The Earth's atmosphere is a complex physical system, and predicting its behavior is crucial for environmental and meteorological purposes. However, even the best theoretical models run on modern computers are insufficient to make precise predictions. In this problem, we will attempt to understand some of the basic atmospheric phenomena based on simple models. You might need the following constants: the intensity of sunlight at Earth (the total solar irradiance) $F_s=1370\text{ W/m}^2$, the molar mass of water $\mu_{\text{H}_2\text{O}}\approx18\text{ g/mol}$, the average molar mass of air $\mu_{\text{air}}\approx29\text{ g/mol}$, and the Stefan-Boltzmann constant $\sigma=5.67\times10^{-8} \text{ W}/(\text{m}^2 \text{K}^4)$. All gases in this problem can be treated as ideal gases. Assume that all air molecules have $5$ degrees of freedom. You may need the following integral:
$$\int_{-\infty}^\infty e^{-ax^2/2} \;dx = \sqrt{\frac{2\pi}{a}}, \hspace{4mm} a> 0.$$
In this section, we study the effect of the atmosphere on the Earth's surface temperature. Assume that Earth and its atmosphere have an albedo $a=0.3$ for solar radiation, which is the reflected fraction of the total incident radiation. You may use this value in all parts of this problem. In addition, assume the Earth radiates as a black body.
Your answer for A.2 should be lower than what you would expect. We now consider adding a thin atmospheric layer at temperature $T_a$, see Figure A.1. The atmospheric layer transmits a net fraction $t_{\text{sw}}$ of the incident solar radiation and a net fraction $t_{\text{lw}}$ of the Earth's thermal radiation. Otherwise, you may treat the atmosphere as a black body.
The infrared radiation emitted by Earth has low energy, incapable of exciting electrons within molecules, but it has the ability to excite the vibrational and rotational modes of the molecules.
Quantum mechanics forbids the vibrational modes of symmetric diatomic molecules, such as nitrogen and oxygen (the most abundant gases in the Earth's atmosphere), to be excited by light. This explains why $\text{N}_2$ and $\text{O}_2$ do not contribute to the greenhouse effect. In general, the absorption of light by molecules is governed by the allowed energy transitions in them. However, the energy of the light absorbed does not have to exactly match the energy gap in the molecule. Suppose that a molecule at rest has a spectral line (an allowed transition) at frequency $f_0$.
For a gas at temperature $T$, the velocity of its molecules is distributed according to Maxwell's distribution. For a molecule of mass $m$, the probability to find a molecule's velocity along one dimension to be between $v$ and $v+dv$ is $p_1(v)dv$, where $p_1(v)$ is a probability distribution function given by
$$p_1(v)=C \exp\left(-\frac{mv^2}{2k_BT}\right)$$
$C$ is a normalisation constant ensuring the probabilities add up to one, and $k_B$ is the Boltzmann constant.
Consider a small cylindrical mass of air at height $z$ above the ground. The pressure and mass density of air at that height are $p(z)$ and $\rho(z)$, respectively, see Figure C.1. Assume a uniform downward gravitational field $g$ and that the pressure on the Earth's surface is $p_o$.
In a real atmosphere, the temperature is not constant but changes with height. The rate of decrease of temperature with height $\Gamma(z) = -dT/dz$ is called the lapse rate. Consider a small mass of air rising adiabatically in the atmosphere such that it remains at mechanical equilibrium with its surrounding.
To analyse the stability of an atmosphere, we imagine starting from an equilibrium state, and then perturbing a small mass of air and analyse its response. Consider a small air mass initially in equilibrium with the surrounding air at height $z$ and temperature $T$. It is then adiabatically displaced vertically by a displacement $\delta z_0$. Assume that throughout the motion, the air parcel always has the same pressure as the surrounding air at the same height. The surrounding atmosphere is unaltered and has a different lapse rate $\Gamma$. Neglect viscosity.
Even though water constitutes a small portion of the atmosphere, it has a significant role in climate science. It is responsible for precipitation, and it is the most significant greenhouse gas. The phase of water depends on what temperature and pressure the water system is at, depicted on a $p-T$ phase diagram, see Figure D.1. When the pressure and temperature lie on the coexistence curve (phase boundary), both liquid and vapour water can be present in the system. The slope of the coexistence curve is given by the Clausius-Clapeyron equation:
$$\frac{dp_s}{dT}=\frac{\Delta S}{\Delta V}$$
where $p_s$ is the saturation pressure (the pressure at the phase transition), and $\Delta S$ and $\Delta V$ are the changes in entropy and volume across the phase transitions, respectively. Treat water vapour as an ideal gas.
Now we consider a `moist' air mass that rises adiabatically starting from a temperature $T_i$. The mass mixing ratio of water vapour (the ratio of the mass of water vapour to the mass of air) is $\phi$. Take the air mass to have a molar specific heat capacity at constant pressure $c_p$. The universal gas constant is $R=8.31 \text{ J}/(\text{mol} \text{ K})$.
Under suitable atmospheric conditions, a bright ring appears around the Sun which is called a halo. Halos are caused by ice crystals present in the upper troposphere. One interesting feature about halos is that they always appear at a specific angle relative to the direction of the Sun.
The most common type of halo forms when tiny ice crystals take the shape of regular hexagonal prisms. Light from the Sun falls onto randomly oriented ice crystals drifting in the atmosphere and scatters in various directions. However, in certain specific directions, the intensity of the refracted light is maximal, and this determines the angle at which the bright ring appears.
Consider a hexagonal ice prism whose six-fold symmetry axis is perpendicular to the direction of the Sun's rays. Investigate a light ray that refracts through two rectangular faces of the prism indicated in Figure E.2. Due to the random orientation of the ice crystals, the light strikes the crystal faces at varying incidence angles $\alpha$.