Absorption of Microwaves in the Atmosphere

V. L. Ginzburg

From the standpoint of the application of microwaves for communications, television, etc., the question of their absorption in the atmosphere is of great importance. We have already dealt with this problem in some detail earlier in a review on radiospectroscopy[^1], but now we shall return to it briefly in connection with the appearance of a number of new data.[^2–^5]

The absorption of radio waves in an atmosphere free of dust, water droplets, etc., occurs only in oxygen and in water vapor. This is explained by the fact that the molecules of all the other gases entering into the composition of the atmosphere in any appreciable amounts have no permanent moments1, whereas the water molecule has a permanent electric moment, and the oxygen molecule a permanent magnetic moment.

The ground term of the $\mathrm{O}_2$ molecule has the structure ${}^3\Sigma_g^-$, and its rotational levels are characterized by the quantum number of the total angular momentum $J = 1, 2, 3\ldots$ and by the rotational quantum number $k = 1, 3, 5, 7\ldots$ (the values $k = 0, 2, 4\ldots$ are forbidden by the Pauli principle). Interaction with rotation leads to multiplet splitting—the separation of a term with given $k$ into three terms with $k = J, J \pm 1$. According to the theory, the energy of all the terms is determined by formula (I,20)2. The experimental values for the multiplet splitting agree well with the theoretical ones,[^2] and therefore in Table I we give only the theoretical values calculated from formula (I,20). In Table I, $\gamma_k/c$, $\gamma_{k\pm1}/c$ are the wave numbers corresponding to the terms with $J = k, k \pm 1$.

The selection rules for magnetic radiation allow transitions with $\Delta J = 0, \pm 1$ and $\Delta k = 0, \pm 2, \pm 4\ldots$ For the $\mathrm{O}_2$ molecule, transitions with

Table 1

Multiplet splitting of oxygen

\(|\Delta k|>2\), to which wavelengths \(\lambda<0.16\ \mathrm{cm}\) correspond, are inactive in radiation.^2 Conversely, in the case \(\Delta k=0\) there is an entire band consisting of lines with \(\Delta J=\pm1\) (transitions from terms \(J=k+1\) to

Fig. 1. Multiplet splitting of various rotational levels of the molecule \(\mathrm{O}_2\) in the ground electronic-vibrational state.

a term with \(J=k\), and conversely) and located in the region \(\lambda\sim0.5\ \mathrm{cm}\). The values of the wave numbers \(\left(\dfrac{\nu_k}{c}-\dfrac{\nu_{k\pm1}}{c}\right)\) corresponding to this band are given in Table 1, as are approximate values of \(\lambda\). Transitions \(J=k+1\leftrightarrow J=k-1\) are forbidden. The arrangement of the terms is clear from Fig. 1.

From the entire band there falls out only one transition \(\dfrac{\nu_1}{c}-\dfrac{\nu_0}{c}\), to which there corresponds the wavelength \(\lambda = 0.25\ \text{cm}\).

Absorption lines have a finite width, due to collisions and, in general, to interaction between molecules (Doppler broadening of lines, associated with the motion of molecules, is significant only at pressures \(p < 0.1\ \text{mm Hg}\)). At atmospheric pressure the width of the lines is quite considerable; as will be seen below, the width \(\dfrac{\Delta \nu}{c} > 0.02\ \text{cm}^{-1}\). Therefore, in the case of \(\mathrm{O_2}\), at atmospheric pressure the individual lines are almost unresolved, and there is a continuous absorption band. As indicated in \({}^{1}\), a theory of line width valid for large values of \(\Delta \nu\) and for distances from the line center \((\nu=\nu_0)\) by an amount \(\nu-\nu_0\) comparable with the distance between lines has not yet been developed. Some very rough considerations \({}^{6}\), however, lead to a theoretical formula for the absorption coefficient \(\chi\), suitable when many lines and arbitrary values of \(\nu_1\), \(\nu_0\), and \(\Delta \nu\) are taken into account. Without dwelling further on the theory, we give here the corresponding formula \({}^{6,2}\):

\[ \left. \begin{aligned} \chi &= 10^{6}\cdot \log_{10} e \cdot \left(\frac{8\pi^{3}\nu N}{3hc}\right) \frac{\sum_{ij} |p_{ij}|^{2} f(\nu_{ij},\nu)\, e^{-E_j/kT}} {\sum_j e^{-E_j/kT}}, \\[4pt] f(\nu_{ij},\nu) &= \frac{\nu}{\pi \nu_{ij}} \left[ \frac{\Delta \nu}{(\nu_{ij}-\nu)^2+(\Delta \nu)^2} + \frac{\Delta \nu}{(\nu_{ij}+\nu)^2+(\Delta \nu)^2} \right], \end{aligned} \right\} \tag{1} \]

where \(p_{ij}\) is the matrix element of the electric (or magnetic) dipole moment, \(\nu_{ij}=\dfrac{E_i-E_j}{h}\), \(E_i\) and \(E_j\) are the energies of states \(i\) and \(j\), \(N\) is the concentration of absorbing molecules, \(T\) is the temperature, \(\nu\) is the frequency of the incident radiation, and the factor \(10^{6}\cdot \log_{10} e = 0.4343\cdot 10^{6}\) is introduced so that the coefficient \(\chi\) be expressed in decibels per kilometer. If this factor is absent, then \(\chi\) is measured in \(\text{cm}^{-1}\). The value \(\chi=10\ \text{db/km}\) corresponds to a decrease in the intensity of the radiation by a factor of 10 over a path of \(1\ \text{km}\).

Taking into account that, in the radio range at ordinary temperatures,

\[ h|\nu_{ij}| \ll kT, \]

one may put

\[ \nu_{ij}e^{-E_j/kT}+\nu_{ji}e^{-E_i/kT} = \frac{h\nu_{ij}^{2}}{2kT} \left(e^{-E_j/kT}+e^{-E_i/kT}\right), \]

and formula (1) assumes the form

\[ \chi = 10^{6}\log_{10} e\cdot \frac{ 8\pi^{3}\nu N \sum_{ij} |p_{ij}|^{2}\nu_{ij} f(\nu_{ij},\nu)e^{-E_j/kT} }{ 6ckT \sum_j e^{-E_j/kT} }. \tag{2} \]

The advantage of formula (2) as compared with (1) consists in the fact that all terms in it are of the same sign (positive). In the optical region, and at reduced pressure also in the radio range, the inequalities may often be regarded as satisfied

\[ \nu \gg |\nu-\nu_{ij}|,\qquad \nu \gg \Delta\nu . \tag{3} \]

In this case

\[ f(\nu_{ij},\nu)=\frac{1}{\pi}\left(\frac{\Delta\nu}{(|\nu_{ij}|-\nu)^2+(\Delta\nu)^2}\right), \]

and formula (2) goes over into formula (I,4); here it must be taken into account that the number of molecules in state 1 is equal to

\[ \frac{g_1Ne^{-E_1/kT}}{\sum_j e^{-E_j/kT}}, \]

where \(g_1\) is the statistical weight of state 1.

In the sums over \(i\) and \(j\) in (1) and (2) there is also a term with \(i=j\). If \(p_{ii}\ne 0\), then such a diagonal term leads to nonresonance absorption \((\nu_{ii}=0)\), which, under certain assumptions, is equivalent to the well-known Debye absorption of dipolar molecules

\[ \varkappa=\frac{8\pi^2Np^2\nu^2\Delta\nu}{3ckT\left(\nu^2+(\Delta\nu)^2\right)}, \tag{4} \]

where \(p\) is the permanent electric or magnetic moment of the molecule. The mechanism of nonresonance absorption is connected with the fact that, because of collisions, the moment \(p\) is essentially not constant; in the absence of collisions \(\Delta\nu\) in (4) is equal to zero and \(\varkappa=0\).

It is necessary once again to emphasize that formulas (1), (2), (4) cannot be regarded as reliably substantiated theoretically, as is the case, for example, for formula (I,4), valid under conditions (3). Further, the line width \(\Delta\nu\) still remains unknown \([\)under conditions (3) \(\varkappa(|\nu_{ij}|+\Delta\nu)=\frac{1}{2}\varkappa(|\nu_{ij}|)]\). At not too high pressures the usual mechanism of line broadening as a result of collisions may be regarded as valid. In this case (see \(^{1}\))

\[ \Delta\nu=\frac{1}{2\pi\tau} =\frac{1}{2\pi}\sqrt{2\pi d^2N\bar v}; \qquad \bar v=\sqrt{\frac{8kT}{\pi M}}, \tag{5} \]

where \(\tau\) is the mean free time, and the transition to the concrete form of \(\tau\) in the second part of the formula refers to the case of collisions of particles of equal mass \(M\) (\(d\) is the effective diameter of a particle or, more precisely, the sum of the radii of the two colliding particles). The value of \(d\) is of the order of the gas-kinetic diameter \(d_0\) and, generally speaking, larger than it. For air \(\pi d_0^2\simeq 3\cdot 10^{-15}\) (from experiments on electron collisions

with \(O_2\) and \(N_2\) molecules) and, thus, the gas-kinetic value of \(\Delta\nu\) at atmospheric pressure \((N=2.7\cdot 10^{19})\) and \(T=300^\circ K\) is equal to

\[ \left. \begin{array}{l} \Delta\nu_0 \simeq 8\cdot 10^8\ \mathrm{sec}^{-1},\\[4pt] \dfrac{\Delta\nu_0}{c} \simeq 0.025\ \mathrm{cm}^{-1}. \end{array} \right\} \tag{6} \]

According to (5), the width \(\Delta\nu\) is proportional to \(N\) or to the gas pressure, since the gas may be regarded as ideal (higher pressures are excluded, since at them the mechanism of collisional broadening becomes inapplicable).

To calculate \(\varkappa\) by formula (2), one must substitute into it the values \(E_i\), \(E_j\), \(\nu_{ij}\), and \(|p_{ij}|^2\), known from spectroscopy; as for the value of \(\Delta\nu\), it can in fact be determined only from experiment. The calculations, however, can be carried out by assigning a series of reasonable values of \(\Delta\nu\).

For oxygen the values of \(\nu_{ij}\) are determined by Table I, the values of \(p_{ij}\) are known from theory, and the factor \(\sum e^{-E/kT}\) is calculated from spectroscopic data. In the general case it is, of course, pointless to write out \(\varkappa\), since the corresponding expression contains all the frequencies given in Table I. However, at some distance from the center of the absorption band it may be assumed that all frequencies \(\nu_{ij}/c=2\ \mathrm{cm}^{-1}\). As a result, for air (21% \(O_2\)) at \(T=293^\circ\) and a pressure of \(760\ \mathrm{mm\ Hg}\) we have

\[ \varkappa =0.34\left(\frac{\nu}{c}\right)^2 \left[ \frac{\Delta\nu/c}{(2-\nu/c)^2+(\Delta\nu/c)^2} + \frac{\Delta\nu/c}{(2+\nu/c)^2+(\Delta\nu/c)^2} + \right. \]

\[ \left. + \frac{\Delta\nu/c}{(\nu/c)^2+(\Delta\nu/c)^2} \right] \ \mathrm{db/km}. \tag{7} \]

Formula (7) is applicable when \(\dfrac{\nu}{c}>4.5\ \mathrm{cm}^{-1}\) or \(\dfrac{\nu}{c}<1.5\ \mathrm{cm}^{-1}\); the first of these conditions is not replaced by the condition \(\dfrac{\nu}{c}>2.5\ \mathrm{cm}^{-1}\) because of the presence of the line \((\nu_1-\nu_0)\) with \(\dfrac{\nu}{c}\simeq 4\ \mathrm{cm}^{-1}\) (see Table I).

Calculations of \(\varkappa\) by formula (2) in the region \(\lambda\sim 0.5\ \mathrm{cm}\) are presented in Fig. 2 as applied to air (21% \(O_2\)) at \(T=300^\circ\), \(p=760\ \mathrm{mm\ Hg}\), and under the assumptions that \(\Delta\nu=0.02,\ 0.05,\) and \(0.1\ \mathrm{cm}^{-1}\). Circles in this figure indicate the experimental points (a table of experimental values of \(\varkappa\) is given in¹ Table III). From the available experimental data it is not yet possible to determine the value of \(\Delta\nu/c\) with any accuracy (for an analysis of the experimental data see²). At present the best choice should be considered \((\Delta\nu/c)=0.02\ \mathrm{cm}^{-1}\). Let us note that, within the accuracy of the experiment, collisions of \(O_2\) molecules with \(O_2\) and of \(O_2\) with \(N_2\) lead to the same broadening. Therefore, the absorption of radio waves by oxygen in air pro-

proportional to the content of \(O_2\). This circumstance is taken into account in formula (7) and in all calculations (Fig. 2, etc.).

The value \((\Delta \nu/c)=0.02\ \mathrm{cm}^{-1}\) agrees well with the gas-kinetic value (6). For \((\Delta \nu/c)=0.02\) the fine structure of the absorption band is still somewhat resolved, whereas for \((\Delta \nu)=0.05\ \mathrm{cm}^{-1}\) this is no longer the case (see Fig. 2). Table II gives the calculated values of \(x\) in db/km for air at \(T=293^\circ\) and atmospheric pressure.

In Table II, for \((\Delta \nu/c)=0.02\), the values of \(x\) in the absorption band itself

Table II

Values of \(x\), in db/km, for air

Fig. 2. Absorption in dry air \((21\%\,O_2)\) at \(T=300^\circ\), \(p=760\ \mathrm{mm\ Hg}\). The circles indicate experimental values.

are given within known limits because of insufficiently precise knowledge of the natural frequencies of the lines listed in Table I (Table I gives calculated values differing from the experimental values in the second digit; see \(^{2}\)). Table II does not take into account the presence of a weak line at \((\nu/c)=4\ \mathrm{cm}^{-1}\); this line is shown in Fig. 3, in which the solid curve refers to absorption by water vapor (see below), and the dotted curve to absorption by oxygen. Absorption in air due to oxygen at \(\lambda>10\ \mathrm{cm}\) is shown in Fig. 4.

In this region practically all the absorption is caused by the third, nonresonant term in formula (7). This term coincides with the semi-

Fig. 3. Theoretical absorption curves in the atmosphere at \(T = 293^\circ\) and \(p = 760\) mm Hg. The dashed curve refers to oxygen \(\left[(\Delta \nu / c)=0.02\ \mathrm{cm}^{-1}\right]\). The solid curve is absorption by water vapor \(\left[\rho = 7.5\ \mathrm{g/m^3},\ (\Delta \nu / c)=0.1\ \mathrm{cm}^{-1}\right]\).

Fig. 4. Absorption (theoretical) in dry air at \(T = 293^\circ\) and \(p = 760\) mm Hg.

… obtained from formula (4), if one sets

\[ p^2=\frac{8}{3}\left(\frac{eh}{4\pi mc}\right)^2, \]

as must be done for oxygen (see²). From Figs. 2, 3, 4 it is apparent that the absorption of microwaves in the atmosphere due to oxygen is, in certain regions, very large. Thus, at the maximum for \(\lambda \simeq 0.5\ \mathrm{cm}\), \(\varkappa \simeq 15\ \mathrm{db/km}\), and the wave intensity falls by a factor of 10 over a path of only \(2/3\ \mathrm{km}\). For \(\lambda>0.7\ \mathrm{cm}\), \(\varkappa<0.1\ \mathrm{db/km}\), and the absorption is noticeable only over paths of many tens of kilometers; finally, for \(\lambda>3\ \mathrm{cm}\), \(\varkappa \leqslant 0.01\text{--}0.02\) (see Table II), and the wave is attenuated by a factor of 10 over a path of 500–1000 km. For convenience, part of the curve of Fig. 3 is reproduced in somewhat greater detail in Fig. 5 (see⁵; the value of \(\Delta\nu/c\) in this figure is greater than \(0.02\ \mathrm{cm}^{-1}\) and is not indicated exactly).

Fig. 5. Absorption in the atmosphere (\(p=760\ \mathrm{mm\ Hg}\)) due to oxygen (curve \(a\), 21% \(O_2\)) and water vapor (curve \(b\); \(\rho=10\ \mathrm{g/m^3}\)). The solid portions of the curves are constructed on the basis of experimental data.

The absorption of radio waves by water vapor in the centimeter region is due to the radiative transition of the molecule

\[ (-+5_{-1})\to(+-6_{-5}), \]

to which corresponds the wavelength \(\lambda=1.3481\ \mathrm{cm}\). All the other transitions for \(H_2O\) lie in the millimeter range: the longest-wavelength transition \((++2_2)\to(--3_{-2})\) corresponds to \(\lambda\simeq 1.62\ \mathrm{mm}\).

In the case of the water molecule, in contrast to the oxygen molecule, the diagonal matrix elements \(p_{ii}=0\), and thus there is no nonresonant absorption. The absorption caused by the single line \(\lambda\simeq 1.35\ \mathrm{cm}\), according to formula (2), into which the corresponding numerical values for water at \(T=293^\circ\) have been substituted, is equal to³

\[ \varkappa_{.35} = \frac{\alpha\nu^2}{c^2} \left[ \frac{\Delta\nu/c} {\left(\dfrac{1}{1.348}-\dfrac{\nu}{c}\right)^2+ \left(\dfrac{\Delta\nu}{c}\right)^2} + \frac{\Delta\nu/c} {\left(\dfrac{1}{1.348}+\dfrac{\nu}{c}\right)^2+ \left(\dfrac{\Delta\nu}{c}\right)^2} \right], \tag{8} \]

\[ \alpha=0.00350\rho=2.63\,q, \]

where \(\rho\) is the number of grams of water vapor in \(1\,m^{3}\) of air and \(q\) is the relative concentration of water molecules in the air \(\left(q=\dfrac{N_{\mathrm{H_2O}}}{2.7\cdot 10^{19}}\right)\). Unlike oxygen, in the case of water it is also necessary to take into account the influence of a number of other strong lines located at shorter wavelengths. Calculation by formula (2) gives, for this absorption from “distant” lines, the value\(^3\)

\[ \varkappa_{\mathrm{dist}}=0{,}0116\,\rho\,\frac{\Delta\nu\cdot \nu^{2}}{c^{3}} =8{,}7q\,\frac{\Delta\nu\cdot \nu^{2}}{c^{3}} . \tag{9} \]

Some experimental values of \(\varkappa/\rho\) for \(\mathrm{H_2O}\) vapor in air are given\(^1\) in Table II, § 6. Here the values of \(\varkappa/\rho\) depend on \(\rho\), since collisions of \(\mathrm{H_2O}-\mathrm{H_2O}\) molecules have a cross section approximately 5 times larger than in the case of collisions \(\mathrm{H_2O}-\mathrm{O_2}\) or \(\mathrm{H_2O}-\mathrm{N_2}\) (see \({}^{1,3}\)). Accordingly,

\[ \text{for } \rho\to 0 \qquad \frac{\Delta\nu}{c}=0{,}087\ \mathrm{cm}^{-1}, \]

\[ \text{for } \rho=18\ g/m^{3} \qquad \frac{\Delta\nu}{c}=0{,}094\ \mathrm{cm}^{-1}, \]

\[ \text{and for } \rho=50\ g/m^{3} \qquad \frac{\Delta\nu}{c}=0{,}107\ \mathrm{cm}^{-1} \]

(the air pressure is atmospheric throughout). The values of \(\varkappa/\rho\) calculated by formulas (8) and (9) are given in Table III for \(T=293^\circ\) and \((\Delta\nu/c)=0{,}1\ \mathrm{cm}^{-1}\). The complete theoretical value is \(\varkappa=\varkappa_{1,35}+\varkappa_{\mathrm{dist}}\). Comparison with experiment shows\({}^{13,1}\) that formula (8) is valid, whereas the experimental value of \(\varkappa_{\mathrm{dist}}\) is 4–5 times greater than the value (9). This fact already reveals the previously mentioned unreliability of formulas (1)—(2) far from the center of the line.

Table III

Values of \(\varkappa/\rho\) in \(db\cdot m^{3}/km\cdot g\) at \((\Delta\nu/c)=0{,}1\ \mathrm{cm}^{-1}\)

In the last column of Table III there is given the semi-empirical value

\[ (\chi/\rho)_{\text{se}}=\frac{\chi_{1.35}+4.5\chi_{\text{res}}}{\rho} \]

(for the shortest waves the values are not given, since there were no measurements in this region).

The experimental values of \(\chi/\rho\), as indicated, agree fairly well with the value \((\chi/\rho)_{\text{se}}\). Thus, in the experiment for \(\lambda=1.27\ \text{cm}\), for all \(\rho\), \((\chi/\rho)_{\text{exp}}=0.0230\) (see Table II), while the value \((\chi/\rho)_{\text{se}}=0.0215\); for a number of points the difference is still smaller (see Table IV).

At \(T=291^\circ\) (\(18^\circ\text{C}\)) the density of saturated vapor is \(15\ \text{g}/\text{m}^3\), and thus at a relative humidity of \(66\%\) the atmosphere contains \(10\ \text{g}/\text{m}^3\) of water vapor. At \(\rho=10\ \text{g}/\text{m}^3\), at the absorption maximum \((\lambda \simeq 1.33\ \text{cm})\), \(\chi=0.25\ \text{db}/\text{km}\), and thus the wave intensity decreases by a factor of 10 over a path of \(40\ \text{km}\). Experimental values of \(\chi\) for \(\rho=10\ \text{g}/\text{m}^3\) are given in Table IV, where in the third column are also indicated the values obtained from the formula \(\chi_{\text{se}}=\chi_{1.35}+4.5\chi_{\text{res}}\) for \((\Delta\nu/c)=0.1\ \text{cm}^{-1}\).

Table IV

Values of \(\chi\) in db/km at \(\rho=10\ \text{g}/\text{m}^3\)

Fig. 6. Theoretical curve for absorption by water vapor in the millimeter range for \(\rho=7.5\ \text{g}/\text{m}^3\) and \(\Delta\nu=0.11\ \text{cm}^{-1}\).

Absorption by water vapor is also shown by the solid line in Fig. 3 for \(T=293^\circ\) and \(\rho=7.5\ \text{g}/\text{m}^3\), which corresponds to a content of \(1\%\) water molecules in the atmosphere. In this case the formula used was

\[ \chi=\chi_{1.35}+\chi_{\text{res}}\quad \text{with}\quad \frac{\Delta\nu}{c}=0.1\ \text{cm}^{-1}. \]

In Fig. 5 the absorption is also given for the assumption \(\rho = 10\ \mathrm{g/m^3}\); the assumptions on which the calculations are based are not indicated\(^5\). However, since part of the curve coincides with the experimental data, the curve corresponds to the value \(\chi_{\text{пэ}}\).

It should not be forgotten that in a number of cases (especially in the tropics) the density of vapor in the atmosphere may be considerably higher than \(10\ \mathrm{g/m^3}\).

The absorption by water vapor in the atmosphere in the millimeter range is clear from Fig. 6, constructed\(^3\) for \(\rho = 7.5\ \mathrm{g/m^3}\) and

\[ \frac{\Delta \nu}{c}=0.11\ \mathrm{cm}^{-1}. \]

The total absorption of oxygen and water vapor is, obviously, determined simply by the sum of the values of \(\chi\) for \(\mathrm{O_2}\) and \(\mathrm{H_2O}\).

Above it was assumed that there is no condensed water in the atmosphere, which is not the case during rain, snowfall, in fog, and in clouds. The absorption by condensed water, which, as is clear from what follows, for \(\lambda < 10\ \mathrm{cm}\) may be appreciable, is due to true absorption of energy in droplets and to scattering of radiation by these droplets; since the scattered waves go mainly to the side, scattering leads to an attenuation of the transmitted wave. For waves \(\lambda < 10\ \mathrm{cm}\) the losses in water (connected with the imaginary part of the complex dielectric constant) are very large, and for small droplets (fog, clouds) the dominant effect is true absorption\(^*\). Scattering (in the centimeter range) need be taken into account only for raindrops. We shall not dwell on this range of questions in any detail, but give in Fig. 7 a graph of the values of \(\chi\) available in\(^5\).

Fig. 7. Absorption in rain, fog, and clouds. Solid curves refer to absorption in rain: \((a)\)—for rain giving precipitation \(h=0.25\ \mathrm{mm/hour}\) (very light rain); \((b)\)—for rain with \(h=1\ \mathrm{mm/hour}\) (light rain); \((c)\)—for rain with \(h=4\ \mathrm{mm/hour}\) (moderate rain); \((d)\)—for a shower \(h=16\ \mathrm{mm/hour}\). Dashed curves indicate absorption in fog and clouds: \((e)\)—density of condensed water \(\rho = 0.032\ \mathrm{g/m^3}\) (visibility \(600\ \mathrm{m}\)); \((f)\)—\(\rho = 0.32\ \mathrm{g/m^3}\) (visibility \(120\ \mathrm{m}\)); \((g)\)—\(\rho = 2.3\ \mathrm{g/m^3}\) (visibility \(30\ \mathrm{m}\)).

\(^*\) Droplets may be regarded as small when their diameter is much smaller than \(\lambda/n\), where \(\lambda\) is the wavelength in vacuum and \(n\) is the refractive index of water for the given wavelength.

This graph was constructed on the basis of numerous as-yet unpublished calculations and, apparently, is quite reliable. Experimental verification confirms\(^5\) the data of Fig. 7. Rain is characterized in this figure by the precipitation it gives per hour. For fog and clouds, the density of condensed water in \(1\ \mathrm{m}^3\) of air is given. For comparison, approximate distances are also indicated (in the caption to the figure) at which, in the corresponding fog or cloud, objects can be seen with the eye.

CITED LITERATURE

1. V. L. Ginzburg, Radiospectroscopy of Molecules, UFN, 31, 320 (1947).
2. J. H. van Vleck, Absorption of Microwaves in Oxygen, Phys. Rev., 71, 413 (1947).
3. J. H. van Vleck, Absorption of Microwaves in Water Vapor, Phys. Rev., 71, 425 (1947).
4. G. W. King, R. M. Hainer and P. C. Cross, Expected Absorption of Microwaves by Water and Certain Related Molecules, Phys. Rev., 71, 433 (1947).
5. L. N. Ridenour (editor), Radar System Engineering, § 2, 15, N. Y. and London (1947).
6. J. H. van Vleck and V. F. Weisskopf, On the Shape of Lines Broadened as a Result of Collisions, Rev. Mod. Phys., 17, 227 (1945).

1. Radiative transitions of the rotational type, which are under discussion in the radio range, are possible only when the molecule has a permanent electric or magnetic moment. ↩

2. Formulas with the numeral I refer to review.[^1] ↩