# An oddity?

1. Apr 18, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Consider 10 He atoms as a system

Thermodynamically, the state of maximum entropy hence the most likely macrostate for the system is 5 He to be excited and 5 in the ground state.

But lower energies are more favourable so there should be more probability for the He to be in the ground state? Or more than 5 in the ground state. However, I don't know the physics behind this last claim. Can someone help out here?

2. Apr 18, 2007

### Mentz114

Interesting. If there's just enough energy around to raise 5 heliums from the ground state, and the next He level is not resonant with these quanta, and all the energy is in the He, it's the only macrostate (?). In a realistic scenario one atom grabbing all the energy has a non-zero probablity.

Lots of ifs, but you need to describe the environment in more detail, because the amount of energy available is important.

Last edited: Apr 18, 2007
3. Apr 18, 2007

### malawi_glenn

doesent that follow quite straight forward fron Boltzmann energy-distribution?

4. Apr 18, 2007

### pivoxa15

What is the Boltzmann energy-distribution? I only know the Boltzmann distribution.

Lets assume the whole universe consists of 10 He atoms. Then the total energy of the universe would determine the macrostate of the 10 atoms? So in this way there is only one entropy hence entropy will not increase over time?

Last edited: Apr 18, 2007
5. Apr 18, 2007

### malawi_glenn

Well there you have it then..

6. Apr 18, 2007

### pivoxa15

So we are talking about a microcanonical ensemble.

So in this case we list all the microstates with this constant E but each is as likely to happen as the other so there is no most likely microstate.

7. Apr 18, 2007

### Mentz114

Having only 2 energy levels means there's only one way to distribute the energy. Give yourself 3 levels and the numbers are more interesting.

Last edited: Apr 18, 2007
8. Apr 19, 2007

### pivoxa15

But there are 10 atoms. So each atom has two ways. The thing is if the energy in the universe is constant than the number of excited atoms must be constant hence one macrostate. And each microstate is equally as probable. So there is only one entropy value in this universe without it increasing or decreasing.

However in the real universe the energy is still constant but there are many different macrostates hence entropy increase. How do we reconcile my hypothetical siutation and the real universe? Is it because in the hypothetical universe there is only one type of atom. Whereas in the real universe there many different atoms hence different energy levels per atom i.e 3 or more. Hence if our universe were literally filled with only He atoms than there would only be one entropy value? Hence one macrostate?

9. Apr 19, 2007

### Mentz114

No, it's the energy distribution which is unrealistic. If your universe was not expanding or contracting it would have constant entropy. I don't know what level you're at with this but hyperPhysics has some good explanations.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/disbex.html

Real Helium atoms have more than 2 energy levels. Our cosmos is complicated and the amount of available energy may not be fixed for all time.

10. Apr 19, 2007

### pivoxa15

Does your last statement imply energy is not conserved? Or leaking into multiverses?

Why does an expansion or contraction of the universe imply entropy change? So entroy increase or decrease dosen't just involve the excitation level of the helium but also the expansion or contraction of the universe?

Last edited: Apr 19, 2007
11. Apr 19, 2007

### Mentz114

I don't know the answer to any of your four questions. But I was referring to available energy.

I think that as entropy increases the amount of available energy decreases. If this is not the case I'm sure someone will come in and correct me.

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