How to Show Average Energy at Low Temperatures?

So in the "keep terms up to first order" we just drop the x^2.In summary, when using the first order expansion (1+x)^-1=1-x for x<<1, the average energy of a system with two non-degenerate energy levels E1 and E2, where E2>E1>0 and at temperature T, is simplified to E1+deltaE*e^(-B*deltaE) where deltaE= E2 -E1 and B=1/kT (k=Boltzmann constant) in the limit that kT<<E2-E1. This is because for small enough values of x, x^2 can be ignored and does not have a noticeable effect on the equation
  • #1
shayan825
16
0
A system has two non-degenerate energy levels E1 and E2, where E2>E1>0. The system is at tempreture T. The Average energy of the system is = E1+E2e^(-B*deltaE) / 1+e^(-B*deltaE) where deltaE= E2 -E1 and B=1/kT (k=Boltzmann constant). show that for very low temperatures kT<<deltaE, average energy= E1+deltaE*e^(-B*deltaE).

hint: use the first order expansion (1+x)^-1=1-x for x<<1
keep terms up to first order only

Here is what I get:

I multiplied 1-e^(-B*deltaE) by E1+E2e^(-B*deltaE) based on (1+x)^-1=1-x and I get E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE), but it is not the answer

Help would be appreciated
 
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  • #2
You are told: $$\bar E = \frac{E_1+E_2 e^{-(E_2-E_1)/kT}}{1-e^{-(E_2-E_1)/kT}}$$ ... and you need to find this in the limit that ##kT<<E_2-E_1##

You did: $$(1-e^{-(E_2-E_1)/kT})(E_1+E_2 e^{-(E_2-E_1)/kT})$$ because ##(1+x)^{-1}\simeq 1-x## for ##x<<1## ... ??!

What are you treating as "x" in that expression?
What is your reasoning?
 
  • #3
Simon Bridge said:
You are told: $$\bar E = \frac{E_1+E_2 e^{-(E_2-E_1)/kT}}{1-e^{-(E_2-E_1)/kT}}$$ ... and you need to find this in the limit that ##kT<<E_2-E_1##

You did: $$(1-e^{-(E_2-E_1)/kT})(E_1+E_2 e^{-(E_2-E_1)/kT})$$ because ##(1+x)^{-1}\simeq 1-x## for ##x<<1## ... ??!

What are you treating as "x" in that expression?
What is your reasoning?

I did it based on the equation (1+x)^-1=1-x. Therefore, (1+e^-BdeltaE)^-1 is equal to 1-e^-BdeltaE. The denominator is (1+e^-BdeltaE) and not (1-e^-BdeltaE) . I made a mistake, sorry.
 
  • #4
shayan825 said:
keep terms up to first order only
This is the part you forgot.
 
  • #5
DrClaude said:
This is the part you forgot.

what should I do? could you please help?
 
  • #6
You have
$$
\frac{f(x)}{1+x} \approx (1-x) f(x)
$$
Then do the multiplication, and keep only terms up to first order.
 
  • #7
hint: the "order" is determined by the power of x.

I did it based on the equation (1+x)^-1=1-x. Therefore, (1+e^-BdeltaE)^-1 is equal to 1-e^-BdeltaE.
So x=e^-BdeltaE right?

Your answer was:
E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

... in terms of x, that is: E1 + deltaE x + E2 x^2

The answer you want is:
E1+deltaE*e^(-B*deltaE) = E1 + deltaE x

compare.
 
  • #8
Simon Bridge said:
hint: the "order" is determined by the power of x.


So x=e^-BdeltaE right?

Your answer was:
E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

... in terms of x, that is: E1 + deltaE x + E2 x^2

The answer you want is:
E1+deltaE*e^(-B*deltaE) = E1 + deltaE x

compare.

So, I just take E2 x^2 from the equation? or there is more to it? Thank you for your help
 
  • #9
Yep: that's all there is to it.

The idea is that if x is small enough to make the binomial approximation (which is what you did), then x^2 is way wayy too small to have a noticeable effect. It is safe to just ignore it. i.e. imagine x~0.00001... at that sort of scale the equations x^2+x+1 and x+1 are hard to tell apart.
 

FAQ: How to Show Average Energy at Low Temperatures?

What is the Boltzmann distribution problem?

The Boltzmann distribution problem is a mathematical concept that describes the distribution of particles in a system at a given temperature. It is named after the Austrian physicist Ludwig Boltzmann and is commonly used in statistical mechanics and thermodynamics.

What does the Boltzmann distribution equation represent?

The Boltzmann distribution equation represents the probability of a particle being in a particular energy state within a system. It takes into account the energy of the system, the temperature, and the number of particles present.

What are the assumptions made in the Boltzmann distribution problem?

The Boltzmann distribution problem assumes that the particles in the system are in thermal equilibrium, meaning they have reached a steady state and are not exchanging energy with the surroundings. It also assumes that the particles are non-interacting, meaning they do not interact with each other.

How is the Boltzmann distribution problem used in real-world applications?

The Boltzmann distribution problem is used in various fields of science, including chemistry, physics, and engineering. It is used to understand the behavior of particles in a system, such as the distribution of energy levels in a gas or the movement of molecules in a chemical reaction. It is also used in statistical mechanics to predict the properties of materials and systems at the molecular level.

What are some limitations of the Boltzmann distribution problem?

One limitation of the Boltzmann distribution problem is that it assumes a system is in thermal equilibrium, which may not always be the case in real-world situations. It also does not account for the effects of quantum mechanics, which can be important at very small scales. Additionally, it assumes that all particles have the same properties, which may not always be true in complex systems.

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