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In principle yes, gravitational waves dissipate energy into ordinary matter they pass through. However the coupling is extremely weak (that gravitational waves

In principle yes, gravitational waves dissipate energy into ordinary matter they pass through. However the coupling is extremely weak (that gravitational waves are so hard to detect is testament to this). In fact this weak coupling is what makes GWs so interesting for observational astronomy: They propagate from the source to our detectors virtually unchanged. (This is in contrast to EM radiation, which is very easy to scatter.) For example the farthest we can see back with EM radiation is about 200k years after the Big Bang, when the plasma of the early universe recombined into neutral hydrogen. By contrast gravitational waves can see back to the Big Bang itself, so it is a truly unique source of information as compared with light.

HappMacDonald 3y ago
I would rate the weak coupling as a far second or third point of interest behind linear signal fall-off vs inverse square for most other kinds of signal such as electromagnetic waves. Gamma ray burst is twice as far away? It's four times dimmer. A thousand times as far? A million times more dim. Gravitational wave signal from <event> is twice as far away? Makes it twice as hard to detect. A thousand times as far away? Only a thousand times as hard to detect.
- marsten 3y ago
  To be clear, gravitational waves follow the same inverse square law as EM radiation, in terms of power. This is after all just a statement of conservation of energy. The linear drop-off you're referring to is when we look at it in terms of field strength (in this case the spacetime strain). Since power is proportional to field squared, this implies a linear drop-off in the field. It just so happens that for GWs it's easier to detect the field, whereas for (most) EM radiation it's easier to detect power. There are field-detection methods for EM radiation as well, which are useful for weak signals. Homodyne and heterodyne detection are good examples.

HappMacDonald 3y ago

I would rate the weak coupling as a far second or third point of interest behind linear signal fall-off vs inverse square for most other kinds of signal such as electromagnetic waves. Gamma ray burst is twice as far away? It's four times dimmer. A thousand times as far? A million times more dim. Gravitational wave signal from <event> is twice as far away? Makes it twice as hard to detect. A thousand times as far away? Only a thousand times as hard to detect.

marsten 3y ago
To be clear, gravitational waves follow the same inverse square law as EM radiation, in terms of power. This is after all just a statement of conservation of energy. The linear drop-off you're referring to is when we look at it in terms of field strength (in this case the spacetime strain). Since power is proportional to field squared, this implies a linear drop-off in the field. It just so happens that for GWs it's easier to detect the field, whereas for (most) EM radiation it's easier to detect power. There are field-detection methods for EM radiation as well, which are useful for weak signals. Homodyne and heterodyne detection are good examples.