Talk:Inexact differential

message: What *is* an inexact differential. I believe I was the person that put that redlink up a long while ago in Entropy (thermodynamic views), and I had hoped that someone would fill that space with an explanation. I still have no idea what the inexact differential symbol δ *means*. Does anyone know? Fresheneesz 09:34, 5 May 2006 (UTC)
replies: text: Yes, loosely it means that the function may not exist; also in thermodynamics it means that that function is not a state function and that it is path dependent.  This is just from memory, I may be off on a few technical points.  We are working to build the inexact differential article.  Help if you want.  Thanks:--Sadi Carnot 13:54, 25 July 2006 (UTC)
thread_title: Inexact differential?

message: Please do not attmept to merge this article into exact differential.  These are seperate but related terms.  See: Talk:First law of thermodynamics for example.  Thanks:--Sadi Carnot 05:48, 25 July 2006 (UTC)
thread_title: No merging

message: I'm not sure how to make a re-direct, so could someone make a redirect from imperfect differential to here?  Over in this locale we call it an "imperfect differential".  Fephisto 21:36, 24 August 2006 (UTC)
replies: text: to make a redirect, create a new page, in which you type #REDIRECT inexact differential
. Mct mht 14:08, 26 August 2006 (UTC)
thread_title: Imperfect Differential

message: it says: The symbol Inexact differential symbol.PNG (d-bar), or δ (in the modern sense)
what is the diff between the two? What is meant by "in the modern sense"? I was taught the d-bar symbol in my physical chemistry course...  —Preceding unsigned comment added by Mikejones2255 (talk • contribs) 03:34, 17 December 2009 (UTC)
thread_title:  d-bar or delta? 

message: I have (almost) rewritten the article to make it more accessible to non-engineers (I am one myself though).
I have shortened the description part (it was way too long and thereby confusing). Also the part with integrals was quite confusing:
:\int_a^b \delta f \ne f(b) - f(a),
This expression really does not make sense ("it is not even wrong"). Assuming that a and b are the initial and final states of the system, the point of the inexact differential is that you cannot get a unique primitive f. Rewriting in variables of the first principle of thermodynamics:
:Q \ne \int_{U_1}^{U_2} \delta Q
The reason this does not make any sense is that we cannot figure out how Q changes with the state U without knowing W. You can phrase it like there is an unsaturated degree of freedom, I suppose.
This concept is really simple, but it seems that thermodynamics textbooks are written by some scribes that love keeping everything difficult to understand. 193.175.53.21 (talk) 11:11, 4 August 2010 (UTC)
thread_title:  Rewrite 

message: I don't see the relevance of the Vanuatuan movie to this topic. It seems nothing more than a slightly comical (lewd) video, with a post-hoc rationalisation.  — Preceding unsigned comment added by 129.67.105.62 (talk) 13:18, 9 June 2011 (UTC)
thread_title:  Vanuatuan? 

message: This article looks weird. What is the exact mathematical definition of inexact differentia? ––<Span xml:lang="zh-Hant-CN" lang="zh-Hant">虞海</Span> (Yú Hǎi) <Big>✍</Big> 17:45, 17 January 2012 (UTC)
replies: text: Let me have another go at explaining it. In thermodynamics, U is a scalar potential, while Q and W are not. This means that any vector field that is a gradient of U is a conservative vector field, while vector fields that are gradients of Q and W are not conservative. When we call differentials of Q and W inexact, and use a \delta or đ symbol for them in place of a d, this expresses the fact that any integral of them will be path dependent because their vector fields are not conservative. Dezaxa (talk) 05:34, 15 April 2014 (UTC)
replies: text: This isn't quite right.  You can't say that "vector fields that are gradients of Q and W are not conservative," because all vector fields that are gradients of anything are conservative.  The point is that it doesn't even make sense to talk about taking the gradient of Q or W.  And the real point is that it doesn't make sense to talk about differential amounts of heat or work either.  So to answer your question, Yu Hai, there is no exact mathematical definition of "inexact differential" (pun intended?), the people who developed thermodynamics invented their own version of mathematics which you can convince yourself kind of makes sense as long as you don't think too hard about it.  Ryan ChemE (talk) 17:20, 4 February 2015 (UTC)
thread_title:  Definition? 

message: It seems to me that the temperature as an integrating factor is a neutral factor in this business. What makes the entropy an exact differential is the specification of the path of integration: a reversible path (Q_rev). With the specification of a path all inexact differentials become exact. Am I totally missing the point or am I right? 109.93.162.246 (talk) 11:31, 4 January 2014 (UTC)
replies: text: I think you are right. While \delta Q is inexact, dQ_\text{rev} is exact, because a reversible path has been specified. Dezaxa (talk) 05:47, 15 April 2014 (UTC)
thread_title:  Integrating Factor ? 

message: There are a few issues with the "Heat and work" section under "Examples."  The first sentence, that a fire requires heat, fuel, and an oxidizing agent, is more related to the "fire triangle" concept in fire safety, and I think is off topic here.  Also the discussion regarding activation energy is misleading.  The activation energy of a reaction has nothing to do with energy flows into or out of the system.  Activation energy simply allows one to relate the speed of a reaction with the system's temperature, and any reaction will occur (however slowly) at any finite temperature, so long as it is thermodynamically favorable.  Heat is not consumed in a reaction to overcome the activation energy, it would only be transferred to maintain the temperature of a reaction with a positive enthalpy change of reaction.  The reason you need to add heat (or work to generate friction) is because you need to raise the temperature high enough to allow the reaction to proceed at a high enough rate to sustain itself.
The physical law relating reactions and activation energy to temperature is the Arrhenius equation: k=Ae^{-E_a/RT} where k is the reaction rate constant.  Finally, the last sentence about integration needing to account for the path taken - what integration?  What is being integrated here, and with respect to what? Ryan ChemE (talk) 14:00, 27 June 2015 (UTC)
thread_title:  Problems with heat and work example