An inquiry regarding a some derivation in Statistical Mechanics book by Huang My question is from Huang's second edition page 98.
For those who don't have the book at hand, I'll quote the derivation which I don't understand.
$u(\vec{r},t) = \langle v \rangle$.

We should then have three independent conservation theorems. For $\chi=m$ we have immediately $$\partial_t(mn)+
\partial_{x_i}
\langle mnv_i \rangle=0$$ or, introducing the mass density $$\rho(\vec{r},t)\equiv mn(\vec{r},t)$$ we obtain $$(5.15)\partial_t \rho + \nabla \cdot (\rho u)=0.$$ Next we put $\chi=mv_i$, obtaining $$(5.16)\partial_t \langle \rho v_i \rangle + \partial_{x_j}\langle \rho v_iv_j \rangle -\frac{1}{m}\rho F_i =0 $$ To reduce this further let us write $$\langle v_iv_j \rangle = \ldots = \langle (v_i-u_i)(v_j-u_j) \rangle +u_i u_j$$ Substituting this into (5.16) we obtain $$(5.17) \rho (\partial_t u_i +u_j\partial_{x_j}u_i)= 1/m \rho F_i- \partial_{x_j} \langle \rho (v_i-u_i)(v_j-u_j) \rangle$$

How was the last identity, eq (5.17) been derived from the previous identity, I don't see it, should it be easy, right?
 A: First I will give the intuition of both equations, and then I will say how to go from one to the other. If you just want the derivation, skip to the derivation section.
Inuition
First let me rewrite
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j}\langle \rho v_iv_j \rangle -\frac{1}{m}\rho F_i =0 $$
as
$$\partial_t \langle \rho v_i \rangle  =\frac{1}{m}\rho F_i - \partial_{x_j}\langle \rho v_iv_j \rangle. $$
Look it at this way, it says the change in momentum density at a point is equal to the force applied plus a term equal to the gradient in the covariance of the velocity. The intuition for this second is that it describes the momentum at a point changing because high momentum fluid is flowing into the point. 
Now let's look at the second expresssion.
$$\rho (\partial_t u_i +u_j\partial_{x_j}u_i)= 1/m \rho F_i- \partial_{x_j} \langle \rho (v_i-u_i)(v_j-u_j) \rangle$$
Here we are taking a piece of mass and following its location and looking at how its velocity changes at its location. The kinematic expression for the change in velocity of this piece of mass will have two terms. One term comes from the change in velocity at a constant point. The other term comes from the fact that even if the velocity at each point is constant, the velocity of an object will change if it moves from a region of high speed to low speed. The rate of velocity change from this is given by $v_i \partial_{x_i} v_j$. Now dynamically, there at two things which would cause the velocity of a piece of mass to change. One is a force, the other has to do that even at a single point, there is a spread in velocities, so particles from other locations will defuse to the location where we are looking. If diffusion is higher on one side than on another, we will see a net flux of particles from the side of higher diffusion and this will affect the average velocity where we are looking. 
Derivation
To derive the second relation from the first. We will procede in five steps: We will


*

*Move the $F$ term to the right hand side,

*Plug in the second to last equation and put the covariance term on the right hand side

*Use the continuity equation (5.15)

*Use the product rule on sum derivatives of products.

*Simplify by cancelling terms


Step one
First step one: moving the $F$ term. We go from  
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j}\langle \rho v_iv_j \rangle -\frac{1}{m}\rho F_i =0 $$
to
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j}\langle \rho v_iv_j \rangle  =\frac{1}{m}\rho F_i $$
Step Two
Now step two. Using the covariance equation $\langle v_iv_j \rangle =  \langle (v_i-u_i)(v_j-u_j) \rangle +u_i u_j$ we get
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j}\langle \rho v_iv_j \rangle  =\frac{1}{m}\rho F_i $$
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j} \rho \left(\langle (v_i-u_i)(v_j-u_j) \rangle +u_i u_j \right) =\frac{1}{m}\rho F_i $$
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j} \rho  u_i u_j  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
Step Three
For step three we will need to use the continuity equation $\partial_t \rho + \nabla \cdot (\rho u)=0.$ We will apply this equation to the $\partial_t \langle \rho v_i \rangle$ term. First lets notice that 
$$\partial_t \langle \rho v_i \rangle = \partial_t \rho \langle v_i \rangle = \partial_t \rho u_i =\rho\partial_t  u_i +u_i \partial_t \rho.  $$
Now applying the continuity equation to the second term on the right hand side, we get 
$$\partial_t \langle \rho v_i \rangle = \rho\partial_t  u_i - u_i \partial_{x_j} \rho u_j.$$
Plugging this equation in, we go from
$$\partial_t \langle \rho v_i \rangle + \partial_{x_j} \rho  u_i u_j  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
to 
$$\rho\partial_t  u_i - u_i \partial_{x_j} \rho u_j + \partial_{x_j} \rho  u_i u_j  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
Step Four
Now we will use the product rule to expand $\partial_{x_j} \rho  u_i u_j$ to $u_i \partial_{x_j} \rho   u_j + \rho  u_j \partial_{x_j}   u_i.$ This takes us from 
$$\rho\partial_t  u_i - u_i \partial_{x_j} \rho u_j + \partial_{x_j} \rho  u_i u_j  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
to
$$\rho\partial_t  u_i - u_i \partial_{x_j} \rho u_j + u_i \partial_{x_j} \rho   u_j + \rho  u_j \partial_{x_j}   u_i  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
Step Five
In step five, the two $u_i \partial_{x_j} \rho u_j$ terms cancel, and we can factor out a $\rho$. We go from
$$\rho\partial_t  u_i - u_i \partial_{x_j} \rho u_j + u_i \partial_{x_j} \rho   u_j + \rho  u_j \partial_{x_j}   u_i  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
to 
$$\rho\partial_t  u_i + \rho  u_j \partial_{x_j}   u_i  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
to
$$\rho\left(\partial_t  u_i +  u_j \partial_{x_j}   u_i \right)  =\frac{1}{m}\rho F_i -\partial_{x_j}\rho \langle (v_i-u_i)(v_j-u_j) \rangle $$
And this is the equation you wanted.
A: We should understand the meaning of $\left<A\right> = \frac{\int{du A\cdot f}}{\int du f}$, where $f$ is full distribution function, thus $f=f(r,u,t)$, and $\rho(r,t) = \int{du f(r,u,t)}$. So using $(5.16)$ and the fact that $\frac{\partial \rho}{\partial t}=-\partial_{x_{j}}{\rho u_{j}}$
$$
(\partial_{t}{\rho})u_{i}+\rho \partial_{t}\left<v_{i}\right> + \partial_{x_{j}}\rho\left<v_{i}v_{j} \right>= 0 
$$
$$
(\partial_{t}{\rho})u_{i}+\rho \partial_{t}\left<v_{i}\right> + \partial_{x_{j}}\rho\left<(v_{i}-u_{i})(v_{j}-u_{j}) \right>+\partial_{x_{j}}(\rho u_{i}u_{j})= 0 
$$
$$
-(\partial_{x_{j}}\rho u_{j})u_{i}+\rho \partial_{t}\left<v_{i}\right> + \partial_{x_{j}}\rho\left<(v_{i}-u_{i})(v_{j}-u_{j}) \right>+\partial_{x_{j}}(\rho u_{i}u_{j})= 0 
$$
It lead us to $(5.17)$
