I Rewriting of Gibbs Free Energy in Peksin (Equations 13.35/13.36)

[thatboi]

Hey all,
On page 446 in Peskin, he provides 2 different ways of writing the Gibbs Free Energy:
$$\textbf{G}(M,t) = M^{1+\delta}h(tM^{-1/\beta})$$, and $$\textbf{G}(M,t) = t^{\beta(1+\delta)}f(Mt^{-\beta})$$ where $h$ and $f$ are some initial condition functions that have a smooth limit as $t\rightarrow 0$. My question is how to see that these 2 equations are equivalent. I figure there is a relation that goes like $\rho_{m} = m^2/M^2$ and $\rho_{m}\sim t$ where the first equivalence is from pg. 436 and the second relation is from pg. 445 but I still cannot seem to make it work.

 

[vanhees71]

It's just that $M \sim |t|^{\beta}$ (cf. p. 441, Eq. (13.7)).

 

[thatboi]

It's just that $M \sim |t|^{\beta}$ (cf. p. 441, Eq. (13.7)).

But I thought the point of 13.35 and 13.36 were to derive 13.7 so we couldn't assume that relation a priori?

 

[TSny]

I don't know anything about the subject material in chapter 13 of Peskin. But maybe it's just some mathematical manipulations to show the equivalence of $$\textbf{G}(M,t) = M^{1+\delta}h(tM^{-1/\beta})$$ and $$\textbf{G}(M,t) = t^{\beta(1+\delta)}f(Mt^{-\beta}).$$

Define the function $H(x)$ by the relation $h(x) \equiv x^p H(x)$, where $p$ is a number that we will specify shortly. Then, $$\textbf{G}(M,t) = M^{1+\delta}h(tM^{-1/\beta}) = M^{1+\delta}\left(tM^{-1/\beta}\right)^p H(tM^{-1/\beta}). $$ Let $p = \beta(1+\delta)$ so that $$\textbf{G}(M,t) = M^{[(1+\delta)-(1+\delta)]} t^{\beta(1+\delta)} H(tM^{-1/\beta}) = t^{\beta(1+\delta)} H(tM^{-1/\beta}).$$ Hence, $$\textbf{G}(M,t) = t^{\beta(1+\delta)} H(tM^{-1/\beta}) =t^{\beta(1+\delta)} H\left[\left(Mt^{-\beta}\right)^{-1/\beta}\right] = t^{\beta(1+\delta)} f(Mt^{-\beta})$$ where $f(Mt^{-\beta}) \equiv H\left[\left(Mt^{-\beta}\right)^{-1/\beta}\right]$.

Does this work?

 

[George Jones]

On page 446 in Peskin, he provides 2 different ways of writing the Gibbs Free Energy

You mean Peskin and Schroeder. Peskin is an altogether different book.

 

[Demystifier]

Peskin is an altogether different book.

Concepts of Elementary Particle Physics.

 

[malawi_glenn]

And Schroeder is Thermal physics ;)

 

[thatboi]

I don't know anything about the subject material in chapter 13 of Peskin. But maybe it's just some mathematical manipulations to show the equivalence of $$\textbf{G}(M,t) = M^{1+\delta}h(tM^{-1/\beta})$$ and $$\textbf{G}(M,t) = t^{\beta(1+\delta)}f(Mt^{-\beta}).$$

Define the function $H(x)$ by the relation $h(x) \equiv x^p H(x)$, where $p$ is a number that we will specify shortly. Then, $$\textbf{G}(M,t) = M^{1+\delta}h(tM^{-1/\beta}) = M^{1+\delta}\left(tM^{-1/\beta}\right)^p H(tM^{-1/\beta}). $$ Let $p = \beta(1+\delta)$ so that $$\textbf{G}(M,t) = M^{[(1+\delta)-(1+\delta)]} t^{\beta(1+\delta)} H(tM^{-1/\beta}) = t^{\beta(1+\delta)} H(tM^{-1/\beta}).$$ Hence, $$\textbf{G}(M,t) = t^{\beta(1+\delta)} H(tM^{-1/\beta}) =t^{\beta(1+\delta)} H\left[\left(Mt^{-\beta}\right)^{-1/\beta}\right] = t^{\beta(1+\delta)} f(Mt^{-\beta})$$ where $f(Mt^{-\beta}) \equiv H\left[\left(Mt^{-\beta}\right)^{-1/\beta}\right]$.

Does this work?

Yes great! I see now thanks a lot.

 

[Demystifier]

And Schroeder is Thermal physics ;)

This explains why his QFT book talks about Gibbs free energy.

 

[vanhees71]

QFT and many-body physics is so closely related that it is no surprise, if you find at least one chapter on it also in vacuum-QFT textbooks ;-).

 

[Demystifier]

QFT and many-body physics is so closely related that it is no surprise, if you find at least one chapter on it also in vacuum-QFT textbooks ;-).

I'm used to partition function $Z$ and closely related Helmholtz free energy $F$ (the relation is $Z=e^{-\beta F}$) in particle QFT, but Gibbs free energy in particle QFT looks quite exotic to me. The Minkowski version of $-\beta F$ is $iW[J]$ which generates only connected Feynman diagrams. What kind of diagrams are generated by the Minkowski version of Gibbs free energy?

 

[vanhees71]

That's of course true, but which thermodynamical potential you address is not that important. They are all related by Legendre transformations.

 

[Demystifier]

That's of course true, but which thermodynamical potential you address is not that important. They are all related by Legendre transformations.

Sometimes it's important. For example the Jarzynski equality in non-equilibrium statistical physics
$$\left\langle e^{-\beta W}\right\rangle = e^{-\beta \Delta F}$$
is valid for $F$, but, as far as I know, there is no analog for other thermodynamic potentials.

The $F$ in thermodynamics is in many ways analogous to the Lagrangian in mechanics, and Lagrangian has some special properties which other "energy" functions in mechanics don't have.