Conceptual Difficulty with Stat Mech

In summary: The second system could have any energy with relative probability given by the Boltzmann factor, just like the first system.
  • #1
Euclid
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0
In statistical mechanics, we define

[tex] \frac{1}{\tau}=\left( \frac{\partial \sigma}{\partial U} \right)_N[/tex]

This formula gives the temperature as a function of the energy of the system and N. So knoweldge of U and N determines the temperature of the system. Conversely, at least when the above equation can be inverted, knowledge of the temperature and N determines the energy exactly.

This is contrasted with the ensemble average: [tex] U = \tau^2 \frac{\partial \log Z}{\partial \tau} [/tex]. This seems to imply that knowledge of the temperature and N only determines the ensemble average energy, whereas before, the energy was determined exactly.

For concreteness, consider a system of N spin 1/2 particles. In the large N limit,

[tex] \frac{1}{\tau}= \frac{-U}{Nm^2B^2}[/tex]

whereas

[tex] <U> = -2mBN \sinh (mB/\tau)[/tex].

How is it that these two formulas are not in conflict with each other?

Edit: I just realized that the second formula is actually an exact expression and in the appropriate limit reduces to the first. However, I am still confused about this. Why is averaging necessary in one case, while not in the other?
 
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  • #2
an instantaneous energy and the average energy are not the same thing.

your first energy, the one involving the derivative of the ln of the partition function, is infact an average (although you didn't explicitly write it this way for some reason).

This seems to imply that knowledge of the temperature and N only determines the ensemble average energy, whereas before, the energy was determined exactly.

Don't forget the partition function! Without that you wouldn't know how the energy is actually partitioned amongst the possible states, and therefore wouldn't know the average energy!
 
  • #3
I am aware of the distinction between the energy of a state and the ensemble average energy of a system. In fact, that's basically the cause of my confusion. In the definition of temperature, there no statistical averaging going on, yet you get the same relationship between T, U and N as you do when you take averages, as in the second formula.

I think about this as follows. Given a system S in thermal equilibrium with a reservoir R at temperature T, it follows from the 1st L.T. that S is at temperture T. Therefore, from the relation [tex] 1/T = \partial_U \sigma(U,N)[/tex] (assuming I can invert this relation to find U in terms of T) we know the energy of the system S. Now on the other hand, we are taught that there is no definite relationship between these quantities. The system S could have any energy with relative probability given by the Boltzmann factor. All we can know is the average energy. It seems to me that there is a genuine conflict here. Am I wrong?
 
  • #4
but you are in the canonical ensemble (constant N,V,T) so the energy will vary..i'm afraid i don't understand your question, can you phrase it differently?
 
  • #5
I think I figured out my problem (and now that I have it is quite clear why you didn't understand my question). The problem was that I was thinking of two systems in thermal equilibrium as necessarily having equal temperatures, while in principle, equal temperature is only the most likely outcome.
 

Related to Conceptual Difficulty with Stat Mech

1. What is conceptual difficulty with statistical mechanics?

Conceptual difficulty with statistical mechanics refers to the challenges that students and researchers may face in understanding and applying the fundamental principles and concepts of this branch of physics. It involves comprehending abstract concepts, mathematical equations, and their physical significance.

2. What are some common misconceptions about statistical mechanics?

Some common misconceptions about statistical mechanics include thinking that it is solely based on probability or that it is only applicable to large systems. Another misconception is that statistical mechanics is only concerned with equilibrium systems, when in reality it can also be applied to non-equilibrium systems.

3. How can one overcome conceptual difficulty with statistical mechanics?

To overcome conceptual difficulty with statistical mechanics, it is important to have a solid understanding of the underlying mathematical concepts and to practice solving problems. It may also be helpful to seek out additional resources, such as textbooks or online tutorials, and to work with a mentor or study group.

4. What are some real-world applications of statistical mechanics?

Statistical mechanics has a wide range of real-world applications, including in thermodynamics, material science, and quantum mechanics. It is used to understand and predict the behavior of gases, liquids, and solids, as well as complex systems such as biological systems and networks.

5. How does statistical mechanics relate to other branches of physics?

Statistical mechanics is closely related to other branches of physics, particularly thermodynamics and quantum mechanics. It uses concepts and principles from these fields to describe the behavior of systems at the microscopic level. It also provides a bridge between the macroscopic and microscopic worlds, allowing for a deeper understanding of the physical world.

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