How to derive Eq. (4.26) from Eq. (4.25) in Birger Bergersen's book (Maier-Saupe model)

[MathematicalPhysicist]

Homework Statement

This is a question from Maier-Saupe model section in the book.

Eq. (4.25) is:
$$(4.25)\ \ \ g=G/N=-k_BT\ln(4\pi\int_0^1 d\mu \exp\{\rho U \beta [(3\mu^2-1)Q-Q^2]\})$$
Eq. (4.26) is:
$$(4.26) \ \ \ g= G/N = -k_B T \ln(4\pi)+\rho U Q^2(1-\frac{2}{5}\beta \rho U)-\frac{8}{105}\beta^2\rho^3 U^3 Q^3 + \frac{4}{175}\beta^3 \rho^4U^4Q^4+\ldots . $$

For the life of me I don't see how to derive it, Birger et al. says it's now straightforward matter to obtain a Taylor expansion of (4.25) in powers of $Q$.

So I assumed I need to expand the exponential i.e with: $\exp(\rho U\beta[(3\mu^2-1)Q-Q^2])=1+\rho U\beta [ (3\mu^2-1)Q-Q^2]+(\rho U \beta [(3\mu^2-1)Q-Q^2])^2/2$ and then to use integration of $\mu$ and expand $\ln(ab)=\ln a + \ln b$, this is how we get the first term in (4.26), but I don't see how to get the other terms.

Can anyone lend some help with this derivation?
Thanks!

Relevant Equations

I posted in the exercise above.

The dictators at physics.stackexchange want to close my post that I post here.

I hope someone can help me with this question, I want to compute this by hand, without Computer algebra software, mainly because I don't know which syntax to use for Mathematica (if you know the syntax, can you give it to me; )

I don't have enough time, please help!

Thanks!

 

[Delta2]

I think your approach to expand the exponential accordind to $e^x=\sum \frac{x^n}{n!}$ where $x=\rho U\beta((3\mu^2-1)Q-Q^2)$ and then do integration of polynomials of $\mu$ seems correct to me however I believe you are not implementing it correctly (except from the first term ). It seems to be a tedious algebraic exercise .

Can you post a screenshot of the book's page cause I think your post might have some typos.

Maybe an expansion of $$\ln(1+y)=-\sum \frac{(-1)^ky^k}{k}$$ has to be done too. (where y is the first term (besides the 1) in the series you get after doing the expansion of the exponential and integrating). This would explain the alternating signs in the powers of Q.

 

[vela]

$$(4.25)\ \ \ g=G/N=-k_BT\ln(4\pi\int_0^1 d\mu \exp\{\rho U \beta [(3\mu^2-1)Q-Q^2]\})$$

You can pull quite a bit out of the integral using the fact that
$$\exp\{\rho U \beta [(3\mu^2-1)Q-Q^2]\} = \exp[-\rho U \beta(Q+Q^2)] \exp[3 \rho U \beta Q \mu^2],$$
so you end up with
$$-k_BT [\ln(4\pi) - \rho U \beta (Q+Q^2) + \ln \int_0^1 d\mu\,e^{3 \rho U \beta Q \mu^2}].$$ Then you could try expanding the exponential as a series and integrate term by term. Finally, expand the log as a series. It looks like the first-order term will cancel like you want then.

 

[MathematicalPhysicist]

OK thank you everyone.

BTW, I am not sure all these uses of the series sums are legitimate from a pure mathematical point of view... but hey... physics is nutz.

 

[MathematicalPhysicist]

I think your approach to expand the exponential accordind to $e^x=\sum \frac{x^n}{n!}$ where $x=\rho U\beta((3\mu^2-1)Q-Q^2)$ and then do integration of polynomials of $\mu$ seems correct to me however I believe you are not implementing it correctly (except from the first term ). It seems to be a tedious algebraic exercise .

Can you post a screenshot of the book's page cause I think your post might have some typos.

Maybe an expansion of $$\ln(1+y)=-\sum \frac{(-1)^ky^k}{k}$$ has to be done too. (where y is the first term (besides the 1) in the series you get after doing the expansion of the exponential and integrating). This would explain the alternating signs in the powers of Q.

I wish I could do the calculation by hand, but unfortunately this dang is inhumane!

 

[MathematicalPhysicist]

s-h-i-t is replaced by dang? LOL...

 

[Delta2]

OK thank you everyone.

BTW, I am not sure all these uses of the series sums are legitimate from a pure mathematical point of view... but hey... physics is nutz.

They are legitimate from mathematical point of view, just when expanding the $\ln(1+y)$ as a series you have take cases $|y|<1$ or $y>1$ cause then the formula for series is abit different.

I wish I could do the calculation by hand, but unfortunately this dang is inhumane!

Just do what @vela suggests (as long as you don't have a typo in the OP in the eq. 4.25, I was thinking along the same lines as @vela but I was afraid you might have a typo there and that the exponent is $…(3\mu^2-1)(Q-Q^2)$). Then all you have to do is series expansion of $e^{-aQ\mu^2}$ ( I hope $a=3\rho U\beta$ is a constant w.r.t $Q$ and $\mu$) and then integrating. And finally series expansion of $\ln(1+y)$ for the proper y.

 

[MathematicalPhysicist]

@Delta2 I might have a typo. Anyhow seems to me hard for a human to do the calculation , I assume it was done by computer algebra system, am I right?

 

[Delta2]

@Delta2 I might have a typo. Anyhow seems to me hard for a human to do the calculation , I assume it was done by computer algebra system, am I right?

I believe it can be done by hand, especially if there is not a typo and we can do the simplification as suggested by @vela . If there is a typo there and the $Q^2$ applies to $\mu^2$ then its going to be aliitle bit harder, might be more than 1 page long.

Also what I believe is another problem here , that we have to deal with a double or more precisely a nested Taylor expansion and i don't see how this can be done unless we truncate one expansion and it is unclear how the author did the truncation. I would personally truncate the expansion involving the integral and keep full the expansion of $\ln$

 

[MathematicalPhysicist]

I believe it can be done by hand, especially if there is not a typo and we can do the simplification as suggested by @vela . If there is a typo there and the $Q^2$ applies to $\mu^2$ then its going to be aliitle bit harder, might be more than 1 page long.

Also what I believe is another problem here , that we have to deal with a double or more precisely a nested Taylor expansion and i don't see how this can be done unless we truncate one expansion and it is unclear how the author did the truncation. I would personally truncate the expansion involving the integral and keep full the expansion of $\ln$

How does @samalkhaiat calls physics - ill defined mathematics...

 

[MathematicalPhysicist]

They are legitimate from mathematical point of view, just when expanding the $\ln(1+y)$ as a series you have take cases $|y|<1$ or $y>1$ cause then the formula for series is abit different.

Just do what @vela suggests (as long as you don't have a typo in the OP in the eq. 4.25, I was thinking along the same lines as @vela but I was afraid you might have a typo there and that the exponent is $…(3\mu^2-1)(Q-Q^2)$). Then all you have to do is series expansion of $e^{-aQ\mu^2}$ ( I hope $a=3\rho U\beta$ is a constant w.r.t $Q$ and $\mu$) and then integrating. And finally series expansion of $\ln(1+y)$ for the proper y.

No, there's no typo I copied it as is.

Believe me one thing I know in my studies is how to copy...

 

[Delta2]

No, there's no typo I copied it as is.

Believe me one thing I know in my studies is how to copy...

ok fine hehe, so proceed as vela suggests in post #3, just that when you expand the exponential take only the first two terms of the series, then integrate from 0 to 1, and you ll have something equal to $1+y(Q)$ where I believe $y(Q)$ will be a single term (not a long sum) and then you can do full series expansion of $\ln(1+y(Q))$ and tell us what you get.

How does @samalkhaiat calls physics - ill defined mathematics...

Sometimes they might get ill defined indeed. But here I don't believe it such the case, its only that we have to do some approximations (in this case truncating the $e^{-aQ\mu^2}=\sum\frac{(-aQ\mu^2)^n}{n!}$ series) sometimes otherwise the result is very complex mathematically.

 

[MathematicalPhysicist]

Sometimes they might get ill defined indeed. But here I don't believe it such the case, its only that we have to do some approximations (in this case truncating the $e^{-aQ\mu^2}=\sum\frac{(-aQ\mu^2)^n}{n!}$ series) sometimes otherwise the result is very complex mathematically.

Aha!
You believe but you don't know for sure since you haven't done the calculation by hand.
I am pretty much sure they let the computer to do the stuff by itself and I didn't see an easy calculation not even in the SM to the second edition of Bergersen's.

I would like to see someone does this computation as rigorous as possible, if it's even possible.

 

[Delta2]

Aha!
You believe but you don't know for sure since you haven't done the calculation by hand.
I am pretty much sure they let the computer to do the stuff by itself and I didn't see an easy calculation not even in the SM to the second edition of Bergersen's.

I would like to see someone does this computation as rigorous as possible, if it's even possible.

Well yes I don't know for sure. By doing the calculation for the first few terms as I outlined in post #12 I see that the constant term agrees, the first power of Q vanishes, but I don't seem to get agreement on the higher than first, powers of Q.

Now it is upon you to do the calculation as follows. Expand fully the exponential, integrate fully to get an an infinite series $1+y(Q)=1+\sum a_n Q^n$ and then use the approximation $\ln(1+y(Q))\approx 1-y(Q)$. if again it doesn't work then I really don't know what's going on here.

You are right that this is not as trivial or straightforward as the book author claims it to be. But the fact that they might be using computer software to do the expansions, it doesn't prevent you of you doing the hard way by hand, putting down some hard work by you.

Finally all I can say more is that is not trivial because it involves nested series expansion, something along the lines of $$\sum a_n (\sum b_k Q^k)^n$$, and I am not so experienced on this type of expansion. Maybe a "big brain" of these forums like @vanhees71 or @Orodruin can offer us his lights upon this not so trivial problem.

 

[MathematicalPhysicist]

Thanks @Delta2 you got to love PF!

 

[MathematicalPhysicist]

I might try to correspond with one of the authors of this book to see what he thinks.
But as I rarely get a response I'll leave it at that!

 

[vela]

Consider this example: $\log e^x = \log \left( 1+ x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right) = a_0 + a_1 x + a_2 x^2 + \cdots$. Expanding the log you get $\log e^x = y - \frac {y^2}{2} + \frac {y^3}{3}$ where $y=x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$.

Note that no power of $y$ will produce a constant term, so we know that $a_0=0$. Now consider $a_1$. If you multiply out $y^n$, the lowest power of $x$ will be $x^n$, so to find $a_1$, we only need to consider $y^1$. By inspection, we get $a_1=1$.

Now consider $a_2$. We're looking for terms involving $x^2$. The $y$ term will contribute $x^2/2!$. We'll also get a contribution from the $y^2$ term:
$$y^2 = \left(x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right) \left(x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right).$$ The only way to get a term proportional to $x^2$ is to multiply the $x$ from the first factor with the $x$ from the second factor. In other words, $y^2 = x^2 + \text{higher orders}$. So the $y^2$ term in the log series contributes $-x^2/2$. We have then $a_2 = (1/2! - 1/2) = 0$.

Now consider $a_3$. The $y$ term will contribute the cubic term $x^3/3!$. From the $y^2$ term, we will get cubic terms from $x$ in one factor multiplied by $x^2$ in the other factor, so the cubic term of $y^2$ is $2 \left(x\cdot\frac{x^2}{2!}\right) = x^3$. Finally, the cubic term of $y^3$ is $x^3$. Hence, we have
$$a_3 x^3 = \frac {x^3}{3!} - \frac{1}{2}x^3 + \frac{1}{3}x^3 = 0,$$ so $a_3 = 0$.

For $a_4$, we have
\begin{align*}
y &\to \frac{x^4}{4!} \\
y^2 &\to 2\left(x \cdot \frac{x^3}{3!}\right) + \left(\frac{x^2}{2!}\right)^2 = \frac 7{12}x^4 \\
y^3 &\to 3\left(x \cdot x \cdot \frac{x^2}{2!}\right) = \frac 32 x^4 \\
y^4 &\to x^4
\end{align*} so
$$a_4 x^4 = \frac{x^4}{4!} - \frac 12 \left(\frac 7{12}x^4\right) + \frac 13\left(\frac 32 x^3\right) - \frac 14 x^4 = 0.$$

 

[MathematicalPhysicist]

@vela I'll read your post later today or tomorrow can't guarantee when since I have an overload of studying and work...

 

[Delta2]

@vela your post shows that it isn't so straightforward to find the series expansion, as the author claims it to be.

 

[vela]

I guess "straightforward" is a matter of perspective. To me (and to the author, I suspect), it's just basic algebra.

 

[MathematicalPhysicist]

I guess "straightforward" is a matter of perspective. To me (and to the author, I suspect), it's just basic algebra.

That's relativity for you!

 

[Delta2]

I guess "straightforward" is a matter of perspective. To me (and to the author, I suspect), it's just basic algebra.

Deep down it might be basic algebra what you doing, however I would not call the whole thing straightforward.

I don't know, maybe I got scared from the nested series expansion and couldn't see this straightforwardly