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Is it possible to deduce the density distribution only from the gravity field around it? Apparently this cannot be done in Newtonian gravity or static cases of GR: in spherical symmetric cases of Newtonian gravity or static cases of GR, the radial density distribution does not alter the gravity fields outside. However, I'm personally not quite familiar with the accelerating cases in GR.

Is is possible to appropriately accelerate any object and obtain its density distribution from the gravity waves around the object? Please state an algorithm or prove not.

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  • $\begingroup$ Well, we cannot purposely accelerate an object enough to make detectable gravitational waves. $\endgroup$ Commented Oct 19 at 20:51
  • $\begingroup$ As a practical matter, this isn’t possible as the previous comment explained. I don’t know whether it’s possible in theory or not, but is there some reason that you think this is of theoretical interest? $\endgroup$ Commented Oct 19 at 21:23
  • $\begingroup$ We can observe neutron star mergers. Can we learn anything about the density distribution from the gravitational waves? $\endgroup$ Commented Oct 20 at 2:02
  • $\begingroup$ @mmesser314 Excellent example! $\endgroup$ Commented Oct 20 at 6:42
  • $\begingroup$ The statement that we can't learn (anything) about a density distribution in Newtonian gravity seems false. Perhaps you could explain better what you mean and why you think that? And when you say accelerated, how is that different from an orbit? $\endgroup$ Commented Oct 20 at 6:48

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In the case of perfect spherical symmetry we can learn nothing about the density profile (other than that it is spherically symmetric). This is a consequence of Newton's shell theorem (or Birkhoff's theorem in GR) that says the potential (or metric) outside all spherically symmetric, static mass distributions of a given total mass are the same and spherically symmetric.

For all other cases, which includes every real object in the universe, there is no exact spherical symmetry and so, for example, an orbiting probe can yield detailed information about the distribution of mass inside the object it is orbiting. This is where much of the information we have about planetary interiors in our Solar System comes from.

Data from orbiting spacecraft can be used to constrain the mass multipole moments - the gravitational potential can be expressed in terms of a series expansion: mass monopole, (no dipole, because no negative masses), quadrupole etc. These in combination with physical modelling constrain the interior structure and density profile (e.g. Mazarico et a. 2014; Ni 2020).

Gravitational waves also contain information about the radial density profile of the (non-black hole) objects in a merging binary system. When the components get close they are distorted by tidal fields and this distortion, which depends on the interior equation of state and density profile, alters the gravitational wave emission and so can be used to model, for example, neutron star interiors (Chatziioannou et al. 2015.

I doubt there are uniquely invertible solutions, even if one were to measure all the multipoles to arbitrarily higher orders with high precision. It is usually the case that other physcal constraints are applied (e.g., hydrostatic equlibrium and an equation of state).

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    $\begingroup$ I think what the question is asking is to what extent does the multipole expansion outside of a bounded mass distribution uniquely determine the distribution. Do you have a citation for your claim that "Only in the case of perfect spherical symmetry can we learn nothing about the density profile (other than that it is spherically symmetric)"? That claims seems surprising to me. $\endgroup$ Commented Oct 20 at 15:29
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    $\begingroup$ @tparkerWhat other situations are there where we can learn nothing about the internal distribution of mass? Or are you arguing that we can learn something in the case of spherically symmetric mass distributions? I make no claim that a mass distribution can be uniquely determined. $\endgroup$ Commented Oct 20 at 15:48
  • $\begingroup$ I'm not entirely sure how to interpret your quoted statement. Is there a more precise delineation of the information that can and can't be learned about the density profile that holds in general, which makes clear what is exceptional about the spherically symmetric case? It is obviously false that for a non-symmetric mass distribution, we can "only learn that it is spherically symmetric". So I'm trying to understand the precise non-trivial mathematical substance of your claim. $\endgroup$ Commented Oct 20 at 15:57
  • $\begingroup$ @tparker I will review later, but I don't think you've understood what I've written. The shell theorem (or Birkhoff's theorem in GR) show that all spherically symmetric mass distributions have the same spherically symmetric exterior potential (or metric). $\endgroup$ Commented Oct 20 at 15:59
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    $\begingroup$ It's also worth noting that the multipole moments of a distribution do not uniquely determine the matter distribution; for an $2^n$-pole, you can scale up the physical size of the configuration by a factor of $\lambda$ and scale down the charge density by a factor of $\lambda^n$ and still get the same multipole moment. $\endgroup$ Commented Oct 20 at 16:12

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