# Earth-like Atmosphere Breaks Second Law of Thermodynamics due to Conduction?

Consider a neutral gas around an Earth-like atmosphere with density, pressure and temperature given by $\rho,\ p$ and $T$ respectively. To keep the system a closed system assume that the atmosphere is far away from any star and no heat is coming from the planet below. The equation of motion for the gas is given by: $$\rho\frac{D \vec{v}}{D t}=-\nabla p - \rho\nabla\Phi,$$ where $\Phi$ is the gravitational potential energy, given by: $$\Phi = -\frac{GM}{r+R},$$ where $R$ is the radius of the planet and $r$ is the distance from the surface of the planet. Now we assume that the atmosphere is at steady-state and axisymmetric and so only depends on $r$. The equation of motion reduces to: $$\frac{dp}{dr} = -\rho\frac{d\Phi}{dr}.$$ It is also assumed that the energy is spread evenly throughout the atmosphere such that: $$\frac{d}{dr}\left(\rho\Phi+\frac{p}{\gamma -1}\right)=0,$$ where $\gamma=5/3$ is the ratio of specific heats and we assume our gas is a monatomic ideal gas. The solution to the above two equations is: $$\rho=A|\Phi|^{1/(\gamma -1)-1},$$ $$p=A(\gamma-1)|\Phi|^{1/(\gamma-1)}+B,$$ where $A$ and $B$ are constants of integration. The key aspect to the solution is that there is a gradient in temperature. Therefore, conduction would act to smooth out the temperature gradients. But this would mean the energy is no longer evenly distributed so surely this means the entropy has decreased? Does this violate the second law of thermodynamics?

• So the question is: "I set up a situation where the atmosphere is hot in some places and cold in others. How is it possible that heat flows?" It basically answers itself. You can have heat flow, because you didn't start with uniform temperature. – knzhou Sep 17 '18 at 15:10

Secondly by focusing on the energy density you are assuming that all types of energy "count equally" for the entropy. This is, however, not true. In particular particles can have kinetic in 3 directions (we can tell you have 3 degrees of freedom because $\gamma = \frac{5}{2}$) but only 1 direction effects the potential energy. This means that there are 3 times as many ways for a particular volume of gas to have a certain internal energy than for it to the same potential energy and so internal energy counts 3 times for the entropy. This means that you expect more energy in internal energy than potential energy.
• Thank you for the answer. Am I correct in thinking that:$$\frac{d}{dr}\left(\rho\Phi+\frac{3p}{\gamma-1}\right)=0,$$ would give maximum entropy? Any pointers to books etc. where I can find out more would be greatly appreciated also. – Peanutlex Sep 17 '18 at 15:16
• No it wouldn't because it is still only considering the entropy distribution and not the full state of the gas. As the previous comment said the correct condition for maximising entropy is that the temperature across the gas is constant (note that this does not hold for the real atmosphere as it is not in thermodynamic equilibrium and so not in the maximum entropy configuration). The constant temperature condition allows you to use the equation of state to relate the pressure and density. For and ideal gas this reduces to $p/\rho = const$ – By Symmetry Sep 17 '18 at 15:53