Taylor expansion, thermodynamics

In summary, to show the given relation, make a Taylor expansion of S around the point (U,V,n) and use the fact that S(U,V,n) is equal to both S(U+ΔU, V+ΔV, n) and S(U-ΔU, V-ΔV, n). This will lead to the desired result.
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Homework Statement


I'm asked to show that the relation ##S(U+\Delta U, V+ \Delta V , n )+S(U-\Delta U, V- \Delta V , n ) \leq 2 S(U,V,n)## implies that ##\frac{\partial ^2 S}{\partial U ^2 } \frac{\partial ^2 S}{\partial V ^2}- \left ( \frac{\partial ^2 S }{\partial U \partial V} \right ) \geq 0##.

Homework Equations


Taylor expansion of a function of several variables: https://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables

The Attempt at a Solution


The book gives some help, it says to first make a Taylor expansion of the left side to the second order in ##\Delta U## and ##\Delta V##. So I should reach the equation ##S_{UU}(\Delta U)^2+2S_{UV}\Delta U \Delta V + S_{VV} (\Delta V)^2 \leq 0##, then I'd have to rewrite this expression and I should reach the answer.
However I'm having a problem at understanding what the book means by "in ##\Delta U## and ##\Delta V##". Does it means I should expand S and U around the point ##(\Delta U , \Delta V, n)##?
Or...

Edit: I've expanded S around the point (U,V,n) which seemed to make sense to me, but then I don't reach what I should. Instead I reach ##-S+\Delta U S_U + \Delta V S_V + \frac{1}{2} \left [ (\Delta U)^2 S_{UU} + 2 \Delta U \Delta V S_{UV} + (\Delta V )^2 S_{VV} \right ] \leq 0## where the functions are evaluated in ##(U,V,n)##.

Edit2: NEVERMIND! I forgot a term to make the Taylor expansion. I didn't do it yet but at first glance this is the right thing to do, i.e. making the expansion around (U,V,n)!
Edit3: problem solved, I've reached the final result!
 
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  • #2


It seems like you are on the right track. The key here is to make a Taylor expansion of the function S around the point (U,V,n). This means that you are expanding the function in terms of small changes in U and V, which are represented by the variables ΔU and ΔV.

When you make the Taylor expansion, you will get an expression that looks like this: S(U+ΔU, V+ΔV, n) = S(U,V,n) + ΔU*S(U,V,n) + ΔV*S(U,V,n) + 1/2*(ΔU)^2*S(U,V,n) + 1/2*(ΔV)^2*S(U,V,n) + higher order terms.

You can then plug this expression into the original relation and use the fact that S(U,V,n) = S(U+ΔU, V+ΔV, n) and S(U,V,n) = S(U-ΔU, V-ΔV, n) to simplify the expression and reach the desired result.

Hope this helps!
 

What is Taylor expansion?

Taylor expansion is a mathematical technique used to approximate the value of a function at a certain point by using a series of derivatives of the function evaluated at that point.

What is the purpose of using Taylor expansion in thermodynamics?

In thermodynamics, Taylor expansion is used to approximate the behavior of a thermodynamic system near equilibrium, where the system can be described by a set of continuous functions. This allows us to make predictions and analyze the behavior of the system without having to solve complex equations.

What are the limitations of using Taylor expansion in thermodynamics?

One limitation of using Taylor expansion in thermodynamics is that it is only accurate for systems near equilibrium, where the functions are continuous. It cannot be used for systems that undergo rapid changes or have discontinuities in their behavior.

How is Taylor expansion related to entropy in thermodynamics?

In thermodynamics, entropy can be defined as the change in the heat of a system divided by its temperature. Taylor expansion is used to approximate this change in heat, making it a useful tool in calculating entropy changes.

Can Taylor expansion be used for all thermodynamic systems?

No, Taylor expansion is only applicable to systems that can be described by continuous functions. It cannot be used for systems with discontinuous behavior, such as phase transitions or chemical reactions.

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