Everage energy of a gas of quantom SHO

[Jason Gomez]

Homework Statement

[/U]
The average energy of a gas of quantum SHO is

$$Eav= \sum_{n=0}^{\infty}n\hbar\omega e^(-n\hbar\omega/kT)\div \sum_{n=0}^{\infty}e^(-n\hbar\omega/kT)$$

can be solved to be

$$Eav=\hbar\omega\div \left \{ e^\left ( \hbar\omega/kt \right ) \right \}-1$$

make use of the following two sums, true when [x]<1:

$$\sum_{n=o}^{\infty }x^n=1/(1-x)$$

$$\sum_{n=0}^{\infty}nx^n=x/(1-x)^2$$

Homework Equations

I tried dividing the two equations, and I think that is the right course of action, but I am not sure what to do with the summations, what I finally get looks like this:

$$\left \{\sum_{n=0}^{\infty} x^n\div \sum_{n=0}^{\infty}nx^n\right \}= 1/(x-x^2)$$

I am not sure, but can I just say:

$$n=1/\left ( x-x^2 \right )$$
b]3. The Attempt at a Solution [/b]

 

[Dickfore]

$$\left \{\sum_{n=0}^{\infty} x^n\div \sum_{n=0}^{\infty}nx^n\right \}= 1/(x-x^2)$$

This is wrong. Use the provided formulas to plug in both of the fractions and do step by step simplification to find the correct expression. then, you will need to identify what x is in terms of the parameters given in the problem.

 

[Jason Gomez]

Thank you, I assume you mean plug in the summation equations that I am to make use of into the Eav equations provided, but I am not sure how. I think I may be making this problem harder than what it is, but I just can't rap my brain around it

 

[Dickfore]

You have a sum in the numerator and denominator of the expression. Both of them are evaluated for you in the statement of the problem.

Note:

The "$\div$" sign means division which is the same as a fraction line.

 

[paranormal]

I would approach this problem by first understanding the physical meaning of the equations. The average energy of a gas of quantum SHO is a measure of the average amount of energy possessed by each particle in the gas. This energy is dependent on the temperature (T) of the gas and the frequency (ω) of the SHO. The first equation provided is a mathematical representation of this average energy, which takes into account the different energy levels (n) of the SHO.

To solve for the average energy, we can use the two summation equations provided. By dividing the two equations, we can simplify the expression to:

Eav= \left \{\sum_{n=0}^{\infty} x^n\div \sum_{n=0}^{\infty}nx^n\right \}= 1/(x-x^2)

We can then use the definition of the geometric series (x^n) to rewrite this expression as:

Eav=1/(x-x^2)=1/\left ( 1-x \right )-1

Substituting x=e^(-\hbar\omega/kT) into this expression, we get:

Eav=1/\left ( 1-e^(-\hbar\omega/kT) \right )-1

Finally, using the definition of the geometric series again, we can simplify this expression to:

Eav=\hbar\omega\div \left \{ e^\left ( \hbar\omega/kt \right ) \right \}-1

This is the final expression for the average energy of a gas of quantum SHO. It is important to note that this expression is only valid when x<1, which implies that the temperature of the gas must be greater than absolute zero. This makes sense, since at absolute zero, the particles would have no energy and the SHO would not be oscillating. Therefore, this expression is only valid at non-zero temperatures.

In conclusion, the average energy of a gas of quantum SHO can be solved using the two summation equations provided, and the final expression is dependent on the temperature and frequency of the SHO. This can be a useful tool for understanding the behavior of gases at the quantum level.