# Stress-energy tensor of point particle when the trajectory is a transcendental equation?

I'm working through Carroll's GR book, and Problem 7.8 is not coming together. I'm missing something idiotically simple, but I'm not sure if I can cleanly write a stress-energy tensor for a point particle with the given path: $$r = \frac{2b}{1+ \cos \theta}$$ and $$t = \sqrt{\frac{2b^3}{M}} \left(\tan \frac{\theta}{2} + \frac{1}{3} \tan^3 \frac{\theta}{2} \right)$$ I have nice relationships for the velocities! $$\dot{\theta} = \sqrt{\frac{M}{2b^3}} (1 + \cos \theta)^2$$ and $$\dot{r} = \sqrt{\frac{2M}{b}} \sin \theta$$ But I need a $T^{00} ({\bf x},t)$ so I can get the quadrupole moment and hence the gravitational waves.

Typically I'd have something like $$T^{00} = M \delta(x^3) \delta(x^2) \left[ \delta(x^1 + (9Mt^2/8)^{1/3} ) + \delta( x^1 - (9Mt^2/8)^{1/3}) \right]$$ But I can't get the $x^i$ in terms of $t$.

Can I cleanly parametrize this system in terms of $\dot{r}$ and $\dot{\theta}$?

What am I missing?

Edit: Here is pretty much word for word the question as printed;

"The gravitational analog of bremsstrahlung radiation is produced when two masses scatter off each other. Consider what happens when a small mass $m$ scatters off a larger mass $M$ with impact parameter $b$ and total energy $E=0$. Take $M \gg m$ and $M/b \ll 1$. The motion of the small mass can be described by Newtonian physics, since $M/b \ll 1$. If the orbit lies in the $(x,y)$ plane and if the large mass sits at $(x,y,z)=(0,0,0)$, calculate the gravitational wave amplitude for both polarizations at $(x,y,z)=(0,0,r)$. Since the motion is not periodic, the gravitational waves will be burst-like and composed of mass different frequencies. On physical grounds, what do you expect the dominant frequency to be? Estimate the total energy radiated by the system. How does this compare to the peak kinetic energy of the small mass?"

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Can you state the metric and so forth for those of us who don't have that book? As written, your question can be answered only by someone who has a copy handy. – DarenW Apr 29 '13 at 1:07
I'm not sure if the information I provided is equivalent to whatever you had in mind for using the metric for the configuration, but I updated the question so that it's as complete as you can get. – user2053414 Apr 29 '13 at 1:14