Particle Current Density in GR TL;DR
Why is there a measure in the definition of particle current density?
Background 
I found myself studying kinetic physics in general relativity, which is fairly well documented in this livingreviews article. Kinetic physics, in this case, refers to switching from "particle description" into a "phase space description" and formulating general relativistic equations there. The only difference between these two descriptions is that one can think of "phase space" as a sort of probability distribution, and it can be used to derive macroscopic (large scale) equations with various approximations in some sense easier.
Question In the notes given above, they simply "define" particle current density (non-numbered equation) as: 
$N^a=-\int_{\mathcal{R}^3} f \sqrt{|g_{ab}|} p^a \frac{dp^1 dp^2 dp^3}{p_0}$
For which $N^0$ is supposedly the "general relativistic equivalent" of number density. I find myself disagreeing with this definition (in particular I am fixated on the term $|g_{ab}|$), as I tried solving it for Schwarzschild geometry. 
For Schwarzschild geometry:
$N^0=\int_{\mathcal{R}^3} f \cdot (r^2 \sin\theta) \cdot p^0 \frac{dp^1 dp^2 dp^3}{E}$, where $E$ is the covariant energy. 
Now I do not understand why there is a factor $\sqrt{|g_{ab}|}$ describing the volume element, because in this case I do not see how $N^0$ can describe density.
The way I understand each term: 


*

*$f(x^\mu, p^\mu)$ describes the probability density 

*$(r^2 \sin\theta)$ describes the effect of infinitesimal volume element

*$p^0$ describes the time dilation in GR

*$E$ I dont really understand why this is here but it seems to cancel $p^0$ when in local lorentz frame so why not.


Now obviously $(r^2 \sin\theta)$ means multiplication by a volume element (easiest to understand as $dV = r^2 \sin\theta dr d\theta d\phi$ in spherical coordinates, forgetting GR for a while). 
If $f$ already describes density, why is there an additional volume element?
Definitions


*

*$f$ is the phase space density

*$\sqrt{|g_{ab}|}|$ is the square root of determinant

*$p^a$ is the 4-momentum of a given particle.

 A: This definition of $N^\alpha$ looks a bit fishy to me to be honest: In "On the Topology and Future Stability of the Universe" Sec. 7.1.1 (available on google books) it is defined as: $$N^\alpha=\int_{P_\xi} f p^\alpha \mu_{_{P_\xi}}.$$ The description of $\mu_{_{P_\xi}}$ in that book sounds overly complicated to me but in sec. 5.3 "Collisionless Matter" of Numerical Relativity of T. W. Baumgarte there are several realtions for $\mu$ to $d^3V_p$ and  $d^4V_p$ depending on what one understands under $f$:
$$N^\alpha=\int p^\alpha \frac {F} m d^4V_p=\int p^\alpha \frac F {p^t} \frac{dmdp_1dp_2dp_3}{\sqrt{-g}},$$
with $F$ as number density in phase space/ distribution function defined by: $$F(x^\alpha,p_\beta)=\frac{dN}{d^3V_x d^4V_p}.$$
Here with a lorentzian metric $(-+++)$ and $$d^4V_p=\frac{-dp_t dp_1dp_2dp_3}{\sqrt{-g}}=\frac{m}{p^t}\frac{dm dp_1dp_2dp_3}{\sqrt{-g}},$$ 
$$d^3V_x=\frac{p^t} m \sqrt{-g} dx_1dx_2dx_3.$$
The $p^t$ in the denominator comes from the substitution $dp_t=-\frac{m}{p^t}dm.$ 
If one only describes particles with identical rest mass $m_0$ one can define $f$ with:
$$F(x^\alpha,p_\beta)=f(x^\alpha,p_i)\delta(m-m_0)\Leftrightarrow f(x^\alpha,p_i)=\frac{dN}{d^3V_x d^3V_p}$$
and $N^\alpha$ becomes:
$$N^\alpha=\int p^\alpha \frac f {p^t} \frac{dp_1dp_2dp_3}{\sqrt{-g}}.$$
A kind of disclaimer at the end: I "derived" the equations for $N^\alpha$ by looking at the expressions for the energy momentum tensor (given in Baumgarte's book) but this should be fine since the ony difference seems to be one $p^\beta$. For a detailed explanation I would recommend reading that section 5.3.
So so summarize I would say $\sqrt{-g}$ should be in the denominator at least if one defines $F$ and $f$ the way I presented.
