Curved spacetime point particle Lagrangian density This is probably trivially related to the question: 
Action for a point particle in a curved spacetime , but am a bit unsure how to write it as a Lagrangian density.
In curved spacetime the action is related to the Lagrangian density by:
$$ S = \int \sqrt{-g} \, \mathrm{d}^4 x \ \mathcal{L}(x^\mu)$$
The simpilest way I can think of describing a point mass taking a path through space time would be something like:
$$ \mathcal{L}(x^\mu) = -m\int d\alpha \ \delta^4(x^\mu-s^\mu(\alpha))$$
Where $\alpha$ is parameterizing a path $s^\mu(\alpha)$ through spacetime.  Is that the correct Lagrangian density?
Can someone show me how to manipulate these mathematical structures to check that this simplifies somehow to the known action for a free particle:
$$ S = - m \int d\tau $$
or how it relates to what was written in the other question?
$$\mathcal S =- m \int\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$$
 A: I didn't find so elegant expression as your ansatz but let me try. A classical
Lagrangian $L\left(  t,q^{i},v^{i}\right)  $ for a particle is a function of
seven variables: time $t$, particle coordinates $q^{i}$ and speed $v^{i}$ (I
use Latin labels for three-dimensional indices while Greek-indexed components
are for 4-metric). Since we are going to talk about massive particles only,
the time can be used to parameterize particle trajectory, hence using the
action
\begin{equation}\tag{1}
\begin{split}
S&=-m\int d\tau\\
&=-m\int\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}\\
&=\int dt\,L,
\end{split}
\end{equation}
we find the classical Lagrangian:
\begin{equation}
\begin{split}
L\left(  t,q^{i},v^{i}\right)  &=-m\frac{d\tau}{dt}\\
&=-m\sqrt{g_{00}\left(
t,q\right)  +2g_{i0}\left(  t,q\right)  v^{i}+g_{ij}\left(  t,q\right)
v^{i}v^{j}},
\end{split}
\end{equation}
where I explicitly showed the arguments of the metric tensor. It is easy to
represent $L$ as a formal 3-fold integral:
\begin{equation}
L\left(  t,q^{i},v^{i}\right)  =-m\int d^{3}x\,\delta^{\left(  3\right)
}\left(  q^{i}-x^{i}\right)  \sqrt{g_{00}\left(  \hat{x}\right)
+2g_{i0}\left(  \hat{x}\right)  v^{i}+g_{ij}\left(  \hat{x}\right)  v^{i}
v^{j}},
\end{equation}
where $g\left(  \hat{x}\right)  =g\left(  t,x\right)  $. Therefore the action
(1) takes the form:
\begin{align}
S  &  =-m\int d^{4}x\,\,\delta^{\left(  3\right)  }\left(  q^{i}-x^{i}\right)
\sqrt{g_{00}\left(  \hat{x}\right)  +2g_{i0}\left(  \hat{x}\right)
v^{i}+g_{ij}\left(  \hat{x}\right)  v^{i}v^{j}}\\
&  =-m\int d^{4}x\sqrt{-g\left(  \hat{x}\right)  }\,\left\{  \frac{\delta^{(  3)}\left(  q^{i}
-x^{i}\right)}{\left[  -g\left(
\hat{x}\right)  \right]  ^{1/2}}  \sqrt{g_{00}\left(  \hat{x}\right)  +2g_{i0}\left(  \hat
{x}\right)  v^{i}+g_{ij}\left(  \hat{x}\right)  v^{i}v^{j}}\right\}
,\nonumber
\end{align}
where $g=\det g_{\mu\nu}$. The strange quantity in the curly brackets above
can be called the Lagrangian density.
Let me represent this density in a more pleasant way. First, I would like to
use ADM-like notations (ADM stands for Arnowitt, Deser, and Misner):
\begin{equation}
\gamma_{ij}=-g_{ij}+\frac{g_{i0}g_{j0}}{g_{00}},\qquad g_{i}=-\frac{g_{i0}
}{g_{00}},\qquad h^{2}=g_{00},
\end{equation}
thus the Lagrangian density has the form:
\begin{equation}\tag{2}
\mathcal{L}\left(  \hat{x},q^{i},v^{i}\right)  =\left(  \frac{h^{2}}
{-g}\right)  ^{1/2}\delta^{\left(  3\right)  }\left(  q^{i}-x^{i}\right)
\left[  \left(  1-g_{i}v^{i}\right)  ^{2}-h^{-2}\gamma_{ij}v^{i}v^{j}\right]
^{1/2}.
\end{equation}
Let me start the explanation with the matrix $\hat{\gamma}$. Suppose, there are
two points $A$ and $B$ separated by $dx$. If you send a light signal from
point $A$ to point $B$ and back, then you can define the spatial distance $dl$
between the two points as $cT/2$, where $c$ is the speed of light and $T$ is the time
interval between signal sending and signal receiving. It is easy to show
that the element of spatial distance such defined is $dl^{2}=\gamma_{ij}
dx^{i}dx^{j}$. Therefore the 3-tensor $\gamma_{ij}$ accounts for spatial
geometry. You can consider it as a kind of induced 3-metric. In fact,
$\gamma_{ij}$ is simply the inverse matrix of the three-dimensional \ part of
the contravariant metric:
\begin{equation}
-g^{in}\gamma_{nj}=\delta_{j}^{i},
\end{equation}
It is also easy to show:
\begin{equation}
-g^{in}\gamma_{nj}=\delta_{j}^{i},\qquad-g=g_{00}\gamma,\qquad\Rightarrow
\qquad\left(  \frac{g_{00}}{-g}\right)  ^{1/2}=\gamma^{-1/2}=\sqrt{-\det
g^{mn}}.
\end{equation}
Therefore, in the quantity $h^{-2}\gamma_{ij}v^{i}v^{j}=\gamma_{ij}
dq^{i}dq^{j}/(h\,dt)^{2}=\left(  \Delta q\right)  ^{2}/(h\,dt)^{2}$ in the
Lagrangian density (2), $\Delta q$ is the real element of spatial
distance along particle's trajectory. 
The quantity $\Delta t=h\,dt=\sqrt
{g_{00}}dt$ defines the proper time for the given point in space,
i.e.
\begin{equation}
\left.  d\tau\right\vert_{dx^{i}=0}=g_{00}\,dt,
\end{equation}
hence $\sqrt{g_{00}}dt$ is the actual time interval in the comoving reference
frame of the particle.
It looks like that $\left(  \Delta q\right)  ^{2}/(h\,dt)^{2}$ is the real
particle speed which can be measured by an external observer, but it is not
completely true. There is also so called a synchronization correction. It is
closely connected to the procedure of synchronizing clocks located at
different points in space. One can show (see, e.g., Landau, Lifshitz, vol. II,
«The classical theory of fields») that if the clocks are synchronized by the passing light signal (as considered above) the time interval $dt$ in the point $A$ should be corrected by $\Delta t=-g_{i}dx^{i}$ in the point $B$. For example, if you choose the closed contour in the space with $g_{i}\neq0$ and try to synchronize all clocks along
the contour, you will find the time difference, which would be recorded upon
returning to the starting point is:
\begin{equation}
\Delta t=-\oint g_{i}dx^{i}.
\end{equation}
Taking into account all above, we can define the quantity
\begin{equation}
V^{i}=\frac{dq^{i}}{(1-g_{i}v^{i})\sqrt{h}dt}=\frac{dq^{i}}{\sqrt{h}\left(
dt-g_{i}dq^{i}\right)  },
\end{equation}
which should be calculated along particle's trajectory. Therefore
\begin{equation}
\frac{h^{-2}\gamma_{ij}v^{i}v^{j}}{\left(  1-g_{i}v^{i}\right)  ^{2}}
=\gamma_{ij}V^{i}V^{j}=V^{2} 
\end{equation}
is the square of the actual particle speed measured by properly calibrated
rulers and proper time determined by clocks synchronized along the trajectory
of the particle. Finally, the Lagrangian density takes the form:
\begin{equation}
\mathcal{L}\left(  \hat{x},q^{i},v^{i}\right)  =-m\,\sqrt{1-V^{2}}\left[
\gamma^{-1/2}\,\delta^{\left(  3\right)  }\left(  q^{i}-x^{i}\right)  \left(
1-g_{i}v^{i}\right)  \right]  ,
\end{equation}
where $\gamma^{-1/2}\,\delta^{\left(  3\right)  }\left(  q^{i}-x^{i}\right)
\times\left(  1-g_{i}v^{i}\right)  $ is correctly defined three-dimensional
Delta-function supplied by the time-synchronizing correction.
The conception of the speed $V^{i}$ is very useful. For example, suppose there
is a stationary metric $g_{\mu\nu}\left(  x^{i}\right)  $ («stationary» means $g_{\mu\nu}$ does not depend on time), then particle's energy, which conserved along the trajectory, is as follows:
\begin{equation}
E=\frac{m\sqrt{g_{00}}}{\sqrt{1-V^{2}}}.
\end{equation}
A: your Lagrangian is almost correct. But you also need to use a mass term that is conserved, which won't be the case of the mass term with your Lagrangian.
If you use: $$ \mathcal{L}_m =  \sum_p m_p \gamma_p^{-1}(-g)^{-1/2}\delta^{(3)}(x^j-x^j_p(\tau_p)),$$
where $\gamma_p=dx^0/cd\tau_p$ the Lorentz factor, $u^\mu_p=dx^\mu_p/cd\tau$ the 4-velocity of the particle (such that $u^\mu u_\mu=-1$) and $x^j_p(\tau_p)$ the trajectory of the p-th particle, then you have all what you need. Indeed, you can directly compute that it reduces to $$S=-\sum_p m_p \int d \tau_p,$$
where $m_p$ is conserved.
Note that this is intimately related to the so-called conserved density $\rho^*$ that is often used in practical applications of general relativity (e.g. in cellestial mechanics). $\rho^*=\sqrt{-g} \gamma_p \rho$, where $\rho$ is the density appearing in the stress-energy tensor. Then, one has the "Newtonian" conservation of the so-called conserved density (i.e. $\partial_0 \rho^*+\partial_i(\rho^* v^i)=0$, with $v^i = u^i/\gamma_p$). (You can derive it from the conservation equation $\nabla_\sigma(\rho u^\sigma)=0$).
