8
$\begingroup$

Only detectors above a certain sensitivity can measure with a high enough signal to noise ratio (SNR) to successfully pick out gravitational waves using matched filtering.

However, once we have discovered signals using LIGO detectors, can we gain anything from looking at the less sensitive or more noisy detectors such as GEO600? Once we know where the signal should be, and what it should look like, is there any way we can get more insight from the noisier data?

In pure signal processing terms: Suppose the same signal is transmitted on two channels, one with a very high SNR and one with a SNR so low that the signal can't be detected. Must the low SNR data simply be thrown out or are we able to combine it with the strong one to better resolve the detected signal?

$\endgroup$

1 Answer 1

2
$\begingroup$

It's a question of what is the most efficient route to new knowledge, or more secure knowledge. There is no need to throw any data away, but one has to make a judgement on what is most practical way ahead. For example, if the SNR is $N$ times lower then, in general, one will have to acquire $N^2$ times more data so that after averaging you get the same precision. However, this requires that something else does not intervene to make the averaging not work, and in practice there may be systematic effects which means the average will be off.

The judgement one must make, therefore, is whether it is better to spend your time improving an instrument such as LIGO, VIRGO and similar, or analyzing its data more rapidly, or increasing the proportion of time for which it is operational, or whether to let other people do that while you meanwhile get all the data you can from some less precise instrument. I don't know how the numbers will pan out in this example: that would require detailed technical knowledge of the limits of each detector. But the general rule is that detectors of less sensitivity or accuracy do have a role to play as long as they are not a lot less sensitive or they can make up for this by the kind of stability which allows for long averaging times. (And of course long averaging is not available for a brief gravitational wave burst.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.