Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitational waves, compensate for the Earth's movement

  1. Feb 15, 2016 #1
    How is it possible to compensate for LIGO's movement in the recent measurement of Gravitational Waves?

    I can understand that we are able to measure something to a tiny fraction of a proton when the subject and observer are both moving relative to each other through space but how is it possible when measuring Gravitational Waves. The waves are traveling across the universe, so the subject and observer are not moving relative to each other. LIGO is moving in so many complicated directions and at speed. We have earths rotational speed at the latitude of observation with earth's axis wobble, earths rotation and trajectory about the sun, the suns rotation and trajectory about the Milky way etcetera. Surely these compounded arcs of movement, speeds and precise angles at the moment of measuring a Gravitational Wave are not known exactly and would have to be calculated from older measurements that were not take to the same level of accuracy as the LIGO experiment is working at.

    Can this movement really be worked out accurately enough to give an absolute reading to such a small degree?

    How is it done?
  2. jcsd
  3. Feb 15, 2016 #2


    User Avatar
    Gold Member

    LIGO does not measure movement or distance relative to the source, it measures movement and distance within itself.

    The gravitational wave sweeps through the Earth, distorting it. The Earth (and everything else) flexes.

    This is Earth, as a gravity waves passes through it:

    This changes the distance between LIGO's emitter and detector.

    This is LIGO (simplistically - actually changes along both axes, as per Earth diagram above):
    See how the horizontal beam changes length?

    When you set up a laser beam so that it interferes with itself, even extremely tiny changes in the length will result in the two beams falling out-of-phase with each other.

    This is a terrible diagram to represent that's really happening, but it's all I could find:
    They set up the laser to reflect back upon itself so that it perfectly self-destructs, resulting in a flat line output. If the length of the beam changes by a tiny fraction, it no longer self-destructs, and instead a weak signal is detected.

    Last edited: Feb 15, 2016
  4. Feb 15, 2016 #3

    You're quite right in thinking there has to be some compensation for the motion of the detectors, but perversely it is not the complex and rapid motion of the detectors through the universe that cause the most headaches. Those movements can be fairly easily factored into the calculations because they are well known and don't change by much.

    It is the earth-bound local effects that cause the biggest problems in cleaning up the signal. Things like a truck hitting a pothole, a minor earthquake, a wave breaking on a nearby beach will all cause noise on the signal from any LIGO detector. In other words, it's the same sort of noise you'd expect to find on any VERY sensitive seismograph.

    The signal from each detector is cleaned up by filters that take these factors into account and the result is then compared with the output from the other detector at the other end of the USA. If a matching signal is received at the second detector then you can more or less rule out local noise.

    The difference in time of detection of the signal provides a clue as to which direction it came from. It also provides a sanity check because the time of detection can only differ by a maximum of the time it would take light to travel between the two detectors.

    With only two LIGOs operating at present you aren't going to get pinpoint accuracy on the direction of the source - you need three LIGOs to start seriously triangulating on the origin of the signal.
  5. Feb 16, 2016 #4


    User Avatar
    Science Advisor
    Gold Member

    Probably 4 or 5 detectors to be precise. That does not diminish the accomplishment.
  6. Feb 16, 2016 #5
    Thank you for your answers. It is a phenomenal accomplishment that all these things can be compensated for and filtered out to the required level of accuracy.

    @DaveC426913 Thanks for your diagrams. What I was trying to understand is that, in your Gif with the horizontal yellow beam, the beam should actually be rotating through the wave which would effect the beam length during the measurement.
  7. Feb 16, 2016 #6
    The detected waveform passed the detector in 0.2 seconds. The earth is essentially stationary in that timeframe.

    Edit: Here is the measured signal, look at the timescale.

  8. Feb 16, 2016 #7
    @Lord Crc
    >The earth is essentially stationary in that time frame.

    How can you say that? If the earth is traveling 67062 miles per hour around the sun in 0.2 we would have moved 0.3725 miles approx which is massive if your measuring something to a fraction of a proton. That's not even counting 1180 kilometers/hr earth rotation or movement of the solar system itself.
  9. Feb 16, 2016 #8
    Poor choice of words. The motion you mention is essentially constant linear motion on the timescale of the detection event. Thus it shouldn't matter, as it would just cause a constant disortion of the signal. At least that's my take on it.
  10. Feb 16, 2016 #9
    Thanks for your input.

    SF Cookie agrees that compensation for the movement of the detectors through the universe is necessary but says the movement can be fairly easily factored into the calculations. What's nagging my thinking is that even if calculating the arc movements of the earth and solar system is done using very accurate numbers the distances involved are so large that even the smallest approximation or rounding in the calculation would mean compensation would not be accurate to the required precision.

    For example when it's stated that the Milky Way Galaxy moving 1.3 million miles per hour through the universe that not exact especially when dealing with split seconds and fractions of a proton distances. Surely our exact movement relative to the waves is unknown.

    Sort of like trying to match up the theory of relativity with quantum theory.
  11. Feb 16, 2016 #10
    Martin, I don't think the LIGO team tries to compensate for the motion of the Solar system or our galaxy. We observe the gravitational waves in our own frame of reference - even if it is not really inertial - and the waves move at the speed of light (as far as we know) in our frame of reference, measured against our proper time. I hope someone can confirm or infirm this, as I am half guessing here -- gravitational waves push the abstraction to a new level ;)
  12. Feb 17, 2016 #11


    User Avatar
    Gold Member

    Gravitational waves, like light, move at the speed of light - In ALL frames of reference - including ours.
    Our movement around the sun and the motion of the galaxy are irrelevant.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook