2
$\begingroup$

Maxwell Boltzmann distribution function (MBDF) has the form $$f(v)=n(\frac{m}{2\pi k_BT})^{\frac{3}{2}} exp(-\frac{mv^2}{2k_BT})$$ [Basic Space Plasma Physics by Rudolf A. Treumann & Wolfgang Baumjohann]. The shifted MBDF has the form $$f(v)=n(\frac{m}{2\pi k_BT})^{\frac{3}{2}} exp(-\frac{m(v-v0)^2}{2k_BT}).$$

This is shown in the below figure for ms= 1.660539040 $10^{-27}$, kb=1.380 $10^{-23}$, T=100, v0=2000, wherein thick lines give the MBDF and dashed one gives the shifted MBDF. MBDF and shifted MBDF..

As far as the beams in plasma are concerned, I believe that one should see a figure of the form: Beam DF Kindly correct me if I am wrong.

What is the mathematical expression that describes the above figure?

$\endgroup$

1 Answer 1

2
$\begingroup$

Since the only plasmas where that bump-on-tail velocity distribution function (VDF) can exist are those that are either weakly collisional or collisionless, it is perfectly okay to add VDFs. That is, those plasmas are neither in thermodynamic or thermal equilibrium, so there is nothing wrong with adding two VDFs as they are not a single temperature (e.g., see https://physics.stackexchange.com/a/268594/59023 or https://physics.stackexchange.com/a/375611/59023).

The general bi-Maxwellian VDF of species $s$ is given by: $$ f_{s}\left( v_{\parallel}, v_{\perp} \right) = \frac{ n_{s} }{ \pi^{3/2} \ V_{T \parallel, s} \ V_{T \perp, s}^{2} } \ exp\left[ - \left( \frac{ v_{\parallel} - v_{o, \parallel, s} }{ V_{T \parallel, s} } \right)^{2} - \left( \frac{ v_{\perp} - v_{o, \perp, s} }{ V_{T \perp, s} } \right)^{2} \right] \tag{0} $$ where $\parallel$($\perp$) refer to directions parallel(perpendicular) with respect to a quasi-static magnetic field, $\mathbf{B}_{o}$, $V_{T_{j, s}}$ is the $j^{th}$ thermal speed (actually the most probable speed), $v_{o, j, s}$ is the $j^{th}$ component of the bulk drift velocity of the distribution (i.e., from the 1st velocity moment), and $n_{s}$ is the number density or zeroth velocity moment of species $s$.

Typically beams (i.e., the little bump in your merged VDF) are not isotropic, so it is common to have a VDF for the core and a VDF for the beam. It is okay to use a form like Equation 0 for both, each with different densities and thermal speeds. The non-equilibrium nature of the plasma allows two such VDFs to effectively stream past each other. They tend to excite instabilities that lead to fluctuations like Langmuir waves.

You could get a little fancier and use a bi-kappa VDF. The bi-kappa distribution function is given by: $$ f_{s}\left( v_{\parallel}, v_{\perp} \right) = A_{s} \left[ 1 + \left( \frac{ v_{\parallel} - v_{o, \parallel, s} }{ \sqrt{ \kappa_{s} - 3/2 } \ \theta_{\parallel, s} } \right)^{2} + \left( \frac{ v_{\perp} - v_{o, \perp, s} }{ \sqrt{ \kappa - 3/2 } \ \theta_{\perp, s} } \right)^{2} \right]^{- \left( \kappa_{s} + 1 \right) } \tag{1} $$ where the amplitude is given by: $$ A_{s} = \left( \frac{ n_{s} \ \Gamma\left( \kappa_{s} + 1 \right) }{ \left( \pi \left( \kappa_{s} - 3/2 \right) \right)^{3/2} \ \theta_{\parallel, s} \ \theta_{\perp, s}^{2} \ \Gamma\left( \kappa_{s} - 1/2 \right) } \right) \tag{2} $$ and where $\theta_{j, s}$ is the $j^{th}$ thermal speed (also the most probable speed), $\Gamma(x)$ is the complete gamma function, and $\kappa_{s}$ is the kappa index and can be any value larger than 3/2.

Further we can show that the average temperature is just given by: $$ T = \frac{ 1 }{ 3 } \left( T_{\parallel} + 2 \ T_{\perp} \right) \tag{3} $$ if we assume a gyrotropic distribution (i.e., shows symmetry about $\mathbf{B}_{o}$ so that the two perpendicular components of a diagonalized pressure tensor are equal).

$\endgroup$
3
  • $\begingroup$ thanks for providing many insights. My query, again, is will Equation 0 will give the required plot? I tied with the above parameters, but I could regenerate the plot. $\endgroup$
    – sreeraj t
    Commented Jun 27, 2019 at 1:16
  • 1
    $\begingroup$ You need to sum two of them and then take a 1D cut because Equations 0 and 1 are 2D functions. $\endgroup$ Commented Jun 28, 2019 at 15:50
  • $\begingroup$ again, thanks for the tip. Now, I could regenerate the plot just by summing both equation which is given in my question. And I do not understand what is meant by 1D cut? Any thought about that? $\endgroup$
    – sreeraj t
    Commented Jul 1, 2019 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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