Can you please explain “In a real search we would grid over the parameters and calculate the SNR time series for each one.” How do we achieve this in the least computationally expensive way? How do you account for unaligned spins or rotating black holes in actual analyses?
@AmbicaG Thank you for the question. I’m sorry for the slow reply. At some point, the best source of information for questions like these are methods papers about CBC searches. For example:
But, briefly, the searches use a grid over a limited number of parameters. Typically, a 3 parameter grid covering the masses of the two black holes and an aligned spin parameter. In most cases, we think this grid is robust enough to catch most binary black hole systems with only some losses. A non-aligned system is still detectable in most cases, but with a less-than-optimal SNR. A second set of analyses, known as parameter estimation, studies each detected system with a larger, 15 dimensional parameter space to measure the values of the spins, positions, and other parameters.
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Apart from the dip in the value of Chi squared at the time of the signal there is a general increase until the dip and subsequent decrease after the dip in L1,H1, not so prominent in V1. Whats the reason for this?
@AmbicaG Far away from the signal, chi-squared should be close to 1. The gradual rise in chi-squared is because times close to the merger have some signal power, but the power does not fit the template due to the template being in the wrong place. In other words, the gradual rise is indicating times when a signal is present, but NOT fitting the template well.
FWIW, here’s the explanation in the tutorial:
Also, the values climb just around this minima. This occurs because the template is starting to slide against the true signal in the data but is not perfectly aligned with it.
Hi! Is there a reason that in the notebook Tuto_2.3_Signal_consistency_and_significance.ipynb when calculating cmass we use m.median1d("mass1") instead of m.mass1? Or are these the same?
@zsembes Thank you for the question!
I checked, and I agree that m.mass1 and m.median1d("mass1") return the same result in this tutorial. So, I think you could use either one.
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