Physical Origin of Gravitational Mass Defect For a static distribution of matter, the mass (let me call it mass1) is
$$m=\frac{4\pi}{c^2}\int^a_0T^0_0r^2dr$$
this is integrating over $4\pi r^2dr$. But then it is said that the element of spatial volume of metric should be $dV=4\pi r^2 \sqrt{g_{00}}dr$. So if it is integrated over the spatial volume, the mass is different (let me call it mass2).
Based on what I read so far, mass1 is the one used to calculate the gravitational radius and thus all the equations of motion. Then what is the meaning of mass2? It is said that the difference between mass1 and mass2 is to stabilize the star. Can someone give more physical insight to these two quantities?  
 A: Mass in General Relativity (GR) is a non trivial concept. When talking about a "mass" one needs to be care full or let me rephrase precise about what one is revering to.
For the following I will discuss the concept of "mass" in case of a spherical symmetric, static body. The line element of static, spherical symmetric space time can be written as: $$ds^2=-e^{\nu(r)}dt^2+e^{\lambda(r)}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2.$$
In free space $T_{\mu\nu}=0$ one can solve Einstein's field equation for that line element/metric which leads to the exterior Schwarzschild solution (ESS). One can get the ESS just by the symmetry arguments, the vanishing of the energy-momentum tensor and asymptotic flatness. In my notation the ESS reads:
\begin{align}
e^{\nu(r)}&=1-\frac{2M_{ESS}}{r},\\
e^{\lambda(r)}&=\left(1-\frac{2M_{ESS}}{r}\right)^{-1}.
\end{align}
Up until now $M_{ESS}$ is just a constant of integration. Without further discussion we can not identify $M_{ESS}$ as a mass; we only know it has the dimension of a mass but so has $2M_{ESS}\equiv R_s$. So there are (to my knowledge) at least two good reasons to call $M_{ESS}$ gravitational mass:


*

*$M_{ESS}$ is the single parameter governing the space time curvature and it is the mass felt by a quasi Newtonian observer at large distances.

*$M_{ESS}$ is the single parameter governing the general relativistic gravitational acceleration $\kappa$ is given by $$\kappa=-\frac 12 e^{(\nu(r)-\lambda(r))/2}\nu'(r)=-\frac{M_{ESS}}{r^2}.$$ For a definition of this radial acceleration I can refer to [Ø.    Grøn,1985]. This is a fully general relativistic expression and from pure coincidence it looks the same like the Newtonian gravitational acceleration.


In case of the ESS it makes sense to call $M_{ESS}$ "gravitational mass".
BUT the ESS says nothing about space with non vanishing $T_{\mu\nu}$. In the current setting lets consider only an ideal fluid as contribution to $T_{\mu\nu}$ so $T_{\mu}{}^{\nu}=\mathrm{diag}(-\epsilon(r),p(r),p(r),p(r)).$ In that case the TOV equations describe the space time curvature in regions where $P(r)$ and $\epsilon(r)$ are not zero. Lets call this region interior with $r\in[0,R]$, where $R$ is the stellar radius. At the surface $r=R$ the solutions of the TOV equations for $\nu(r)$ and $\lambda(r)$ must match the ones of the ESS. And one can show that inside the star $$e^\lambda(r)=(1-m(r)/r)^{-1},$$ with $$m(r)=4\pi\int_0^r \epsilon(\tilde r) \tilde r^2 d\tilde r$$ and the matching condition $m(R)=M_{ESS}$.
So maybe now to the point of what is $m(r)$. Ok at $r=R$ $m(R)=M_{ESS}$ holds and so at the stellar surface $m(R)$ has all the meanings of $M_{ESS}$ BUT for $r\neq R$ we do not know anything about $m(r)$ at the current point we can not say anything about $m(r)$ at $r\neq R$ because inside the star we introduced it just as a variable to handle the TOV equations. Inside the star $m(r)$ has no real meaning. It is often called "enclosed mass" but that is only in that sense true that once this quantity $m(r)$ is integrated from centrum to surface it is the gravitational mass.
When looking at it is is a completely odd expression: it is an integral over energy density but without the proper volume element of GR and why do we only integrate over the energy density and not over the trace of $T_{\mu\nu}$. 
The general expression for "active gravitational mass" $M_G$ for a static configuration in GR is the one for the "total enclosed energy" $U=M_{TW}$. I will give an expression for the total enclosed energy which is the active gravitational mass in Tolman-Whittaker form:
$$M_G(R)=M_{TW}(R)=U(R)=4\pi\int_0^R (-T_0^0+T_1^1+T_2^2+T_3^3)e^{(\nu(r)+\lambda(r))/2} r^2 dr.$$
This is the integral over the trace of the energy momentum tensor with the proper GR volume element. It corresponds to the total internal energy and the active gravitational mass enclosed in a sphere of radius $R$ and similar expressions can be given for non spherical but static configurations. BUT how does this expression make sense in case of the TOV equations: I just claimed that $m(R)=M_{ESS}$ is the gravitational mass of the star. So what now? Well $$m(R)=4\pi\int_0^R \epsilon(r) r^2 d r$$ and 
$$M_G(R)=M_{TW}(R)=U(R)=4\pi\int_0^R (\epsilon(r)+3P(r))e^{(\nu(r)+\lambda(r))/2} r^2 dr $$ look extremely different but they are the same. I claim $m(R)=M_G(R)=M_{TW}(R)=U(R)=M_{ESS}(R)$ in case of the TOV equations. This is a highly non trivial result which can be obtained by integrating $M_{TW}$ by parts using the TOV equations and its specific boundary conditions at $r=0$ and $r=R$. This holds only for $m(R)$ not for $m(r)$ with $r\neq R$ so again $m(r)$ has only a real meaning at $R$. $M_{TW}(r)=U(r)$ on the other hand has proper meaning at $r\neq R$: it really is the enclosed total energy up to a radius $r$. Plus one can show that even inside the star $$\kappa=-\frac{m_{TW}(R)}{r^2},$$ which again is a non trivial result of the TOV equations.
So this integral over just $\epsilon(r)$ without the proper volume element only has real meaning in case of the TOV equations and in principle only at $r=R$. It is a freak accident that this integral gives the gravitational mass: there are arguments for it in different books but in principle it is not so oblivious result of the TOV differential equations and boundary conditions. In the general case $M_{TW}$ would be the correct expression for the active gravitational mass and total energy.
In case of the TOV equation or deformed stars there are a lot of other masses defined by various integrals: proper mass or baryonic mass would be two examples.
A: To put it simply, you would like to see $\ m_1 \ $ as the total mass of the star minus the (classical) binding gravitational energy; this is given (classically) by
$$
U =  - \frac 35 \frac{G m_{\text{TOT}}^2}{r}.
$$
Of course, bound states have negative energies. So the idea is that the gravitational field of our star is generated by a lower mass. As a first, very shoddy, approximation, one would write something like: 
$$
m_1 =  m_{\text{TOT}} -  \frac 35  \frac{G m_{\text{TOT}}^2}{rc^2}.
$$
This doesn't go far, however, because the metric is different near the star. I'm assuming you are reading Landau-Lifshitz vol. 2, chapter 100: in the Schwarzschild solution you consider a metric in which the time and the radius agree with a flat metric at infinity. Now they write this solution in a very convenient form, which they derive from theoretical principles of simmetry and simplicity. This gives them the classical Schwarzschild metric
$$
ds^2 = \big( 1 - \frac{r_g}{r} \big) c^2dt^2 - \big( 1 - \frac{r_g}{r} \big)^{-1} dr^2  - r^2\big(\sin^2\theta \  d\phi^2 + d \theta ^2  \big).  
$$
What we have to understand here is that $\ r \ $ is not really the radius, that is the distance from the center, and $\ m \ $ is not really the mass: they are parameters which are calculated exploting the radial simmetry and the requirement that the metric is asymptotically flat (form. (100,12)). So, this $\ m \ $ (your $\ m_1 \ $) is the mass you would find measuring the gravitational field at great distance. In the book they get this through long calculations, hiding the idea  a little bit. 
So, this $\ r \ $ has the property that a sphere centered in the star has area $\ 4 \pi r^2, \ $ and this $\ m \ $ can be obtained by this confusing integral:
$$
m =  \int_0^a \frac{1}{c^2} T_0^0 4 \pi r^2 dr = \frac{4 \pi}{c^2} \int_0^a  \cal E r^2 dr.
$$    
In the book it's written that the "total mass," your $\ m_2 \ $  would be
$$
m_\text{TOT} = \frac{4 \pi}{c^2} \int_0^a  \cal E \frac{r^2 dr}{\sqrt{ 1 - \frac{r_g}{r}}} > m,
$$
BUT 1) the metric inside the star does not need to have an horizon at $\ r_g: \ $ in fact in reality the real internal metric would be smooth; 2) this would not calculate the energy of the gravitational field: intuitively, space is curved like the bottom of a cup near the star (for example see MTW \ "Gravitation,\ " page 614, fig. 23.1 ) so you see that the radius $\ r \ $ is like the euclidean radius measured from outside spacetime (like the radius of a parallel seen as a circle in 3D space  as opposed to the radius measured along the meridian). With a bit of differential geometry you should see that the given integral measures something like the total mass-energy of the star minus  the energy of the gravitational field.
