How thick is the atmosphere? A derivation of the Boltzmann distribution | Hacker News Reader

How thick is the atmosphere? A derivation of the Boltzmann distribution

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101 pointssimulate20 comments

Note that while the temperature rockets up to 1000C in the upper atmosphere (thermosphere), it's offset by the decrease in density: almost a thousand-fold between 100-150 km.

High temperature is not the same as hot. Those particles have a high temperature which means they're moving around a lot (crude estimate from conservation of energy: v = √kT/m). When you touch a "hot" thing, the heat you feel is all those particles bumping into your body. So in the upper atmosphere those particles might be very excited, but the low density means there isn't sufficient heat transfer to burn you.

EDIT: There are other interesting considerations. Rocket controllers use precise atmospheric models to know when to throttle down/up during a launch. There is a point called maximum dynamic pressure (max Q) where the stress on the airframe is maximised due to the speed of the vehicle and local air pressure. On Shuttle launches you hear the controllers call out "Go at throttle up" to let the pilot know that it's safe to punch the engines. https://www.youtube.com/watch?v=Em-Krwbn25A

Unless you're moving faster than possible for your particles to return to equilibrium. (either via conduction or via radiation) After all, heat transfer is a large scale statistical phenomenon.

So, if you're moving fast, despite the low density you're hitting a lot of (anisotropic) fast moving particles. This is a difference between static friction and dynamic collisions.

This is why space ships still need heat shields bigger than expected by just friction alone.

Edit: yes it is related to maximum dynamic pressure, you could call it maximum heat differential.

Also fun: "The Kármán line... represents the boundary between Earth's atmosphere and outer space.... Theodore von Kármán... was the first person to calculate that the atmosphere around this altitude becomes too thin to support aeronautical flight, since a vehicle at this altitude would have to travel faster than orbital velocity to derive sufficient aerodynamic lift to support itself."

https://www.wikipedia.org/wiki/Kármán_line

Is this actually a well-defined altitude, or does it depend on the properties of the vehicle? (Certainly, the minimum speed to maintain level flight normally does.) The Karman line is near but distinct from the boundary between the thermosphere and the exosphere, which is well-defined: it's where atoms become so thin that their mean free path becomes so long they stop bouncing off each other and just follow parabolic trajectories under gravity.

This description made no sense to me until I read this from the same page:

> The Karman line is therefore the highest altitude at which orbital speed provides sufficient aerodynamic lift to fly in a straight line that doesn't follow the curvature of the Earth's surface.

I'd like to upvote this twice. I love articles that explain scientific principles in a way that I could explain them to my 11 year old daughter. This article is a perfect example, balancing the equations that I want to see with the text that she needs to grasp the concept without understanding the equations. These random never-thought-about-the-physics-behind-that articles are terrific.

You have a gifted daughter.

Statistical mechanics is one of the most underrated Physics disciplines, having been largely replaced with the simpler thermodynamics in many undergrad courses.

If there was one Physical formula for humanity to leave behind, it should be Boltzmann's entropy formula. Conceptually linking microstates to macrostates could lead to a glut of other discoveries, including thermodynamics (think Industrial Revolution), solid-state (semiconductor stuff), and molecular/nanoscale physics.

Aren't statistical physics and thermodynamics complementary? Thermodynamics is about deriving results given an equation of state and statistical physics is a method for finding the equation of state.

Yes, thermodynamics is rooted in empirical results and the use of thermodynamics in industrial age engineering pre-dates the statistical mechanics explanations of why the laws of thermodynamics are true.

What do you mean by “replaced”? Do you have an example of a place and time where statistical mechanics was taught, and not thermodynamics, and later thermodynamics had replaced statistical mechanics in the curriculum?

From my (limited) experience with UK undergrad courses, it's more that stat mech and thermo were taught, but stat mech has been dropped/made optional. It's somewhat understandable, because a lot of interesting things have happening in other areas, e.g. quantum mechanics, astrophysics, etc. and the length of a degree hasn't changed. It's just one of those ironies that stat mech may have been ahead of it's time, and with cheap computing power it becomes interesting again (see molecular dynamics).

Still, when i took stat mech largely by accident (i enjoyed that lecturer), it felt like a lot of things came together which i had previously just accepted in thermo, and other fields, too (e.g. early quantum mechanics, from what basis did they start). Hence my opinion to why it's underrated (you also hear very little of it in text books - it's just not "cool").

That formula is carved on Boltzmsn’s tombstone.

you had me at lagrange multipliers

A similar derivation is given in the first edition of Kittel’s Thermal Physics, as well as in the Quantum Mechanics book by Griffiths IIRC.

Feynman has a similar dicussion starting from the P=nkT equation: http://www.feynmanlectures.caltech.edu/I_40.html

Does Kittel define Shannon entropy?

I'd have to grab the first edition copy from the library but I think not, he does use Lagrange multipliers though.

Should K.E. be labeled as vertical kinetic energy in the first set of equations instead of vertical potential energy?

Yes, you are right. It's a typo.

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