Statistical mechanics average energy

I can point out the error.In summary, the conversation is about finding the average energy per particle using the equation u = (Eo + E1 e^(-B deltaE)) / (1 + e^(-B deltaE)) and the value of B = 1/T. The first term in the equation can be easily derived, but the second term is giving the person trouble. They tried to Taylor expand the exponential, but everything seems to cancel out and appear as before. The person is attempting to derive the solution for the energy as T approaches 0 and infinity from the average energy per particle. They are advised to try the T-> infinity first and use the approximation 1/(1+x) ≈ 1
  • #1
sarahger9
3
0

Homework Statement



average energy per particle u = (Eo + E1 e^(-B deltaE)) / (1 + e^(-B deltaE))
B = 1/T


Homework Equations



Possibly relevant: e^x = 1 + x^2 / 2! + x^3 / 3! ...

The Attempt at a Solution



It tells me the average energy is about u = Eo + (deltaE)e^(-B delatE) as t approaches 0, and u = (1/2)(Eo + E1) - (1/4)B(delataE)^2 as T approaches infinity.

I can easily derive the first term in both of these equations, but the second is giving me some trouble. I tried to Taylor expand the exponential, but everything seems to cancel out and appear as before.
 
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  • #2
sarahger9 said:

Homework Statement



average energy per particle u = (Eo + E1 e^(-B deltaE)) / (1 + e^(-B deltaE))
B = 1/T


Homework Equations



Possibly relevant: e^x = 1 + x^2 / 2! + x^3 / 3! ...

The Attempt at a Solution



It tells me the average energy is about u = Eo + (deltaE)e^(-B delatE) as t approaches 0, and u = (1/2)(Eo + E1) - (1/4)B(delataE)^2 as T approaches infinity.

I can easily derive the first term in both of these equations, but the second is giving me some trouble. I tried to Taylor expand the exponential, but everything seems to cancel out and appear as before.


What is the question?
 
  • #3
Sorry, I am attempting to derive the solution that was given to me, the energy as T approaches 0 and infinity from the average energy per particle
 
  • #4
sarahger9 said:
Sorry, I am attempting to derive the solution that was given to me, the energy as T approaches 0 and infinity from the average energy per particle

You might want to try the T-> infinity first. I got their answer. Just Taylor expand. And you will also need to use that

[tex]\frac{1}{1+x} \approx 1-x [/tex]

If you don't get it, post your steps
 

1. What is statistical mechanics average energy?

Statistical mechanics average energy is a concept in physics that describes the average amount of energy possessed by a system of particles. It is a statistical average that takes into account the distribution of energy among the particles in the system.

2. How is statistical mechanics average energy calculated?

Statistical mechanics average energy can be calculated using the Boltzmann distribution, which takes into account the temperature and energy levels of the particles in the system. It is also possible to calculate it using the partition function, which describes the probability of a given energy state.

3. What is the significance of statistical mechanics average energy?

The average energy of a system is an important quantity in statistical mechanics because it allows us to make predictions about the behavior of the system. It can also provide insight into the thermodynamic properties of the system, such as its heat capacity and entropy.

4. How does statistical mechanics average energy relate to thermodynamics?

Statistical mechanics average energy is closely related to thermodynamics, as it provides a microscopic explanation for the macroscopic properties of a system. It allows us to understand how the energy is distributed among the particles and how this affects the overall behavior of the system.

5. Can statistical mechanics average energy be applied to all systems?

Statistical mechanics average energy can be applied to a wide range of systems, from simple gases to more complex systems such as solids and liquids. However, it may not accurately describe systems that are very small or very large, or those that are not in thermal equilibrium.

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