A Phys. Rev. Letter by M Wilkinson on large deviation analysis of rapid-onset rainfall

The following is a record of some notes I made on the main mathematical developments in Michael Wilkinson’s 2016 Physical Review Letter on the large deviation analysis of rapid-onset rain showers (reference: Wilkinson, M., 2016, Large Deviation Analysis of Rapid Onset of Rain Showers, Phys. Rev. Lett. 116, 018501). This publication will be referred to as MW2016 herein. The published Letter has an accompanying Supplement which will also be referred to here. The Supplement provides additional details of some intricate derivations underlying key results presented in the main Letter.

One of the main contributions of MW2016 is that it shows that large deviation theory is the correct framework for analysing the implications of a collector-drop model of rapid rain formation which was discussed in a previous paper by Kostinski and Shaw (reference: Kostinski, A., Shaw, R., 2005, Fluctuations and Luck in Droplet Growth by Coalescence, Bull. Am. Meteorol. Soc. 86, 235). In this collector-drop model, a runaway droplet falls and coalesces at a rapidly increasing rate with many smaller water particles. The collision rates are assumed to increase according to a simple power law function of the number of collisions. In addition to formulating a volume fraction cut-off mechanism for raindrop size, MW2016 also successfully uses large deviation analysis in conjunction with asymptotic approximations to study the link between the time scale for rapid onset of rain showers and the number of water particle collisions needed for raindrop formation. In the present article, I provide a mathematical review of the key points in MW2016 and also detail some errata in the published Letter which were identified in the review process.

Mathematical review

In the collector-drop model, a runaway droplet falls and collides with a large number $\mathcal{N}$ of small droplets of radius $a_0$ according to an inhomogeneous Poisson process. If the random variable $t_n$ represents the inter-collision time leading up to the $n$ -th collision, the time for a droplet to experience runaway growth is then the random variable

$T = \sum_{n=1}^{\mathcal{N}} t_n \quad \quad \quad \quad \quad (1)$

The inter-collision times $t_n$ are independent exponentially distributed random variables with

$P_n(t_n) = R_n \exp(-R_n t_n) \quad \quad \quad \quad \quad (2)$

where $R_n$ is the number of collisions per second, i.e., the collision rate, by the time of the $n$ -th collision. The collision rate $R_n$ is modelled as

$R_n = R_1 f(n) \quad \quad \quad \quad \quad (3)$

where

$R_1 = \epsilon N_0 \pi(a_0 + a_1)^2 \alpha (a_1^2 - a_0^2) \quad \quad \quad \quad \quad (4)$

is the collision rate of a drop of radius $a_1$ with droplets of radius $a_0$ . Here, $\epsilon$ is the coalescence efficiency, $N_0$ is the number density of microscopic water droplets per $m^3$ , $\pi(a_0 + a_1)^2$ is the effective cross-section area and $\alpha (a_1^2 - a_0^2)$ is the relative velocity. The constant $\alpha$ is a constant of proportionality between terminal velocity and the radius squared of a water particle.

The function $f(n)$ in (3) characterises the dependence of $R_n$ on the number of collisions. In the Kostinski and Shaw 2005 paper referenced above, $f(n) = n^2$ , but we can have $f(n) = n^{4/3}$ when $\epsilon \approx 1$ and $a_n \gg a_1$ . MW2016 treats $f(n)$ as a simple power law, so that

$R_n = R_1 n^{\gamma} \quad \quad \quad \quad \quad (5)$

The average inter-collision time leading to the $n$ -th collision is then given by

$\langle t_n \rangle \equiv \tau_n = \frac{1}{R_n} = \frac{1}{R_1} n^{-\gamma} \quad \quad \quad \quad \quad (6)$

and the mean time for explosive growth is obtained as

$\langle T \rangle = \sum_{n=1}^{\mathcal{N}} \tau_n = \frac{1}{R_1} \sum_{n=1}^{\mathcal{N}} n^{-\gamma} \quad \quad \quad \quad \quad (7)$

When $\gamma > 1$ , the mean time for explosive growth converges as $\mathcal{N} \rightarrow \infty$ , giving

$\lim_{\mathcal{N} \rightarrow \infty} \langle T \rangle = \frac{1}{R_1} \zeta(\gamma) \quad \quad \quad \quad \quad (8)$

where $\zeta(\gamma)$ is Riemann’s zeta function.

A cut-off mechanism for raindrop size is obtained by considering the liquid water content of a cloud, expressed as a volume fraction $\Phi_l$ . Since $N_0$ is the number of water particles of radius $a_0$ per $m^3$ , and each particle is a sphere with volume $\frac{4}{3} \pi a_0^3$ , the volume fraction of liquid water per $m^3$ of cloud is given by

$\Phi_l \approx N_0 \cdot \frac{4}{3} \pi a_0^3 \quad \quad \quad \quad \quad (9)$

For example, if each droplet is initially of radius $a_0 = 10 \mu m = 10^{-5}m$ , and there are $N_0 \approx 10^9$ such droplets per $m^3$ of cloud, then $\Phi_l \approx 10^{-6}$ .

Using (2), the fraction of droplets undergoing runaway growth between times $t$ and $t + \delta t$ is given by

$P(t) (t + \delta t - t) = P(t) \delta t \quad \quad \quad \quad \quad (10)$

For example, if $P(t) \delta t \approx 10^{-6}$ , then `one in a million’ water droplets will undergo runaway growth between times $t$ and $t + \delta t$ .

If the volume of each runaway droplet increases by a factor of $\mathcal{N}$ , the volume fraction of liquid water removed from the cloud per $m^3$ over the time interval of length $\delta t$ will then be

$\mathcal{N} \Phi_l P(t) \delta t \quad \quad \quad \quad \quad (11)$

For example, if $\mathcal{N} = 10^3$ , $\Phi_l \approx 10^{-6}$ and $P(t) \delta t \approx 10^{-6}$ , then the volume fraction of liquid water content per $m^3$ is reduced by $10^3 \cdot 10^{-6} \cdot 10^{-6} = 10^{-9}$ over the time interval of length $\delta t$ . This represents a $\frac{10^{-9}}{10^{-6}}\cdot 10^2 = 0.1$ per cent reduction in $\Phi_l$ .

As $\delta t \rightarrow 0$ , the instantaneous change of the liquid water content of a cloud per $m^3$ due to runaway growth of droplets is thus

$\frac{d \Phi_l}{d t} = -\mathcal{N} \Phi_l P(t) \quad \quad \quad \quad \quad (12)$

Therefore, the instantaneous percentage rate of decline of the volume fraction is

$-\frac{d \ln \Phi_l}{d t} = \mathcal{N} P(t) \quad \quad \quad \quad \quad (13)$

Integrating (13) over a time interval then gives a percentage loss $\mu$ in that interval:

$\mu = \mathcal{N}\int_0^t dt^{\prime} P(t^{\prime}) \quad \quad \quad \quad \quad (14)$

The onset of a shower is determined by the criterion that $\mu$ be a significant percentage of the liquid water content of a cloud that is removed by raindrop formation as a result of $\mathcal{N}$ particle collisions. We are now interested in finding the time scale $t^{*}$ over which this significant reduction in $\Phi_l$ occurs. In accordance with (14), this $t^{*}$ satisfies

$\mathcal{N^{*}}\int_0^{t^{*}} dt^{\prime} P(t^{\prime}) = 1 \quad \quad \quad \quad \quad (15)$

where $\mathcal{N^{*}} \equiv \mathcal{N}/\mu$ .

To make progress from here, MW2016 approximates $P(t)$ in (15) using the large deviation principle, which is explained in detail in, e.g., an article by Touchette (reference: Touchette, H., 2009, The large deviation approach to statistical mechanics, Phys. Rep. 478, 1). With $T$ defined as in (1) and $\langle T \rangle$ given by (7), we implement the large deviation principle here by approximating the probability $P(T)dT$ at some value $\bar{T}$ in the small-value tail of the distribution of $T$ as

$P(\bar{T})d\bar{T} = P(T \in [\bar{T}, \bar{T} + d\bar{T}])$

where $P(\bar{T})$ is given by

$P(\bar{T}) = \frac{1}{\langle T \rangle} \exp[-J(\tau)] \quad \quad \quad \quad \quad (16)$

with $\tau \equiv \bar{T}/\langle T \rangle$ . The function $J(\tau)$ in (16) is referred to as an entropy function or rate function in large deviation theory, and is defined below. Using (16) in (15) we get the condition for the onset of rainfall as

$\mathcal{N^{*}} \int_0^{t^{*}} \frac{d \bar{T}}{\langle T \rangle} \exp \bigg[-J\bigg(\frac{\bar{T}}{\langle T \rangle}\bigg)\bigg] = \mathcal{N^{*}} \int_0^{t^{*}} d \tau^{\prime} \exp [-J(\tau^{\prime})] = 1 \quad \quad \quad \quad \quad (17)$

with $\tau^{\prime} \equiv \bar{T}/\langle T \rangle$ . A saddle-point approximation for the integral here at the maximum value of the integrand, $\exp[-J(\tau^{*})]$ , with $\tau^{*} \equiv \bar{T}^{\ast}/\langle T \rangle$ , then gives the condition for the onset of rainfall as

$\mathcal{N}^{*} \exp[-J(\tau^{*})] = 1 \quad \quad \quad \quad \quad (18)$

Taking logs, we can write the condition alternatively as

$\ln \mathcal{N}^{*} = J(\tau^{*}) \quad \quad \quad \quad \quad (19)$

What is required now is to find an expression for $\tau^{*}$ in terms of $\ln \mathcal{N}^{*}$ , using (19). To obtain this, it is necessary to find an expression for the rate function $J(\tau)$ for the random sum defined by (1) and (2).

Consider a single inter-collision time $t_n$ with density (2). Following the approach in the article by Touchette cited above for the purposes of the large deviation analysis here, we obtain the log of the Laplace transform of the random variable $t_n$ , often referred to as its scaled cumulant generating function, as

$\lambda_n (k) = -\ln \langle \text{e}^{-kt_n} \rangle = -\ln\bigg(\int_0^{\infty} d t_n \text{e}^{-kt_n} R_n \text{e}^{-R_n t_n}\bigg) = \ln\bigg(1 + \frac{k}{R_n}\bigg) \quad \quad \quad \quad \quad (20)$

To ensure the logarithm on the right-hand side is always defined, we need to restrict the admissible values of $k$ to the interval $(-R_1, \infty)$ . Inter-collision times are assumed to be independent, so for a sum $T$ of inter-collision times such as (1), we get the log of the Laplace transform of $T$ as

$\lambda(k) = -\ln \langle \text{e}^{-kT} \rangle$

$= -\ln \bigg \langle \prod_{n=1}^{\mathcal{N}} \text{e}^{-kt_n} \bigg \rangle = -\ln \prod_{n=1}^{\mathcal{N}}\langle \text{e}^{-kt_n} \rangle = \sum_{n=1}^{\mathcal{N}} \ln\bigg(1 + \frac{k}{R_n}\bigg) \quad \quad \quad \quad \quad (21)$

This is differentiable with respect to $k$ for all $k \in (-R_1, \infty)$ , so by the Gartner-Ellis Theorem of large deviation analysis, discussed in Touchette, the rate function in (16) is given by

$J(\tau) = \sup_k \{\lambda(k) - k\tau\} \quad \quad \quad \quad \quad (22)$

The expression on the right-hand side of (22) is referred to as the Fenchel-Legendre transform of $\lambda(k)$ .

Note that the expression in (22) is the negative of the expression usually given in large deviation theory. This is because the outer function in the formulation of the log of the Laplace transform in (20) above is the negative of the expression used, e.g., in Touchette, and this formulation also has the parameter $k$ negated. The alternative formulation would be

$\lambda_n(k) = \ln \langle \text{e}^{kt_n} \rangle = \ln\bigg(\int_0^{\infty} d t_n \text{e}^{kt_n} R_n \text{e}^{-R_n t_n}\bigg) = -\ln\bigg(1 - \frac{k}{R_n}\bigg)$

and the associated entropy function for this would then be the more conventional

$J(\tau) = \sup_k \{k\tau - \lambda(k)\}$

In this case, the admissible values of $k$ would be those in the interval $(-\infty, R_1)$ . Essentially, the signs of $\lambda(k)$ and $k$ have been negated in the formulation in MW2016, as this is more convenient for the analysis.

Differentiating the bracketed expression in (22) with respect to $k$ and setting equal to zero we obtain

$\lambda^{\prime}(k) - \tau = 0 \quad \quad \quad \quad \quad (23)$

But

$\lambda^{\prime}(k) = \sum_{n=1}^{\mathcal{N}} \frac{\big(\frac{1}{R_n}\big)}{1 + \frac{k}{R_n}} = \sum_{n=1}^{\mathcal{N}} \frac{1}{R_n + k} \quad \quad \quad \quad \quad (24)$

Putting this in (23) we get

$\tau = \sum_{n=1}^{\mathcal{N}} \frac{1}{R_n + k} \quad \quad \quad \quad \quad (25)$

This implicitly gives an optimal value of $k$ as a function of $\tau$ , say $k^{*}(\tau)$ . If this could be obtained explicitly, inserting it into the right-hand side of (22) would give the required rate function $J(\tau)$ :

$J(\tau) = \lambda(k^{*}(\tau)) - k^{*}(\tau) \cdot \tau \quad \quad \quad \quad \quad (26)$

However, in practice it is not possible to obtain an exact expression for $J(\tau)$ in this way. Instead, MW2016 obtains asymptotic expressions for $J(\tau)$ which are valid for small $\tau$ . This is achieved by reparameterising $J(\tau)$ using a scaled variable $\kappa$ , defined as

$\kappa = \frac{k^{*}}{R_1} \quad \quad \quad \quad \quad (27)$

From (7) and (8) we get, as $\mathcal{N} \rightarrow \infty$ ,

$\tau = \frac{\bar{T}}{\langle T \rangle} = \frac{R_1 \bar{T}}{\zeta(\gamma)} \quad \quad \quad \quad \quad (28)$

Using (5), (27) and (28) in (25) we get

$\tau(\kappa) = \frac{R_1}{\zeta(\gamma)} \sum_{n=1}^{\infty} \frac{1}{R_1 n^{\gamma} + R_1 \kappa} = \frac{1}{\zeta(\gamma)} \sum_{n=1}^{\infty} \frac{1}{\kappa + n^{\gamma}} \quad \quad \quad \quad \quad (29)$

The expression on the right-hand side of (29) appears in equation (18) in MW2016.

We now write (2.21) in reparameterised form as $\mathcal{N} \rightarrow \infty$ as

$S(\kappa) = \sum_{n=1}^{\infty} \ln \bigg(1 + \frac{R_1 \kappa}{R_1 n^{\gamma}}\bigg) = \sum_{n=1}^{\infty} \ln(1 + \kappa n^{-\gamma}) \quad \quad \quad \quad \quad (30)$

We also write

$k^{*}(\tau) \cdot \tau = \frac{R_1 \kappa \zeta(\gamma) \tau}{R_1} = \zeta(\gamma) \kappa \tau \quad \quad \quad \quad \quad (31)$

Then we re-express (26) in the form

$J(\kappa)= S(\kappa) - \zeta(\gamma) \kappa \tau(\kappa) \quad \quad \quad \quad \quad (32)$

This appears as equation (19) in MW2016.

Following the approach in the Supplement to MW2016, the next stage in the development is to obtain an asymptotic expression for the sum in (30) as $\kappa \rightarrow \infty$ . (It is assumed from now on that $R_1 = 1$ , to simplify the expressions). Upon differentiating this, we will then obtain a corresponding asymptotic expression for $\tau$ in accordance with (29), which can be inverted to obtain a small $\tau$ approximation for $\kappa$ , and subsequently a small $\tau$ approximation for $J(k)$ . Finally, this will be used in (19) to obtain an expression for $\tau^{*}$ in terms of $\ln\mathcal{N}^{*}$ .

Begin by approximating the sum in (30) as an integral. Observe that

$\int_0^{\infty} dn \ln(1 + \kappa n^{-\gamma})$

$= \int_0^1 dn \ln(1 + \kappa n^{-\gamma}) + \int_1^2 dn \ln(1 + \kappa n^{-\gamma}) + \int_2^3 dn \ln(1 + \kappa n^{-\gamma}) + \cdots$

$= \int_1^0 -dx \ln(1 + \kappa (1-x)^{-\gamma})$

$+ \int_1^0 -dx \ln(1 + \kappa (2-x)^{-\gamma}) + \int_1^0 -dx \ln(1 + \kappa (3-x)^{-\gamma}) + \cdots$

$= \sum_{n=1}^{\infty} \int_0^1 dx \ln(1 + \kappa (n-x)^{-\gamma})$

(The third line is obtained by making the change of variable $n = \overline{m} - x$ in each integral in the second line, with $\overline{m}$ representing the fixed positive integer in the upper limit, so $dn = -dx$ , and $x = 0$ when $n = \overline{m}$ whereas $x = 1$ when $n = \overline{m} - 1$ ). Therefore

$\sum_{n=1}^{\infty} \ln(1 + \kappa n^{-\gamma}) \equiv \int_0^{\infty} dn \ln(1 + \kappa n^{-\gamma})$

$+ \bigg\{\sum_{n=1}^{\infty} \ln(1 + \kappa n^{-\gamma}) - \int_0^{\infty} dn \ln(1 + \kappa n^{-\gamma})\bigg\}$

$= \int_0^{\infty} dn \ln(1 + \kappa n^{-\gamma}) + \sum_{n=1}^{\infty} \int_0^1 dx \bigg \{ \ln(1 + \kappa n^{-\gamma}) - \ln(1 + \kappa (n-x)^{-\gamma})\bigg\}$

In MW2016 this is written as

$S = S_0 + \sum_{n=1}^{\infty} \triangle S_n \quad \quad \quad \quad \quad (33)$

where

$S = \sum_{n=1}^{\infty} \ln(1 + \kappa n^{-\gamma})$

$S_0 = \int_0^{\infty} dn \ln(1 + \kappa n^{-\gamma})$

$\triangle S_n = \int_0^1 dx \bigg \{\ln(1 + \kappa n^{-\gamma}) - \ln(1 + \kappa (n-x)^{-\gamma})\bigg\} \quad \quad \quad \quad \quad (34)$

Write $m = n - x$ . Then as $x \rightarrow 0$ we have

$\frac{\ln(1 + \kappa n^{-\gamma}) - \ln(1 + \kappa (n-x)^{-\gamma})}{x}$

$= \frac{\ln(1 + \kappa (m+x)^{-\gamma}) - \ln(1 + \kappa m^{-\gamma})}{x}$

$\rightarrow \frac{\partial \ln(1 + \kappa n^{-\gamma})}{\partial n} = \frac{-\gamma \kappa}{n(\kappa + n^{\gamma})}$

Therefore, when $n \gg 1$ , we can approximate each $\triangle S_n$ by using

$\triangle S_n = \int_0^1 dx x \bigg\{\frac{\ln(1 + \kappa n^{-\gamma}) - \ln(1 + \kappa (n-x)^{-\gamma})}{x}\bigg\}$

$\sim \frac{\partial \ln(1 + \kappa n^{-\gamma})}{\partial n} \int_0^1 dx x$

$= \frac{-\gamma \kappa}{2n(\kappa + n^{\gamma})} \quad \quad \quad \quad \quad (35)$

When $\kappa n^{-\gamma} \gg 1$ but $n$ is not necessarily large, we can write

$\triangle S_n \sim \int_0^1 dx \ln[\kappa n^{-\gamma}(1 + n^{\gamma}/\kappa)] - \int_0^1 dx \ln[\kappa (n-x)^{-\gamma}(1 + (n - x)^{\gamma}/\kappa]$

$= -\gamma \int_0^1 dx \{ \ln(n) - \ln(n-x)\} + O(\kappa^{-1}) \quad \quad \quad \quad \quad (36)$

and we can ignore the $O(\kappa^{-1})$ component as $\kappa \rightarrow \infty$ . Next, make the change of variable $y = n - x$ in the integral in (36). Then when $x = 0$ , $y = n$ , when $x = 1$ , $y = n-1$ , and $dy = -dx$ . We get

$\triangle S_n \sim \gamma \int_n^{n-1} dy \{\ln(n) - \ln(y)\}$

$= -\gamma \int_{n-1}^n dy \{\ln(n) - \ln(y)\}$

$= -\gamma\bigg[(n-1) \ln\bigg(\frac{n-1}{n}\bigg)+ 1\bigg] \quad \quad \quad \quad \quad (37)$

In order to evaluate $S$ using (33), MW2016 splits the summation over $\triangle S_n$ into two parts: choose an integer $M$ such that $\kappa M^{-\gamma} \gg 1$ and let $M \rightarrow \infty$ as $\kappa \rightarrow \infty$ . Use (35) for $\triangle S_n$ when $n > M$ , and (37) when $n \leq M$ . We get

$S \sim S_0 - \gamma \sum_{n=1}^M \bigg[(n-1) \ln\bigg(\frac{n-1}{n}\bigg)+ 1\bigg] - \gamma \kappa \sum_{n = M+1}^{\infty} \frac{1}{2n(\kappa + n^{\gamma})} \quad \quad \quad \quad \quad (38)$

We now approximate the second summation using an integral:

$- \gamma \kappa \sum_{n = M+1}^{\infty} \frac{1}{2n(\kappa + n^{\gamma})} \sim \frac{-\gamma \kappa}{2} \int_M^{\infty} dn \frac{1}{n(\kappa + n^{\gamma})} = \frac{-\gamma}{2} \int_M^{\infty} dn \frac{1}{n\bigg(1 + \frac{n^{\gamma}}{\kappa}\bigg)} \quad \quad \quad \quad \quad (39)$

Make the change of variable $y = n^{\gamma}/\kappa$ . Then $dy = \frac{1}{\kappa} \gamma n^{\gamma - 1} dn$ , so $dn = \frac{1}{y} \frac{n}{\gamma} dy$ , and when $n = M$ , $y = M^{\gamma}/\kappa$ , while when $n = \infty$ , $y = \infty$ . The integral in (39) becomes

$\frac{-\gamma}{2} \int_{M^{\gamma}/\kappa}^{\infty} \frac{1}{y} \frac{n}{\gamma} dy \frac{1}{n(1+y)}$

$= -\frac{1}{2} \int_{M^{\gamma}/\kappa}^{\infty} dy \frac{1}{y(1+y)}$

$= -\frac{1}{2} \int_{M^{\gamma}/\kappa}^{\infty} dy \bigg \{ \frac{1}{y} - \frac{1}{y+1} \bigg \}$

$= \frac{1}{2} \ln\bigg(\frac{M^{\gamma}}{\kappa}\bigg) - \frac{1}{2} \ln\bigg(\frac{M^{\gamma}}{\kappa} + 1\bigg)$

$= \frac{1}{2} \ln \bigg(\frac{\frac{M^{\gamma}}{\kappa}}{\frac{M^{\gamma}}{\kappa}+ 1}\bigg)$

$= -\frac{1}{2} \ln(1 + \kappa M^{-\gamma})$

$= -\frac{1}{2} \ln \bigg[\kappa M^{-\gamma} \bigg(1 + \frac{1}{\kappa M^{-\gamma}}\bigg)\bigg]$

$= -\frac{1}{2} \ln(\kappa) - \frac{1}{2} \ln(M^{-\gamma}) - \frac{1}{2} \ln \bigg(1 + \frac{1}{\kappa M^{-\gamma}}\bigg)$

$= -\frac{1}{2} \ln(\kappa) + \frac{\gamma}{2} \ln(M) + O(\kappa^{-1})$

Putting this in (38) we get

$S \sim S_0 - \gamma \sum_{n=1}^M \bigg[(n-1) \ln\bigg(\frac{n-1}{n}\bigg)+ 1\bigg] -\frac{1}{2} \ln(\kappa) + \frac{\gamma}{2} \ln(M) + O(\kappa^{-1}) \quad \quad \quad \quad \quad (40)$

We can write this as

$S \sim S_0 - \gamma \bigg\{\sum_{n=1}^M \bigg[(n-1) \ln\bigg(\frac{n-1}{n}\bigg)+ 1\bigg] - \frac{1}{2} \ln(M)\bigg\}$

$-\frac{1}{2} \ln(\kappa) + O(\kappa^{-1}) \quad \quad \quad \quad \quad (41)$

Taking the limit as $M \rightarrow \infty$ , the term in curly brackets in (41) converges to a constant $C$ , and we get

$S \sim S_0 - \frac{1}{2} \ln(\kappa) - \gamma C + O(\kappa^{-1}) \quad \quad \quad \quad \quad (42)$

This is equation [10] in the Supplement to MW2016.

From (29), we see that differentiating $S(\kappa)$ in (30) gives a time $T(\kappa) = \zeta(\gamma) \tau(\kappa)$ . The corresponding asymptotic expression for $T(\kappa)$ is obtained by differentiating (42). We get

$\frac{\partial S}{\partial \kappa} \sim \int_0^{\infty} dn \frac{1}{\kappa} \bigg(\frac{1}{1 + \frac{n^{\gamma}}{\kappa}}\bigg) - \frac{1}{2\kappa} \quad \quad \quad \quad \quad (43)$

In the integral in (43), make the change of variable $x = \frac{n^{\gamma}}{\kappa}$ , so that $(\kappa x)^{1/\gamma} = n$ . Then

$dn = \frac{1}{\gamma} (\kappa x)^{\frac{1}{\gamma} -1} \cdot \kappa dx = \frac{1}{\gamma} \kappa^{-\frac{\gamma -1}{\gamma}} \cdot x^{-\frac{\gamma -1}{\gamma}} \cdot \kappa dx$

The integral in (43) then becomes

$\kappa^{-\frac{\gamma -1}{\gamma}} \cdot \frac{1}{\gamma} \int_0^{\infty} dx \frac{x^{-\frac{\gamma -1}{\gamma}}}{1+x}$

so we get

$\frac{\partial S}{\partial \kappa} = T(\kappa) \sim A(\gamma) \cdot \kappa^{-\frac{\gamma -1}{\gamma}} - \frac{1}{2\kappa} \quad \quad \quad \quad \quad (44)$

with

$A(\gamma) \equiv \frac{1}{\gamma} \int_0^{\infty} dx \frac{x^{-\frac{\gamma -1}{\gamma}}}{1+x} \quad \quad \quad \quad \quad (45)$

(MW2016 notes that the integral in (45) is actually the beta function $B\big(1-\frac{1}{\gamma}, \frac{1}{\gamma}\big)$ ). Integrating the first term on the right-hand side of (44), which is $\frac{\partial S_0}{\partial \kappa}$ , gives

$S_0 = \gamma A(\gamma) \kappa^{\frac{1}{\gamma}} \quad \quad \quad \quad \quad (46)$

Substituting for $S_0$ in (42) with the expression in (46) gives

$S \sim \gamma A(\gamma) \kappa^{\frac{1}{\gamma}} - \frac{1}{2} \ln(\kappa) - \gamma C + O(\kappa^{-1}) \quad \quad \quad \quad \quad (47)$

This is equation (21) in MW2016.

To obtain an explicit expression for the rate function $J(\tau)$ , we must now invert (44) to express the parameter $\kappa$ in terms of the time $T$ , and hence in terms of the dimensionless time given by $\tau = \frac{T}{\zeta(\gamma)}$ . To this end, observe that from (44) we get

$T \sim A \kappa^{\frac{1}{\gamma}} \kappa^{-1} - \frac{1}{2} \kappa^{-1} = \kappa^{-1} \bigg(A \kappa^{\frac{1}{\gamma}} - \frac{1}{2}\bigg) \quad \quad \quad \quad \quad (48)$

Therefore

$\kappa T \sim A \kappa^{\frac{1}{\gamma}}$

$\kappa \sim \bigg(\frac{T}{A}\bigg)^{-\frac{\gamma}{\gamma-1}} = \bigg(\frac{A}{\zeta(\gamma)}\bigg)^{\frac{\gamma}{\gamma-1}}\bigg(\frac{T}{\zeta(\gamma)}\bigg)^{-\frac{\gamma}{\gamma-1}} = b\tau^{-\frac{\gamma}{\gamma-1}} \quad \quad \quad \quad \quad (49)$

with $b \equiv \big(\frac{A}{\zeta(\gamma)}\big)^{\frac{\gamma}{\gamma-1}}$ . Now, using (47) and (48), the rate function is given by

$J(\tau) = S(\kappa^{*}) - T(\kappa^{*})\kappa^{*}$

$= \gamma A(\gamma) \kappa^{\frac{1}{\gamma}} - \frac{1}{2} \ln(\kappa) - \gamma C -A \kappa^{\frac{1}{\gamma}} + \frac{1}{2}$

$= (\gamma - 1) A\kappa^{\frac{1}{\gamma}} - \frac{1}{2} \ln(\kappa) - \bigg(\gamma C - \frac{1}{2}\bigg) \quad \quad \quad \quad \quad (50)$

Using (49) in (50) we get

$J(\tau) = (\gamma - 1) A b^{\frac{1}{\gamma}} \tau^{-\frac{1}{\gamma-1}} + \frac{1}{2} \frac{\gamma}{\gamma - 1} \ln(\tau) + D \quad \quad \quad \quad \quad (51)$

where $D$ is another constant.

We are now finally in a position to obtain the result given in (27) in MW2016. From (51) we see that to leading order we have

$J(\tau^{*}) \propto [\tau^{*}]^{-\frac{1}{\gamma-1}}$

so from (19) we obtain

$\tau^{*} \propto [\ln \mathcal{N}^{*}]^{-(\gamma - 1)} \quad \quad \quad \quad \quad (52)$

Errata in MW216

In the course of working with MW2016, some errata have come to light which I will record here.

In a private written communication with Michael Wilkinson, I pointed out that an error was made in the Supplement to MW2016 which details the asymptotic analysis. The issue arises in the Supplement when combining equations (17) and (19) there to get equation (20), which is then presented as equation (24) in the main MW2016 text. Equation (17) for $\kappa$ in the Supplement has a leading order term involving the dimensionless time variable $\tau$ with an exponent of the form $-\gamma/(\gamma-1)$ . Using (17) in equation (19) for the entropy, we should get the leading order $\tau$ term appearing with exponent $-1/(\gamma-1)$ , because $\kappa$ in (19) appears raised to the power $1/\gamma$ . However, the exponent is incorrectly given as $-\gamma/(\gamma-1)$ . This error affected the time scale relationship reported in (27) in the main MW2016 text. Combining the expression for the entropy with the condition (12) in the main MW2016 text for the onset of a shower, $\tau$ should be related to $\ln \mathcal{N}^{\ast}$ with an exponent $-(\gamma-1)$ on the latter, i.e., the reciprocal of $-1/(\gamma-1)$ . However, the exponent that is actually reported in equation (27) of MW2016 is $-(\gamma-1)/\gamma$ .

In an online meeting with Michael Wilkinson, I also pointed out that the formula for the constant $C$ given as equation (11) in the Supplement to MW2016 cannot be correct, and the formula is incompatible with the numerical value claimed for $C$ in equation (12) in the Supplement. For example, the expression involving natural logarithm terms in (11) must be negative, but the numerical value given in (12) is positive. Using numerical calculations in MAPLE, I pointed out that the following variant of the formula in (11) produces a numerical value close to the one reported in (12), as $M$ becomes large:

$C = 1 + \bigg\{\sum_{n=1}^M \bigg[(n-1) \ln\bigg(\frac{n-1}{n}\bigg)+ 1\bigg] - \frac{1}{2} \ln(M)\bigg\}$

The expression in curly brackets appears in equation (41) in my review above.

Finally, in another private written communication with Michael Wilkinson I pointed out that the symbol $T$ is being used inconsistently in MW2016. In equation (4) in MW2016, $T$ is defined as a random variable, i.e., a sum of random inter-collision times for a runaway raindrop. However, the same symbol $T$ is used in the entropy function in equation (7) and in the related equation (16) in MW2016, where $T$ is no longer a random variable, but rather an exogenous bound on values of the random variable $T$ . The distinction between the two becomes a delicate matter in some discussions, so it would have been better to use something like an overbar to distinguish between the random variable $T$ and the related symbol $\bar{T}$ , for example in expressing the large deviation principle $P(T \in [\bar{T}, \bar{T}+d\bar{T}]) \approx \exp(-J(\bar{T}))d\bar{T}$ .

A Phys. Rev. Letter by M Wilkinson on large deviation analysis of rapid-onset rainfall

Published by Dr Christian P. H. Salas

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