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I know that $δ$ sometimes represents the Dirac delta function but in my book it states "Suppose that equilibrium has been established Then a slight change in the position of the piston should not change the free energy since it is at a minimum that is $δA=0$" but in terms of this what exactly does it mean?

Qmechanic
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Astronomical
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2 Answers2

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I know that $\delta$ represents the Dirac delta function but in my book it states "Suppose that equilibrium has been established Then a slight change in the position of the piston should not change the free energy since it is at a minimum that is δA=0" but in terms of this what exactly does it mean.

Here, the symbol $\delta$ is being used to indicate "change." You can read $\delta A$ as the "change in Free Energy."

When a function like $A$ is at a maximum or a minimum (or saddle point) the function is stationary, meaning it doesn't change at first order when its function arguments change.

Since the Free Energy $A$ is a function of $T$, $V$, and $N$, we can take $$\delta A =0 $$ as effectively equivalent to: $$ \frac{\partial A}{\partial T} = 0\; $$ $$ \frac{\partial A}{\partial V} = 0\; $$ $$ \frac{\partial A}{\partial N} = 0\; $$

This is because, at first order: $$ \delta A = \frac{\partial A}{\partial N}\delta N + \frac{\partial A}{\partial V}\delta V + \frac{\partial A}{\partial T}\delta T\;. $$

hft
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  • This makes a lot of sense thank you so much! – Astronomical Sep 18 '22 at 18:21
  • no problem, you're welcome – hft Sep 18 '22 at 18:22
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    +1. It might be worth to point out that often the symbol $\mathrm d$ is used instead of $\delta$. – Tobias Fünke Sep 18 '22 at 18:24
  • I don't think it is worth it at this point, since there already is another answer (the question seems to be a duplicate) that gets into the different between $d$ $\delta$ and $\Delta$. Probably not worth it to expand this answer too. – hft Sep 18 '22 at 19:02
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That $\delta$ is not a Dirac delta, but the symbol for indicating a differential form that is an inexact differential, i.e. a differential whose integration depends on the path of integration,

$\displaystyle \int_{\ell^1_{A\rightarrow B}} \delta f \ne \displaystyle \int_{\ell^2_{A\rightarrow B}} \delta f \qquad$ in general, for different integration paths $\ell^i_{A\rightarrow B}$ with the same extreme points $A$, $B$

and not only on the values of a function (its primitive) at the extreme points of the integration path, like exact differentials

$\displaystyle \int_{\ell_{A\rightarrow B}} d f = f(B) - f(A) \qquad $ for every integration path with extreme points $A$ and $B$.

basics
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    I don't think that the $\delta$ here means inexact differential, tho, due to its appearance with $A$, the free energy, which is a state function. I guess it should mean that to first order, the free energy does not change. – Tobias Fünke Sep 18 '22 at 18:00
  • I really don't know. Probably it only means a variation of a function, without caring about if it's an exact or an inexact differential at all. For sure it's not a Dirac's delta – basics Sep 18 '22 at 18:11
  • Im sorry im still confused does it just mean a "slight variation"? because in another part of my book it says " if the system is in equilibrium with the given T and P(T) the Gibbs potential of this state must be at a minimum. That is for any parameters other than T and P are varied slightly we must have δG=0." so is this sentence just saying that if even if we change Temperature and Pressure slightly the change in the Gibbs free energy is still 0 – Astronomical Sep 18 '22 at 18:19
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    @goforhenry A "slight variation" can be taken to mean variation at "first order" with respect to the function's arguments. This means that the first order partial derivatives of the function are all zero at that point. You can see my answer for the equations that correspond to these words. – hft Sep 18 '22 at 18:21