# How do compute an energy momentum tensor, given some equations of motion

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.
$$T^{\mu \nu} = (\rho + p) U^\mu U^\nu + p g^{\mu \nu}$$
where:
$$c = G = 1$$ 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
where:
$$\delta$$ 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.

• +1 Ottimo ! (very good in Italian), but I think you meant cylindrical coordinates – magma Dec 16 '18 at 21:10
• 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... – John Doe Dec 16 '18 at 22:47
• @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. – Michele Grosso Dec 17 '18 at 16:43
• @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 $dx dy$. – Michele Grosso Dec 17 '18 at 16:53