This problem can be found in a paper called "Gravitational Radiation From Point Masses In A Keplerian Orbit", but I do not have access to this, so cannot see how to do it.

I have been given a problem where we have two (point-like, non-relativistic) masses $m_1,m_2$, which are in some kind of binary orbit. They each trace out an ellipse (of different sizes), and we are given two formulae / equations of motion:

  • One for the distance between the two masses, $r(t)$, for all time (given in terms of the eccentricity)

  • One for the angular velocity $\dot\phi(t)$ in terms of $r(t)$

Then I am asked to compute the energy momentum tensor for this system.

I don't know exactly where to start. I had some ideas - it would be helpful if I knew a particular action to apply a variation to to obtain the energy-momentum tensor, but I don't know which action to use. I don't know what form of metric to use either (if I did, I could perhaps compute the Ricci tensor and Ricci scalar, then use Einstein's equations to find $T_{\mu\nu}$). But again, I don't have this starting point.

Any ideas or hints?


Let us consider the energy-momentum tensor $T^{\mu \nu}$ of a perfect fluid. Then we specialize to the configuration in question.
$T^{\mu \nu} = (\rho + p) U^\mu U^\nu + p g^{\mu \nu}$
$c = G = 1$ natural units
$\rho$ energy density in the rest frame
$p$ pressure in the rest frame
$U^\mu$ four-velocity
$g^{\mu \nu}$ inverse metric tensor

A nonrelativistic system means:
1) Energy density close to mass density in the rest frame
2) Pressure negligible. In fact $p = (1/3) v^2 \rho$ where the Newtonian velocity $v$ is negligible compared to $1$ (speed of light in natural units).
3) $U^\mu$ close to $(1, \vec v)$. Again the spatial part is negligible compared to the time component.

Hence the energy-momentum tensor of a binary system with point-like nonrelativistic masses $m_1$ and $m_2$ can be approximated in a polar coordinates system $(t, r, \theta, z)$ centered in one of the focal points as
$T^{t t} = m_1 (1/r) \delta(r - r_1) \delta (\theta - \theta_1) \delta (z) + m_2 (1/r) \delta(r - r_2) \delta (\theta - \theta_2) \delta (z)$
Other components negligible
$\delta$ Dirac delta function
The factor $(1/r)$ allows for the unity when integrating the $\delta$ function in polar coordinates
$r_1, \theta_1, r_2, \theta_2$ as functions of time $t$ are the laws of motion of the masses and are given

Note: The delta functions allow for a straightforward integration to get the quadrupole moment which in turn allows for the gravitational perturbation.

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  • $\begingroup$ +1 Ottimo ! (very good in Italian), but I think you meant cylindrical coordinates $\endgroup$ – magma Dec 16 '18 at 21:10
  • $\begingroup$ Thank you for this, it mostly makes sense now. One question I have is where exactly does the factor of $1/r$ come from, relative to the original equation for $T_{\mu \nu}$? Is it because $\rho$ is the mass density rather just the mass? I'm not convinced by that part... $\endgroup$ – John Doe Dec 16 '18 at 22:47
  • $\begingroup$ @magma. Yes, in three dimensional space it is called cylindrical coordinate system. I had in mind the plane of the orbit, that is why I stated polar. $\endgroup$ – Michele Grosso Dec 17 '18 at 16:43
  • 1
    $\begingroup$ @John Doe. Yes, $\rho$ is the mass density that you have to integrate in the plane of the orbit. The infinitesimal area in a plane in polar coordinates $(r, \theta)$ is $r dr d\theta$. You have to assume the factor $1/r$ to offset the $r$ of the infinitesimal area. In this way the integration of the $\delta$ function gives the mass $m$. Of course, if you use cartesian coordinates $(x, y)$ in the plane of the orbit you do not need any factor, as the infinitesimal area is $dx dy$. $\endgroup$ – Michele Grosso Dec 17 '18 at 16:53

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