# what it would mean to have a high temperature

by aaaa202
Tags: temperature
 P: 870 So I have been thinking about what it would mean to have a high temperature. This means the the change in entropy per change in energy is very low. But what does this mean combinatorically? For instance, take a canonical ensemble of N atoms and U energy. When is temperature high? I have feeling that it must be when U>>N but I'm not sure why. Maybe it's wrong. Can anyone explain what a high temperature corresponds to in terms of how many atoms you have compared to how much energy?
 PF Patron P: 2,895 Temperature has nothing to do with the number of atoms, it has to do with the average energy per ion. The entropy change, per heat added, also has nothing to do with the number of atoms, surprisingly. This is because the entropy of the parts of a system add when you add more parts, but the change in entropy, per heat added, doesn't care how many parts there are-- fewer parts means more energy is being added per part, and more parts means less gets added per part, so the number of parts ends up not mattering at all.
 P: 870 What you mean that it doesn't care how many atoms there are. If you have 10 atoms sharing 1000 energy units compared to 100 atoms sharing 1000 energy units, then there is more energy per atom in the first case. Which system has the highest temperature?
P: 32

## what it would mean to have a high temperature

is temperature defined only as a thermodynamic equilibrium quantity? and if so, do you must have a gigantic heat sink attached to the system that you are interested in (i.e. be in equilibrium with a huge heat sink)? And if so, does it matter how many atoms you have in your specific system?
P: 786
 Quote by aaaa202 So I have been thinking about what it would mean to have a high temperature. This means the the change in entropy per change in energy is very low. But what does this mean combinatorically? For instance, take a canonical ensemble of N atoms and U energy. When is temperature high? I have feeling that it must be when U>>N but I'm not sure why. Maybe it's wrong. Can anyone explain what a high temperature corresponds to in terms of how many atoms you have compared to how much energy?
Ken G is right. Think of temperature as proportional to the average kinetic energy per particle. That means temperature doesn't depend on the number of particles. Temperature is equal to 2E/3k where E is the average kinetic energy of a particle and k is Boltzmann's constant.

If the average kinetic energy of a particle is high, the temperature is high. Usually, energy levels for a particle have energy $\epsilon n^2$ where n=1,2,3,4... and $\epsilon$ is the ground state energy. So temperature is high when $U/N>>\epsilon$ or, in other words, $n>>1$

When the temperature is low, things get complicated, but that's usually around 1 Kelvin, or less.
 P: 870 I don't get it. Say 1000 particles share 1000J. Then the average energy per particle is 1J/particle. Say you have 100 particles sharing 1000J. Then the average energy per particle is 10J. So the average kinetic energy is 10 times as high and so is the temperature... What's wrong saying this? Its a closed system..
P: 786
 Quote by aaaa202 I don't get it. Say 1000 particles share 1000J. Then the average energy per particle is 1J/particle. Say you have 100 particles sharing 1000J. Then the average energy per particle is 10J. So the average kinetic energy is 10 times as high and so is the temperature... What's wrong saying this? Its a closed system..
There's nothing wrong with it. Except for minor quibbles, there is nothing wrong with it.
 P: 870 So then we conclude that the temperature is dependent on the number of atoms, since more atoms = higher energy density. I think you've mistaken what I said.
P: 786
 Quote by aaaa202 So then we conclude that the temperature is dependent on the number of atoms, since more atoms = higher energy density. I think you've mistaken what I said.
For a gas, lets say E=total kinetic energy, N=number of particles, T=temperature. Then

$$E=\tfrac{3}{2}N\,k\,T[/tex/ where k is Boltzmann's constant. The energy per particle will be [tex]E/N=\tfrac{3}{2}kT$$

More atoms means lower energy per particle, if you keep E constant. If you keep T constant, then energy per particle stays the same. In other words, if you keep total energy constant, then, yes, the temperature depends on the total number of particles. The more particles, the lower the temperature.
 P: 870 Okay.. I think all this has been of no relevance to what I was looking for. So let me be very concrete. I recently did an exercise, where you had a ferromagnet of 3 atoms in a B-field. If you wrote the partition function you found that the most probable state at T=0 was the lowest energy state - just like the boltzmann distribution. At high temperatures you however found that all the states of the magnet were equally probable. Can someone tell me how this happens? Is this when there is a very high energy-density in the system (the ferromagnet and the B field)? What does a high energy density do for the combinatorics? And lastly another question: How can you even view the B-field as a resevoir like the boltzmann distribution? You must be able to, because it's under this ensemble-resevoir idea that the whole partition function idea is concretized as far as I can see.
P: 786
 Quote by aaaa202 Okay.. I think all this has been of no relevance to what I was looking for. So let me be very concrete. I recently did an exercise, where you had a ferromagnet of 3 atoms in a B-field. If you wrote the partition function you found that the most probable state at T=0 was the lowest energy state - just like the boltzmann distribution. At high temperatures you however found that all the states of the magnet were equally probable. Can someone tell me how this happens? Is this when there is a very high energy-density in the system (the ferromagnet and the B field)? What does a high energy density do for the combinatorics? And lastly another question: How can you even view the B-field as a resevoir like the boltzmann distribution? You must be able to, because it's under this ensemble-resevoir idea that the whole partition function idea is concretized as far as I can see.
Rather than have me look it up, what are the energy states of the atoms? Can you characterize them somehow?
 P: 870 The exact numbers don't matter so let's just say they're: -2J,-1J,0,1J,2J a) How is it possible to view a B-field as a resevoir? b) What would a high temperature for a system like this mean? - high energy density?
P: 786
 Quote by aaaa202 The exact numbers don't matter so let's just say they're: -2J,-1J,0,1J,2J a) How is it possible to view a B-field as a resevoir? b) What would a high temperature for a system like this mean? - high energy density?
Ok, so there are five energy levels total. That makes a difference. You don't mean J for joules, energy is always positive, so I expect those are something like angular momentum states. The energies are something like $\epsilon_i=2\epsilon, \epsilon, 0, \epsilon, 2\epsilon$. The Boltzmann distribution will give $$\frac{N_i}{N}=\frac{e^{-\epsilon_n/kT}}{Z}$$ where $$Z=\sum_{n=-2}^2e^{-\epsilon_n/kT}$$ Just to see how things go, you can set $\epsilon=1, k=1$. If you plug in the numbers, You can see for T=1/1000, almost everything is in the 0 state, but at T=1000, they are evenly spread out among all five states.

The total energy is $$E=\sum_{n=-2}^2\epsilon_n e^{-\epsilon_n/kT}$$ When E/N is very small compared to $\epsilon$ then most particles will be in the ground state. When E/N is much larger than $\epsilon$ then the energy will be almost evenly distributed among the five states. The highest E/N you can possibly have is then $(2+1+0+1+2)\epsilon/5=1.2\epsilon$.

The fact that you have only a limited number of energy states makes all the difference.
 P: 743 Combinatorics, huh? Well think of it this way: you have n degrees of freedom (n is very large, but finite), and k units of energy (assuming every degree of freedom is quantized with the same steps for simplicity), then the number of ways of arranging the energy is just n choose n+k-1. The entropy is just ln (n+k-1 choose k). Adding one unit of energy will increase the entropy to ln C(n+k,k). A high temperature means that adding a unit of energy increases the entropy very little. Combinatorically, that's when k is already pretty large. The number of combinations increases more than linearly with energy, but less than exponentially, so the logarithm will increase less than linearly with increasing k.
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