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Should entropy and temperature be dual?
In Classical Continuum Mechanics, the entropy(s) and temperature(T) are considered to be dual in the second thermodynamics principle,
by define a relation between internal energy(e) and Helmhotz free energy(ψ): ψ = e - s * T . As you can find it in many CM books, such as [2008][J.N.Reddy]Introduction to Continuum Mechanics.
Here is my questions:
1. Is the transformation between internal energy and Helmhotz free energy a Legendre Transform?
2. From aspects of thermodynamics or heat transfer, what's the proof of keeping this duality?
3. For constitutive models considering effects of high temperature and gradient temperature, should we remain the duality?
Any comments are welcome!
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Re: Should entropy and temperature be dual?
Take a look at Krishna Garikipati's lecture series, particularly https://www.youtube.com/watch?v=xROHGSp3088 and some of the adjacent lectures.
-- Biswajit
Remarkable explanation!
Thanks Biswajit,
You suggested a good lecture video.
But I was still wondering that how do we know the slope/derivative, internal energy and entropy, will equal to the temperature?
Re: Temperature and entropy
For equilibrium thermodynamics, no matter how high the temperature, standard arguments apply. See for example http://www.feynmanlectures.caltech.edu/I_44.html, http://www.feynmanlectures.caltech.edu/I_45.html, and http://www.feynmanlectures.caltech.edu/I_46.html.
For systems that are not close to equilibrium, see http://www.physik.uni-augsburg.de/theo1/hanggi/Casas.pdf. I'm by no means an expert on the topic but others on iMechanica might be.
-- Biswajit
entropy
Why does the derivative of the internal energy with respect to the entropy equal to the temperature?
Don't we usually define that the change of entropy equals to the heat over the temperature? ΔS = Q_rev / T
Re: dU/dS == T ?
To quote Feyman's lectures: "The heat Q put into the system, plus the work W done on the system, is the increase in the energy U of the system; the latter energy is sometimes called the internal energy". This comes from the conservation of energy.
dU = dQ + dW
From 2nd law, dQ = T dS
=> dU = T dS + dW
dW is mechanical work and is typically assumed not to be affected by a change in entropy, i.e., dW/dS = 0
Take the limit as dS ->0 and you get the relation that's normally used. (d means an increment)
-- Biswajit
Hi Biswajit,
Hi Biswajit,
Thanks a lot for your explanation and online book. I think you're right.
So the answers to my three questions will be:
1. Yes, it is a Legendre Transform; to transfer variable entropy in internal energy to temperature in Helmhotz free energy.
2. If the mechanical work is typically not be affected by a change in entropy; just as you specified.
3. For thermodynamics, no matter the temperature or gradient of temperature will affect the duality, as long as 2 is satisfied.