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 $0$.


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

  • 1
    $\begingroup$ Yes, you are correct in assuming $\nabla_{v}(Xf_{s})$ Thank you for the working out. It is clear. $\endgroup$ – hahahasan Feb 8 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.