# Taking moments of the Vlasov equation

Given the following term:

$$\nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]$$

where $$\textbf{E}$$ and $$\textbf{B}$$ are the electric and magnetic fields respectively experienced by a particle travelling with velocity $$\textbf{v}$$ and $$f_{s}$$ is the distribution function of particle velocities (assumed to be Maxwellian). I want to know how simply multiplying this expression by some arbitrary polynomial function, $$X(\textbf{v})$$ and integrating over velocity will give:

$$- \frac{e}{m_{s}} \int (\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s} d^{3}v$$

According to T.J.M. Boyd & J.J.Sanderson in the book Physics of Plasmas one needs to use integration by parts and use the limit:

$$\lim_{|\textbf{v}|\to\infty} (Xf_{s})=0$$

I am struggling to see how this follows and would appreciate a detailed explanation with as few a step skipped as reasonable.

I am assuming $$\nabla_{v}(Xf_{s})$$...?

Integration by parts:

$$- \frac{e}{m_{s}} \int \underbrace{(\textbf{E} + \textbf{v} \times \textbf{B})}_{f} \cdot \underbrace{\nabla_{v}Xf_{s}}_{g'} d^{3}v =$$

$$\propto \underbrace{(\textbf{E} + \textbf{v} \times \textbf{B})}_{f} \cdot \underbrace{Xf_{s}}_{g} \Biggr|_{\text{limits}} - \int \underbrace{\nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]}_{f'}\cdot \underbrace{Xf_s}_g.$$

The limits will be $$v$$ from $$0$$ to $$\infty$$, where $$v = |\mathbf{v}|$$ and its angular part is just a constant.

I know that $$\lim_{|\textbf{v}|\to\infty} (Xf_{s})=0$$ and I want:

$$Xf_{s} \Biggr|_{\text{limits}} = (Xf_s)\Biggr|_{\infty} - (Xf_s)\Biggr|_{0} = 0 - 0,$$

because $$f_s \propto v^2 =0$$ at $$0$$.

So:

$$- \frac{e}{m_{s}} \int (\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s}d^{3}v = - \int \nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]\cdot Xf_s.$$

• Yes, you are correct in assuming $\nabla_{v}(Xf_{s})$ Thank you for the working out. It is clear. – hahahasan Feb 8 at 13:10