Part 1: How do gravitational waves decay with distance?

My Newtonian instincts tell me that gravitational waves would likely decay in the same way that Newtonian gravity decays, but at this point I've realized how misleading Newtonian gravity can be.

My guess would be that for a gravitational wave with an initial amplitude $a$ that traveled distance $r$ would have an amplitude of $r^{-2}a$. doubling the distance means one quarter the amplitude, triple the distance means one ninth the amplitude, and so on. I would also guess that the energy of gravitational waves is related to their amplitude in a similar way to kinetic waves.

Part 2: Does the decay of gravitational waves depend on time?

Let's say I take two gravitational wave pulses. One passes through empty space and the other passes an object that distorts space-time in a measurable way, such as a black hole.

If I measure them after they have traveled the same distance, would one have a higher amplitude or contain more energy than the other? Would this be due to time being warped through the warping of space-time? Again, the distance they traveled is the same so I imagine the only difference would be the time it took to cross that distance.

Please keep in mind that I have no experience with the math side of general relativity or special relativity. I tried searching for answers online but I couldn't find anything that a physics enthusiast with no university level physics courses could understand.

  • $\begingroup$ on part 2, one term to look up is "superradiant scattering". A wave (of any sort, including a gravitational wave) passing a rotating black hole can be amplified $\endgroup$ Oct 24, 2019 at 11:06

1 Answer 1


Part 1: no, the amplitude decays as 1/r. The power or energy of the wave decays like the square of that, like 1/r^2 (one over r squared). This is the same as in electromagnetic waves, the gravitational wave amplitude (in General Relativity) components obey the same equations as that for electromagnetic waves for the electric or magnetic field. The reason is that although forces go like 1 over r squared, that can be also used to see how the electric field (or Newtonian gravitational field) changes with distance r, but it is true for STATIC or slowly changing fields, not true for radiating fields, eg, propagating waves.

To figure out propagating waves you need to figure out how the time change of the fields (electric, magnetic or gravitational, they give the same results) change. You need to derive the wave equations for those. It then comes out that the fields go like 1/r. The fields are the amplitudes, A. The power or energy in all cases those cases goes like A squared. That holds true for radially propagating electromagnetic fields and gravitational fields. For gravitational fields it is the metric of the spacetime in Einstein's General Relativity equations that define the fields. For gravitational waves, in the linearized case, the waves are represented as small changes, with amplitude A, of the background metric. In both cases it is the square of A that is the power or energy (and actually it is power or energy density, per unit area), which then goes like 1/r squared.

It is easy why it must be so. Realize that those energies are conserved as they propagate. So the total energy at a distance r, summed around the whole spherical wavefront (yes, consider them radiating away radially symmetrically) is then the power or energy density, multiplied by the area of a sphere. Since area = 4 x pi x r^2, the energy or power density must go like 1/r^2, so the total power or energy is the same at any spherical distance r. It's energy conservation that leads to that behavior.

Please be aware that these statements for the gravitational field are true for linearized gravity, i.e. For Einstein's equations, but where the waves or disturbances in the gravitational field, the metric of spacetime, is small. For very strong gravity and gravity changes things get much more complicated because Einstein's equations are nonlinear, gravity interacts with itself, and gravity is also be a source of a gravitational field. Things get complicated and you need some understanding of General Relativity to understand what that really means. If you are interested, there's plenty good online lectures and summaries (and bad ones also, careful).

Part 2: the decay of gravitational wave amplitude or power or energy density. Does it depend on time?

It might (remember that in General Relativity you can change coordinates and mix space and time)@ but in the linearized Einstein equations it does not, if you stick with the flat coordinate system as the reference frame. It'll depend on distance. And yes, then there is so called lensing, which is waves passing close to bodies some of the gravity and distort spacetime (in your terms, all of those together really just create the curvature of spacetime), and thus space near them, and thus a bending of the waves like lenses do. So waves going near them will bend, and typically travel a longer distance (but it all depends on the overall distribution of mass around their overall path), and arrive delayed. We can measure the so called lensing delays for electromagnetic waves and actually used it to measure the mass densities around various galaxies and galaxy clusters. We can even measure the mass densities of dark matter around them. We have not yet observed gravitational wave lensing, but it is expected and will be used to measure more details from bodies or objects that are not visible through electromagnetism.

So, yes gravitational waves will vary their amplitude and power with the longer lensing distances, and be delayed. The effects will be small, and the amplitude changes will be much harder to detect accurately than the timing delays. The space based gravitational interferometer to be launched in the next few years will be able to see the delays, not sure about the distance variations through the amplitude changes, but time delays determines distance anyway so we'll measure those distance changes also.


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.