Entropy Confusion: Feynman's Explanation and My Question

In summary, Feynman discusses an example of putting a hot stone at temperature T_1 into cold water at temperature T_2 causing a flow of heat \Delta Q between the two. The entropy change is given to be \Delta S = \frac{\Delta Q}{T_2} - \frac{\Delta Q}{T_1}. If either of the above occurred then you would have to take these into account when calculating the entropy.
  • #36
You should have, because you are completely wrong. Find me *any* discussion of statistical mechanics in this:

In what way?

True chemical engineers also need/use mechanical engineering theory of pipes pumps and fittings, structural engineering theory of pressure vessels and structures, and so on and so forth. So there is much of this in chem eng literature.

This just proves my point about how much overlap there is between disciplines.

But all this would be at nought without the theory of the chemicals and their reactions that go into these plants.

It is often said in textbooks on physical chemistry that thermodynamics (read classical here) define what reactions are possible, but tell us nothing about the rates of these reactions. The reaction may be thermodynamically feasible, but so slow as to be unusable.

For instance glass is soluble in pure water.
The catch is that the rate of solution is measured on the geological timescale.

The mathematics of these rates is definitely the province of statistical mechanics. I amsure you will find lots of reaction rate information in the references you mention amongst others.

Physical Chemists also use a slightly different notation when they discuss classical thermodynamics - it has much to commend it.

This is simply labelling some of the variables with subscripts to indicate the conditions, so for instance rather than using

[tex]\Delta Q\quad or\quad q[/tex]

[tex]\Delta {Q_v}\quad or\quad {q_v}[/tex] or [tex]\Delta {Q_p}\quad or\quad {q_p}[/tex]

are used to indicate conditions of constant volume or pressure.
This helps ensure the appropriate equations are used in calculating quantities such as enthalpy, entropy, free energy etc.

There is another entropy thread concurrent with this one where we are working through rather better without all this squabbling.

I had though my " attitude" one of evenhandedness to both CT and SM as both have their place, both supply answers unavailable to the other and both concur where they overlap.
 
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  • #37
mr. vodka said:
Hm, I've come to the realization I probably know too little to have enough insight for decent arguments.

But Andy Resnick, let me ask you one thing. SM has a rigid definition of entropy that in essence allows you to calculate the entropy of any situation (in practice, one might be lacking knowledge/skill to actually do it). But more importantly, and basically my question, how is entropy defined in classical thermodynamics? I was taught that dS = dQ/T viewed from a reversible process, but this definition is incredibly limited, so I assume there is a more general one out there that everybody has neglected to teach me. Could you tell me?

EDIT: re-reading my post, it may sound as though I am sniping, but that wasn't meant that way, it was/is an honest question

Having questions is always good :)

Entropy is defined by the second law. That may not sound helpful, but recall the second law, properly written, is a statement on the rate of entropy *production* during a process.

SM is a theory of equilibrium states, nothing more. Thus, the "definition" of entropy in SM (k ln (W)) can only correspond to thermo*statics*. In thermodynamics, the entropy is related to the maximum amount of heat it is possible to generate during a process. Specifically,

[tex]T \dot{S} \geq Q[/tex]

It's important to recall that Q is not the 'heat' but the *rate* of heating, as seen from conservation of energy:

[tex] \dot{E} = W + Q[/tex]

Similarly, W is the *rate* of work (not the partition function). In thermostatics and SM, the time derivatives are gotten rid of, and differentials added (often times very confusingly).

The problem of thermodynamics, from this point forward, is first how to determine the constitutive functionals W, S, and F (the free energy) and second, to derive thermostatics from the thermodynamic functionals previously assigned.

Let me emphasize again that just because one has a "rigid" definition of thermostatic functionals, one *cannot* extrapolate to thermodynamics. k ln W is not a 'more fundamental' definition of entropy, since the domain of validity of SM is limited to only equilibrium states.

Does this help?
 
  • #38
Count Iblis said:
Statistical mechanics as used in practice (like evaluating partition functions or doing monte carlo simulations), may not be all that useful to engineers who use thermodynamics. But that doesn't mean that the foundations of thermodynamics lie firmly within the realm of what we in theoretical physics call "statistical mechanics".

Note that statistical mechanics can also be the study of chaos theory, far out of equilibrium phenomena etc. etc.

Look, I've tried to be as explicit as possible: SM hold for equilibrium, which is thermo*statics*. Thermo*dynamics* lies outside the domain of validity of SM. Parroting the word salads of others is not going to help you when you leave the comfort of previously worked examples and venture forth on your own.
 
  • #39
Studiot said:
In what way?

<snip>

I note that your overlong post neglected to contradict my claim.
 
  • #40
Studiot said:
In what way?
But all this would be at nought without the theory of the chemicals and their reactions that go into these plants.

It is often said in textbooks on physical chemistry that thermodynamics (read classical here) define what reactions are possible, but tell us nothing about the rates of these reactions. The reaction may be thermodynamically feasible, but so slow as to be unusable.

For instance glass is soluble in pure water.
The catch is that the rate of solution is measured on the geological timescale.

The mathematics of these rates is definitely the province of statistical mechanics. I amsure you will find lots of reaction rate information in the references you mention amongst others.

No, I do not think this is correct. Statistical mechanical models have certainly helped provide some insight into the microscopic phenomena that account for measured reaction rates, and in the very simplest cases, statistical mechanical models can provide accurate numerical estimates for reaction rates. Note however, that in order to do this, the shape of the entire potential energy surface for the reaction must be known (which is an intractable problem for systems larger than a few atoms), and even then one must assume a "quasi-equilibrium" between the reactants and the activated complexes at the tops of the reaction barriers. Microscopic theories of chemical kinetics are highly approximate in practice, and rely on empirical constants and constraints for accurate work.

The most fundamental equation in chemical kinetics, the Arrhenius equation, is an empirically derived formula, and the field remains firmly grounded in using empirical models for reaction rates. Chemical engineers certainly rely primarily on empirical data and models when determining things like how scaling up a chemical reactor will affect dynamic quantities like heat and material transport, and how those will then affect reaction rates. In almost all cases, the microscopic details of what is going on at the molecular scale is far too low-level and far too "fuzzy" to be worth considering.

In any case, these microscopic theories are not well-enough understood or developed to be useful to a chemical engineer in the first place. Try telling them, "I have a model that describes in detail the molecular scale processes that must be contributing to the large-scale process you are trying to model." They might say, "Great! How accurately can it predict changes in reaction rates over such-and-such a range of temperatures and pressures." If one then responded, "It does really well, and is never off by more than an order of magnitude or so.", then one would be lucky if all they do is laugh in one's face. For their purposes, such a large uncertainty is almost certainly completely untenable.
 
  • #41
I note that your overlong post neglected to contradict my claim.

What exactly do you mean by that?
 
  • #42
At first I thought to myself

Perhaps they teach a different brand of reaction kinetics in American universities from that put forward in Oxford and Cambridge in the Uk.

But no, wait, I find on Page 274 of what is probably the foremost American text on the subject of physical chemistry by Moore, a chapter entitled
'Collision theory of gas reactions', followed by statistical mechanics maths.

So I see all is OK and it is just that some at PhysicsForums have not read this book or its brothers.

For the record

The link between chemical kinetics and statistical mechanics is to do with the probabilities of two molecules meeting.
This must obviously happen for those two molecules to react chemically.

This application of SM is obviously different from the application Andy et al are describing, so can be expected to have a different appearance.

In particular they are not necessarily about equilibrium. SM can be and indeed is employed in non equilibrium situations.

Mix 1 mole of sodium hydroxide with 1 mole of hydrochloric acid.
You have a definite non equilibrium situation.

I wasn't thinking of the Arrhenius equation when I mentioned SM, but it is true you can get to it from there.
However this is not the fundamental equation of chemical kinetics; it merely describes the temperature dependence of what is known as the 'rate constant'. This, of course, is only useful with simple order reactions that have a single rate constant.
 
  • #43
Studiot said:
At first I thought to myself

Perhaps they teach a different brand of reaction kinetics in American universities from that put forward in Oxford and Cambridge in the Uk.

But no, wait, I find on Page 274 of what is probably the foremost American text on the subject of physical chemistry by Moore, a chapter entitled
'Collision theory of gas reactions', followed by statistical mechanics maths.

So I see all is OK and it is just that some at PhysicsForums have not read this book or its brothers.

Nothing you wrote there provides any refutation or counter-examples for anything I have written. I assure you that my understanding of chemical kinetics and SM is just fine :wink:. Go back and read my post carefully, and you will see that I was not talking about the *existence* of microscopic theories of chemical reactions, I was talking about their *accuracy* when applied to real, non-ideal experimental systems of non-trivial complexity. Everything in that section on "collision theory of gas reactions" provides a highly-idealized conceptualization of the microscopic phenomena. This is immensely useful for building an intuitive qualitative understanding of chemical reactions, but it does not provide quantitatively correct descriptions for any but the most simple reactions.

For the record

The link between chemical kinetics and statistical mechanics is to do with the probabilities of two molecules meeting.
This must obviously happen for those two molecules to react chemically.

That is not really a "link" at all ... it is as you say, obvious, and a necessary condition. However, it says nothing at all about what happens *after* the molecules come together. What is the probability of reaction? How does it depend on the quantum states and/or relative orientation of the molecules in the collision? How does the energy flow between the internal degrees of freedom of the molecules during the reaction, and how does that affect the reaction probability? Those are the phenomena we try to understand in chemical kinetics and dynamics (at least in the U.S. :wink:) ... and while they are intensely interesting (to me at least), they have not led to accurate general theories of chemical reactions, even for gas phase systems of moderate complexity.

In particular they are not necessarily about equilibrium. SM can be and indeed is employed in non equilibrium situations.

Please give an example (what you have written below does not qualify).

Mix 1 mole of sodium hydroxide with 1 mole of hydrochloric acid.
You have a definite non equilibrium situation.

Ok, now write down a detailed description of the microscopic chemical processes involved in that reaction from the first-principles equations of stat. mech. (don't forget to take into account the effects of the solvent molecules!) Then use those results to come up with a quantitative prediction of the reaction rate and compare it to what was observed experimentally. I am afraid that you will find this is simply not possible.

I wasn't thinking of the Arrhenius equation when I mentioned SM, but it is true you can get to it from there.

However this is not the fundamental equation of chemical kinetics; it merely describes the temperature dependence of what is known as the 'rate constant'. This, of course, is only useful with simple order reactions that have a single rate constant.

What does any of that have to do with our discussion? Show me any equation describing chemical kinetics that is both accurate, and derived from first-principles using only the microscopic properties of the chemical system (i.e. the potential energy surface). You will find that, where such equations exist, they are typically applicable only to fairly simple chemical systems (i.e. unimolecular dissociation of a gas phase molecule). I am not aware of any such equations that can be extended to general cases in the gas phase, or any that apply for chemical reactions in condensed phases, at least not without modification with empirically derived expressions or constants.
 
  • #44
Andy Resnick said:
Look, I've tried to be as explicit as possible: SM hold for equilibrium, which is thermo*statics*. Thermo*dynamics* lies outside the domain of validity of SM. Parroting the word salads of others is not going to help you when you leave the comfort of previously worked examples and venture forth on your own.

This is very misleading. What you wrote about entropy e.g. was wrong. In statistical mechanics we do have a definition of entropy for non-equilibrium systems. Also, note that thermodyamics is strictly spreaking always "thermostatics" as F. Reif explains in detail in his book.

The second law does not define entropy at all. Rather one has a general definition of entropy applicable for general systems that does not assume thermal equilibrium. Then the second law can be derived under some assumptions (like the equal prior probability assumption and time reversibility).

When we do thermodynamics what we actually do is approach a complicated situation as closely as we can from within thermostatics. The typical undergraduate textbook method is to consider only initial and final states which are accuratley describred by thermostatics.

What engineers do in practice when considering the actual dynamics of a system in terms of fluid velocity field, temperature and pressure distribution, is also strictly speaking approaching the real problem from within thermostatics. What happens here is that you add a lot of external variables in the description of the system. So, you just use a different coarse graining level and make statistical assumptions about those desgrees of freedom that you do not keep in your description.

Those assumptions do not have to be that it is in equilibrium, you can e.g. describe the situation using some Boltzmann distribution function whose evolution is given by a collision integral. But it should be clear that whenever you describe a system consisting of 10^23 degrees of freedom formally in terms of formulas that can be described in, say, 100 bits, you are necessarily making statistical assumptions about all those degrees of freedom.
 
  • #45
Studiot said:
What exactly do you mean by that?

You claimed "The whole of chemical reaction dynamics, and therefore chemical engineering, is built on statistical mechanics. Only the simplest reactions have linear dynamics."

I provided 3 major chemical engineering references, and SM is nowhere to be found in them. Now you provide a text on physical chemistry, and make some more claims (which I cannot substantiate; amazon does not allow me to see the table of contents). However, other texts on physical chemistry that I can see the TOC (Atkins is one, Silbey is another) again have *nothing* on SM.

Now you claim SM can be applied to nonequilibrium situations, and again, do not provide a reference. You do not see how kinetics is a linearized situation.

Just as I can't make you eat your vegetables, I can't make you learn. I can, however, try and prevent you from spreading misinformation to others.
 
  • #46
Count Iblis said:
<snip> In statistical mechanics we do have a definition of entropy for non-equilibrium systems. <snip>

I would be very interested in seeing this; I have not seen it before. Can you supply a reference?
 
  • #47
Count Iblis said:
<snip>

When we do thermodynamics what we actually do is approach a complicated situation as closely as we can from within thermostatics.

Are you kidding me? This is a joke, right?

Count Iblis said:
<snip>What engineers do in practice when considering the actual dynamics of a system in terms of fluid velocity field, temperature and pressure distribution, is also strictly speaking approaching the real problem from within thermostatics.

Can you provide a reference for this outlandish statement?

Count Iblis said:
Those assumptions do not have to be that it is in equilibrium, you can e.g. describe the situation using some Boltzmann distribution function whose evolution is given by a collision integral. But it should be clear that whenever you describe a system consisting of 10^23 degrees of freedom formally in terms of formulas that can be described in, say, 100 bits, you are necessarily making statistical assumptions about all those degrees of freedom.

You are completely changing the subject here. I never claimed one cannot describe a large number of discrete particles using statistical methods; I claim that a statistical description does *not* underlie thermodynamics. SM is *not* 'more fundamental' than thermodynamics. Thermodynamics does not 'emerge' from SM.
 
  • #48
The most fundamental equation in chemical kinetics, the Arrhenius equation, is an empirically derived formula,

I wasn't thinking of the Arrhenius equation when I mentioned SM,

What does any of that have to do with our discussion?

Please give an example (what you have written below does not qualify).

I cannot hold a rational discussion with someone who introduces something, then appears to deny it's relevance.

Or appears to deny that if you mix some of the (nearly) stongest acid with some of the (nearly) strongest alkali you have a non equilibrium situation.
 
  • #49
Just as I can't make you eat your vegetables, I can't make you learn. I can, however, try and prevent you from spreading misinformation to others.

I can't imagine why you are intent on being so downright rude, especially when most of what I say supports you.

However cast first the beam from thine own eye.

However, other texts on physical chemistry that I can see the TOC (Atkins is one, Silbey is another) again have *nothing* on SM.

1) I specified American textbooks, not UK ones.

2) Atkins is an excellent book: My copy has two chapters about Statistical Thermodynamics, Ch19 entitled the concepts and Ch20 entitled the machinery.

3) I do not know Silbey so cannot comment.

4) I did provide a gentle comment on your bibliography in reply to your question, which is more than you did for mine when I provided an example of a non equilibrium situation in the chemical reaction.

Do you deny that standard chemical reaction kinetics can be applied to this reaction?
 
  • #50
Studiot said:
I cannot hold a rational discussion with someone who introduces something, then appears to deny it's relevance.

As I explained in context when I introduced it, the Arrhenius equation was derived empirically, rather than from first principles. That was my point, and you failed to address it.

Or appears to deny that if you mix some of the (nearly) stongest acid with some of the (nearly) strongest alkali you have a non equilibrium situation.

What are you talking about? How could you possibly get that from what I wrote? Of course it's not at equilibrium! I challenged you to use stat mech to derive the rate constant.

My point is that, while chemical kinetics can be rationalized qualitatively in terms of stat mech, you can't work the other way and derive quantitatively correct expressions from first principles...empirical adjustments are required.
 
  • #51
Andy Resnick said:
Are you kidding me? This is a joke, right?



Can you provide a reference for this outlandish statement?



You are completely changing the subject here. I never claimed one cannot describe a large number of discrete particles using statistical methods; I claim that a statistical description does *not* underlie thermodynamics. SM is *not* 'more fundamental' than thermodynamics. Thermodynamics does not 'emerge' from SM.

I suspect we are using different definitions of SM, and what it means for X to be the foundation of Y. My definitions have nothing to do with how scientists and engineers work in the subject areas in practice. What I'm saying is similar to the statement that physics is the foundation of chemistry and biology. This does not imply that chemist use the Schrödinger equation to calculate the properties of molecules. The fact that they don't doesn't mean that physics is not a foundation of chemistry.
 
  • #52
SpectraCat

What are you talking about? How could you possibly get that from what I wrote? Of course it's not at equilibrium! I challenged you to use stat mech to derive the rate constant.

OK well I stated that SM methods are also applicable to non equilibrium situations and offered an example

In particular they are not necessarily about equilibrium. SM can be and indeed is employed in non equilibrium situations.

Mix 1 mole of sodium hydroxide with 1 mole of hydrochloric acid.
You have a definite non equilibrium situation

Your response was to state that my example was inadequate (did not qualify) and to ask for, I presume another one.

Please give an example (what you have written below does not qualify).

Now addressing the second part of your statement above.
Why do you keep harping on about the 'rate constant' when I presume you know that it applies strictly to equilibrium situations?
Are you actually asking me to set up and solve the relevant differential equations (which involve this constant) that do lead to the actual rate of reaction?


I wonder if, as you yourself observed so well in your post #29 that we are working on different interpretations of Statistical Mechanics?

Taking Andy’s definition form his post #31 (I agree with this)

Statistical Mechanics: (classical or quantum) mechanics of a large number of discrete particles.

I think my example qualifies as it is the average action of some 1024 molecules.

This is not about partition functions, energy surfaces or whatever – though we could discuss those.
 
  • #53
Count Iblis said:
I suspect we are using different definitions of SM, and what it means for X to be the foundation of Y. My definitions have nothing to do with how scientists and engineers work in the subject areas in practice. What I'm saying is similar to the statement that physics is the foundation of chemistry and biology. This does not imply that chemist use the Schrödinger equation to calculate the properties of molecules. The fact that they don't doesn't mean that physics is not a foundation of chemistry.

That's all well and good to have an idiosyncratic point of view...which you do not divulge except for an opaque metaphor... but it makes meaningful communication impossible.
 
  • #54
Studiot said:
SpectraCat



OK well I stated that SM methods are also applicable to non equilibrium situations and offered an example

<snip>

That was a poor example. To wit; you offered an example from *kinetics*. Kinetics is steady-state.

The reason steady-state can be modeled in SM is because it can be transformed to appear like *equilibrium*. Kinetics is a linearized theory- steady-state conditions are a linear condition.

You never responded to my example: calculate Q(t) for the OP's problem.
 
  • #55
Studiot said:
SpectraCat



OK well I stated that SM methods are also applicable to non equilibrium situations and offered an example



Your response was to state that my example was inadequate (did not qualify) and to ask for, I presume another one.



Now addressing the second part of your statement above.
Why do you keep harping on about the 'rate constant' when I presume you know that it applies strictly to equilibrium situations?
Are you actually asking me to set up and solve the relevant differential equations (which involve this constant) that do lead to the actual rate of reaction?


I wonder if, as you yourself observed so well in your post #29 that we are working on different interpretations of Statistical Mechanics?

Taking Andy’s definition form his post #31 (I agree with this)



I think my example qualifies as it is the average action of some 1024 molecules.

This is not about partition functions, energy surfaces or whatever – though we could discuss those.

You haven't said anything about "average action of molecules" until this very last post, and you still haven't said anything detailed about how your example has anything to do with the discussion at hand, which as I understand it, has two facets: first, whether or not thermodynamics can be derived from statistical mechanics alone, and second, whether or not chemical kinetics can be derived from statistical mechanics alone. I have largely stayed out of the first discussion, but I have been participating in the second.

My point all along has been that *if* statistical mechanics by itself were enough to explain all of chemical kinetics, then one should be able to take any system at any point in time, like your example of 1M HCl mixed with 1M NaOH at the moment of mixing, and using just the molecular-scale description of the system, write down equations for the time-evolution of the chemical species in the system so that its state at any other time could be accurately predicted, including any environmental effects such as changes in temperature, solvent composition, ionic strength, etc. To my knowledge, this cannot be done in the general case, starting from statistical mechanics. Of course I don't want you to actually do this, I just wanted to illustrate the difficulty (impossibility) of the task.

Your point is taken with regard to my use of the term "rate constant" in the second post .. notice that in my original post, I asked you to describe the "reaction rate", which is what I meant in my second post as well.

I don't think this has anything to do with us having "different definitions of SM" ... I too basically agree with what Andy wrote in post #31, although I would also stipulate that SM as applied often necessarily involves significant simplifications/idealizations of the underlying molecular physics (e.g. approximation of molecular vibrations as uncoupled harmonic oscillators), which can lead to significant deviations with respect to real systems.
 
  • #56
You haven't said anything about "average action of molecules" until this very last post

My understanding of the average action of a large numebr of particles or molecules is that this is what is meant by a statistical approach.

and you still haven't said anything detailed about how your example has anything to do with the discussion at hand, which as I understand it, has two facets: first, whether or not thermodynamics can be derived from statistical mechanics alone, and second, whether or not chemical kinetics can be derived from statistical mechanics alone

No indeed I didn't say much about the CT v SM debate. I felt others had already said more than enough about that issue already.

In fact I watched this thread for a while before saying anything at all. When I did, others were already discussing the original issue ( which was neither of the above) in minute detail. My aim was to extend the horizons somewhat and point out there are wider aspects, although remaining relevant to the original OP. I was not intending to do his homework for him.

I agree my claim about chemical engineering to be rather grandiloquent, but it was only a throwaway statement tacked on the end of something solid, not a bone to be latched onto.

That was a poor example. To wit; you offered an example from *kinetics*. Kinetics is steady-state.

The reason steady-state can be modeled in SM is because it can be transformed to appear like *equilibrium*. Kinetics is a linearized theory- steady-state conditions are a linear condition.

I really have no idea what you mean by this so would be grateful for an explanation.

In my book chemical kinetics is a single or system of differential equations of rate of change of concentration v time. These may be linear or non linear; obviously linear is preferable but many have to be solved numerically. these are all derived from the statistical observation that the rate of reaction (=rate of change of concentration) is proportional to some function of the concentration, linear if you are lucky.

However I don't see whether being linear or not has any bearing on the state of equilibrium or non equilibrium. I notice this 'requirement' has been dropped for SM but I don't understand what you mean by 'steady state' ?
 
  • #57
Enough. Both of you (Studiot and Count Iblis) have oscillated between making grandiose claims (e.g., SM is a more fundamental theory than thermodynamics) while offering the most threadbare meager evidence (10% of a single book, kinetic theory, silence), and constantly switching the topic around.

I don't have a problem with students that do not know things; I do have a problem with your apparent inability to google "non equlibrium statistical mechanics". If you did, you would see that there is quite a lot of research on the subject- did you think I am the first person to point out deficiencies in SM?- and honestly, I would have been thrilled to have a discussion about mode coupling theory, the Fokker-Planck equation, Smoluchouski equation, and the fluctuation-dissipation theorem.

Neither of you have done any effort to learn anything new. All I see is the constant re-packaging of dried-up undergraduate elementary concepts held up as exemplars of physical theory. This is, frankly, an insult to those of us who actually do research.

I do not have the time or effort to teach the willfully ignorant. Maybe after you have done some learning we can try again.
 
  • #58
Andy Resnick said:
I haven't given anyone a chance to respond, but I feel like I have been having this argument for over a month and not getting anywhere.

So, here's a simple challenge: solve this, and I will reconsider my earlier claims.

Using SM, solve either Q(t) or T(t) in the OP's question. If SM is truly a more fundamental theory than thermodynamics, this should be trivial.

Perhaps you should specify what exactly you have in mind when you say "fundamental", because I get the impression it is not what everyone else means when they say fundamental. It seems that you feel that a theory is more fundamental if it is more useful; this is not generally what people have in mind when they use that word.

Generally, a "fundamental theory" is taken to mean the "more real theory", in the sense that it is a more accurate description of Nature in principle. This does not necessarily mean it makes more accurate predictions. String Theory, were it to somehow be verified, would be considered the most fundamental theory, but no one would every dream of trying to calculate the trajectory of a baseball using it.

SM is considered more fundamental than Thermodynamics because it starts from a microscopic perspective and works up to the macroscopic quantities. Thermodynamics, on the other hand, is phenomenological, and having disposed of the need to have a microscopic description it is able to describe phenomena for which the microscopic picture is either unknown or exceedingly complicated. Similarly, Phenomenological Landau Theory is often more useful than starting with a microscopic Hamiltonian, but it wouldn't generally be considered "more fundamental".

So, if SM cannot be used to calculate your requested quantities, that says nothing of its "fundamental-ness" in the usual use of the term. So, what exactly do you mean by "fundamental"?
 
  • #59
This will be the second time I have quoted the Bible in this thread.

Despite suffering rudeness and dismissive arrogance by the Science Advisor I have decided to turn the other cheek and walk away, as advised in the introductory sticky notices.
 
  • #60
Studiot said:
This will be the second time I have quoted the Bible in this thread.

Despite suffering rudeness and dismissive arrogance by the Science Advisor I have decided to turn the other cheek and walk away, as advised in the introductory sticky notices.

Yes, and also the entropy in this thread is just too high to have a meaningful discussion. And now that Mute has backed my general position, I think this is a good point to stop this discussion in this thread. Perhaps we can start another thread about the foundations of thermodynamics, statistical mechanics etc. etc.
 
<h2>1. What is entropy confusion?</h2><p>Entropy confusion refers to the misunderstanding and misinterpretation of the concept of entropy, particularly in relation to the second law of thermodynamics. It is often attributed to physicist Richard Feynman's explanation of entropy, which has been misinterpreted by many.</p><h2>2. What is Feynman's explanation of entropy?</h2><p>Feynman's explanation of entropy is that it is a measure of the amount of disorder or randomness in a system. He used the example of a jigsaw puzzle to illustrate this concept, stating that there are many more ways for the pieces to be disordered than for them to be ordered, and therefore, a disordered state is more likely.</p><h2>3. What is the second law of thermodynamics?</h2><p>The second law of thermodynamics states that the total entropy of a closed system will never decrease over time. In other words, the disorder or randomness of a closed system will always increase or remain the same.</p><h2>4. How is entropy related to the second law of thermodynamics?</h2><p>Entropy is directly related to the second law of thermodynamics because it is a measure of the disorder or randomness in a system, and the second law states that this disorder will always increase or remain the same. This means that the entropy of a closed system will always increase or remain the same over time.</p><h2>5. What is the significance of understanding entropy confusion?</h2><p>Understanding entropy confusion is important because it allows us to have a better understanding of the second law of thermodynamics and its implications. It also helps us to avoid common misconceptions and misinterpretations of this concept, which can lead to incorrect conclusions and misunderstandings in various fields of science.</p>

1. What is entropy confusion?

Entropy confusion refers to the misunderstanding and misinterpretation of the concept of entropy, particularly in relation to the second law of thermodynamics. It is often attributed to physicist Richard Feynman's explanation of entropy, which has been misinterpreted by many.

2. What is Feynman's explanation of entropy?

Feynman's explanation of entropy is that it is a measure of the amount of disorder or randomness in a system. He used the example of a jigsaw puzzle to illustrate this concept, stating that there are many more ways for the pieces to be disordered than for them to be ordered, and therefore, a disordered state is more likely.

3. What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will never decrease over time. In other words, the disorder or randomness of a closed system will always increase or remain the same.

4. How is entropy related to the second law of thermodynamics?

Entropy is directly related to the second law of thermodynamics because it is a measure of the disorder or randomness in a system, and the second law states that this disorder will always increase or remain the same. This means that the entropy of a closed system will always increase or remain the same over time.

5. What is the significance of understanding entropy confusion?

Understanding entropy confusion is important because it allows us to have a better understanding of the second law of thermodynamics and its implications. It also helps us to avoid common misconceptions and misinterpretations of this concept, which can lead to incorrect conclusions and misunderstandings in various fields of science.

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