θ (r,z) = ∑ Θ (z)J (k r) (8.83 ) rz s,rz 1 s s
∑ Θ (r,z) = Θ (z)J (k r), (8.84 ) zz s,zz 0 s s
making clear that the first boundary condition is satisfied. In more detail, we have
( ) ( ) ---μ--- ksz --μ---- −ksz Θs,rz(z)∕ksμ = 2Ns − λ + 2μ Ds e − 2Ms − λ + 2μCs e + λ-+-μ-k z (C e− ksz − D eksz) (8.85 ) λ + 2μ s s s
ksz −ksz λ-+-μ-- ( −ksz ksz) Θs,zz(z)∕ksμ = (Ds − 2Ns )e − (Cs + 2Ms )e − λ + 2μ ksz Cse + Dse . (8.86 )
The boundary conditions on the faces provide four equations allowing one to determine the constants Cs, Ds,Ms, and Ns. We have
2 C = -1--4ps(1 −-σ-)--1 −-qs +-2qsxs- (8.87 ) s πa2 ksY (1 − qs)2 − 4qsx2s
1 4ps(1 − σ2) qs(1 − qs + 2xs) Ds = ---2-------------------2-------2 (8.88 ) πa ksY (1 − qs) − 4qsxs
1 ps(1 + σ )2qsx2s + (1 − 2σ)(1 − qs + 2qsxs) Ms = − --2-------------------------2-------2-------- (8.89 ) πa ksY (1 − qs) − 4qsxs
-1--psqs(1-+-σ-)2x2s-−-(1-−-2σ)(1-−-qs +-2xs) Ns = − πa2 ksY (1 − qs)2 − 4qsx2s , (8.90 )
with the notation xs ≡ ksh and qs ≡ exp(− 2xs). This was found by [5]. At this point, [28Jump To The Next Citation Point] pointed out that the component of spatial frequency zero of the pressure was not taken into account (recall that the ζs are the nonzero solutions of J1(ζ) = 0). The preceding displacement vector has a zero average on the strained face. One must now consider the resulting force acting on the body under the uniform pressure
1 p0 = ---2, (8.91 ) πa
producing a force of 1 N after integration on the disk. But this force produces an acceleration, which should be added to the Navier–Cauchy equations (8.67View Equation) (recall that the mirrors of GW interferometers are practically free in the longitudinal degree of freedom in the detection band). This effect can be taken into account by an extra displacement of the form
{ δ1ur(r,z) = p0σY[-r(1 − z∕h) ] δ u (r,z) = p σr2 − -1(z − z2∕2h) . (8.92 ) 1 z 0 Y2h Y
This extra displacement contributes only axial stress
δ1Θzz = − p0(1 − z∕h), (8.93 )
all other extra stress components being null. The equilibrium equations remain satisfied for
1 N 1 N ∂zδ1Θzz = p0∕h = ρ × -----2 = ρ × ----- = ρ¨z (8.94 ) ρh πa Mass
as remarked by [28Jump To The Next Citation Point]. Now, the sum of the displacement (8.66View Equation) and of the extra displacement (8.92View Equation) satisfies all boundary conditions except the vanishing of the radial stress on the edge. We have for the FB component of the radial stress
′ Θs,rr(z) = λ(ksAs (z) + ∂zBs (z))J0(ζs) + 2μksAsJ 1(ζs). (8.95 )
But, due to the fact that J ′(ζs) = J0(ζs) 1, and after substituting the explicit expressions of As and Bs, we get
p J (ζ ) [ Θs,rr(z) = − p0-----s--0-s------ (1 − qs + 2qsxs(1 + xs))e−ksz − qs(1 − qs + 2xs(1 − xs))eksz (1 − qs)2 − 4qsx2s ] − kszqs(1 − qs + 2xs )eksz − ksz(1 − qs + 2qsxs)e− ksz. (8.96 )
It is numerically easy to check that this function of z is not very different from linear. It has a vanishing average. Therefore, it is possible to find an approximate solution using the Saint-Venant principle once more. We add to our displacement one more extra displacement giving a linear edge stress compensating the preceding. The second extra displacement is of the form
{ 1−σ δ2ur(r,z) = -Y-(ω0 + ω1z )r 2 (8.97 ) δ2uz(r,z) = − 2Yσ(ω0 + ω1z∕2 )z − 1−Y-σω1r2 ,
such that the equilibrium equations (8.67View Equation) are identically satisfied. It contributes only radial stress, thus leaving unchanged the boundary conditions, except for a radial contribution
δ2Θrr(z) = ω0 + ω1z. (8.98 )
As usual, we require a minimum residual stress on the edge, which amounts to having null resulting mean force and torque on the edge. If we define
∫ h I = 1- Θ (a,z )dz (8.99 ) 0 h 0 rr
∫ h I1 = -1- Θrr(a,z)zdz (8.100 ) h2 0
then the values of ω0 and ω1 are
ω0 = 6I1 − 4I0 (8.101 )
ω = (6I − 12I )∕h. (8.102 ) 1 0 1
The explicit expression (8.96View Equation) allows one to compute I0 and I1. One finds
I0 = 0 (8.103 )
I = p S (8.104 ) 1 0
with
a2-∑ psJ0(ζs) S = h2 ζ2 (8.105 ) s s
and
ω = 6Sp , ω = − 12Sp ∕h. (8.106 ) 0 0 1 0
In summary, the total displacement is now
⃗utot(r,z) = ⃗u (r,z) + Δ ⃗u(r,z) (8.107 )
with
{ ∑ ur(r,z) = As(z)J1(ksr) uz(r,z) = ∑ Bs(z)J0(ksr) (8.108 )
and
Δ ⃗u(r,z) = δ1⃗u(r,z) + δ2⃗u(r,z) { -σ 1−σ- = p0Yσ(12− z∕h)1-+ Y (ω20 + ω1z2)σr 2 1−-σ- 2 (8.109 ) p02Y r ∕h − Y (z − z ∕2h) − Y (ω0z + ω1z ∕2 ) − 2Y ω1r

8.4.3 Strain energy

The strain tensor is now of the form

E (r,z) = E (r,z) + ΔE (z), (8.110 ) ij 0,ij ij
where the component E0,ij is computed from ⃗u, whereas the component ΔEij is computed from Δ ⃗u. Now the strain energy density w (r,z) is defined by
1 Y [ ( w (r,z) = ----------------- σE (r,z)2 + (1 − 2σ) Err(r,z)2 2(1 + σ)(1 − 2σ) )] +E ϕϕ(r,z)2 + Ezz(r,z)2 + 2Erz(r,z)2 (8.111 )
∑ E (r,z) ≡ iEii(r,z) being the trace of the strain tensor. Thus, integrated strain energy U, i.e., our target, is
∫ 2π ∫ h ∫ a U = dϕ dz rdrw (r,z). (8.112 ) 0 0 0
The squares of the strain tensor components obviously contain, in general, squares of the main strain, squares of the extra strains and cross products. However, in the r integral, cross products vanish, so that the total internal energy is the sum of two contributions
U = U0 + ΔU. (8.113 )
These can be computed separately. We have
2 ∑ 2 2 2 U0 = 1 −-σ-- J-0(ζs)ps -1-−-qs-+-4qsxs-. (8.114 ) πaY s>0 ζs (1 − qs)2 − 4qsx2s
The dimension of U is J N−2. And for the second contribution, we have
[ ] a2 ( h)4 ( h)2 ΔU = ----3-- -- + 12σξ -- + 72 (1 − σ)ξ2 (8.115 ) 6 πh Y a a
with
∑ ξ ≡ psJ0(ζs)∕ζ2s. (8.116 ) s>0
For our three reference examples, using Virgo mirrors, we get, with a loss angle of 10–6,

8.4.4 Explicit displacement and strain tensor

In Section 9, we shall need the explicit expressions of the displacement vector and particularly of the trace of the strain tensor. We have, after the preceding calculations for the FB components of the main displacement,

ps(1-+-σ) { 2 − ksz 2 ksz As (z) = πa ζ YΔ − [(1 − 2σ)Qs + 2qsxs]e + [(1 − 2σ )Rs − 2qsxs]e s[ s −k z k z]} +ksz Qse s + Rse s (8.133 )
with the notation
2 2 ks ≡ ζs∕a, xs ≡ ksh, qs ≡ exp(− 2xs), Δs ≡ (1 − qs) − 4qsxs, Qs ≡ 1 − qs + 2qsxs, Rs ≡ qs(1 − qq + 2xs). (8.134 )
In the same way
ps(1-+-σ-){ 2 −ksz 2 ksz Bs(z) = πaζsY Δs [2(1 − σ)Qs − 2qsxs]e + [2(1 − σ)Rs + 2qsxs]e [ − ksz ksz]} +ksz Qse − Rse (8.135 )
The z derivative of Bs(z) is needed as well,
p (1 + σ) { ∂zBs (z) = -s-2----- − [(1 − 2σ)Qs − 2qsx2s]e− ksz + [(1 − 2σ )Rs + 2qsx2s]eksz πa Y[Δs ]} − ksz Qse −ksz + Rseksz (8.136 )
and the combination is still
{ } 1-[∂ A (z) − k B (z)] = (1-+-σ)ps (2qx2 − k zQ )e−ksz − (2qx2 − k zR )eksz . (8.137 ) 2 z s s s πa2Y Δs s s s s s s s s
For the extra displacement we have
1 Δur (r,z) = πa2Y--[c0 + c1z]r (8.138 )
1 [ ] Δuz (r,z) = --2--- c2r2 + c3z + c4z2 (8.139 ) πa Y
with the notation
c0 = σ + 6(1 − σ)S, c1 = − [σ + 12 (1 − σ )S]∕h, c2 = − 1c1, 2 c3 = − (1 + 12 σS ), c4 = (1 + 24σS )∕2h (8.140 )
This allows one to plot the (virtually) deformed solid (see Figure 59View Image) in our three examples. We have amplified the displacement by a large factor, to give a better idea of the shape.
View Image

Figure 59: Virtually deformed mirror under beam pressure. (1 N integrated pressure) by modes having 1 ppm clipping losses. For more clarity, the displacements have been amplified by a factor of 7 × 107.

We find the FB components of the main strain tensor to be

∑ Err (r,z) = ksAs (z)J′1(ksr) (8.141 ) s
∑ E ϕϕ(r, z) = ksAs (z)J1(ksr) (8.142 ) s ksr
∑ Ezz(r,z) = ∂zBsJ0(ksr) (8.143 ) s
1-∑ Erz (r,z) = 2 (∂zAs − ksBs )J1(ksr) (8.144 ) s
This gives, in particular, the FB component of the trace of the strain tensor
∑ E (r,z) = Es(z)J0(ksr) (8.145 )
with
E (z) = − 2(1-+-σ)(1-−-2σ-)ps-[Q e−ksz − R eksz] . (8.146 ) s πa2Y Δs s s
And for the extra contributions, we get
ΔErr (z) = --1---(c0 + c1z) (8.147 ) πa2Y
ΔE (z) = --1---(c + c z) (8.148 ) ϕϕ πa2Y 0 1
--1--- ΔEzz (z ) = πa2Y (c3 + 2c4z) (8.149 )
ΔErz = 0 (8.150 )
ΔE (z) = ---1--(2c + c + 2(c + c )z) πa2Y 0 3 1 4 --1---[ ] = − (1 − 2σ)πa2Y 1 − 12S − (1 − 24S )z∕h (8.151 )
This allows us to map the energy density in the material (see Figures 60View Image, 61View Image, and 62View Image).
View Image

Figure 60: Distribution of strain energy in the mirror substrate (LG0,0 w = 2 cm, logarithmic scale)
View Image

Figure 61: Distribution of strain energy in the mirror substrate (Flat mode b = 9.1 cm, logarithmic scale)
View Image

Figure 62: Distribution of strain energy in the mirror substrate (LG5,5 w = 3.5 cm, logarithmic scale)

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