In a set of notes about cosmology, I have found the following claim:
The 0th moment of the Vlasov equation yields the continuity equation. For that, upon integrating over the momentum, we have to integrate the last term by parts and use that the potential is independent of the momentum. Taking the first moment and using the continuity equation yields the Euler equation.
I have tried several times to do the calculations that are described, to no avail. The starting point seems to be the Vlasov equation that was previously derived for the cosmological fluid, which is:
$$\dfrac{\partial f}{\partial\tau}+\dfrac{\vec{p}}{ma}\dfrac{\partial f}{\partial\vec{x}}-am\vec{\nabla}\phi\cdot\dfrac{\partial f}{\partial\vec{p}}=0$$
where $a$ is the scale factor, $\tau$ denotes conformal time, $\vec{\nabla}\equiv\vec{\nabla}_{\vec{x}}$ is the gradient with respect to comoving coordinates $\vec{x}$, and $f=f(\vec{x},\vec{p},\tau)$ is the distribution function.
The author is using the following definitions for the density, mean streaming velocity and velocity dispersion of the fluid, respectively:
$$\rho(\vec{x},\tau)=\dfrac{m}{a^3}\int d^3p\ f(\vec{x},\vec{p},\tau)$$
$$v_i(\vec{x},\tau)=\dfrac{\displaystyle\int d^3p\ \dfrac{p_i}{am}\ f(\vec{x},\vec{p},\tau)}{\displaystyle\int d^3p\ f(\vec{x},\vec{p},\tau)}$$
$$\sigma_{ij}(\vec{x},\tau)=\dfrac{\displaystyle\int d^3p\ \dfrac{p_i}{am}\dfrac{p_j}{am}\ f(\vec{x},\vec{p},\tau)}{\displaystyle\int d^3p\ f(\vec{x},\vec{p},\tau)}-v_i(\vec{x})v_j(\vec{x})$$
and the equations that we want to obtain by taking moments of the Vlasov equation are:
- Continuity equation:
$$\delta'+\vec{\nabla}\cdot[\vec{v}(1+\delta)]$$
- Euler equation:
$$v'_i+\mathcal{H}v_i+\vec{v}\cdot\vec{\nabla}v_i=-\nabla_i\phi-\dfrac{1}{\rho}\nabla_i(\rho\sigma_{ij})$$
where $\delta$ is the density contrast, defined as $\rho=\bar{\rho}(1+\delta)$, and $\mathcal{H}=a'/a\equiv (1/a)\partial_\tau a$.
I have attempted to follow the steps described in the text, by integrating the Vlasov equation over the space of momenta:
$$\int d^3 p\ \dfrac{\partial f}{\partial\tau}+\int d^3 p\ \dfrac{\vec{p}}{ma}\dfrac{\partial f}{\partial\vec{x}}-\int d^3 p\ am\vec{\nabla}\phi\ \dfrac{\partial f}{\partial\vec{p}}=0$$
but I am lost and don't know how to proceed. For instance, I could manipulate the first term in the following way:
$$\int d^3 p\dfrac{\partial f}{\partial\tau}=\dfrac{\partial}{\partial\tau}\int d^3p\ f=\dfrac{\partial}{\partial\tau}\bigg(\dfrac{a^3}{m}\underset{=\rho(\vec{x},\tau)}{\underbrace{\dfrac{m}{a^3}\int d^3p\ f}}\ \bigg)=$$
$$=\dfrac{\partial}{\partial\tau}\bigg(\dfrac{a^3}{m}\rho(\vec{x},\tau)\bigg)=\dfrac{1}{m}\dfrac{\partial}{\partial\tau}[a^3\rho(\vec{x},\tau)]=\dfrac{1}{m}(3a^2a'\rho+a^3\rho')$$
However, this seems to take me nowhere. Any help or advice, please?
Best Answer
Probably the first thing to note is that the spatial and momenta coordinates are independent, so $\nabla_\mathbf{x}\mathbf{p}=0$ and $\nabla_\mathbf{p}\mathbf{x}=0$. For the terms, it's probably easiest to take each one independently and reduce it. My background is more on the plasma physics side than cosmology, so the Vlasov equation I'm familiar with is, $$\partial_tf+\mathbf{v}\cdot\nabla_xf+\mathbf{a}\cdot\nabla_vf=0$$ where $\mathbf{a}=\dot{\mathbf{p}}$ and all constants are set to unity for simplicity. For the first moment, we're multiply by $\mathbf{v}$ and integrating over all velocity-space, $$\int\mathbf{v}\partial_tf\,\mathrm{d}\mathbf{v}+\int\mathbf{v}\left(\mathbf{v}\cdot\nabla_x\right)f\,\mathrm{d}\mathbf{v}+\int\mathbf{v}\left(\mathbf{a}\cdot\nabla_v\right)f\,\mathrm{d}\mathbf{v}=0$$ So let's look at the first term: $$ \int\mathbf{v}\partial_tf\,\mathrm{d}\mathbf{v} = \partial_t\int\mathbf{v}f\,\mathrm{d}\mathbf{v} = \partial_tn\mathbf{u}$$ where we used the definition of $n\mathbf{u}$ from the 0th moment case.
For the second term, we use the independent coordinate aspect to re-write it as, $$\int\mathbf{v}\left(\mathbf{v}\cdot\nabla_x\right)f\,\mathrm{d}\mathbf{v}=\int\nabla_x\cdot\left(\mathbf{v}\mathbf{v}f\right)\,\mathrm{d}\mathbf{v}=\nabla_x\cdot\int\mathbf{v}\mathbf{v}f\,\mathrm{d}\mathbf{v}$$ This last equality is then $n$ times the average of $\mathbf{v}\mathbf{v}$: $$\nabla_x\cdot\int\mathbf{v}\mathbf{v}f\,\mathrm{d}\mathbf{v}=\nabla_x\cdot\left(n\langle\mathbf{v}\mathbf{v}\rangle\right)$$ We can express the velocity $\mathbf{v}$ as the sum of a mean field, $\mathbf{u}$, and a thermal fluctiation, $\mathbf{w}$, then we get $$\nabla_x\cdot\int\mathbf{v}\mathbf{v}f\,\mathrm{d}\mathbf{v}=\nabla_x\cdot\left(n\langle\mathbf{u}\mathbf{u}\rangle+n\langle\mathbf{w}\mathbf{w}\rangle\right)=\nabla_x\cdot n\mathbf{u}\mathbf{u}+\nabla_x\cdot\sigma$$
Lastly, the final term requires using the derivative of three products, $$\nabla\cdot(abc)=(\nabla a)bc+a(\nabla b)c+ab\nabla c,$$ to re-write the integral as, $$\int \mathbf{v}\left(\mathbf{a}\cdot\nabla_v\right)f\,\mathrm{d}\mathbf{v}=\int\nabla_v\cdot\left(f\mathbf{v}\mathbf{a}\right)\,\mathrm{d}\mathbf{v}-\int f\mathbf{v}\left(\nabla_v\cdot\mathbf{a}\right)\,\mathrm{d}\mathbf{v}-\int f\mathbf{a}\cdot\nabla_v\mathbf{v}\,\mathrm{d}\mathbf{v}$$ By a similar argument to the 0th moment case, the first two terms on the right hand side are zero (i.e., the divergence theorem says the velocity goes as $v^2$ which means $f\to0$ faster than $\mathbf{S}\to\infty$). This leaves, $$\int \mathbf{v}\left(\mathbf{a}\cdot\nabla_v\right)f\,\mathrm{d}\mathbf{v}=-\int f\mathbf{a}\cdot\nabla_v\mathbf{v}\,\mathrm{d}\mathbf{v}=-n\mathbf{a}$$
The three terms then combine to form the momentum conservation equation. If you are careful with your coefficients (i.e., $a$'s and $\nabla\phi$), you should still be able to derive the same equations.