- #31
Aero51
- 545
- 10
Here's my take on the problem. Lunch turned into dinner which turned into drinks which turn into forgetting.
Here is my proof. Its really not mine but is paraphrased from Hills "An introduction to statistical thermodynamics" We will first start out with the assumption that the lotto balls are bouncing around in a plenum with a single opening. A ball is "selected" when it is sucked out from this opening. Let's assume that each ball is its own system, denoted Bj, with a state Ej, corresponding to (in this case) the balls kinetic energy. We will not define the limits of the possible states, however we will say that any system can be in any available potential state. The number of systems with a state by nj.
It follows that the following conditions must be satisfied
\sum n_{j} = s number of systems found in state E1...E2... etc [1]
\sum n_{j} E_{j} = E_{t} total sypersystem energy [2]
The total number of states found in the supersystem, St is computed by the formula:
S_t = s! / ((\prod_{j} n_{j})!) [3]
It follows that the probability of observing a a specific state:
P_{j} = \overline{n_{j}}/S [4]
and therefore
\overline{E}=\sum P_{j}E_{j} [5]
We want to calculate the most probable distribution, or the most likely observed n (n*) for our supersystem. To do this we take the natural log of the distribution St and utilize LeGrange multipliers to calculate the maximum. Although the number of balls B1, B2, ...Bn, is finite, it is safe to assume that their accessible (but not necessary probable) energy levels are close to ∞. This allows us to apply Sterlings Approximation, which yields the following equation constrained by the our first two equations:
\partial / \partial (n_{j}) [Ln(S_{t}) - \alpha \sum n_{j} - \beta \sum n_{j} E_{j}] = 0 [6]
which results in:
n^{*}_j = S_t e^{-\alpha}e^{-\beta E_j} [7]
Substituting this equation into our first two leads to the following
e^{\alpha} = \sum e^{-\beta E_j} [8]
\overline{E}= \sum (E_j e^{-\beta E_j}) / \sum e^{-\beta E_j} [9]
This also leads to the following result
P_j = n^*_j = e^{-\beta E(j)}/(\sum e^{-\beta E(j)}) [10]
I probably did a bad job paraphrasing this derivation, but again its not mine. It is sound, but my explanation may not be. I could have just skipped to the probability function and written below to get my point across.
This where I come in and mess everything up:
Recalling that the energy of each ball is defined by the basic kinetic energy equation:
E_j = 1/2 m V_{j}^2 = 1/2 m (V_x^2 +V_y^2+V_z^2)_{j} [11]
Which leads to the probability function given a velocity:
P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j}) [12]
And recalling that the ball needs to be sucked into the tube to be "selected", leads to the notion that only a certain combination of Vx, Vy Vz for each ball will permit selection (IE a lotto ball velocity parallel to the orifice would not result in its selection). I do not want to sit down and figure this specific number out so I will not.
Further more, with these velocities comes a spatial constriction, (a location that the ball must be in order to be "selected, which is proportional to the ratio of the orifice and ball diameters).
The obvious conclusion is that each ball must have both a certain position and a certain velocity to be selected. Let Pposition denote the probability of being in a certain position.
P_{velocity}\cap P_{position} = P_{velocity} P_{position} [13]
since we'll say velocity and position are independent of each other. This leads to the following expression for selection.
P_{selection}= P_{velocity} P_{position} = P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j}) P_{position} [14]
As was mentioned before there was actually a set of velocities and positions which had to be met in order for the ball to be "selected" This leads to the total probability of being selected:
P_{total selection}= \sum_{vx,vy,vz}P_{velocity} \sum_{x,y,z}P_{position}
Where the sums denote the permissible velocities and positions. I can't prove this isn't 1/n, but if you can prove it is show me :)
Here is my proof. Its really not mine but is paraphrased from Hills "An introduction to statistical thermodynamics" We will first start out with the assumption that the lotto balls are bouncing around in a plenum with a single opening. A ball is "selected" when it is sucked out from this opening. Let's assume that each ball is its own system, denoted Bj, with a state Ej, corresponding to (in this case) the balls kinetic energy. We will not define the limits of the possible states, however we will say that any system can be in any available potential state. The number of systems with a state by nj.
It follows that the following conditions must be satisfied
\sum n_{j} = s number of systems found in state E1...E2... etc [1]
\sum n_{j} E_{j} = E_{t} total sypersystem energy [2]
The total number of states found in the supersystem, St is computed by the formula:
S_t = s! / ((\prod_{j} n_{j})!) [3]
It follows that the probability of observing a a specific state:
P_{j} = \overline{n_{j}}/S [4]
and therefore
\overline{E}=\sum P_{j}E_{j} [5]
We want to calculate the most probable distribution, or the most likely observed n (n*) for our supersystem. To do this we take the natural log of the distribution St and utilize LeGrange multipliers to calculate the maximum. Although the number of balls B1, B2, ...Bn, is finite, it is safe to assume that their accessible (but not necessary probable) energy levels are close to ∞. This allows us to apply Sterlings Approximation, which yields the following equation constrained by the our first two equations:
\partial / \partial (n_{j}) [Ln(S_{t}) - \alpha \sum n_{j} - \beta \sum n_{j} E_{j}] = 0 [6]
which results in:
n^{*}_j = S_t e^{-\alpha}e^{-\beta E_j} [7]
Substituting this equation into our first two leads to the following
e^{\alpha} = \sum e^{-\beta E_j} [8]
\overline{E}= \sum (E_j e^{-\beta E_j}) / \sum e^{-\beta E_j} [9]
This also leads to the following result
P_j = n^*_j = e^{-\beta E(j)}/(\sum e^{-\beta E(j)}) [10]
I probably did a bad job paraphrasing this derivation, but again its not mine. It is sound, but my explanation may not be. I could have just skipped to the probability function and written below to get my point across.
This where I come in and mess everything up:
Recalling that the energy of each ball is defined by the basic kinetic energy equation:
E_j = 1/2 m V_{j}^2 = 1/2 m (V_x^2 +V_y^2+V_z^2)_{j} [11]
Which leads to the probability function given a velocity:
P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j}) [12]
And recalling that the ball needs to be sucked into the tube to be "selected", leads to the notion that only a certain combination of Vx, Vy Vz for each ball will permit selection (IE a lotto ball velocity parallel to the orifice would not result in its selection). I do not want to sit down and figure this specific number out so I will not.
Further more, with these velocities comes a spatial constriction, (a location that the ball must be in order to be "selected, which is proportional to the ratio of the orifice and ball diameters).
The obvious conclusion is that each ball must have both a certain position and a certain velocity to be selected. Let Pposition denote the probability of being in a certain position.
P_{velocity}\cap P_{position} = P_{velocity} P_{position} [13]
since we'll say velocity and position are independent of each other. This leads to the following expression for selection.
P_{selection}= P_{velocity} P_{position} = P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j}) P_{position} [14]
As was mentioned before there was actually a set of velocities and positions which had to be met in order for the ball to be "selected" This leads to the total probability of being selected:
P_{total selection}= \sum_{vx,vy,vz}P_{velocity} \sum_{x,y,z}P_{position}
Where the sums denote the permissible velocities and positions. I can't prove this isn't 1/n, but if you can prove it is show me :)
