Apply the Legendre Transformation to the Entropy S as a function of E

In summary, the conversation is about the Sackur-Tetrode equation and using Legendre transformations to obtain a new potential with different independent parameters. The participants discuss the representation of ds and finding a new potential, F, in terms of dT, dV, and dN. They also mention the relationship between d(Ts) and s.
  • #1
GravityX
19
1
Homework Statement
Apply Legendre Transformation to Entropy
Relevant Equations
##g(m)=f(x(m))-m*x(m)## and ##x(m)=(f')^-1(m)##
Hi,

Unfortunately I am not getting anywhere with task three, I don't know exactly what to show

Bildschirmfoto 2022-11-28 um 16.30.54.png

Shall I now show that from ##S(T,V,N)## using Legendre I then get ##S(E,V,N)## and thus obtain the Sackur-Tetrode equation?
 
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  • #2
You should rather think in terms of differentials. The "natural independent parameters" for ##S## are ##(E,V,N)##. Now you want to get another potential with the "natural independent parameters" ##(T,V,N)##. So first write down the differential ##\mathrm{d} S## in terms of ##(E,V,N)## and then think about, how to Legendre transform to a new potential with the other set of independent parameters.
 
  • #3
Thanks vanhees71 for your help 👍

I have now represented ##ds## as follows

$$ds=\frac{\partial S}{\partial E}dE+\frac{\partial S}{\partial V}dV+\frac{\partial S}{\partial N}dN$$

$$ds=\frac{1}{T}dE+\frac{P}{T}dV-\frac{\mu }{T}dN$$

Now I would just have to get rid of the ##dE## or rather I would have to express ##dE## with the help of ##dT##, ##dV##, ##dN##, right?
 
  • #4
Write it in the form of ##\mathrm{d} E=...## then find a new potential, ##F## ("free energy"), such that instead of a differnetial with ##\mathrm{d} s##, ##\mathrm{d} V## and ##\mathrm{d}N## you get one with ##\mathrm{d} T##, ##\mathrm{d}V## and ##\mathrm{d}V##. Note that
$$\mathrm{d}(Ts)=s \mathrm{d} T + T \mathrm{d} s!$$
 
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Likes GravityX
  • #5
Thanks for your help vanhees71 👍 I think I got it now 😀
 

FAQ: Apply the Legendre Transformation to the Entropy S as a function of E

What is the Legendre transformation?

The Legendre transformation is a mathematical operation used to convert one function into another. It is commonly used in thermodynamics to convert functions of extensive variables (such as energy or volume) into functions of intensive variables (such as temperature or pressure).

How is the Legendre transformation applied to entropy as a function of energy?

In thermodynamics, entropy (S) is a function of energy (E) and temperature (T). The Legendre transformation can be applied to this relationship to obtain a new function, S' as a function of T and a new variable, the conjugate of energy, which is called the thermodynamic potential or free energy (F). This is expressed as S'(T,F) = S(E) - ET.

What is the significance of applying the Legendre transformation to entropy?

Applying the Legendre transformation to entropy allows us to express the thermodynamic properties of a system in terms of intensive variables, which are often easier to measure and manipulate in experiments. It also helps in simplifying calculations and understanding the relationships between different thermodynamic quantities.

Can the Legendre transformation be applied to other thermodynamic functions?

Yes, the Legendre transformation can be applied to any function that is a function of extensive variables. For example, it can be applied to the internal energy (U) to obtain the Helmholtz free energy (A), or to the volume (V) to obtain the Gibbs free energy (G).

Are there any limitations to applying the Legendre transformation?

The Legendre transformation is only applicable to functions that are convex and have a unique global minimum. This means that it cannot be applied to functions that have multiple minima or are not continuous. It is also important to ensure that the transformed function is physically meaningful and satisfies the laws of thermodynamics.

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