Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I understand that the 2nd law of thermodynamics roughly states that, if you have a body (or a gas in a chamber) that is hot at one end and cold on the other, the heat will always flow from the hot to the cold part and to get the opposite effect, one has to put energy in from the outside (through a machine or something).

Now, I don't understand why this fact cannot be explained just through probabilities (of the velocities of the gas molecules, say).

It would seem to me that it is simply very, very, very unlikely that faster moving molecules all end up in (approximately) one spot at any time.

But from all the fuzz about the 2nd law, I'm led to believe that there has to be more behind it than probability. So where am I wrong? Why is the second law beyond probability? How is the 2nd law tested? (so that one can rule out simple probability?)

ps.: I haven't yet had a course on probability theory. So my understanding of it is limited.

share|improve this question
    
I would suggest picking up a basic undergrad statistical mechanics book (Kittel and Kroemer's Thermal Physics is a nice one) and looking at the first chapter or so -- it basically covers all that you're asking about. –  j.c. Nov 29 '10 at 0:29
1  
Actually, as the answers will tell you, the second law is just probability. It states that the entropy of an isolated system does not decrease. This is just a precise way of saying that a system tends to move into its most probable state. [Entropy ~ log(probability)] What you've heard about heat flow is not the 2nd law itself, just a consequence of it. –  David Z Nov 29 '10 at 1:38

4 Answers 4

up vote 7 down vote accepted

I assure you, it is all probability AND statistics.

Well you see when you say "a gas is at 300 Kelvin", then it does not mean all the molecules in the gas are "at 300 Kelvin", it rather states an average. It represents the total behavior of the gas compared to anything at any other temperature. So there are actually molecules with more kinetic energy, and those with less kinetic energy, interacting with each other and the environment they are in, however due to the massive number of collisions and interactions, the result with highest probability(i.e. same temperature box&gas stays at the same temperature) is the macroscopic outcome.

Probability, combined with statistics is a very powerful tool to represent macroscopic nature, and I suggest you to take probabilistic maths and statistical thermodynamics courses to further investigate such issues.

share|improve this answer
    
Well, then my concerns seem to have been unjustified and one can indeed explain these macroscopic effects through probabilities. Thanks for clearing that up. –  Sam Nov 29 '10 at 7:40

It's very, very unlikely that heat should flow from cold to hot, for values of "very very unlikely" involving extremely large powers of ten. We're talking monkeys-with-typewriters-writing-Shakespeare kind of unlikely.

There's a very good, largely nonmathematical discussion of the second law and entropy in Sean Carroll's From Eternity to Here, which discusses entropy in detail in the context of the "arrow of time." It's a very good treatment of the subject, and well worth reading.

share|improve this answer
    
Monkeys-typewriting-Shakespeare is rather an example of an opposite thing, that given enough time/monkeys it is almost sure that Shakespeare work will be reproduced. –  mbq Nov 29 '10 at 12:18
2  
And given enough time, it is also true that a high-entropy system will eventually fluctuate into a lower-entropy state. In both cases, the amount of time required to be sure of seeing it is astronomical. –  Chad Orzel Nov 29 '10 at 12:31
    
Penrose also discusses this in his new book---Cycles of Time--This paper gives a sort of condensed version:accelconf.web.cern.ch/AccelCo...S/THESPA01.PDF I guess, though, that I wonder why a cosmological explanation of low entropy initial condition is really necessary, but I seem to be in a minority. Isn't Boltzmann's definition enough? –  Gordon Jan 27 '11 at 7:02
    
Sorry, this link should work:docs.google.com/… –  Gordon Jan 27 '11 at 7:12

I also found that with modern computers with about a days effort, I could write a 2D kinetic particle simulation code. Collide two particles, and conservation of energy and momentum leaves only a single undetrmined quantity, which I simply set by using a random number generator. Throw in a few hundred or thousand particles, and you can see how the system evolves, and what sort of statistical properties emerge. This allows someone who isn't really good at math, but good at programming to design numerical experiments. I postulate that if these tools had been available when the subject was first researched that if would have developed much quicker.

share|improve this answer
    
Any link towards your code ? –  Frédéric Grosshans Jun 6 '12 at 14:54

I belief that when the 2nd law of thermodynamics is stated carefully in the right way it becomes a total no-brainer, i.e., just plain obvious.

The only concept you have to know about is that of macro- and micro-states:

A micro-state is a state that characterizes position/velocity of each and every molecule in your gas, or, more general, the exact state of the system.

A macro state is a collection of many micro-states that share a common characteristic. For example, there are many ways you can arrange the molecules of a gas and still have the same volume, temperature and pressure. The number of micro-states that make up a given macro-state is called multiplicity

The 2nd law of thermodynamics simply states that over the course of time, multiplicity will increase: Macro-states with more micro-states are more probable than macro-states with fewer micro-states.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.