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I can't understand some steps in obtaining the collision term in the Boltzmann equation for plasma. For the first time it was made by L.D. Landau in his article "The kinetic equation in the case of Coulomb interaction" (Zh. Eksper. i Teoret. Fiz. , 7 : 2 (1937) pp. 203–209)*.

I also can't find proper literature, it often happens that authors use an approximate approach with a cross section and get approximate expression without an accurate factor, which is important for me. So I try to read Landau's article and understand his actions.

*English version can be find here: https://www.sciencedirect.com/science/article/pii/B9780080105864500298

Question

In general, I'm not able to catch the series expansion of the collision probability $\omega(p,p',\tilde{p},\tilde{p'})$ at $\tilde{p}=p,\ \tilde{p'}=p'$, where:
$p,\ \tilde{p}$ - momentums of the first particle before and after collision,
$p',\ \tilde{p'}$ - momentums of the second particle before and after collision,
$\tilde{p_i}=p_i+\mathit{\Delta_i}\ ,\ \tilde{p'}=p_i'+\mathit{\Delta_i'}\ $, and $\mathit{\Delta},\ \mathit{\Delta'}$ are considered as small parameters.

My Attempt

There is how I would do this (let me use $p_{\alpha}, p_{\beta}$ instead of $p, p'$):
$$\omega(p_{\alpha},p_{\beta},\tilde{p_{\alpha}},\tilde{p_{\beta}})\approx \omega(p_{\alpha},p_{\beta},p_{\alpha},p_{\beta})+\frac{\partial \omega}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i+\frac{\partial \omega}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i$$

It'll help to write the collision term $J_{\alpha\beta}=\iiint \omega(p_{\alpha},p_{\beta},\tilde{p_{\alpha}},\tilde{p_{\beta}})(\tilde{f_{\alpha}}\tilde{f_{\beta}}-f_{\alpha}f_{\beta})d^3p_{\beta}d^3\tilde{p_{\alpha}}d^3\tilde{p_{\beta}}$ as:
$$J_{\alpha\beta}\approx \iiint \omega(p_{\alpha},p_{\beta},p_{\alpha},p_{\beta})\left(f_{\alpha}\frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i + f_{\beta}\frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i \right)d^3p_{\beta}d^3\tilde{p_{\alpha}}d^3\tilde{p_{\beta}} + \iiint \left [ \left ( \frac{\partial \omega}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i+\frac{\partial \omega}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i \right )\left(f_{\alpha}\frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i + f_{\beta}\frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i \right) + \omega\left ( f_{\alpha}\frac{\partial^2 f_{\beta}}{\partial \tilde{p_{\beta}}_i \partial \tilde{p_{\beta}}_j}\frac{\mathit{\Delta_{\beta}}_i \mathit{\Delta_{\beta}}_j}{2} + f_{\beta}\frac{\partial^2 f_{\alpha}}{\partial \tilde{p_{\alpha}}_i \partial \tilde{p_{\alpha}}_j}\frac{\mathit{\Delta_{\alpha}}_i \mathit{\Delta_{\alpha}}_j}{2} + \frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i} \frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_j}\mathit{\Delta_{\alpha}}_i\mathit{\Delta_{\beta}}_j \right ) \right ]d^3p_{\beta}d^3\tilde{p_{\alpha}}d^3\tilde{p_{\beta}}\ ,$$ where $\tilde{f}=f(r,\tilde{p},t),\ f=f(r,p,t)$. The first integral is zero, because the integrand is odd relatively to the point $\tilde{p_{\alpha}}=p_{\alpha}$ (or $\tilde{p_{\beta}}=p_{\beta}$).

Landau integrates by parts a portion of the second integral over $d^3p_{\beta}$: $$\iiint \frac{\partial \omega}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i \left(f_{\alpha}\frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i + f_{\beta}\frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i \right)d^3p_{\beta}d^3\tilde{p_{\alpha}}d^3\tilde{p_{\beta}}$$ But there are different variables in the derivation $\frac{\partial}{\partial \tilde{p_{\beta}}}$ and the differential $d^3p_{\beta}$ (with and without tilde). Moreover, $\frac{\partial \omega}{\partial \tilde{p_{\beta}}_i}$ is only a function of $p_{\alpha},p_{\beta}$, so I can't integrate over $d^3\tilde{p_{\beta}}$. But there are only my notes here, and I guess, I totally misunderstand something. Landau does it in the different way, which confuses me.

Landau's approach

1 step

He writes $\omega$ as: $$\omega \left( \frac{p_{\alpha}+\tilde{p_{\alpha}}}{2},\frac{p_{\beta}+\tilde{p_{\beta}}}{2},\tilde{p_{\alpha}}-p_{\alpha},\tilde{p_{\beta}}-p_{\beta} \right) = \omega \left( p_{\alpha}+\frac{\mathit{\Delta_{\alpha}}}{2},p_{\beta}+\frac{\mathit{\Delta_{\beta}}}{2},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right)$$ and states the following:

($\omega$ should of course be expanded only with respect to $\mathit{\Delta_i}$, appearing in $p_i+\mathit{\Delta_i}/2$ and $p_i'+\mathit{\Delta_i}'/2$)

This is the first thing, which I don't understand.

I would do so: $$\omega(\xi_{\alpha}^1,\xi_{\beta}^1,\eta_{\alpha}^1,\eta_{\beta}^1)\approx \omega(\xi_{\alpha}^0,\xi_{\beta}^0,\eta_{\alpha}^0,\eta_{\beta}^0)+\frac{\partial \omega}{\partial \xi_{\alpha i}}\frac{\mathit{\Delta_{\alpha}}_i}{2}+\frac{\partial \omega}{\partial \eta_{\alpha i}}\mathit{\Delta_{\alpha}}_i + \frac{\partial \omega}{\partial \xi_{\beta i}}\frac{\mathit{\Delta_{\beta }}_i}{2}+\frac{\partial \omega}{\partial \eta_{\beta i}}\mathit{\Delta_{\beta }}_i$$ where $(\xi_{\alpha}^0,\xi_{\beta}^0,\eta_{\alpha}^0,\eta_{\beta}^0) = (p_{\alpha},p_{\beta},0,0)$,

$(\xi_{\alpha}^1,\xi_{\beta}^1,\eta_{\alpha}^1,\eta_{\beta}^1) = (p_{\alpha}+\mathit{\Delta_{\alpha}}/2,\ p_{\beta}+\mathit{\Delta_{\beta}}/2,\ \mathit{\Delta_{\alpha}},\ \mathit{\Delta_{\beta}})$
and $(\xi_{\alpha},\xi_{\beta},\eta_{\alpha},\eta_{\beta}) = \left( \frac{p_{\alpha}+\tilde{p_{\alpha}}}{2},\frac{p_{\beta}+\tilde{p_{\beta}}}{2},\tilde{p_{\alpha}}-p_{\alpha},\tilde{p_{\beta}}-p_{\beta} \right)$.

It seems legit, because: $$\left ( \frac{\partial \omega}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \left ( \frac{\partial \xi}{\partial \tilde{p}} \right )_p + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \left ( \frac{\partial \eta}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \cdot \frac{1}{2} + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \cdot 1$$

2 step

He writes the series as: $$\omega \left( p_{\alpha}+\frac{\mathit{\Delta_{\alpha}}}{2},p_{\beta}+\frac{\mathit{\Delta_{\beta}}}{2},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right) = \omega \left( p_{\alpha},p_{\beta},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right) + \frac{\partial \omega}{\partial p_{\alpha i}}\frac{\mathit{\Delta_{\alpha}}_i}{2} + \frac{\partial \omega}{\partial p_{\beta i}}\frac{\mathit{\Delta_{\beta}}_i}{2}$$

It leads to several questions. Why are there:

  • $\omega \left( p_{\alpha},p_{\beta},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right)$ instead of $\omega (p_{\alpha},p_{\beta},0,0)$;
  • $\frac{\partial \omega}{\partial p_i}$ instead of $\frac{\partial \omega}{\partial \tilde{p_i}}$ (no tilde instead of tilde);
  • $\frac{\mathit{\Delta}_i}{2}$ instead of $\mathit{\Delta}_i$?

3 step

He integrates by parts over $d^3p_{\beta}$: $$\iiint \frac{\partial \omega}{\partial p_{\beta i}}\frac{\mathit{\Delta_{\beta}}_i}{2} \left(f_{\alpha}\frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i + f_{\beta}\frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i \right)d^3p_{\beta}d^3\tilde{p_{\alpha}}d^3\tilde{p_{\beta}}$$ which is legit if the previous steps are legit.

Excuse me for the long description. Could you please help me with the problem? Or maybe advice good literature?

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It seems that I found a solution. But I still can't understand why Landau writes $\omega \left( p_{\alpha},p_{\beta},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right)$ instead of $\omega (p_{\alpha},p_{\beta},0,0)$. Maybe he really wanted to make the first integral zero, because he used that $\omega \left( p_{\alpha},p_{\beta},\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}} \right)$ is odd function relatively to $\mathit{\Delta_{\alpha}},\mathit{\Delta_{\beta}}$, but forgot that it's not possible to get such expression. Anyway, here are my thoughts.

We should use the following fact: $$\omega(p_{\alpha},p_{\beta},\tilde{p_{\alpha}},\tilde{p_{\beta}})=\omega(\tilde{p_{\alpha}},\tilde{p_{\beta}},p_{\alpha},p_{\beta})$$ It means that the probability of the direct process ($p_{\alpha},p_{\beta}\rightarrow \tilde{p_{\alpha}},\tilde{p_{\beta}}$) is equal to the probability of the reverse process ($\tilde{p_{\alpha}},\tilde{p_{\beta}}\rightarrow p_{\alpha},p_{\beta}$). Therefore: $$\frac{\partial \omega}{\partial \tilde{p_{\alpha}}}=\frac{\partial \omega}{\partial p_{\alpha}},\quad \frac{\partial \omega}{\partial \tilde{p_{\beta}}}=\frac{\partial \omega}{\partial p_{\beta}}$$ Or in general form: $$\left ( \frac{\partial \omega}{\partial \tilde{p}} \right )_p =\left ( \frac{\partial \omega}{\partial p} \right )_{\tilde{p}} \tag{1}$$ Recall the derivatives expression through $\xi,\eta$: $$\left ( \frac{\partial \omega}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \left ( \frac{\partial \xi}{\partial \tilde{p}} \right )_p + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \left ( \frac{\partial \eta}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \cdot \frac{1}{2} + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \cdot 1$$ $$\left ( \frac{\partial \omega}{\partial p} \right )_{\tilde{p}} = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \left ( \frac{\partial \xi}{\partial \tilde{p}} \right )_{\tilde{p}} + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \left ( \frac{\partial \eta}{\partial \tilde{p}} \right )_{\tilde{p}} = \left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} \cdot \frac{1}{2} + \left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi} \cdot (-1)$$ So, (1) is true if $\left ( \frac{\partial \omega}{\partial \eta} \right )_{\xi}=0$.

Let's write the derivatives also through $p,\mathit{\Delta}$: $$\left ( \frac{\partial \omega}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}} \left ( \frac{\partial p}{\partial \tilde{p}} \right )_p + \left ( \frac{\partial \omega}{\partial \mathit{\Delta}} \right )_p \left ( \frac{\partial \mathit{\Delta}}{\partial \tilde{p}} \right )_p = \left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}} \cdot 0 + \left ( \frac{\partial \omega}{\partial \mathit{\Delta}} \right )_p \cdot 1 \tag{2}$$ $$\left ( \frac{\partial \omega}{\partial p} \right )_{\tilde{p}} = \left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}} \left ( \frac{\partial p}{\partial p} \right )_{\tilde{p}} + \left ( \frac{\partial \omega}{\partial \mathit{\Delta}} \right )_p \left ( \frac{\partial \mathit{\Delta}}{\partial p} \right )_{\tilde{p}} = \left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}} \cdot 1 + \left ( \frac{\partial \omega}{\partial \mathit{\Delta}} \right )_p \cdot (-1) \tag{3}$$ Using (1), (2), (3) we get: $$2\left ( \frac{\partial \omega}{\partial p} \right )_{\tilde{p}}=\left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}}$$ Finally, we can obtain: $$\left ( \frac{\partial \omega}{\partial \tilde{p}} \right )_p = \frac{1}{2}\left ( \frac{\partial \omega}{\partial \xi} \right )_{\eta} = \left ( \frac{\partial \omega}{\partial p} \right )_{\tilde{p}} = \frac{1}{2} \left ( \frac{\partial \omega}{\partial p} \right )_{\mathit{\Delta}}$$ $$\omega(p_{\alpha},p_{\beta},\tilde{p_{\alpha}},\tilde{p_{\beta}})\approx \omega(p_{\alpha},p_{\beta},p_{\alpha},p_{\beta})+\frac{\partial \omega}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i+\frac{\partial \omega}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i = \omega(p_{\alpha},p_{\beta},p_{\alpha},p_{\beta}) + \left ( \frac{\partial \omega}{\partial p_{\alpha i}} \right )_{\mathit{\Delta_{\alpha}}}\frac{\mathit{\Delta_{\alpha}}_i}{2} + \left ( \frac{\partial \omega}{\partial p_{\beta i}} \right )_{\mathit{\Delta_{\beta}}}\frac{\mathit{\Delta_{\beta}}_i}{2}$$ Now the integration $$\iiint \frac{\partial \omega}{\partial p_{\beta i}}\frac{\mathit{\Delta_{\beta}}_i}{2} \left(f_{\alpha}\frac{\partial f_{\beta}}{\partial \tilde{p_{\beta}}_i}\mathit{\Delta_{\beta}}_i + f_{\beta}\frac{\partial f_{\alpha}}{\partial \tilde{p_{\alpha}}_i}\mathit{\Delta_{\alpha}}_i \right)d^3p_{\beta}d^3\mathit{\Delta_{\alpha}}d^3\mathit{\Delta_{\beta}}$$ is legit.

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