Let us study a system with several particles: odd number of 2D particles, aligned parallel to the detector, spaced every 2*r*. The detector is placed symmetrically in the middle. We will publish these conditions later. An identical probability distribution gives the probability of locating each particle. Although we use Gaussian distributions, any probability distribution works, because (int _{-infty}^{infty}P(x)dx=1). In this way, we do not become attached to any particular form of wave packet. Again, to simplify the calculations, we project 2D particles onto the plane of the detector to work with 1D distributions. Figure 3 shows such a configuration for (N=9) particles. The solid red line marks the detector and the dotted red lines are the limits of the visibility tunnel.

We define the transmittance of the dilute gas cloud *TR* as proposed in Ref.^{11}. The transmittance is the probability that a photon that the detector would have detected in the absence of a cloud passes unabsorbed through the whole *NOT*-cloud of elements and is detected by the detector. Collisions with individual particles are independent, so we can think of this process as a Markov chain:

$$begin{aligned} TR = prod _{n=1}^N big ( 1-G,P(o_n) big ), end{aligned}$$

(seven)

where (G,P(o_n)) is the probability that the photon is absorbed by *not*-th gas molecule, which is compensated by (on) of the detector, see Eq. (5) in Ref.^{11}.

Now we take advantage of periodicity. Identical pieces (of the same shape and number) of the probability distribution escape and flow into the tunnel of visibility. Thus, we can “unfold” a single distribution periodically instead of considering all distributions in one place. Then we virtually “shift” the detector *NOT* times (per a period of 2*r*) and take the product of all its positions. This way we can replace (o_n=r(2n-N-1)/2), and eq. (7) becomes:

$$begin{aligned} TR = prod _{n=1}^N left( 1-G,Pleft( rfrac{2n-N-1}{2}right) right) . end{aligned}$$

(8)

Figure 4 illustrates this idea. A distribution is divided into several pieces, periodically every 2*r*. Values of *not*, *oh*, (P_v(o)), (G,P_v(o)) and (1-G,P_v(o)) are superimposed for each part for convenience.

As expected, all probabilities (P_v(o)) sum up to 1, which means that the analyzed interval contains the entire particle. Probability does not “leak” laterally. We interpret this as a conserved mass in the system.

The following applies to (G=const):

$$begin{aligned} sum _{n=1}^{N} P_v(o_n)=1 Rightarrow sum _{n=1}^{N} G,P_v(o_n)=G=const . end{aligned}$$

(9)

Transmission is the product of (1-G,P_v(o_n)), see eq. (seven). The sum of its components is always constant. (sum (1-G,P_v(o_n))=const) As shown above. However, the constant sum does not guarantee that the product is constant:

$$begin{aligned} sum _{n=1}^{N} a_n=sum _{n=1}^{N} b_n not Rightarrow prod _{n=1}^{N} a_n=prod _{n=1}^{N} b_n. end{aligned}$$

(ten)

This shows that even *for closed conserved-mass systems, the transmission may change* because the conservation of mass depends on a sum (of certain elements), but the transmission depends on a product (of the same elements). In general, the transmission depends on how the distributions are divided. This division depends on (i) the shapes of the probability distributions and (ii) the width of the detector.

The shape of the normal distribution depends only on its standard deviation. detector size *r* significantly influences the values of the individual components of the product for one or the other (r,sim,stdev) Where (rinfluencing the product and, finally, the measured transmission.

In the classical case, for well localized particles (ideal gas) and a macroscopic detector, we have (stdev,ll,r). This way, any nonzero probability of a particle always arrives at a piece making all the elements of the product of Eq. (7) equal to 1, except one element. The single element less than 1 determines the value of the entire product. The product does not change when changing *r* because there will always be only one such element. Thus, in this case, the detector size cannot affect the transmission measurement. This explains why we do not observe any dependence of the transmission on the detector size in classical systems.

This analysis applies to any number of particles (*NOT*). Even for *NOT* the (on) substitution leading to Eq. (8) should be slightly different.

Figure 5 shows the dependence of transmittance on particle standard deviation for measurement with a fixed-size detector. The following section describes the details of the graph.

### Dense or inhomogeneous clouds

If there is much more than one particle per detector area (as in any real-world configuration), then we repeat the above reasoning several times in the following way. We divide the gas cloud into enough parts that each of them statistically contains only one particle per “detection area”. We calculate the (partial) transmission for each of these parts independently. From the property of independence of the probability of absorption by individual gas particles, we calculate the product of the partial transmissions, obtaining the total transmission of the whole cloud.

The same approach works for analyzing inhomogeneous gas clouds. One has to divide such a cloud into homogeneous parts, calculate the (partial) transmissions separately and take their product to obtain the total transmission.

Alternatively, we can do the trick of adjusting the constant *g*. We can put it equal to (1-TR_{cl})where (TR_{cl}) is the classical transmittance of the cloud. We then take a set of “artificial” particles distributed exactly every 2*r* as requested above. Such an artificial particle represents all the real particles present in the visibility tunnel of a given piece. Remember, however, that the spread (eg standard deviation) of this artificial particle is the same as that of any cloud particle. That is, we do not sum the masses of individual particles to calculate the propagation speed. This last method is very efficient for numerical calculations.

### 3d cloud

For a three-dimensional gas cloud, it must first be projected onto the plane of the detector. For such a 2D model, we require to distribute the particles evenly: one particle per detection zone. The simple way to analyze 2D is to consider the normal distribution and a square detector with side equal to 2*r*. For such a system: (i) an analytical solution is available, see (11) and (18) in Ref.^{11} and (ii) the square shape of the detector makes it possible to cover the whole plane with adjacent detectors. Then we can perform the same periodic reasoning given above for the 1D model.

An arbitrary shape of the detector makes reasoning more difficult and changes the quantitative equations. However, this is still possible because it requires only a finite area of the detector. However, qualitatively, the presented principle of the dependence of the transmission on the detector area is valid.

. Transmittance of gases ultradilutes revisited scientific reports