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I am trying to sample Maxwell-Jüttner distribution using the Sobol method as described in Zenitani Loading relativistic Maxwell distributions in particle simulations (2015). Equation (2) in the paper assumes $m=c=1$ and then goes on to describe how to sample the four-velocity. I have a few questions regarding the Sobol algorithm:

  1. If $m$ and $c$, are not taken to be unity, how will that affect the value of generated $u$? I can see how this will affect $f(u)$, but does $u$ also need to be scaled?
  2. In order to get $v$, which the normal three velocity vector, do we just need to use the relation $u = \gamma v$, and express $v$ in terms of $u$?

I am attaching my distribution plot for reference.

enter image description here

Prav001
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    Related: https://physics.stackexchange.com/q/298469/25301 – Kyle Kanos Aug 27 '19 at 19:32
  • @KyleKanos Yeah I saw that post, and I don't have any problem as such in sampling. I am just confused about proper normalization, as to how should I scale my u(and therefore v), and f(u) when the mass and speed of light is not taken to be unity. – Prav001 Aug 27 '19 at 22:24
  • I linked a related query since it's about the same paper in both. I wasn't trying to suggest it answers your own question. – Kyle Kanos Aug 27 '19 at 22:45

1 Answers1

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OP's equations are \begin{align} f(\mathbf{u})\,\mathrm{d}^3u&=\frac{N}{4\pi m^2cTK_2(mc^2/T)}\exp\left(-\frac{\gamma mc^2}{T}\right)\,\mathrm{d}^3u\tag{1} \\ \Rightarrow f(u)\,\mathrm{d}u&=\frac{N}{TK_2(1/T)}\exp\left(-\frac{\sqrt{1+u^2}}{T}\right)u^2\,\mathrm{d}u\tag{2} \end{align} where $\mathbf{u}=\gamma\mathbf{v}$ is the spatial components of the 4-velocity, $\gamma$ the Lorentz factor, $N=\int f(\mathbf{u})\,\mathrm{d}u$ is the total number density, $T$ the temperature and $K_2(x)$ the modified Bessel function. The $4\pi$ factor is dropped in (2) due to use of spherical coordinates and it is plain to see that, $$ u=\gamma v\implies \gamma=\sqrt{1+u^2/c^2} $$

Since $u$ is the velocity, then $m\neq1$ should be straight-forward re-insertion. Additionally, since $u=\gamma v$, then there should be no changes to $u$, only to $f(u)$. Lastly, indeed you should invert the relation between $u$ and $v$ to get the 3-velocity, though the work is done for you in Zenitani's paper (and above): $$ v_i=\frac{u_i}{\gamma}=u_i\cdot\left(1+u_i u^i\right)^{-1/2} $$

In my opinion, it may not be the best of ideas to not use $c=1$, since using such scaling is intended for convenience (in both notation and in coding) in relativistic settings. Since the Maxwell-Jüttner distribution describes the relativistic distribution of velocities, it probably would be better to stick with $c=1$.

Kyle Kanos
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  • Thanks for the answer. The reason why I am reluctant to use c= 1 is that because this sampling is part of a bigger code, and in other places, we have used c= 3e10. When I tried rescaling f(u) by $\frac{1}{(mc)^3}$ and replacing u by u/c in the definition of gamma, I am not scaled version of my original plot. New plot: https://imgur.com/WYY6cYA Something tells me I need to rescale u too. – Prav001 Aug 29 '19 at 19:00
  • I've always been taught that you should keep units out of codes, rescaling for visualizing only (maybe not even then). $u$ will have same units as $v$, of so I don't think additional factors of $c$ is needed. I don't know why you're scaling $f(u)$ by $1/(mc)^3$, though, that's not what's called for... – Kyle Kanos Aug 29 '19 at 19:22
  • The $\frac{1}{(mc)^3}$ factor comes from equation (1). Eqn. 2 is obtained by moving to spherical coordinates and setting m=c=1. So I just rescaled f(u) in Eqn (2) by replacing N by $N * \frac{1}{(mc)^3}$ and $ \sqrt {1+u^2}$ as $ \sqrt {1+\frac{u^2}{c^2}}$. I think I will stick with m=c=1 to avoid any confusion. – Prav001 Aug 29 '19 at 21:16
  • I'm still confused. There's a clear $m^2c$ in the denominator in (1) & nothing else, so you've got a few too many factors floating there. There's also a factor of $mc^2$ in the Bessel fcn & another in the exponent, have you accounted for those? – Kyle Kanos Aug 29 '19 at 21:54
  • Sorry for the confusion, but I am not using T, rather I have defined a dimensionless temperature $\theta = \frac{T}{mc^2}$. So $T = mc^2 \theta $ and I can effectively replace T in exponential and the Bessel function in Eqn 1 by $\theta$. So equation 1 becomes $ f(u) = \frac{N}{4 \pi (mc)^3 \theta K_{2}(\frac{1}{\theta})}* exp(- \frac{\gamma}{\theta}) $ I am using $\theta$ = 0.4 – Prav001 Aug 29 '19 at 22:14