How is the Mathematical Conversion between Δ and dX Expressed?

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Discussion Overview

The discussion revolves around the mathematical conversion between the symbols Δ (delta) and dX in the context of thermodynamics, specifically relating to the expressions for entropy and heat transfer. Participants explore the implications of these conversions in both integral and differential forms.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes the visual understanding of the integral representation of heat transfer and entropy, questioning how to express the conversion mathematically.
  • Another participant suggests that integrating dQ/T leads to dS, and thus ∫1/T dQ = ∫ dS, which can be expressed as ∫ dS = ΔS.
  • A different viewpoint emphasizes that the integral of dQ/T in a reversible system is path-independent, suggesting that this defines a differentiable function related to entropy.
  • One participant mentions that dQ is not an exact 1-form and requires an integrating factor (1/T) to become exact, referencing Caratheodory's work in thermodynamics and differential geometry.
  • There is a contention regarding the interpretation of ∫ dS, with some participants asserting it equals S, while others argue it represents ΔS, indicating a distinction between definite and indefinite integrals.
  • Participants clarify that ΔS represents the change in S between the endpoints of the integral, leading to further discussion on the nature of the integral being definite versus indefinite.

Areas of Agreement / Disagreement

Participants express differing interpretations of the relationship between ΔS and S, particularly regarding the nature of the integrals involved. There is no consensus on the mathematical expression of the conversion between delta and dX, and the discussion remains unresolved.

Contextual Notes

Participants highlight the importance of distinguishing between definite and indefinite integrals, which affects the interpretation of ΔS and S. There are also references to the need for integrating factors in certain contexts, which may not be universally accepted or understood among participants.

mraptor
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hi,

Reading a book on thermodynamics and the guy often uses something like this :

∫1/T dQ = ΔS

and then he says "this in differential form" :

dQ/T = dS

I kind of get the idea visually that one slice of "integral" will be dQ and you can think of it this way.
But my question is how do you mathematically express this conversion between delta <==> dX (both ways).
Some Examples would be nice.

thank you
 
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hi mraptor! :smile:
mraptor said:
I kind of get the idea visually that one slice of "integral" will be dQ and you can think of it this way.
But my question is how do you mathematically express this conversion between delta <==> dX (both ways).

integrate dQ/T = dS, and obviously you get …

∫1/T dQ = ∫ dS :wink:

then write ∫ dS = ∆S :smile:
 
While I do not know much Physics

The integral of dQ/t in a reversible system is independent of path. This means that the integral defines a differentiable function on the thermodynamic phase space. This function I think is called entropy and its derivative is dQ/T.

The mathematical fact that you want is that when you have an integral that is path independent - AKA conservative - starting from any point one get a well defined function by integrating along any path from the starting point to another point. You should try to prove for yourself that this function can be differentiated.
 
In other words, dQ is not an exact 1-form, it needs an integrating factor (1/T) to make it exact. Caratheodory re-wrote known (as of 1910) thermodynamics from the point of view of differential geometry (as known to him).
 
tiny-tim said:
hi mraptor! :smile:integrate dQ/T = dS, and obviously you get …

∫1/T dQ = ∫ dS :wink:

then write ∫ dS = ∆S :smile:

I thought :

∫ dS = S

not ΔS ! that is what I can't get..
and also what about the reverse conversion..
 
mraptor said:
I thought :

∫ dS = S

not ΔS ! that is what I can't get..

ΔS is the change in S between the endpoints of the integral.
 
pasmith said:
ΔS is the change in S between the endpoints of the integral.

So you are saying that he is implying that he is doing definite-integral, rather than indefinite-integral..
That is why he is getting ΔS, rather than S
 
mraptor said:
… he is doing definite-integral, rather than indefinite-integral..
That is why he is getting ΔS, rather than S

yup! :biggrin:
 

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