3.3 Thermal distortions in the steady state

We assume from now on that the temperature field is axisymmetric. Due to the temperature field, at the same time, thermal expansion of the material causes a change of the shape of the mirror and a distortion of the reflecting face. If we call ⃗u (r,z ) the displacement vector, i.e., the difference between the coordinates of a given atom before and after heating, the classical relevant quantities are the strain tensor E ij and the stress tensor Θij. In the presence of a temperature field T, the two tensors are related by the generalized Hooke law for isotropic media via the Lamé coefficients λ, μ:
Θij = δij(λE − νT ) + 2μEij, (3.87 )
where ν is the stress temperature modulus and E the trace of the strain tensor. δij is the Kronecker tensor. In cylindrical coordinates (r,ϕ,z), assuming cylindrical symmetry, the displacement vector has only two components, ur (r,z ) and uz(r,z). Then the strain tensor has four components:
∂u u (r,z) Err(r,z) = ---r(r,z) , Eϕϕ(r,z) =--r---- , ∂r [ r ] ∂uz- 1- ∂ur- ∂uz- Ezz(r,z) = ∂z (r,z) , Erz = 2 ∂z (r,z) + ∂r (r,z ) . (3.88 )

The relation (3.87View Equation) is, in detail,

( || Θrr = − νT + λE + 2μErr { Θ ϕϕ = − νT + λE + 2μE ϕϕ | Θ = − νT + λE + 2μE , (3.89 ) |( zz zz Θrz = 2μErz
where, as above, T (r,z) is the excess temperature field given by the generic FB expansion
∑ T (r,z) = Ts(z)J0(ζsr∕a). (3.90 ) s
The stress tensor must obey the homogeneous divergence equation in the absence of external applied forces (static equilibrium), i.e., a special case of the Navier–Cauchy equations,
{ ∂rΘrr + (Θrr − Θ ϕϕ)∕r + ∂zΘrz = 0 (∂ + 1∕r)Θ + ∂ Θ = 0 . (3.91 ) r rz z zz

Moreover, the following boundary conditions must be satisfied:

Θzz(r,±h ∕2) = 0, Θrz(a,z) = 0, Θrz (r,±h ∕2) = 0, Θrr (a, z) = 0. (3.92 )
It is more convenient, at the end of the calculations, to express the results in terms of the Poisson ratio σ and Young’s modulus Y, using the correspondence
------Y-σ------- ---Y----- ----ν---- λ = (1 + σ)(1 − 2σ ), μ = 2(1 + σ ), 2(λ + μ ) = α(1 + σ), (3.93 )
where α is the linear thermal expansion coefficient.

3.3.1 Thermal expansion from thermalization on the coating

In the case of a heat source localized on the reflecting face, so that the temperature field T (r,z) is known, it is possible to satisfy Equation (3.91View Equation) and all but one of the boundary conditions in Equation (3.92View Equation) by a displacement vector of the form

∫ r u (r,z) = ----ν----1- T (r′,z )r ′dr′ (3.94 ) r 2 (λ + μ )r 0
[ ∫ z ] ---ν----- ′ ′ uz(r,z) = 2(λ + μ) −h∕2T (r,z)dz + Φ(r) , (3.95 )
where Φ(r) is a function to be determined. It can be checked that Θzz = 0. The equilibrium equations then reduce to
{ ∂zΘrz = 0 (∂ + 1∕r )Θ = 0 . (3.96 ) r rz

We have

[ ∫ ] ∂Θrz νμ ∂T 1 r ∂2T ′ ′ ′ -----(r,z) = --------- ---(r,z ) +-- ---2 (r ,z)rdr . (3.97 ) ∂z 2(λ + μ ) ∂r r 0 ∂z
But let us recall that T is harmonic [see Equation (3.1View Equation)], so that
2 ( ) ∂-T- = − 1-∂-- r∂T- (3.98 ) ∂z2 r ∂r ∂r
and consequently the first equilibrium equation (3.96View Equation) is identically satisfied
∂Θrz-(r,z) = 0. (3.99 ) ∂z
Now we have
νμ [ 1 ∂T ] (∂r + 1∕r )Θrz =--------- Φ ′′(r) + -Φ ′(r) + ---(r,− h∕2) . (3.100 ) 2(λ + μ ) r ∂z
Therefore, in order to satisfy the second equilibrium equation (3.96View Equation), we take
∫ r ′∫ r′ Φ(r) = − dr- ∂T-(r′′,− h ∕2)r′′dr′′ + C, (3.101 ) 0 r′ 0 ∂z
where C is an arbitrary constant. The stress component Θrz is now explicitly known
νμ [∫ z ∂T 1 ∫ r( ∂T ∂T ) ] Θrz (r,z) = --------- ---(r,z′)dz ′ +-- ---(r′,z) − ---(r′,− h ∕2) r′dr ′ (3.102 ) 2(λ + μ) −h∕2 ∂r r 0 ∂z ∂z
making clear that Θrz (r,− h ∕2) = 0, and since it has been shown that ∂zΘrz (r,z ) = 0, we have, simply,
Θ = 0. (3.103 ) rz
Two more boundary conditions are satisfied. At this point, the only remaining unsatisfied condition is the vanishing of Θrr on the edge. Indeed, we have
∫ r Θrr(r,z) = − -ν-μ--1- T (r′,z)r′dr′ (3.104 ) λ + μ r2 0
or explicitly
( ) --νμ--∑ -a- ζsr Θrr(r,z) = − λ + μ Ts(z)ζsrJ1 a , (3.105 ) s>0
so that
ν μ ∑ J1(ζs) Θrr(a,z) = − λ-+-μ- Ts(z)--ζ---. (3.106 ) s>0 s
It is easy to check numerically that the function Θ (a, z) rr is almost linear for any type of beam. Therefore, it can be almost cancelled by an opposite linear stress on the edge. Such a stress can be induced by the following extra displacement vector:
δur(r,z) = ---λ-+-2μ--- (ω0r + ω1rz) (3.107 ) 2 μ(3λ + 2μ)
δuz(r,z) = − ----λ------(ω0z + ω1z2∕2) − ---λ-+-2μ---ω1r2, (3.108 ) μ(3λ + 2μ ) 4μ (3 λ + 2μ)
where ω0 and ω1 are arbitrary constants. One can check that this vector firstly satisfies the equilibrium equations (3.91View Equation), secondly has identically null stress components Θrz and Θzz, and finally produces a radial edge stress:
δΘrr (a,z) = ω0 + ω1z. (3.109 )
The constants ω 0 and ω 1 can now be chosen in order to minimize the quadratic error:
∫ h∕2 2 Q(ω0, ω1) = [Θrr(a,z) + ω0 + ω1z ] dz. (3.110 ) −h∕2
After using the classical mean-squares formulas, this cancels the mean force and the mean torque on the edge:
∫ h∕2 ω0 = − 1- Θrr(a,z )dz (3.111 ) h −h∕2
∫ h∕2 ω1 = − 12- Θrr(a,z)zdz. (3.112 ) h3 − h∕2
The complete displacement ⃗ U ≡ ⃗u + δ⃗u now satisfies the Navier–Cauchy equations, all constraints on the faces, and induces null mean force and torque on the edge. Owing to the principle of de Saint-Venant [38], we can conclude that the displacement is correct almost everywhere in the bulk material, except possibly near the edge. But any effective optical beam has a vanishing intensity near the edge in order to prevent diffraction losses, so that the solution U⃗ is relevant for our purpose. If we use Young’s modulus Y, the Poisson ratio σ and the linear thermal expansion coefficient α instead of λ,μ,ν, we find
αY χ 𝜖P ∑ J0(ζs)sinh γs ω0 = -πKh---- ps--ζ3----d----, (3.113 ) s s 1,s
with γs ≡ ζsh ∕2a. The FB coefficients ps are identical to the p0,s computed in Section 3.1.3 and
∑ ω1 = − 12-αY-χ𝜖aP- ps J0(ζs)γscosh-γs −-sinh-γs. (3.114 ) πKh3 s>0 ζs4 d2,s

With the same notation, we have, explicitly, the components of the displacement:

∑ u (r,z) = α(1 + σ)a Ts(z)J (ζ r∕a) (3.115 ) r ζs 1 s s>0
1 − σ δur(r,z) = ------(ω0 + ω1z)r (3.116 ) Y
∑ [ ( ) ] uz(r,z) = α-(1 +-σ)𝜖P- ps -1--+ sinh(ζsz∕a) − cosh-(ζsz∕a-) J0(ζr∕a) (3.117 ) 2πK s ζs d2,s d1,s d2,s
[ ] δuz (r,z ) = − 2-σ ω0 + 1ω1z z − (1-−-σ-)ω1r2. (3.118 ) Y 2 2Y
The displacement vector is defined up to a constant vector. We have chosen the constant in such a way that the displacement is zero at (r = 0,z = 0). We can see in Figure 23View Image the global deformation of the mirror in the cases Ex1, Ex2 and Ex3.
View Image

Figure 23: Thermal deformation of the mirror under three types of readout beams (1 W absorbed power in the coating and exaggerated by a factor of 2 × 105)

For the deformation of the coating, we have

uz(r,− h ∕2) = α(1 + σ)Φ (r) (3.119 )
or, in detail,
∑ uz (r,− h∕2 ) = Us [1 − J0(ζsr∕a )] (3.120 ) s>0
with
( ) Us = α-(1 +-σ-)𝜖P-ps sinh-γs+ coshγs- . (3.121 ) 2πK ζs d1,s d2,s
(The displacement has now been taken to be zero at the center of the reflecting face), and
δuz(r,− h∕2 ) = − 1-−-σ-ω1r2. (3.122 ) 2Y
The geometrical effects of heating (see Figure 23View Image) are mainly a thermally-induced aberration due to the change of the reflecting surface by u (r,− h∕2 ) + δu (r) z z, then a change of the optical path through the substrate by a quantity
∫ h∕2 ∫ h∕2 δZ (r) = (nR − 1) Ezz(r,z)dz = α(1 + σ)(nR − 1) T (r,z)dz (3.123 ) −h∕2 −h∕2
(nR being the nominal refractive index), which can be directly included in the thermal lens expression [Equation (3.40View Equation)], which has the same dependence on temperature. Note that the Saint-Venant correction contributes a constant (independent on r), so that we can ignore it in a lensing study. Thus, the global thermal lens is identical to the result found in Section 3, up to the correction
-dn dn- dT → dT + α (1 + σ)(nR − 1). (3.124 )
Estimations of the weighted curvature of the distorted surface are obtained with the same technique as in Section 3. For Laguerre–Gauss modes, we obtain
∑ 1 − σ c = − c(LsG )U (sLG)− -----ω (L1G ). (3.125 ) s>0 2Y
For a flat mode, this is
∑ 1 − σ (F) c = − c(sF)U(sF) − ------ω1 , (3.126 ) s>0 2Y
where the coefficients cs have been defined by Equations (3.70View Equation) and (3.75View Equation).
View Image

Figure 24: Deformation of the reflecting coating for three types of readout beams (coating absorption). Dashed line: best parabolic fit 2 ˆuz + δˆuz = cr + d (weighted by the intensity profile) giving the effective curvature radius

We can see in Figure 24View Image the distorted reflecting face of the mirror in our three examples. The results in terms of curvature radius are

We see in the case of axisymmetry that the use of unconventional modes (either flat or high-order LG) allows one to dramatically reduce spurious thermal effects in mirrors to be installed in advanced GW detectors, where high light-power flows are planned. Up to two orders of magnitude can be gained with respect to the present Virgo configuration for thermal lensing, or for thermal deformation of the coating. As in the thermal lens Section 3, we give some results (see Table 4) for LG modes having the same (1 ppm) clipping losses.


Table 4: Curvature radii from thermal expansion due to coating absorption for modes having 1 ppm clipping losses
order (p,q) w [cm] Rc of th. aberr. [km W]
(0,0) 6.65 77
(0,1) 5.56 149
(1,0) 6.06 141
(0,2) 4.93 210
(1,1) 5.23 201
(2,0) 5.65 197
(0,3) 4.49 258
(1,2) 4.70 250
(2,1) 4.97 250
(3,0) 5.35 250
(0,4) 4.17 299
(1,3) 4.32 290
(2,2) 4.52 292
(3,1) 4.76 298
(4,0) 5.11 300
(0,5) 3.91 332
(1,4) 4.03 324
(2,3) 4.18 326
(3,2) 4.36 332
(4,1) 4.58 342
(5,0) 4.91 348

3.3.2 Thermal expansion from internal absorption

When the linear absorption of light through the bulk material results in an internal heat source, the temperature field is no longer harmonic, and we are bound to solve explicitly the thermo-elastic equations (3.91View Equation) and (3.92View Equation). As seen earlier, the case of internal absorption leads to a symmetric temperature field. However, we shall derive the general thermoelastic solution, which will also prove useful in Section 4 below.

The temperature field is assumed to be of the generic form (ks ≡ ζ∕a)

∑ T (r,z) = Ts(z)J0(ksr). (3.127 ) s
We consider a displacement vector of the form
{ ∑ ur(r,z) = ∑ sAs(z)J1(ksr) (3.128 ) uz(r,z) = sBs(z)J0(ksr).
The equilibrium equations (3.91View Equation) reduce to a system of ordinary differential equations,
[ 2 2 ] (λ + 2μ ) ∂zAs − k sAs − (λ + μ)∂z(∂zAs + ksBs ) = − ks νTs(z) (3.129 )
(λ + 2μ)[∂2B − k2B ] + (λ + μ)k (∂ A + k B ) = ν∂ T (z ), (3.130 ) z s s s s z s s s z s
so that, by a basic combination of these two, we get
[ ] ∂2z − k2s (∂zAs + ksBs) = 0, (3.131 )
a solution of which is
∂ A + k B = k C cosh (k z) + k D sinh(k z), (3.132 ) z s s s s s s s s s
where Cs and Ds are arbitrary constants. By substituting in Equation (3.129View Equation) we get
[ ] λ + μ ν ∂2z − k2s As = -------k2s (Cs sinh (ksz) + Ds cosh(ksz)).− ks------Ts (z ) (3.133 ) λ + 2μ λ + 2μ
A solution of which is
[ ] As (z) = Ms sinh(ksz) + Ns cosh(ksz) + --λ-+-μ--- Csksz cosh (ksz) + Dsksz sinh (ksz ) 2(λ + 2μ) ν 𝒯s − ----------, (3.134 ) ks(λ + 2μ)
where Ms and Ns are two arbitrary constants and 𝒯s (z ) is a special solution of
[∂2 − k2 ]𝒯 (z) = T (z). (3.135 ) z s s s
We can now find Bs from Equation (3.132View Equation)
Bs(z) = [Cs − Ms ]cosh(ksz) + [Ds(z) − Ns (s)]sinh(ksz ) [ ( ) + --λ-+-μ--- Cs cosh(ksz) + kszsinh(ksz) 2(λ + 2μ ) ( )] ν ∂ 𝒯s +Ds sinh (ksz ) + ksz cosh(ksz) + -----------(z). (3.136 ) λ + 2μ ∂z
The boundary conditions,
Θrz (±h ∕2) = Θzz(±h ∕2 ) = 0, (3.137 )
lead to the system (we return to the Poisson ratio and to the linear thermal expansion coefficient)
∂-𝒯s 4α(1 + σ) ∂z (h∕2) = [γssinh γs − (1 − 2σ )cosh γs]Cs + [γs coshγs − (1 − 2σ) sinh γs]Ds +4 (1 − σ)M cosh γ + 4(1 − σ)N sinh γ (3.138 ) s s s s
∂-𝒯s 4α(1 + σ) ∂z (− h∕2) = [γssinhγs − (1 − 2σ )coshγs]Cs − [γs coshγs − (1 − 2σ) sinh γs]Ds +4 (1 − σ)Ms cosh γs − 4(1 − σ)Ns sinh γs (3.139 )
− 4α (1 + σ)ks𝒯s(h∕2) = [2(1 − σ)sinh γs − γscosh γs]Cs + [2(1 − σ )coshγs − γs sinh γs]Ds − 4(1 − σ)Ms sinh γs − 4(1 − σ)Ns coshγs (3.140 )
− 4α (1 + σ )k 𝒯 (− h ∕2) = − [2 (1 − σ )sinh γ − γ cosh γ ]C + [2(1 − σ)cosh γ − γ sinh γ ]D + s s s s s s s s s s 4(1 − σ)Ms sinh γs − 4(1 − σ )Ns cosh γs (3.141 )
with γ ≡ k h ∕2 ≡ ζ h∕2a s s s. It is easy to combine these equations to find
4α (1 + σ ) Cs = -----′′---(e′ssinh γs − ksos coshγs) (3.142 ) Γs
4α-(1 +-σ) ′ Ds = Γ ′ (o scosh γs − ksessinhγs) (3.143 ) s
α(1 + σ) [ ′ ] Ms = (1-−-σ-)Γ-′′ (2(1 − σ) sinh γs − γscosh γs)es + (γssinh γs − (1 − 2σ) coshγs) ksos (3.144 ) s
[ ] -α(1-+-σ)- ′ Ns = (1 − σ)Γ ′s (2(1 − σ)cosh γs − γssinh γs)os + (γs coshγs − (1 − 2σ) sinh γs)kses (3.145 )
with the notation
Γ ′s ≡ sinhγs coshγs + γs, (3.146 )
′′ Γs ≡ sinh γscosh γs − γs. (3.147 )
We have also used the symmetrized coefficients (even and odd parts)
( | es = 1 (𝒯s(h ∕2) + 𝒯s (− h ∕2)) |{ o = 21(𝒯 (h ∕2) − 𝒯 (− h ∕2)) s′ 21 s s (3.148 ) ||( e s = 2(∂z𝒯s(h∕2 ) + ∂z𝒯s (− h∕2 )) o′s = 12(∂z𝒯s(h∕2 ) − ∂z𝒯s (− h ∕2)).
The radial stress is
∑ ∑ J1(ksr) Θrr(r,z) = Θ(r1r),s(z)J0(ksr) − Θ (r2r),s(z)------- (3.149 ) s s ksr
with
(1) ∂Bs Θ rr,s(z) = − νTs (z ) + (λ + 2μ )ksAs (z) + λ-∂z-(z) (3.150 )
(2) Θrr,s(z) = 2μksAs (z). (3.151 )
It can be checked that Θ(1)(z) rr,s gives a null contribution to the mean force and to the mean torque on the edge, i.e.,
∫ ∫ h∕2 (1) h∕2 (1) Θ rr,s(z)dz = Θ rr,s(z)zdz = 0. (3.152 ) −h∕2 −h∕2
Therefore these two mean moments can be computed with As (z). We have
1 ∫ h∕2 1 + σ 2σ sinh γs -- As (z)dz = α -------′-------[ksessinh γs − o′scoshγs] h −h∕2 1 − σ Γs γs 1 + σ 1 ∫ h∕2 + α --------- 𝒯s(z)dz (3.153 ) 1 − σksh − h∕2
∫ h∕2 ( ) 1- As (z)zdz = α 1 +-σ-2σ- sinh-γs-− cosh γs [e′s sinh γs − ksoscosh γs] h −h∕2 1 − σ Γ ′′s γs ∫ h∕2 + α 1 +-σ-1-- 𝒯s(z)zdz. (3.154 ) 1 − σksh − h∕2

In the special case of bulk absorption, we have, due to symmetry,

os = e ′s = 0 (3.155 )
and
βa2P p [ χγ sinh γ ] es = − ------s- 1 + ---s-----s- (3.156 ) πK ζ4 2d1,s
βaP p o′s = − -----s(sinh γs + γs cosh γs). (3.157 ) πK ζ3

The explicit expression for functions As(z) and Bs(z) is, finally,

[ 1-+-σ-αaβP-- sinh-γs As(z) = αa 1 − σ πK ζ3 1 + Γ kszsinh(ksz) ( s s ) ] − γs-coshγs-−-(1-−-2σ)-sinh-γs + (1 − σ)-χ-- cosh(k z) (3.158 ) Γ s d1,s s
1 + σ αaβP [ sinh γs Bs(z) = αa -----------3 − -------ksz cosh(ksz) 1( − σ πK ζs Γ s ) ] γscosh-γs +-2(1-−-σ-)sinh-γs -χ-- + Γ − (1 − σ)d sinh (ksz) (3.159 ) s 1,s

Two boundary conditions have been forgotten. We need vanishing Θrr and Θrz on the edge r = a. However, a numerical investigation shows that Θ (a, z) rr is practically constant, having an average value of

αaP Y χ ∑ p [ sinh γ ( χ 2σ sinh γ ) ] 𝜃 ≡ ⟨Θrr (a, z)⟩ = − ----------- --s4 1 − ------s (1 − σ)----+ --------s- J0(ζs). (3.160 ) (1 − σ )πK s ζs γs d1,s Γ s
In the same spirit as in the preceding case (Saint-Venant correction) we can add an extra displacement
1 −-σ- δur(r,z) = − Y 𝜃r (3.161 )
2σ δuz (r,z ) = --𝜃z, (3.162 ) Y
which induces null δΘrz and δΘzz extra stresses, trivially satisfies the equilibrium equations, and produces a constant δ Θrr = − 𝜃. Now the Θrz stress component is antisymmetric with respect to z
αY ∑ Ts[ ] Θrz (r,z ) = ------ --- kszcosh (ksz)sinh γs − γs cosh γs sinh (ksz ) J1(ksr), (3.163 ) 1 − σ s Γ s
so that it is zero at z = ±h ∕2 with a vanishing average value on the edge z ∈ [− h∕2,h ∕2]. Moreover, it can be checked that the values taken on the edge are weak compared to other places and other components. Therefore, the sum
⃗ ⃗u + δu, (3.164 )
satisfies exactly the equilibrium equations, exactly the boundary conditions on the faces, and on average on the edge. The displacement vector at z = − h∕2 represents the deformation of the reflecting face. We have
Z (r) = uz (r,− h∕2 ) + δuz (r,− h∕2 )[ ] α(1 +-σ)βP-a-∑ ps- 2sinh-γs -χ-- 2σ- = πK ζ3sinh γs Γ ′ − d J0(ζsr∕a) + Y 𝜃z. (3.165 ) s s s 1,s

One can see in Figure 25View Image the distorted shape of the mirror in three situations. The thermally-induced curvature radius can be computed as usual. (See Figure 26View Image for the profiles of the reflecting surface in three situations and the best fitted paraboloid.) For our three examples, we obtain the following figures

See Table 5 for several other LG modes.


Table 5: Curvature radii from thermal expansion due to bulk absorption for modes having 1 ppm clipping losses
order (p,q) w [cm] Rc of th. aberr. [km W]
(0,0) 6.65 165
(0,1) 5.56 318
(1,0) 6.06 290
(0,2) 4.93 440
(1,1) 5.23 410
(2,0) 5.65 404
(0,3) 4.49 533
(1,2) 4.70 507
(2,1) 4.97 504
(3,0) 5.35 512
(0,4) 4.17 613
(1,3) 4.32 586
(2,2) 4.52 585
(3,1) 4.76 597
(4,0) 5.11 615
(0,5) 3.91 677
(1,4) 4.03 654
(2,3) 4.18 650
(3,2) 4.36 661
(4,1) 4.58 684
(5,0) 4.91 713

In Table 6 we give numerical results for our three examples.

View Image

Figure 25: Thermal expansion of the mirror under three types of readout beams (heating by 1 W internal absorption of light, exaggerated by a factor of 2 × 105)
View Image

Figure 26: Thermal aberration caused by internal heating for 1 W absorbed power. Dashed lines: nearest paraboloid (weighted by the intensity distribution)


Table 6: Thermal aberrations from coating and bulk absorption
results (coating abs.) LG0,0 w = 2 cm flat b = 9.1 cm LG5,5 w = 3.5 cm
Curvature radius 5.8 km W 167 km W 478 km W
Coupling losses 4.3 ×10–2/W2 4.4 ×10–3/W2 6.7 ×10–3/W2
results (bulk abs.) LG0,0 w = 2 cm flat b = 9.1 cm LG5,5 w = 3.5 cm
Curvature radius 22 km W 327 km W 937 km W
Coupling losses 3.0 ×10–3/W2 2.2 ×10–3/W2 1.8 ×10–3/W2


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