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ID Date Author Type Category Subject
16   Tue Jun 22 22:28:09 2021 KojiGeneralDesign specsTest Mass wedge design

The ETM wedge of 0.5deg will allow us to separate the AR reflections. We will be OK with the ITM wedge of 0.5deg too. 0.36 deg for ITM is also OK, but not for the ETM.

- Attachment 1 shows the deflection of the 2128mn and 1418nm beams by the test mass wedge. Here, the wedge angle of 1deg was assumed as a reference. For the other wedge angle, simply multiply the new number (in deg) to the indicated values for the displacement and angle.

- Attachment 2 shows the simplified layout of the test masses for the calculation of the wedge angle. Here the ITM and ETM are supposed to be placed at the center of the in-vacuum tables. Considering the presence of the cryo baffles, we need to isolate the pick-off beam on the BS table. There we can place a black glass (or similar) beam dump to kill the AR reflection. For the ETM trans, the propagation length will be too short for in-vacuum dumping of the AR reflection. We will need to place a beam baffle on the transmon table.

- I've assumed the cavity parameter of L=38m and RoC(ETM)=57m (This yields the Rayleigh range zR=27m). The waist radii (i.e. beam radii at the ITM) for the 2128nm and 1418nm beams are 4.3mm and 3.5mm, while the beam radii at the ETM are 7.4mm and 6.0mm, respectively,

- Attachment 3: Our requirement is that the AR reflection of the ALS (1418nm) beam can be dumped without clipping the main beam.
If we assume the wedge angle of 0.5deg, the opening of the main and AR beams will be (2.462+4.462)*0.5 = 3.46 deg. Assuming the distance from the ETM to the in-air trans baffle is 45" (=1.14m), the separation of the beams will become 69mm. The attached figure shows how big the separation is compared with the beam sizes. I declare that the separation is quite comfortable. As the main and AR beams are distributed on both sides of the optic (i.e. left and right), I suppose that the beams are not clipped by the optical window of the chamber. But this should be checked.
Note that the 6w size for the 2128nm beam is 44mm. Therefore, the first lens for the beam shrinkage needs to be 3" in dia, and even 3" 45deg BS/mirrors are to be used after some amount of beam shrinkage.

- Attachment 4 (Lower): If we assume the same ITM wedge angle of 0.5deg as the ETM, both the POX/POY and the AR beams will have a separation of ~100mm. This is about the maximum acceptable separation to place the POX/POY optics without taking too much space on the BS chamber.

- Attachment 4 (Upper): Just as a trial, the minimum ITM wedge angle of 0.36deg was checked, this gives us the PO beam ~3" separated from the main beam. This is still comfortable to deal with these multiple beams from the ITM/

Attachment 1: wedge.pdf
Attachment 2: Layout.pdf
Attachment 3: ETM.pdf
Attachment 4: ITM.pdf
20   Fri Aug 6 04:34:43 2021 KojiGeneralGeneralTheoretical Cooling Time Limit

I was thinking about how fast we can cool the test mass. No matter how we improve the emissivity of the test mass and the cryostat, there is a theoretical limitation. I wanted to calculate it as a reference to know how good the cooling is in an experiment.

We have a Si test mass of 300K in a blackbody cryostat with a 0K shield. How fast can we cool the test mass?

$m\,C_p(t)\,T'(t) = -\epsilon\,\sigma A\,[T(t)^4 - 0^4]$

$T(0) = T_0$

Then assume the specific heat is linear as

$C_p(t) = c_{p0} T(t)$

The actual Cp follows a nonlinear function (cf Debye model), but this is not a too bad assumption down to ~100K.

Then the differential equation can be analytically solved:

$T(t) = T_0 \left( 1 + t/t_0 \right )^{-1/2}$,

where the characteristic time of t0 is

$t_0 = \frac{m c_{p0}}{2\,\epsilon\,\sigma A\,T_0^2 }$.

Here T_0 is the initial temperature, cp0 is the slope of the specific heat (Cp(T_0) = c_p0 T_0). epsilon is the emissivity of the test mass, sigma is Stefan Boltzmann constant, A is the radiating surface area, and m is the mass of the test mass.

Up to the characteristic time, the cooling is slow. Then the temperature falls sqrt(t) after that.

As the surface-volume ratio m/A becomes bigger for a larger mass, in general, the cooling of the bigger mass requires more time.

For the QIL 4" mass, Mariner 150mm mass, and the Voyager 450mm mass, t0 is 3.8hr, 5.6hr, and 33.7hr respectively.

• If the emissivity is not 1, just the cooling time is expanded by that factor. (i.e. The emissivity of 0.5 takes x2 more time to cool)
• And if the shields are not cooled fast or have a finite temperature in the end, of course, the cooling will require more time.
• 1.25 t0 and 8 t0 tell us how long it takes to reach 200K and 100K.

This is the fundamental limit for radiation cooling. Thus, we have to use conductive cooling if we want to accelerate the cooling further more than this curve.

Attachment 1: cooling_curves.pdf
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