Legendre Transformation: Find f(T,v)

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In summary, the conversation discusses finding a Legendre transformation for the given equation and suggests looking at the thread 'why Legendre Transform' on physicsforums for further information.
  • #1
jesuslovesu
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Homework Statement



du = T ds - p dv

Find a Legendre transformation giving f(T,v)

The Attempt at a Solution


Can anyone check if this is remotely correct?

f(T,v)
[tex]df = \partial f/\partial T dT + \partial f/\partial v dv[/tex]

du = Tds - p dv
u = f - vp
d(f-vp) = Tds + v dp - p dv - v dp
df = Tds - pdv
f(T,v) = U(T) -P(v) ?
 
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  • #2
I'm not sure about the work but these are thermodynamic potentials you're working with, the one given is the internal energy, and if you scroll about halfway down to see them in differential form you'll see you should end up with H, enthalpy. Should give you something else to look for, I don't have my thermo book on me but it worked out those transforms
 
  • #3
jesuslovesu said:

Homework Statement



du = T ds - p dv
Find a Legendre transformation giving f(T,v)

Allthough its rather late in the day to answer this question, you may want to look at the thread called 'why Legendre Transform' in physicsforums.
 

1. What is a Legendre transformation?

A Legendre transformation is a mathematical operation that transforms a function of one variable into a function of a different variable. It is commonly used in thermodynamics and statistical mechanics to relate different thermodynamic potentials.

2. When is a Legendre transformation necessary?

A Legendre transformation is necessary when we want to express a function in terms of its conjugate variable, or when we want to switch between different thermodynamic potentials. It is also used in other areas of physics and mathematics, such as in Lagrangian and Hamiltonian mechanics.

3. How do you find f(T,v) using a Legendre transformation?

To find f(T,v), we first start with a function of two variables, such as the internal energy U(S,v). We then use the following formula: f(T,v) = U - TS, where S is the conjugate variable of T. This transforms the function into a new function of T and v.

4. Can a Legendre transformation be applied to any function?

No, a Legendre transformation can only be applied to functions that are convex. This means that the function must have a positive second derivative. If the function is not convex, the transformation will not be well-defined.

5. What is the physical significance of a Legendre transformation?

The physical significance of a Legendre transformation is that it allows us to switch between different thermodynamic potentials, which are useful in describing the behavior of a system. It also helps in simplifying calculations and finding relationships between different variables in a system.

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