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Entry  Tue Jul 18 09:13:08 2017, Gabriele, General, Measurements, Shear and bulk losses in tantala 7x
    Reply  Wed Jul 19 21:19:14 2017, Gabriele, General, Measurements, Shear and bulk losses in tantala 
Message ID: 369     Entry time: Tue Jul 18 09:13:08 2017     Reply to this: 370
Author: Gabriele 
Type: General 
Category: Measurements 
Subject: Shear and bulk losses in tantala 

S1600525 has been coated in Fort Collins with 480nm of pure tantala. I used the emasured loss angles (after deposition, before annealing) to estimate the shear and bulk loss angles.


First, my COMSOL simulation shows that even if I don’t include the drum-like modes, I still have a significant scatter of shear/bulk energy ratio. The top panel shows indeed the ratio shear/bulk for all the modes I can measure, and the variation is quite large. So, contrary to my expectation, there is some room for fitting here. The bottom panel just shows the usual dilution factors.

Then I tried to fit the total losses in my sample (the substrate is negligible) using four different models:
1) one single loss angle for both bulk and shear, constant in frequency
2) one single loss angle for both bulk and shear, linear in frequency
3) separate bulk and shear loss angles, constant
4) separate bulk and shear loss angels, linear in frequency 
Instead of using Gregg harry's technique (taking pairs of losses together), I simply fit the whole datasets with the assumptions above. I derived the 95% confidence intervals for all parameters. I also weighed each data point with the experimental uncertainty. I’m not sure yet how to compare the performance of the various models and decide which is the best one, since clearly the more parameters I plug into the model, the better the fit gets.
If I use two different loss angles, but constant, I get numbers similar to what Gregg presented at the last Amaldi conference (G1701225), but inverted in bulk and shear. I cross checked that I didn’t do any mistake. Instead, if I allow linear dependency on frequency of bulk and shear, I get a trend similar to the one in Gregg's slides.
My plan is to have this sample annealed today or tomorrow and measure it again before the end of the week. 


One loss angle - constant

\phi = (6.99 \pm 0.05) \times 10^{-4} \mbox{ rad}

One loss angle - linear in frequency

\phi = (6.91 \pm 0.07) \times 10^{-4} +\frac{f-1 \mbox{ kHz}}{1 \mbox{ kHz}} \cdot (3.3 \pm 2.2) \times 10^{-6} \mbox{ rad}

Bulk and shear - constant

\begin{matrix} \phi_{shear} = (6.79 \pm 0.12) \times 10^{-4} \mbox{ rad} \\ \phi_{bulk} = (8.54 \pm 0.98) \times 10^{-4} \mbox{ rad} \end{matrix}

Bulk and shear - linear in frequency

\begin{matrix} \phi_{shear} = (6.9 \pm 0.4) \times 10^{-4} +\frac{f-1 \mbox{ kHz}}{1 \mbox{ kHz}} \cdot (9.9 \pm 7.4) \times 10^{-6} \mbox{ rad} \\ \phi_{bulk} = (6.4 \pm 3.7) \times 10^{-4} +\frac{f-1 \mbox{ kHz}}{1 \mbox{ kHz}} \cdot (-14 \pm 39) \times 10^{-6} \mbox{ rad} \end{matrix}

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