I want to get a Boltzmann distribution of the $v_x$, $v_y$ and $v_z$ velocity components (please, notice that the distribution is one-dimensional). To do so, I need the cumulative distribution function $F(v_x)$. I can get a random number and set it as $F(v_x)$. After that, how can I get $v_x$ with this information? Is there any other way to get $v_x$ using both, the $F(v_x)$ and the Boltzmann distribution itself?
Asked
Active
Viewed 908 times
2 Answers
0
A Boltzmann distribution is system dependent--it depends on the energy eigenstates. Moreover, if the system lacks symmetry, then Vx, Vy, and Vz may have very different distributions. However, if you're talking about an ideal gas--then it's the standard Maxwell-Boltzmann distribution. To get just Vx, you can use the equal-parition property, and rewrite a 1-d distribution accordingly. Or: you can compute a 3D distribution's magnitude and randomly orient it on a sphere (uniform in cos(theta) and phi), taking the x-projection.
To compute Vx from the inverse CDF (also called the PPF, percentile point function): generate uniform random numbers x on [0, 1], then Vx = F^(-1)(x) should be distributed according to the pdf, f(Vx).
JEB
- 42,131
0
Let $f(v_x)$ be the p.d.f. for $x$-component of velocity, and let $g(v)$ be the p.d.f. for total velocity magnitude (Maxwell-Boltzmann distribution) which is given. Then Bayes theorem gives: \begin{align} f(v_x)=\int_0^\infty dv~g(v)h(v_x|v) \end{align} in which $h(v_x|v)$ is the conditional p.d.f. such that $h(v_x|v)\delta v_x$ gives the probability that the $x$-component of velocity lies in the interval $(v_x,v_x+\delta v_x)$ given that its total velocity magnitude is $v$. Once we know $h(v_x|v)$ the problem is solved.
Obviously since we must have $v_x\leq v$, $h(v_x|v)=0$ for $v_x>v$. In what follows we shall deal with the case where $v_x\leq v$.
To derive the form of $h(v_x|v)$ we shall assume isotropy as regards the orientation of the velocity vector (whose length is given to be $v$). This means that if we take our origin at the tail of the velocity vector then the probability that the tip of the vector shall be located on any point on the surface of the sphere of radius $v$ is uniformly distributed over the surface area of the sphere. In other words the probability that the tip of velocity vector shall lie in the area neighborhood $\delta A$ of point $\mathbf{r}$ on the surface of the sphere (of radius $v$) is $(1/4\pi v^2)\delta A$. If $p(\mathbf{r})$ is the p.d.f. such that $p(\mathbf{r})\delta A$ gives the probability that the tip of velocity vector lies in the area neighborhood $\delta A$ of point $\mathbf{r}$ on the surface of the sphere (of radius $v$) then $p(\mathbf{r})=\frac{1}{4\pi v^2}$.
Applying Bayes theorem again we get: \begin{align} h(v_x|v)=\int_A dA~p(\mathbf{r})q(v_x|\mathbf{r},v) \end{align} in which $q(v_x|\mathbf{r},v)$ is the p.d.f. such that $q(v_x|\mathbf{r},v)\delta v_x$ gives the the probability that the $x$-component of velocity lies in the interval $(v_x,v_x+\delta v_x)$ given that its total velocity magnitude is $v$ in direction $\mathbf{r}$. The problem is now simplified to obtaining $q(v_x|\mathbf{r},v)$ which is easy. Let $\hat{\mathbf{x}}$ be the unit vector along +X-axis and $\hat{\mathbf{r}}\equiv\mathbf{r}/|\mathbf{r}|$. Then \begin{align} q(v_x|\mathbf{r},v)=\delta(v_x-|v\hat{\mathbf{r}}\cdot\hat{\mathbf{x}}|) \end{align} in which $\delta ()$ is the Dirac-delta function. The problem is in principle solved. Needless to say, actual integration to obtain $f(v_x)$ is most likely tedious and I shall not attempt it.
Deep
- 6,642