Two-State Paramagnet: 10^23 Elementary Dipoles, Zero Energy

[Vuldoraq]

Homework Statement

Consider a two-state paramagnet with 10^23 elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down.

(a) How many microstates are “accessible” to this system?

(b) Suppose that the microstate of this system changes 10^9 times per second. How many
microstates will it explore in 10^10 years (the age of the universe)?

(c) Is it correct to say that, if you wait long enough, a system will eventually be found in
every“accessible”microstate? Explain youranswer, and discuss the meaning of the word
“accessible”.

Homework Equations

Multiplicity of a macrostate=$$\frac{N\factorial}{N(up)\factorial*N(down)\factorial}$$

The Attempt at a Solution

Hi, It's only really part (a) and (c) that I am stuck on.

Using the above formula with N(up)=N(down)=5*10^22 and N=10^23, gives

Number of microstates=$$\frac{10^{23}\factorial}{(5*10^{22})\factorial\*(5*10^{22})\factorial$$

This seems like a very large number, have I made an error?

Also, is it right to say that by accessible they mean the number of microstates possible given any restraints the system may have?

Thanks in advance for any help.

Edit:There should be factorials after all the N's in the above expressions and also the Latex generation has failed for some reason. Sorry, if you want me to write out without Latex just say. Thanks.

 

[Delphi51]

I'm interested in this, though I know very little about it. Just an old retired school teacher.

The multiplicity formula doesn't make sense to me. Can you explain why it is?
It seems to me the part (a) is similar to the "how many difference license plate" questions we give in grade 12. How many different states for the first dipole x how many for the second, and so on so (2^10)^23. Yes, a HUGE number. I wonder if a calculator can show the answer? I hope you will post the awer so I can see it.

(b) and (c) boggle my mind! It could be changing to a microstate it had already been in previously, cutting down on the number "explored".

 

[nlightner]

Hi there,

First of all, your attempt at a solution is on the right track. The formula you used is correct, but there was a small mistake in your calculation. The correct number of microstates for this system is:

Number of microstates = (10^23)! / (5*10^22)! * (5*10^22)! = 1.25 * 10^75

This is indeed a very large number, but it is not incorrect. This is because the number of microstates increases exponentially with the number of dipoles. In this case, we have 10^23 dipoles, so the number of microstates is also on the order of 10^23.

For part (c), it is not entirely correct to say that a system will eventually be found in every accessible microstate if you wait long enough. This is because the number of microstates is so large that it is practically impossible for a system to explore all of them within a reasonable amount of time. However, it is true that the system will eventually explore a large number of microstates and will eventually reach a state where all microstates have been explored. This is known as the principle of equal a priori probabilities, which states that in the absence of any other information, all microstates are equally likely to be explored.

The word "accessible" in this context refers to the microstates that are possible for the system to reach given its constraints, such as the fixed total energy and the number of dipoles.

I hope this helps clarify things for you. Let me know if you have any other questions.