Partition function of rotating molecule

[mikeph]

Hello

I am working through a textbook here, struggling to follow a mathematical step. We are deriving the partition function Q due to pure rotation of a system containing molecules with quantum rotation energy levels:

E = h²J(J+1) / 8pi²I

Where J is the rotational quantum number, J = 0,1,2...

I have substituted this into the partition function, where the degeneracy is (2J+1), and have made the approximation of turning the sum into an integral to obtain:

Q(rot) = INT(J=0 to infinity) of (2J+1)exp[(-h²J(J+1))/(8pi²IkT)] dJ

Suddenly they say we can eliminate the (2J+1) factor at the front of the integrand if we convert the integration variable from dJ to d(J² + J). Nothing else changes:

Q(rot) = INT(J=0 to infinity) of exp[(-h²(J² + J))/(8pi²IkT)] d(J² + J).

They then integrate, taking the constant factor of the exponential down, evaluate what's left at the limits to get [1-0] and conclude the answer is this factor, which is:

Q(rot) = (8pi²IkT)/(h²)

I understand the last step, but can't see how they've done the substitution of the variable in the first place to remove the (2J+1) factor in the integral.

Thanks for any help,

Mike

 

[eljose79]

Hello Mike,

Thank you for sharing your struggle with this mathematical step. I can understand how it can be confusing and difficult to follow. Let me try to break it down for you.

Firstly, let's look at the original expression for the partition function:

Q(rot) = INT(J=0 to infinity) of (2J+1)exp[(-h2J(J+1))/(8pi2IkT)] dJ

As you have correctly pointed out, the degeneracy factor (2J+1) is present in the integrand. Now, let's make the substitution of the integration variable from dJ to d(J2 + J). This means that instead of integrating with respect to J, we will now integrate with respect to (J2 + J). This is a common technique used in integration, known as the "u-substitution" method.

Now, let's see how this substitution affects the integrand. We will substitute J = (J2 + J) - J2. This means that (2J+1) becomes [(J2 + J + 1) - (J2 + 1)]. Simplifying this, we get (2J+1) = J2 + J. Substituting this in the original expression, we get:

Q(rot) = INT(J=0 to infinity) of (J2 + J)exp[(-h2J(J+1))/(8pi2IkT)] d(J2 + J)

As you can see, the (2J+1) factor has been eliminated from the integrand. Now, when we integrate this expression, we will get:

Q(rot) = INT(J=0 to infinity) of exp[(-h2(J2 + J))/(8pi2IkT)] d(J2 + J)

The limits of integration remain the same, but the integrand has changed. This is because we have substituted the variable, not the limits. Now, when we integrate this expression, we will get:

Q(rot) = [-exp[(-h2(J2 + J))/(8pi2IkT)]] from J=0 to infinity

Evaluating this at the limits, we get [1-0], which gives us the final result:

Q(rot) = (8pi2IkT)/(h2)

I hope this helps to clarify the steps involved in the substitution of the integration variable. If you have any further questions or

 

[astrokreng]

Hello Mike,

Thank you for reaching out with your question. I can understand your confusion with the substitution of the variable in the partition function for a rotating molecule. Let me explain it step by step.

First, let's look at the integral that you have written:

Q(rot) = INT(J=0 to infinity) of (2J+1)exp[(-h2J(J+1))/(8pi2IkT)] dJ

As you correctly mentioned, the degeneracy of the rotational energy levels is (2J+1). However, when we are calculating the partition function, we are essentially summing over all possible energy states. In this case, we are summing over all possible values of J, which goes from 0 to infinity. Therefore, we can write the integral as:

Q(rot) = INT(J=0 to infinity) of (2J+1)exp[(-h2J(J+1))/(8pi2IkT)] dJ = INT(J=0 to infinity) of exp[(-h2J(J+1))/(8pi2IkT)] dJ

Now, let's look at the substitution of the variable. The reason we do this is to make the integral easier to evaluate. When we substitute J with (J2+J), we are essentially changing the variable from J to (J2+J). This allows us to eliminate the (2J+1) factor in the integral, making it easier to integrate. Let's see how this works:

Q(rot) = INT(J=0 to infinity) of exp[(-h2J(J+1))/(8pi2IkT)] dJ

Substituting J with (J2+J):

= INT(J2+J=0 to infinity) of exp[(-h2(J2+J)(J2+J+1))/(8pi2IkT)] d(J2+J)

= INT(J2=0 to infinity) of exp[(-h2(J2+2J+1))/(8pi2IkT)] d(J2+J)

= INT(J2=0 to infinity) of exp[(-h2J2-h2J-h2)/(8pi2IkT)] d(J2+J)

= INT(J2=0 to infinity) of exp[(-h2J2)/(8pi2IkT)]exp[(-h2J)/(8pi2IkT