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Mar 9 at 4:19 comment added Bababeluma @NameYYY interesting
Mar 8 at 23:28 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 19:54 comment added Name YYY @Bababeluma : I checked it previously. Independently of the number of particles, the system ends up in this state.
Mar 8 at 17:46 comment added Bababeluma According to the principle of detailed balance, equilibrium result shouldn't depend on the form of your interaction as long as it sustain channels between every pair in the phase space. Sometimes the balance can't be achieved and you thermalize in some unintuitive subspace. The histogram looks pretty rough, Maybe try N=40000 n=200 that shouldn't be too demanding for your PC
Mar 8 at 17:41 comment added Bababeluma Have you tried bigger N and n? Maybe you should first check that your lineshape doesn't vary with bigger or smaller N and n.
Mar 8 at 14:07 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 13:58 comment added Name YYY @LPZ : I started with a simple simulation that assumes that per each timestep all particles must interact. The results do not depend on the number $N$ (after reaching some threshold $N = N_{\text{max}}$) and on the number of simulated interactions $n$ (after reaching some threshold $n = n_{\text{max}}(N)$. For the plots, I considered $N = 4000$ and $n = 20$.
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Mar 8 at 13:56 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 13:53 comment added LPZ Ok, and out of curiosity (doing the simulations myself), what did you choose for $N$ and how many collisions did you simulate?
Mar 8 at 13:48 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 13:48 comment added Name YYY @LPZ : overall, I mean that the probability does not depend on the specific pair, it is constant. For the given particle, the probability of pairing with the rest of the particles is $1/(N-1)$ (and not $1/N$ as I wrote). If considering all the N particles, you are correct that the probability is $2/N(N-1)$. I will clarify this point, thanks! For the second simulation, I still sample directions uniformly, but when selecting the pairs I associate the weight $\propto 1/(p_{1}\cdot p_{2})$ to each of the pair.
Mar 8 at 13:11 comment added LPZ Why is your probability to choose a pair $1/N$ and not $2/N/(N-1)$? For the second simulation, are you using $p_1\cdot p_2$ to bias the choice the of pair which is not uniform anymore, or are you still choosing uniformly the pair and use it to bias the outcome of the collision which is not isotropic anymore?
Mar 8 at 13:07 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 13:02 comment added Name YYY @MichaelSeifert : I have added the plots showing the initial and final distributions, please see the updates to my question; the binning is for constant bin widths, and the amount of particles is enough to distinguish the behavior of the low-energy tail. I also checked that the 4-momentum is conserved in the collision (in particular, the total final energy is equal to the total initial energy, and the total momentum $\to 0$), as well as verified that the final particles have isotropy in directions.
Mar 8 at 12:55 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 12:44 comment added Michael Seifert Now that my results agree with yours, I'm not sure what could be going wrong. Possibilities include: errors in the collision algorithm (incorrectly distributed polar angles? failure to transform the collisions back into the lab frame?); improper binning techniques (make sure they're all of the same width in $E$); insufficient data to distinguish the exponent of $E$ in the density of states to a sufficient resolution. Without more information about the simulation I'm not sure this can be answered here.
Mar 8 at 12:36 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 12:28 answer added Michael Seifert timeline score: 6
Mar 8 at 12:27 history edited Name YYY CC BY-SA 4.0
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Mar 8 at 12:01 history asked Name YYY CC BY-SA 4.0