How to derive this $\dfrac{dT}{d\tau}$? I am studying the paper "Gravitational field of a particle falling in a Scharzschild geometry analyzed in tensor harmonics" by Zerilli. The author calculates the gravitational radiation emitted by a particle falling along a geodesic into a Schwarzschild black hole.
The stress energy tensor of the particle is:
$$T^{\mu\nu}=m_0 \dfrac{dT}{d\tau} \dfrac{dz^\mu}{dt} \dfrac{dz^\nu}{dt} \dfrac{1}{r^2}\delta(r-R(t)) \delta^{(2)}(\Omega-\Omega(t))$$
where the trajectory of the particle is $z^\mu=\left( T(\tau),R(\tau), \theta(\tau), \phi(\tau) \right)$.
In the appendices the author calculates the expression of the stress energy tensor of a particle falling radially into the black hole with the method of tensor harminics. For expample the 00 component is:
$$A_{lm}^{(0)} = m_0 \dfrac{dT}{d\tau} \left(1-\dfrac{2m}{r}\right)^2 \dfrac{1}{r^2} \delta(r-R(t)) Y_{lm}^*$$
In order to calculate the gravitational radiation emitted one must Fourier transform this expression: the author gives the procedure: 1) multiply by $\exp(i\omega t)dt$, the write $dR=dt/(dR/dt)$ so that the delta is simplified thanks to its properties. At the end of the calculations the author reports:
$$A_{lm}^{(0)}=\dfrac{m_0}{2\pi} \sqrt{\left(l+\dfrac{1}{2}\right) \dfrac{r}{2m}} \dfrac{1}{r^2} \exp(i\omega T(r))$$
The last two expressions give me troubles:

*

*Starting from the Fourier transform: where does the $\dfrac{dT}{d\tau}$ goes? Since it is the time component of the trajectory of the particle I thought that it could be derived by the Lagrangian as Wald does, i.e. from
$$-1=-\left( 1-\dfrac{2m}{r} \right) \dfrac{dT}{d\tau} + \left( 1- \dfrac{2m}{r} \right) \dfrac{dR}{d\tau}$$
from this I can derive the $\dfrac{dT}{d\tau}$ term, but it depends on $\dfrac{dR}{d\tau}$, whose dependence I do not know. Wald derives $\dfrac{dR}{d\tau}$ from the above Lagrangian by using the fact that (through Killing vecotrs) one has $E=\left(1-\dfrac{2m}{r} \right)\dfrac{dT}{d\tau}$, but then I need the energy $E$. From thecontour conditions of Zerilli problem the particle stars at infinity with 0 velocity, so at infinity $E=m_0c^2$, but this doesn't seem to be the way Zerilli calculated the Fourier transform, so I am lost;

*The second problem I have is with the the second equation I have reported: where does the $\left( 1-\dfrac{2m}{r} \right)^2$ term comes from?

 A: Well the paper is exactly about how to obtain such a decomposition. Just reading carefully shows that
$$ T^{00}= \sum_{L M} A^{(0)}_{LM}(r,t) \text{a}_{LM}^{(0)}$$
from eq (A1). So besides the Fourier transform and resolving the delta, one must multiply with ${\rm a}$ and sum over $M$ to get the result you want. Note $M$ is not $m$, so $m$ seems to be the Schwarzschild mass, while $M$ is the index of the spherical harmonic "magnetic number" and $m_0$ the particles mass, avoid being clumsy with the notation to understand better.
Question 1:
The author explicitly says that is the case of a particle falling in starting with "zero velocity", so you do know how $R$ depends on $t$ indirectly. You can solve the geodesic equation for the particle to find its wordline explicitly.
Question 2:
For the second question you only need to follow the metric used and the factors
$$\frac{dz^0}{dt}= \frac{dT(\tau)}{dt}= \frac{d\tau}{d t}\frac{dT}{d\tau}$$
check the metric and you should get the factors $$\left(1-\frac{2m}{r}\right)$$.
You will also get one of those factors from ${\rm a}^{(0)}$, a look at Table I should help.
PD: Also the author is a bit clumsy with using $T$ for the inverse function of $R(t)$ and also for the time component of the path.
A: The equations of motion for a geodesic starting at rest at infinity are very simple, and as usual the key is to start with the constants of motion:
$$ -1 = \frac{d z^\mu}{d\tau}\frac{d z^\nu}{d\tau} g_{\mu\nu} = -\left(1-\frac{2m}{r}\right)\left(\frac{d T}{d\tau}\right)^2 + \left(1-\frac{2m}{r}\right)^{-1}\left(\frac{d R}{d\tau}\right)^2$$
and
$$ 1=\mathcal{E} = -g_{t\mu}\frac{d z^\mu}{d\tau} = \left(1-\frac{2m}{r}\right)\frac{d T}{d\tau}$$
(The specific energy $\mathcal{E}$ is one because the particle starts at rest at infinity.)
These can be solved for $\frac{d T}{d\tau}$ and $\frac{d R}{d\tau}$
\begin{align}
\frac{d T}{d\tau} &= \left(1-\frac{2m}{r}\right)^{-1}\\
\frac{d R}{d\tau} &= -\sqrt{\frac{2m}{r}}
\end{align}
These can be combined to calculate.
$$\frac{d R}{dT} = \frac{\frac{d R}{d\tau}}{\frac{d R}{d\tau}} =  -\sqrt{\frac{2m}{r}}\left(1-\frac{2m}{r}\right) $$
You now have all the ingredients you need to calculate $A_{LM}$. (Also note that you can make your life easy by placing the incoming particle on the pole. This implies that  $A_{LM}=0$ for all $M$ except 0.)
