Timeline for Finding $p(z), \rho(z)$ of an ideal classical gas in a box
Current License: CC BY-SA 3.0
10 events
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May 13, 2015 at 13:23 | comment | added | M. Zeng | this is just like when you flip $N$ coins, you know that very likely you will get roughly half heads and half tails, because for each coin P(head)=P(tail)=1/2 and when the $N$ coins are tossed they are completely independent from each other. | |
May 13, 2015 at 13:21 | history | bounty ended | Trajan | ||
May 13, 2015 at 13:20 | vote | accept | Trajan | ||
May 13, 2015 at 13:20 | comment | added | M. Zeng | Yes, this is a single-particle phase space, meaning each point in this phase space represents a possible state for a single particle and the phase space density (which is proportional to the Boltzmann factor) determines how likely a single particle will be found in that (p,q) state. This single particle approach works because the particles are assumed to be a non-interacting ideal gas, which means all the particles are independent from each other but each of them are governed by the same probability distribution given by the same single-particle phase space density. | |
May 13, 2015 at 11:53 | comment | added | Trajan | Ok I get that but surely the $\rho(z)$ should be integrating over $d^Np$ and $d^Nq$(for $N$ particles) you seem to have only done $1$. For the second part I cannot see why you have just considered the particles above $z$? I cannot see why particles above the height z affect it? | |
May 13, 2015 at 2:12 | comment | added | M. Zeng | @sandstone the $\rho(z)$ is calculated based on the phase-space argument, which in this case already has the M-B distribution included, as you can see the Boltzmann factor inside the integration. The denominator is the total phase-space volume and the numerator is the sub-space volume with fixed $z$ and with all the other degrees of freedom $(x,y,p)$integrated over (you can see that $A$ is just the integration over $x$ and $y$). Consequently, the ratio gives the probability density for a particle being located at height $z$ | |
May 12, 2015 at 10:19 | comment | added | Trajan | Sorry I dont understand your the first half of your solution | |
May 12, 2015 at 10:03 | comment | added | Trajan | Could we have used the Maxwell Boltzmann distribution some how as well? | |
May 11, 2015 at 9:37 | history | edited | M. Zeng | CC BY-SA 3.0 |
added 240 characters in body
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May 11, 2015 at 9:29 | history | answered | M. Zeng | CC BY-SA 3.0 |