Equilibrium

Thermal Pressure

In Earth’s atmosphere, thermal energy is the primary pressure source, producing a force per unit area $$P = nkT$$where $n$ is the number of atmospheric particles per unit volume, $T$ is temperature on the Kelvin scale, and $k = 1.38 \times 10^{-16} {\rm \, erg \, K^{-1}}$ is Boltzmann’s constant.

Thermal pressure can also be represented in terms of mass density $\rho$ as $$P = \frac {\rho k T} {\mu m_p}$$where $m_p$ is a proton’s mass and $\mu m_p$ is the mean mass per atmospheric particle.

Hydrostatic Equilibrium

An atmosphere can remain motionless, in a state known as hydrostatic equilibrium, if pressure forces balance gravitational forces. The difference in pressure force per unit area between the top and bottom of an atmospheric layer of thickness $\Delta r$ is $$\Delta P = \frac {dP} {dr} \, \Delta r$$The area of that layer is $4 \pi r^2$ in a spherical environment, and its mass is $4 \pi r^2 \rho \, \Delta r$. Gravitational forces pulling the layer inward therefore balance the net pressure forces pushing it outward if $$\frac {dP} {dr} = - g \rho = - \frac {\rho v_c^2} {r}$$so that a motionless layer remains in hydrostatic equilibrium.

Gravitational Temperature

Expressing $\rho$ in terms of thermal pressure and temperature and plugging the result into the equation of hydrostatic equilibrium leads to $$\frac {r} {P} \frac {dP} {dr} = - \frac {\mu m_p v_c^2} {kT} = - 2 \frac {T_\varphi} {T}$$ where $$T_\varphi = \frac {\mu m_p v_c^2} {2k}$$is a gravitational temperature that depends on the local circular velocity. Multiplying $T_\varphi$ by $k$ gives the mean kinetic energy of atmospheric particles traveling at the circular velocity of the gravitational potential.

Scale Height

The ratio $T / T_\varphi$ determines the thickness of an equilibrium atmosphere supported by thermal pressure. One can define the pressure scale height of an atmosphere to be $$\lambda_P = \left( - \frac {1} {P} \frac {dP} {dr} \right)^{-1}$$so that $\lambda_P$ is the radial distance over which pressure declines by an amount comparable to its local value.

The thickness of an atmospheric layer divided by its radius then corresponds to $$\frac {\lambda_P} {r} = \frac {1} {2} \frac {T} {T_\varphi}$$

Earth’s atmosphere is thin compared to its radius because its temperature $(T \approx 300 \, {\rm K})$ is much less than the gravitational temperature $(T_\varphi \approx 10^5 \, {\rm K})$, resulting in a pressure scale height $\lambda_P \approx 8.6 \, {\rm km}$.

A massive galaxy’s atmosphere tends to be thick, because its temperature is comparable to $T_\varphi$, making its scale height $\lambda_P$ comparable to its radius.

Other Forms of Pressure

In a galactic atmosphere, thermal pressure is not necessarily the only form of pressure support. Turbulent gas motions can provide some of the resistance to gravity. Magnetic fields and cosmic-ray pressure can also provide some support.

If other forms of pressure support are significant, then a galaxy’s atmosphere can be thick, with scale height $\lambda_P$ comparable to radius $r$, even if the atmosphere’s temperature is substantially lower than the gravitational temperature.