Indistinguishable and Distinguishable Dice Numbers and States

Probability Distributions for Sums of n s-Sided Dies (or sums of n particles with s states each) This user-interactive program shows the probability of the possible sums when n s-sided dies are thrown. It handles both the case where the dies cannot be distinguished from each other and the case where they can be distinguished. The former case (labeled Indist here) is relevant to quantum Bose-Einstein (BE) particles while the latter case is relevant to classical Maxwell-Boltzmann (MB) particles. For particles, the number of dots on the side of the die has some correspondence to the energy level of each die and the sum of the dots of all the dies in the throw would give the total energy of that particular particle state. Since the particles are not distinguishable, only the sum of their energies (dots) is important in describing the macro state of this system. It should be pointed out that indistiguishable objects do not exist unless quantum effects are active so that, for everyday objects, all probability results will be for distinguishable with MB statistics. If particles cannot be distinguished, then the sum of a five die throw resulting in side values (a,b,c,d,e) cannot be counted separately from a similar throw like (b,a,c,d,e). Said another way, for indistinguishable particles, a change in the order in which they fall does not invoke a new result. That means that there are many fewer sums to be counted for indistinguishable particles than for distinguishable ones. And this results in a difference in the relative distribution (probability) function for the two cases. The total number of possible ways that indistinguishable (BE) particle sums can occur is computed by the function:

`ZBE=(n+s-1)!/(n!(s-1)!)`

while for distinguishable (MB) particles we have:

`ZMB=n^s`

The learner can see from the plots (which have the same horizontal sum scale) that the BE distribution has considerably (about 30%) larger variance than the MB distribution. That means that the width of the BE momentum or energy distribution will be greater than the width of the MB distribution. At this point, it is important to point out that the requirement for particles to behave as BE particles is that they have thermal deBroglie wavelength , `lambda_(BT)`, greater than the average particle spacing. An equation for `lambda_(BT)` is `lambda_(BT)=h/sqrt(3MkT)` where h is the Planck constant, M is mass, k is Boltzmann's constant, and T is the Kelvin temperature. As an example, hydrogen molecules, `H_2`, at 1 degree Kelvin have `lambda_(BT)`=1.76 nanometers which means that `H_2` does not have to be a liquid (which has a spacing of 0.36 nanometers) in order to qualify for BE behavior. The learner's conclusion from all this is that particles start behaving as BE particles when their wavelengths are long enough that there is cosiderable overlap between the quantum properties of of neighboring particles. What about the photons (which are well known to be BE particles) in a black body cavity? We know that typical photon wavelength in a 3000 K black body is 1000 nanometers. The BE requirement then is that the photon spacing is less than 1000 nanometers. So what do we expect the photon density to be at 3000 K? What about the much simpler case of photons in a laser or even a passive linear resonator? The criterion here would be that there be one photon for each wave length of light in the laser. Thus a laser that is 1000 wavelength long would need to have at least n=1000 photons inside at any one time. It should be noted here that, laser threshold can be defined as there being an average of at least n=1 photon in the laser resonator. If the wavelength is 1000 nm, then the length, `L`, would have to be less than 0.001 m or 1 mm. The number of photons passing a point in the laser cavity would then be nc/L where c is the speed of light, which turns out to be `3*10^(14)`. Since each photon has energy of about 1 electron volt, the power passing a point is about `3*10^(14)*1.6*10^(-19)=4.8*10^(-5)` Joules per second or about 48 microwatts. Many other more massive elements than `H_2` have to be a lot colder to behave as BE particles and still not be liquid or solid. The ways that the BE distributions are computed is the following for() loops which handles n=3 particles (dies) and s=3 states (sides):
`"for " i=`lo`;i lt `hi`;i++)` die 1
`for " j=i;j lt `hi`;j++)` die 2
`for " k=j;k lt `hi`;k++)` die 3
`"sumList".Add(i+j+k)`;
where `lo` is the number value for the starting state (number of dots) and `hi` is the number value for the ending state. Since dies have dot numbers ranging from 1 to their number of sides, `lo` for dies would be 1 and `hi` would be the number of sides. Note that for the case "BE" the nested sums start at the value of next outer loop. That is all that is needed to avoid repeats of the set of integers, (i,j,k). For the distinguishable case "MB", all of the loops start at the lo value which leads to huge numbers of repeats when the number of particles is high, A recursive algorithm for BE particle group states with any number of particles (dies) and any number of states (sides) is the following: The state (throw) indices are obtained from the data[] array and all of the data[] for the various states are stored in dataA[] for later printing. The sumA[] array stores all of the sums of the indices of the various states. These sums may be considered to be proportional to the total momentum or energy of each state. The second loop, where the recursive call is made, uses the call

`"loopBE"(i, "index" + 1,"nDies","sides")`

A limited number of the state index sets as well as their sums are listed here. The bar chart type plots are the exact data that are obtained from the computer codes given above. The (usually) smooth curves are gaussian type plots of the form

`f(p)=exp[-(p-p0)^2/w^2]`

where p0 is the bin sum value at the center of the curve and w is an expression for the width at the 1/e point of f(p). For the Indist case the expression used for w is w=[(n*s)/4.0]*sqrt(loge(2)) while for the Dist case the expression used for w is w=sqrt(n)/2.0*(s-1); The result is that the width of the Indist case is considerably greater than that of the Dist case especially when s is small. In fact, when s=2, the width of the Indist case is undefined because all its sums are equally probable. The learner has access to both changes in number of particles (dice) and the number of states (dots) on each side of the dies. Please click the links below to see how the Bose-Einstein distribution is derived and is related to the number of particles and states discussed here.

Bose Distribution Derivation Indistinguishable Dice Sums Bose-Einstein Condensate