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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?

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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

  1. Move the $F$ term to the right hand side,
  2. Plug in the second to last equation and put the covariance term on the right hand side
  3. Use the continuity equation (5.15)
  4. Use the product rule on sum derivatives of products.
  5. 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.

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  • $\begingroup$ The reason i didnt think that $\langle \rho v\rangle=\rho \langle v\rangle$ is that i thought the rho should participate in the integration in the expectation value, but you assume you can pull it out of the integral, why is that? $\endgroup$ – MathematicalPhysicist Jan 9 '17 at 16:00
  • $\begingroup$ At each point you have some possibly non-equilibrium velocity distribution with probability density function say $f(\mathbf{v})$. According to the book, $\langle \cdot \rangle$ is defined as the average, at each position, of a quantity, with the average weighted by $f$. But $\rho$ is already the total number of particles at a point (times mass), integrated over $\mathbf{v}$, so $\rho$ does not depend $\mathbf{v}$. Therefore $\rho$ is a constant and can be pulled out of the average. This is in fact stated explicitly (for $n$ instead of $\rho$) at the bottom of page 97. $\endgroup$ – Brian Moths Jan 9 '17 at 16:26
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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 $(5.17)$

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  • $\begingroup$ You forgot about the product rule and $\rho$ does depend on time. $\endgroup$ – Brian Moths Jan 9 '17 at 13:44
  • $\begingroup$ yep just check it, i'm sorry $\endgroup$ – saberfull Jan 9 '17 at 13:48

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