-2
\$\begingroup\$

Looking at some ESA papers I found an unit that I don't know, picometers per square root hertz.

Anybody could tell me what this unit is for?

Thanks!

\$\endgroup\$
  • 7
    \$\begingroup\$ I would personally appreciate more context, e.g. a link to the paper in question. \$\endgroup\$ – Dzarda Jul 20 '15 at 9:51
  • 3
    \$\begingroup\$ It'd help a lot if you provided context of where you found this unit. Just "ESA paper" is very vague. \$\endgroup\$ – Curd Jul 20 '15 at 9:52
  • 2
    \$\begingroup\$ And what is ESA? European Space Agency or what? More context, please! \$\endgroup\$ – Lorenzo Donati supports Monica Jul 20 '15 at 9:54
1
\$\begingroup\$

I found this:

"The main aim of LISA Pathfinder is to demonstrate the concept of low-frequency gravitational wave detection in space. Since the gravitational wave "signal" (also called "ripples in space-time") is caused by the acceleration of massive objects (e.g. black holes mergers), the most interesting frequencies have been calculated to have a period of hundreds of seconds or more. In order to detect such signals, other frequencies must be carefully filtered out. The detection sensitivity is limited by the intrinsic noise of the measurement system in a specific frequency band. For this reason, signal and noise are normally referred to as "spectral density" and all measurement units are specified in "per square root Hertz" at a particular frequency."

Link: http://sci.esa.int/lisa-pathfinder/47856-picometre-precision-demonstrated-by-lisa-pathfinder-tests/

That explains the denominator. As for the picometers, a lot of people switch between wavelength and frequency. So I suppose they are intending picometers to be a wavelength.

In order to give you a bit more detail and specific information, we would need a bit more information as well.

Good luck!

\$\endgroup\$
  • 4
    \$\begingroup\$ No, the picometers does not refer to wavelength. The instrument is ultimately measuring the displacement of some masses that is caused by gravity waves, so naturally, the unit of measurement is length -- picometers, in this case. And the frequency in \$\sqrt{Hz}\$ is not the absolute frequency, but rather the bandwidth of the measurement. Noise power (for white noise) is directly proportional to bandwidth, and power is proportional to the square of the amplitude, so the noise amplitude (voltage, current, displacement, etc.) is proportional to the square root of the bandwidth. \$\endgroup\$ – Dave Tweed Jul 20 '15 at 11:43

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