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

To quantify which of the fit below is the most significant, I did a Bayesian analysis (thanks Rory for the help!).

In brief, I compute the Bayes factors for each of the models considered below. As always in any Bayesian analysis, I had to assume some prior distribution for the fit parameters. I used uniform distributions, between 0 and 20e-4 for the loss angles, and between -100e-6 and 100e-6 for the slope. I checked that the intervals I choose for the priors have only a small influence on the results.

The model that has the highest probability is the one that considers different bulk and shear frequency depent loss angles. The others have the following relative probabilities

One loss angle constant:                       1/13e+13
One loss angle linear in frequency:      1/5.5
Bulk/shear angles constant:                  1/48784
Bulk/shear angles linear in frequency: 1/1

So the constant loss angle models are excluded with large significance. The single frequency dependent loss angle is less probable that the bulk/shear frequency dependent model, but only by a factor of 5.5. According to the literature, this is considered a substantial evidence in favor of frequency dependent bulk/shear loss angles.

Quote:

Results

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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