Equilibrium Statistics -- Euler summation formula

[MathematicalPhysicist]

Homework Statement

In the calculation in high temperatures of $Z_{rot} = (\sum_{j=0}^\infty (2j+1)\exp{j(j+1)\theta_{rot}/T})^N$; they use Euler summation formula:

$$\sum_{n=0}^\infty f(n) = \int_0^\infty f(x)dx+\frac{1}{2}f(0)-\frac{1}{12}f'(0)+\frac{1}{720}f^{(3)}(0)+\ldots$$

for $f(x) = (2x+1)\exp{x(x+1)\theta_{rot}/T}$.

Now they get that: $Z_{rot} = \bigg(T/\theta_{rot}+1/3+\theta_{rot}/(15T)+\ldots \bigg)^N$.

Now as for the third term I did the calculation and I get a minus sign, i.e. I believe it should be: $-\theta_{rot}/(15T)$ instead of $+\theta_{rot}/(15T)$.

I get that the factor that multiplies $\theta_{rot}/T$ is $12/720-1/12$.

Am I right or wrong?

Homework Equations

The Attempt at a Solution

 

[TSny]

Should the argument of the exponential function in $Z_{rot}$ and $f(x)$ have an overall negative sign? If not, I don't see how $\int_0^\infty f(x) dx$ could converge.

 

[MathematicalPhysicist]

@TSny You are correct there should be a minus sign in the argument of the exponential, but in the Solution Manual it doesn't appear.
But in that case I think it should be -1/3 and not +1/3.

 

[Ray Vickson]

Homework Statement

In the calculation in high temperatures of $Z_{rot} = (\sum_{j=0}^\infty (2j+1)\exp{j(j+1)\theta_{rot}/T})^N$; they use Euler summation formula:

$$\sum_{n=0}^\infty f(n) = \int_0^\infty f(x)dx+\frac{1}{2}f(0)-\frac{1}{12}f'(0)+\frac{1}{720}f^{(3)}(0)+\ldots$$

for $f(x) = (2x+1)\exp{x(x+1)\theta_{rot}/T}$.

Now they get that: $Z_{rot} = \bigg(T/\theta_{rot}+1/3+\theta_{rot}/(15T)+\ldots \bigg)^N$.

Now as for the third term I did the calculation and I get a minus sign, i.e. I believe it should be: $-\theta_{rot}/(15T)$ instead of $+\theta_{rot}/(15T)$.

I get that the factor that multiplies $\theta_{rot}/T$ is $12/720-1/12$.

Am I right or wrong?

Homework Equations

The Attempt at a Solution

Please either use parentheses or $e^{a}$ instead of $\exp(a)$. From what you wrote it is impossible to tell whether you want to compute
$\sum (2j+1) e^{j(j+1)} \theta/T$ or $\sum (2j+1) e^{j(j+1) \theta}/T$ or $\sum (2j+1) e^{j(j+1) \theta/T}$. Using "exp" these would be $\sum (2j+1) \exp(j(j+1))\, \theta/T$ or $\sum (2j+1) \exp((j(j+1) \theta) /T$ or $\sum (2j+1) \exp((j(j+1) \theta/T)$. Using a "solidus" fraction $\frac a b$ would be even better"
$$\sum (2j+1) \frac{\exp(j(j+1)) \theta}{T}$$ or $$ \sum (2j+1) \frac{\exp(j(j+1) \theta)}{T}$$ or $$ \sum (2j+1) \exp \left( \frac{j(j+1) \theta}{T} \right)$$
And, of course, there is the issue of the missing "-" sign.

Also: does the Euler summation formula really go from $f^{\prime}(0)$ to $f^{(3)}(0)$, skipping over $f^{\prime \prime}(0)$ altogether?

 

[MathematicalPhysicist]

It should be: $$\sum (2j+1)\exp{\bigg(\frac{j(j+1)\theta_{rot}}{T}\bigg)}$$

 

[Ray Vickson]

It should be: $$\sum (2j+1)\exp{\bigg(\frac{j(j+1)\theta_{rot}}{T}\bigg)}$$

Shouldn't that have a "-" sign in the $\exp(\cdots)$? If not, you have a divergent series (unless $\theta_{rot}/T < 0$).

 

[MathematicalPhysicist]

No, a minus sign is indeed missing in the SM; and as far as I can tell $\theta_{rot}/T>0$, as I wrote this approximation(?) is used for high temperatures, i.e $T \gg \theta_{rot}$.

BTW, you can find this SM scanned in the net; it's in the solution of problem 2.8 in the book by Bergersen's and Plischke's.

 

[TSny]

@TSny You are correct there should be a minus sign in the argument of the exponential, but in the Solution Manual it doesn't appear.
But in that case I think it should be -1/3 and not +1/3.

I get +1/3. It comes from a combination of +1/2 and -1/6. The +1/2 is from (1/2)f(0) and the -1/6 is from -(1/12)f'(0).

 

[MathematicalPhysicist]

@TSny ok I agree it doesn't get changed.

In which case it's indeed $+1/15 (\theta_{rot}/T)$ and not $-1/15 (\theta_{rot}/T)$, and what's wrong in the book is the missing sign in the exponential, am I correct?

 

[TSny]

@TSny ok I agree it doesn't get changed.

In which case it's indeed $+1/15 (\theta_{rot}/T)$ and not $-1/15 (\theta_{rot}/T)$, and what's wrong in the book is the missing sign in the exponential, am I correct?

Yes, I think so.

 

[MathematicalPhysicist]

@TSny , @Ray Vickson after correcting the minus sign in the exponential; I have yet another minus sign problem which I want to correct or ask for your corroboration:

They write in the SM that:

$$E = \frac{\partial}{\partial \beta} \ln Z_{rot} = Nk_B T \frac{1-\frac{\theta_{rot}^2}{15T^2}+\ldots}{1+1/3 \frac{\theta_{rot}}{T}+\frac{\theta_{rot}^2}{15T^2}+\ldots}$$

But in the numerator I get that it should be: $-1+\frac{\theta_{rot}^2}{15T^2}+\ldots$; I believe this expression for $E$ with the derivative is the definition of the internal energy, so I guess again a minus sign is missing here, am I correct?

 

[TSny]

I believe this expression for $E$ with the derivative is the definition of the internal energy, so I guess again a minus sign is missing here, am I correct?

Yes, a minus sign is missing in the expression for $E$.

https://en.wikipedia.org/wiki/Parti...s)#Calculating_the_thermodynamic_total_energy

 

[MathematicalPhysicist]

Yes, a minus sign is missing in the expression for $E$.

https://en.wikipedia.org/wiki/Parti...s)#Calculating_the_thermodynamic_total_energy

Thanks.

There's another thing that get me to scratch my hair metaphorically speaking;

They write that the series expansion of the above $E$ should be: $E_{rot}=Nk_BT(1-\frac{\theta_{rot}}{3T}-\frac{2\theta_{rot}^2}{45T^2}+\ldots)$, but I get that instead of 45 it should be 15;

Where I have expanded:
$$1/(1+1/3 (\theta_{rot}/T)+(\theta_{rot}^2/(15T^2)+\ldots) = 1-\bigg[1/3 (\theta_{rot}/T)+(\theta_{rot}^2/(15T^2)+\ldots\bigg]$$

Have they got it wrong, yet again?
:-D

 

[TSny]

Where I have expanded:
$$1/(1+1/3 (\theta_{rot}/T)+(\theta_{rot}^2/(15T^2)+\ldots) = 1-\bigg[1/3 (\theta_{rot}/T)+(\theta_{rot}^2/(15T^2)+\ldots\bigg]$$

I haven't checked their answer, but you will need to go one order higher in your expansion of the denominator to get all of the $\theta_{rot}^2/T^2$ contributions.

 

[MathematicalPhysicist]

I agree I forgot $\bigg[ 1/3 \theta/T + \theta^2/(15T^2) +\ldots\bigg]^2$.

I get with this that the factor that multiplies $\theta^2/T^2$ is $-2/15+1/9 = (-6+5)/45=-1/45$, is this correct?

 

[TSny]

I agree I forgot $\bigg[ 1/3 \theta/T + \theta^2/(15T^2) +\ldots\bigg]^2$.

I get with this that the factor that multiplies $\theta^2/T^2$ is $-2/15+1/9 = (-6+5)/45=-1/45$, is this correct?

Yes, I think that's right even though it doesn't agree with their factor of $-2/45$ as given in post #13.