Delta function integration to find stress energy for a test particle

Question

I am reading a paper, where the author considers the following integral:

\begin{equation}\tag{1} T^{\mu \nu} = \int_{-\infty}^{\infty} \delta^4 (x-z(\tau)) \frac{dz^{\mu}}{d \tau} \frac{dz^{\nu}}{d \tau} d\tau \end{equation}

where $z^{\mu} = (T(\tau), R(\tau), \Theta(\tau), \Phi(\tau))$ and $\tau$ is the proper time. For context, it describes the stress energy tensor for a particle moving on a geodesic.

Now, they state the following simplification:

\begin{equation} T^{\mu \nu} = \frac{dT}{d \tau} \frac{dz^{\mu}}{dt} \frac{dz^{\nu}}{dt} \frac{\delta(r-R(t))}{r^2} \delta^{(2)} (\Omega - \Omega(t))\tag{2} \end{equation}

where $\Omega = (\theta, \phi) \text{ and } \delta^{(2)} (\Omega) = \delta(cos\theta)\delta(\phi)$.

I struggle to understand this simplification; I do not even see intuitively how $\frac{1}{r^{2}}$ and $cos\theta$ factors appear there. I would really appreciate some help in understanding this.

Answer 1

1. They have integrated out $\tau$ by using $\delta(t-\tau)$;
2. They have used spherical coordinates, which leads to $\delta(r-R(t))/r^2$
3. The other components remain unchanged, thus you have $\delta(\cos(\theta))$ and $\delta(\phi)$.

The general delta function in 4D curved space is $$ \frac{\delta^{4}(x-x_c)}{\sqrt{ |g| }} $$ where $g$ is determinant of metric.

Delt function in spherical coordinates: Since $\sqrt{g}=r^2\sin\theta$, the full delta function without any additional symmetries should be $$ \frac{1}{r^2\sin\theta} \delta( r- r_c) \delta( \theta- \theta_c) \delta( \phi- \phi_c) $$