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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?

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  • $\begingroup$ I don't want to post this an answer, because you already have everything in the question. Take the equation before "but I am .." everything after that I don't get. It's missing a time derivative in the first term. The last term is zero (integration by parts). For the second term move the spatial derivative outside the integral. Final step: look up a continuity equation that is less weird (i.e. in rho not delta) $\endgroup$
    – Bort
    Oct 19, 2023 at 13:19
  • $\begingroup$ @Bort Thank you, you're right, the missing time derivative was a typo. But I don't know what to do with that term, or how to proceed. For example, you say I should move the spatial derivative outside the integral in the second term, so I would get something like: $$\int d^3p\dfrac{\vec{p}}{ma}\dfrac{\partial f}{\partial\vec{x}}=\vec{\nabla}_\vec{x}\cdot\int d^3p\dfrac{\vec{p}}{ma}=\vec{\nabla}_\vec{x}\cdot\int d^3p\ \vec{u}$$ where $\vec{u}=\vec{p}/ma$ is the comoving velocity. But what do I do now with that integral? And how do you integrate the third term by parts? $\endgroup$ Oct 19, 2023 at 16:49
  • $\begingroup$ Could you do it for the case of plasma physics & the E&M force? $\endgroup$
    – Kyle Kanos
    Oct 19, 2023 at 18:06
  • $\begingroup$ @KyleKanos No, I don't know how to do that. I encountered this assumption in a set of particle physics notes and I was curious to know how it's done, since I hate to just believe that equations are true without knowing the explanation behind them, but I never saw the derivation done for the E&M force either. $\endgroup$ Oct 19, 2023 at 21:51

2 Answers 2

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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.

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The various moments of Vlasov's equation are calculated by multiplying the equation by a function $\Phi(\mathbf{p})$ and integrating over $d^3p$:

\begin{equation} \int \Phi(\mathbf{p})\frac{\partial f}{\partial \tau}d^3p + \int \Phi(\mathbf{p})\frac{\mathbf{p}}{ma}\frac{\partial f}{\partial \mathbf{x}}d^3p - \int \Phi(\mathbf{p})am\nabla \phi\cdot \frac{\partial f}{\partial \mathbf{p}} d^3p = 0 \,. \end{equation} Let's consider each term separately.

$1$) $\int \Phi(\mathbf{p})\frac{\partial f}{\partial \tau}d^3p$ \begin{equation} \int \Phi(\mathbf{p})\frac{\partial f}{\partial \tau}d^3p = \frac{\partial}{\partial \tau}\int \Phi(\mathbf{p})fd^3p \end{equation}

$2$) $\int \Phi(\mathbf{p})\frac{\mathbf{p}}{ma}\frac{\partial f}{\partial \mathbf{x}}d^3p$ \begin{equation} \int \Phi(\mathbf{p})\frac{\mathbf{p}}{ma}\frac{\partial f}{\partial \mathbf{x}}d^3p = \frac{\partial}{\partial \mathbf{x}}\int \Phi(\mathbf{p})\frac{\mathbf{p}}{ma}fd^3p \end{equation} $3$) $\int \Phi(\mathbf{p})am\nabla \phi\cdot \frac{\partial f}{\partial \mathbf{p}} d^3p$ \begin{equation} \int \Phi(\mathbf{p})am\nabla \phi\cdot \frac{\partial f}{\partial \mathbf{p}} d^3p = - \nabla \phi \cdot \int a m \frac{\partial \Phi(\mathbf{p})}{\partial \mathbf{p}}f d^3p \end{equation} Now, take $\Phi = \text{const}$. This will yield the $0$-th moment of Vlasov's equation. In particular, if you take $\Phi = \frac{m}{a^3}$, you obtain:

$1$) \begin{equation} \frac{\partial}{\partial \tau}\int \Phi(\mathbf{p})fd^3p = \frac{\partial \rho}{\partial \tau} \end{equation}

$2$) \begin{equation} \frac{\partial}{\partial \mathbf{x}}\int \Phi(\mathbf{p})\frac{\mathbf{p}}{ma}fd^3p = \frac{\partial}{\partial \mathbf{x}} \cdot (\rho\mathbf{v}) \end{equation}

$3$) \begin{equation} \nabla \phi \cdot \int a m \frac{\partial \Phi(\mathbf{p})}{\partial \mathbf{p}} d^3p = 0 \end{equation} This yields \begin{equation} \frac{\partial \rho}{\partial \tau} + \frac{\partial}{\partial\mathbf{x}} \cdot (\rho\mathbf{v}) = 0 \end{equation} which is the standard formulation of a continuity equation, and should correspond to the one you mentioned if I am not mistaken.

If you take $\Phi(\mathbf{p}) \sim \mathbf{p}$, you obtain the $1$-st moment of Vlasov's equation. In particular, with $\Phi(\mathbf{p}) = \frac{\mathbf{p}}{a^3}$ you obtain the following terms:

$1$) \begin{equation} \int \frac{\mathbf{p}}{a^3}\frac{\partial f}{\partial \tau}d^3p = \frac{\partial}{\partial \tau}\int \frac{\mathbf{p}}{a^3}fd^3p = \frac{\partial}{\partial \tau}(a\rho\mathbf{v}) \end{equation}

$2$) \begin{equation} \int \frac{\mathbf{p}}{a^3}\frac{\mathbf{p}}{ma}\frac{\partial f}{\partial \mathbf{x}}d^3p = \frac{\partial }{\partial \mathbf{x}}\int \frac{\mathbf{p}\mathbf{p}}{ma^4}fd^3p = \frac{\partial}{\partial \mathbf{x}}[a\rho(\overleftrightarrow{\sigma} + \mathbf{v}\mathbf{v})] \end{equation}

$3$) \begin{equation} \int \frac{\mathbf{p}}{a^3}am\nabla \phi \cdot \frac{\partial f}{\partial \mathbf{p}}d^3p = -\frac{m}{a^2}\nabla \phi \cdot \int \frac{\partial (\mathbf{p})}{\partial \mathbf{p}} f d^3p = - a \rho \nabla \phi \end{equation}

The equation becomes \begin{equation} \frac{\partial}{\partial \tau}(a\rho\mathbf{v}) + \frac{\partial}{\partial \mathbf{x}}[a\rho(\overleftrightarrow{\sigma} + \mathbf{v}\mathbf{v})] = -a\rho\nabla\phi \end{equation}

Now, developing the first two terms and recalling the vector identity \begin{equation} \nabla\cdot(\rho\mathbf{v}\mathbf{v}) = \mathbf{v}\nabla\cdot(\rho\mathbf{v}) + \rho(\mathbf{v}\cdot\nabla)\mathbf{v} \end{equation}

you can recognise the presence of the left hand side of the continuity equation, which is equal to $0$. Your final result is

\begin{equation} \rho\mathbf{v}\frac{\partial a}{\partial \tau} + \rho a \frac{\partial \mathbf{v}}{\partial \tau} + a\rho(\mathbf{v}\cdot \nabla)\mathbf{v} = -a\rho\nabla\phi - a \nabla\cdot(\rho \overleftrightarrow{\sigma}) \end{equation}

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    $\begingroup$ So your answer to the question, "How do you obtain the momentum conservation equation from the 1st moment of Vlasov's equation?" is the last sentence: verify it yourself, which doesn't seem particularly useful since OP stated they've had issues in the derivation. $\endgroup$
    – Kyle Kanos
    Oct 20, 2023 at 1:07
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    $\begingroup$ I've modified my answer. It should be complete now. $\endgroup$
    – chewbocca
    Oct 20, 2023 at 4:49
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    $\begingroup$ @chewbocca Thank you very much for your answer and for clarifying how to get the Euler equation too. There is something I don't quite get: when you first consider each term separately, for a generic $\Phi$, how did you integrate by parts in (3)? I am assuming that $\vec{\nabla}\phi$ is independent of $\vec{p}$ (is this correct?) and taking $u=\Phi$ and $dv=\partial f/\partial\vec{p}d^3p$, which means that: $$\int d^3p\ \Phi am\vec{\nabla}\phi\dfrac{\partial f}{\partial\vec{p}}=am\vec{\nabla}\phi\bigg(\Phi f-\int d^3p\dfrac{\partial\Phi}{\partial\vec{p}}f\bigg)$$ Why is it wrong? Thanks! $\endgroup$ Oct 20, 2023 at 18:26
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    $\begingroup$ @WildFeather yes, $\nabla \phi$ is independent of $\mathbf{p}$. When you integrate by parts, the first term is $0$ because a properly defined distribution function should be $0$ at infinity. $\endgroup$
    – chewbocca
    Oct 20, 2023 at 19:18
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    $\begingroup$ @WildFeather $\mathbf{p}\mathbf{p}$ is actually a second order tensor which is called a dyadic. The $i,j$ component of such a tensor is written as $(\mathbf{p}\mathbf{p})_{i,j} = p_ip_j$. For instance, if your momentum is in Cartesian coordinates $\mathbf{p} = (p_x, p_y, p_z)$, the tensor $\mathbf{p}\mathbf{p}$ would be \begin{equation} \begin{bmatrix} p_x^2 & p_xp_y & p_xp_z \\p_yp_x & p_y^2 & p_yp_z \\ p_zp_x & p_zp_y & p_z^2\end{bmatrix} \end{equation} $\endgroup$
    – chewbocca
    Oct 25, 2023 at 17:00

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