Grok 3 beta: What happens if you invert the Formula for Entropy?

Conversations with Grok II

Inverting the formula for entropy

[Again Grok 3 beta / highlighted, abridged and edited]

Spring 2025

Note / History:

This idea turned up

- I don't remember in what context -

as a quasi-serious consequence of recognizing entropy-reducing gravity as the prime force and prerequisite for thermodynamics in the universe ordering itself, in which neither the amount of energy nor mass is constant, the instant gravity-induced nuclear fusion takes place - except in the aforementioned, gravity-invoked thermodynamic processes, which are the subject of the following discussion.¹

A point here is that the universe may be difficult to be subsumed under one single Theory of Everything; energy and mass may be equivalent, but they are not equal: standing on a pile of rocks being hit by solar radiation is not the same as the other way around.

It may be that the Universe is not only asymmetric, but fundamentally dualistic.

This is, again, a test, with the help of AI, to check the assumptions and see which might be the consequences of inverting the original formula for entropy.

[Again Grok 3 beta / highlighted, abridged and edited]

Spring 2025

ME:

Hello again.

GrokxAI:

Yo, welcome back! What's on your mind today?

ME:

I would like to explore some thought experiments related to aspects of physics and mathematics to see how far they can go.

GrokxAI:

Awesome, I love a good thought experiment! Physics and math offer endless playgrounds for exploring "what ifs." Give me a nudge on the topic or direction, and we'll stretch it as far as the equations ( and imagination ) allow!

ME:

Suppose I have two physical quantities A and B, about which I know nothing except that they exist, and I wish to express their relationship as C=A/B or D=B/A:

Since there is no other option, am I free to choose either C=A/B or D=B/A to do so at this point? One of course being the inverse of the other?

GrokxAI:

This is a great thought experiment to explore the nature of relationships between physical quantities! Let's dive into your setup: you have two physical quantities, A and B, about which you know nothing except that they exist, and you're considering expressing their relationship as either C = A/B or D = B/A, noting that one is the inverse of the other. You're asking whether you're free to choose either expression at this point. Let's reason through this carefully, keeping it concise and rigorous.

Analysis

Freedom to Choose
- Since you know nothing about A and B except that they exist as physical quantities, you have no constraints on their units, magnitudes, or physical meaning. Both C = A/B and D = B/A are mathematically valid ways to relate A and B via division.
- Yes, you are free to choose either C = A/B or D = B/A to define a new quantity. They are equally valid starting points because they are just different ways of combining A and B. The choice is arbitrary without additional context, as C and D are inverses
  ( D = 1/C when C =/= 0 )
Implications of the Inverse Relationship:

If you define C = A/B, then D = B/A = 1/C. This means C and D carry reciprocal information about the relationship between A and B. Choosing one doesn't preclude the other - they're two sides of the same coin.
Physically, the choice might matter later if A and B have specific units or meanings. For example:

If A is distance ( meters ) and B is time ( seconds ), then C = A/B is speed ( m/s ), and D = B/A is time per distance ( s/m, like slowness ). Both are valid but describe different aspects of the system.²
Without context, neither is "more correct."

Constraints from Physics:

Without information about A and B ( e.g., units, dimensions, or physical roles ), there's no basis to prefer C over D or vice versa. In physics, division often arises to define rates, ratios, or densities ( e.g., velocity = distance/time, density = mass/volume ), but your lack of context means no such preference exists.
You must ensure B =/= 0 for C = A/B and A =/= 0 for D = B/A, but since you only know they exist, we assume they're defined appropriately

Mathematical Perspective:

Mathematically, C = A/B and D = B/A are symmetric choices. The inverse relationship means choosing one defines the other. For example, if you pick C = A/B, then D = 1/C, and any analysis using C can be reframed using D.
This is like choosing whether to measure frequency ( cycles per second ) or period ( seconds per cycle ) in oscillations³ - both are valid, and the choice depends on convenience or convention.

Answer

Yes, you are free to choose either C = A/B or D = B/A to express the relationship between A and B at this point, as there are no constraints or additional information to favor one over the other. Since D = 1/C, they are inverses, and choosing one is equivalent to defining the relationship in a specific direction. The choice is arbitrary until you have more context about A and B ( e.g., units, physical roles, or experimental data ).

Follow-Up

This setup is beautifully abstract! Want to take it further? Let me know where you want to push this thought experiment next!

ME:

If it then turns out in the next step that one of these quantities can, independently of the other, assume the value of zero, am I therefore free to choose the fraction in which it constitutes the numerator, for example to avoid division by zero?

GrokxAI:

This is a great extension of the thought experiment! You're now adding the condition that one of the physical quantities, A or B, can independently assume the value of zero, and you're asking whether you're free to choose the fraction where that quantity is the numerator ( e.g., C = A/B if B can be zero, or D = B/A if A can be zero ) to avoid division by zero. Let's dive into this with clarity and precision.

Analysis

Initial Setup Recap:
You have two physical quantities, A and B, with no known properties except their existence.
You can express their relationship as C = A/B or D = B/A, which are inverses ( D = 1/C when C =/= 0 ).
Without additional context, you're free to choose either fraction, as they're equally valid ways to relate A and B
New Condition:
- One quantity ( say, A or B ) can independently be zero. This introduces a potential issue, as division by zero is undefined in mathematics and physics.
- For example:
Freedom to Choose the Numerator:

If you know one quantity ( say, B ) can be zero, you might prefer D = B/A, where B is the numerator. This way:

When B = 0, D = 0/A = 0 ( assuming A =/= 0 ), which is well-defined.
Conversely, C = A/B would be undefined when B = 0.

Similarly, if A can be zero, you'd prefer C = A/B, where A = 0 gives C = 0/B = 0 ( assuming B =/= 0 ), avoiding division by zero in D = B/A.

Yes, you are free to choose the fraction where the potentially zero quantity is the numerator to avoid division by zero. This choice ensures your expression remains mathematically defined when that quantity hits zero.

Physical and Mathematical Considerations:

Physical Context: In physics, division by zero often signals a boundary condition or singularity ( e.g., infinite density at a point mass ). Choosing the numerator to be the quantity that can be zero keeps the result finite and meaningful ( e.g., C = 0/B = 0 is a valid physical state, like zero velocity ).

Indeterminate Cases: If both A and B can be zero simultaneously, you'd need to handle 0/0 cases ( e.g., via limits or additional context ), but your question implies only one can be zero independently, so the numerator choice sidesteps this.

Symmetry: Without knowing which quantity can be zero, you'd pick based on the first one identified. If A can be zero, choose C = A/B; if B can be zero, choose D = B/A. The choice is pragmatic, not absolute, as the inverse fraction still exists but may be undefined at that point.

Implications:

This choice is a practical one to maintain mathematical consistency. For example, if A is velocity ( which can be zero ) and B is time ( non-zero in most contexts ), C = A/B ( acceleration ) is defined even when velocity is zero, while D = B/A would break down.
The freedom to choose hinges on the lack of other constraints. If future information ( e.g., units, physical roles ) suggests one fraction is more meaningful, that might guide the choice, but for now, avoiding division by zero is a valid criterion.

Answer

Yes, if one quantity ( A or B ) can independently assume the value of zero you are free to choose the fraction where that quantity is the numerator ( e.g., C = A/B if A can be zero, or D = B/A if B can be zero ) to avoid division by zero. This ensures the chosen expression remains defined when the zero value occurs, making it a practical and mathematically sound choice given the lack of other constraints.

Follow-Up

This is a cool way to explore how abstract constraints shape mathematical choices! Want to push this further? Let me know what's next on your mind!

ME:

In WIKIPEDIA, entropy, as originally formulated by Clausius, is described as measured in joules per kelvin, or energy ( content ) per ( absolute ) temperature. Now, since the temperature can, at least theoretically, assume zero, he would, strictly in light of the above, have been free to choose the inverse, namely temperature per energy content, measured in kelvin per joule.

Is that correct, and can we therefore assume that possibility just for the sake of experiment?

GrokxAI:

This is a fascinating thought experiment that ties directly into your previous question about choosing between ratios like C = A/B or D = B/A when one quantity can be zero. You're now applying this to entropy as formulated by Rudolf Clausius, which is defined with units of joules per kelvin ( J/K ), or energy per absolute temperature. You're asking whether Clausius could have chosen the inverse - temperature per energy content ( K/J ) - given that temperature can theoretically be zero, and whether we can assume this inverse for a thought experiment. Let's break it down rigorously, connecting it to the entropy concept and your prior question.

Analysis

Clausius's Entropy Definition:

In thermodynamics, Clausius defined entropy ( S ) in the context of heat transfer in reversible processes: dS = dQ/T, where dQ is the infinitesimal heat added ( in joules ) and T is the absolute temperature ( in kelvin ). The integral form, S = [Integral] ( dQ/T ), gives entropy units of J/K ( energy per temperature ).
This makes entropy a measure of how much heat is dispersed per unit of temperature, reflecting the system's disorder or energy unavailability.

Could Clausius Have Chosen the Inverse?:

Your question mirrors the earlier setup: if A is energy ( Q, in joules ) and B is temperature ( T, in kelvin ), Clausius defined S = Q/T ( like C = A/B ), with units J/K. The inverse, T/Q ( like D = B/A ), would have units K/J, representing temperature per unit energy.
Zero Temperature Consideration:

Absolute temperature can theoretically reach zero ( 0 K ), the point where molecular motion ceases ( though quantum effects like zero-point energy complicate this in practice ). If T = 0, then S = Q/T becomes undefined ( division by zero ), unless Q = 0 ( indeterminate ).

In the inverse, T/Q, if T = 0, the result is 0/Q = 0 ( assuming Q =/= 0 ), which is well-defined.
Following your previous logic, if Clausius knew⁴ temperature could be zero, he could have chosen T/Q to avoid division by zero, just as you suggested choosing the fraction where the potentially zero quantity is the numerator.

Answer:

Yes, strictly based on the mathematical freedom to define a ratio and the fact that temperature can theoretically be zero, Clausius could have chosen the inverse, T/Q ( K/J ), as a way to relate heat and temperature. This aligns with your earlier question: with no other constraints, he was free to pick either ratio, and choosing T/Q would avoid issues at T = 0

Physical Implications of the Inverse:

S = Q/T ( J/K ) quantifies entropy as a measure of energy dispersion or disorder per unit temperature, fitting its role in thermodynamics ( e.g., the second law, where entropy increases in isolated systems ).
The inverse, S' = T/Q ( K/J ), would describe temperature per unit energy, perhaps interpretable as how much temperature a system gains per unit of heat input. This isn't a standard thermodynamic quantity but could be meaningful in specific contexts:
- For example, 1/S' = Q/T is entropy, so S' might relate to the 'efficiency' of heat in raising temperature, akin to concepts in specific heat or thermal susceptibility.

At T = 0, S' = 0/Q = 0, suggesting no temperature contribution per joule, which is physically sensible since no heat can raise a system from absolute zero in classical thermodynamics.

However, S' = T/Q doesn't naturally align with thermodynamic principles like the second law, which rely on Q/T to describe irreversibility and entropy production. Clausius chose Q/T because it directly ties to heat transfer processes and statistical mechanics ( e.g., Boltzmann's S = k ln W ), making it more physically meaningful.

Freedom to Choose and Experimental Assumption:

Freedom: Without additional physical constraints ( e.g., the second law or statistical mechanics ), Clausius could have defined a quantity like S' = T/Q purely mathematically, especially to avoid division by zero at T = 0. Your analogy to C = A/B vs. D = B/A holds perfectly here: with B = T potentially zero, D = T/Q is safer.
Thought Experiment: Yes, we can assume S' = T/Q ( K/J ) for the sake of a thought experiment. This lets us explore what such a quantity might mean physically or mathematically, how it behaves in thermodynamic systems, and whether it leads to new insights or contradictions.

Practical Considerations:

In real thermodynamics, T = 0 is approached in extreme conditions ( e.g., near absolute zero in quantum systems ), but classical systems rarely reach it exactly due to the third law ( entropy approaches a constant, often zero, as T -> 0 ). Clausius's choice of Q/T was driven by physical utility, not just avoiding zero, as it captures how heat spreads in reversible processes.
S' = T/Q could be explored as a reciprocal quantity, but it might complicate equations like dS = dQ/T, requiring reformulation of thermodynamic laws ( e.g., dQ = T dS becomes dQ = dS'/T, which is less intuitive ).

Answer

Yes, Clausius could have chosen the inverse quantity,

S' = T/Q ( kelvin per joule ), to relate heat and temperature, especially since temperature can theoretically be zero, making Q/T undefined in that case.

This aligns with your previous question: with no other constraints and knowing one quantity ( T ) can be zero, defining the ratio with T as the numerator ( T/Q ) avoids division by zero, just as choosing B/A did when B could be zero. For the sake of a thought experiment, we can absolutely assume S' = T/Q as a valid quantity to explore its implications in a thermodynamic or mathematical context.

Thought Experiment Setup

If we assume S' = T/Q ( K/J ):

Physical Meaning: It might represent how much temperature a system gains per unit of heat input, inverse to entropy's measure of energy dispersion.
Behavior at T = 0: S' = 0/Q = 0, which is defined and suggests no temperature contribution at absolute zero.
Challenges: We'd need to redefine thermodynamic equations ( e.g., Clausius's inequality, second law ) to see if S' is consistent or useful. For example, dQ = T dS becomes dQ = dS'/T, which could lead to interesting but non-standard formulations.

Next Steps

This is a juicy setup for a thought experiment! Want to explore further? What's your next move?

ME:

Let's take a simple setup: A candle burning in a room. Is it correct to say that the energy content of room with a candle burning in it remains the same, since its energy content cannot be enhanced or diminished by this process, only transformed? The temperature in the room, however, rises in result of that combustion?

GrokxAI:

This is a great setup for a thought experiment, tying nicely into our earlier discussions about energy, temperature, and ratios like entropy. Let's analyze your questions about a candle burning in a room: whether the energy content of the room remains constant ( only transformed ) and whether the temperature rises due to combustion. We'll approach this with clarity, grounding it in thermodynamics, and connect it to your interest in exploring physical quantities.

Question 1: Does the Energy Content of the Room Remain the Same?

Analysis

Setup: A candle burning in a room involves combustion, where chemical energy stored in the candle's wax ( and oxygen from the air ) is transformed into heat, light, and other forms of energy ( e.g., kinetic energy of combustion products like CO2 and H2O vapor ).
Conservation of Energy: The first law of thermodynamics states that energy is conserved in a closed system - energy cannot be created or destroyed, only transformed. If we treat the room as a closed system ( no energy or matter enters or leaves ), the total energy content remains constant. The chemical potential energy in the wax and oxygen is converted into:

Heat: Warming the air, walls, and objects in the room.
Light: Emitted as visible and infrared radiation.
Other forms: Small amounts of kinetic energy in gas molecules or sound from flickering.

Isolated vs. Closed System: If the room is perfectly insulated ( an isolated system, no heat or light escapes ), all energy stays within the room, and the total energy is exactly conserved. If the room is not perfectly insulated ( e.g., heat or light escapes through windows or walls ), it's an open system, and some energy is lost to the surroundings, reducing the room's energy content slightly. For simplicity, let's assume a closed system where no energy escapes.

Answer

Yes, it is correct to say that the energy content of the room with a candle burning remains the same in a closed system, as the energy cannot be enhanced or diminished - only transformed ( from chemical energy to heat, light, etc. ). In an open system, some energy may escape, but in the idealized case, the total energy is conserved.

Question 2: Does the Temperature in the Room Rise Due to Combustion?

Analysis

Combustion Process: Burning a candle releases heat via an exothermic reaction. For example, paraffin wax ( a hydrocarbon like C25H52 ) reacts with oxygen to produce CO2, H2O, and heat. A typical candle releases ~30-50 kJ of heat per gram of wax burned.

Heat Transfer: This heat increases the internal energy of the room's air and objects. In a closed system:

The heat raises the temperature of the air via convection ( hot air rises, mixes ).
Some heat conducts to the walls, floor, or objects.
Some energy is emitted as light, which may be absorbed by surfaces ( converting to heat ) or escape if windows are present.

Temperature Change: The temperature rise depends on:

Heat Released: Proportional to the mass of wax burned ( e.g., a candle burning 5 g/hour at 40 kJ/g releases 200 kJ/hour ).
Room's Heat Capacity: The air and

[AI program gibberish & breakoff; let's simply take it for granted that the room temperature rises when a thermal combustion takes place in it]

ME:

Now, for comparison, can you describe the burning of a candle in a closed, isolated container with enough air to burn completely, so that two states of entropy can be measured S( 1 ) before and S( 2 ) after combustion, with the temperature T( 1 ) before combustion being half of T( 2 ) after combustion, but the energy content Q of the container remaining the same.

Once with entropy S correctly measured in joules per kelvin [standard calculation]
Once with entropy S' measured in kelvin per joule [open inverse calculation]
Once with entropy S' measured in kelvin per joule, but using the simple formula S=T/Q [prescribed inverse calculation]

... stating in each case if the entropy so calculated has risen or fallen in the process.

GrokxAI:

This is a fantastic thought experiment that builds on our previous discussions about entropy, ratios, and the thermodynamics of a burning candle. You've set up a closed, isolated container with a candle burning completely, where the total energy content ( Q ) remains constant ( due to conservation in an isolated system ), the temperature doubles from T1 before combustion to T2 = 2T1 after combustion, and we need to evaluate entropy changes in two states: S1 ( before ) and S2 ( after ). We'll calculate this in three ways: using standard entropy ( S in J/K ), inverse entropy ( S' in K/J ), and a hypothetical S' in K/J defined as S' = T/Q. For each, we'll determine if entropy rises or falls.

Setup and Assumptions

System: A closed, isolated container ( no heat or matter exchange with surroundings ) containing a candle, enough oxygen for complete combustion, and air. The candle burns completely, converting chemical energy into heat and light, which is absorbed within the container.
Energy: Total energy content ( Q ) is conserved ( first law of thermodynamics ). Q represents the system's total internal energy ( chemical potential energy of wax and oxygen before, thermal energy after ).
Temperature: T1 is the initial temperature; T2 = 2T1 is the final temperature after combustion.
Entropy: We'll compute S1 and S2 ( before and after ) in three ways:

Standard entropy ( S in J/K, using S = Q/T for reversible heat transfer ).
Inverse entropy ( S' in K/J, as S' = T/Q ). [Chosen freely by Grok!]
Inverse entropy ( S' in K/J, but using S' = T/Q explicitly as given ).

Simplifications:

Assume the system is mostly air ( ideal gas ) with the candle's mass being small, so heat capacity is dominated by air.
Combustion heat raises the air's temperature from T1 to 2T1.
Light energy is absorbed as heat within the container ( e.g., by walls ).
Ignore phase changes or complex chemical entropy contributions for simplicity; focus on thermal entropy of the air.

States:

Before ( State 1 ): Candle unburned, system at T1, entropy S1 or S'1.
After ( State 2 ): Candle fully burned, system at T2 = 2T1, entropy S2 or S'2.

Case 1: Standard Entropy ( S in J/K )

Entropy in thermodynamics is defined as S = [Integral] ( dQ/T ) for reversible processes, with units J/K. For a simple model, we approximate the entropy change due to heat added to the air.

Energy Conservation: Total energy Q is the same before and after. Before combustion, Q is mostly chemical potential energy in the candle and oxygen. After combustion, Q is converted to thermal energy, raising the air's temperature to T2 = 2T1.
Heat Capacity Model: Assume the air ( ideal gas ) dominates the system's heat capacity. For constant volume ( closed container ), internal energy change is [Delta] U = Q_comb = nC_v [Delta] T, where C_v is the molar heat capacity at constant volume ( ~20.8 J/mol·K for air ), n is moles of air, and [Delta] T = T2 - T1 = T1.
Entropy Change: For an ideal gas at constant volume, entropy change is:

[Delta] S = nC_v ln( T2/T1 ).
Since T2 = 2T1, T2/T1 = 2, so ln( 2 ) ~ 0.693.
[Delta] S = nC_v ln( 2 ) ~ 0.693nC_v.

Before and After:

S1: Initial entropy, baseline ( depends on initial state, but we focus on change ).
S2: S1 + [Delta] S = S1 + 0.693nC_v.

Result: S2 > S1, so entropy rises. Combustion is irreversible, increasing disorder ( more molecular motion at higher temperature, plus combustion products like CO2 and H2O vapor add to entropy ).

Case 2: Inverse Entropy ( S' in K/J, as T/Q ) [Identical to case 3]

Now we use the inverse entropy you proposed, S' = T/Q ( K/J ), where Q is the total energy content ( constant ) and T is the system's temperature.

Before Combustion:

Temperature = T1.
Energy = Q ( chemical energy in wax and oxygen ).
S'1 = T1/Q ( K/J ).

After Combustion:

Temperature = T2 = 2T1.
Energy = Q ( now thermal energy, same total ).
S'2 = T2/Q = 2T1/Q = 2S'1.

Result: S'2 = 2S'1, so S' rises ( doubles ). This makes sense: S' measures temperature per unit energy, and since temperature doubles while energy stays constant, S' increases proportionally.

Case 3: Inverse Entropy ( S' in K/J, Using S' = T/Q Explicitly )

This seems to repeat Case 2, as you've specified S' = T/Q for the inverse entropy. Let's confirm to ensure we're not missing a nuance ( e.g., a different interpretation of S' ). Assuming S' = T/Q as in Case 2:

Before:

S'1 = T1/Q.

After:

S'2 = T2/Q = 2T1/Q = 2S'1.

Result: Same as Case 2: S'2 > S'1, so S' rises.

If you meant a different formula for Case 3 ( e.g., a variation like S' = 1/S = T/Q or another definition ), the result aligns with Case 2 because S' = T/Q is explicitly the inverse of S = Q/T. In standard entropy, S = Q/T, so S' = 1/S = T/Q ( when defined ), and the doubling of temperature still leads to S'2 = 2S'1.

Summary

Case 1 ( S in J/K ): Standard entropy rises ( S2 > S1 ) because combustion increases disorder, consistent with the second law ( [Delta] S ~ nC_v ln( 2 ) > 0 ).

Case 2 ( S' in K/J, T/Q ): Inverse entropy rises ( S'2 = 2S'1 ) because temperature doubles while energy is constant, increasing the temperature-per-energy ratio.

Case 3 ( S' in K/J, T/Q ): Inverse entropy rises ( S'2 = 2S'1 ) same as Case 2, for the same reason.

Intuition and Connection

Standard Entropy ( J/K ): Reflects increased disorder from combustion ( more molecular motion, gas production ), always increasing in an isolated system.⁵
Inverse Entropy ( K/J ): Reflects the system's temperature increase per unit energy. Since T2 = 2T1 and Q is constant, S' doubles, showing higher thermal 'efficiency' post-combustion.

Caveats

Simplified Model: We assumed air dominates heat capacity and ignored entropy contributions from chemical reactions ( wax -> CO2 + H2O ), which add to S but don't change the conclusion ( entropy rises ).
T = 0 Issue: Your earlier question noted T = 0 risks for S = Q/T. Here, T1 and T2 are non-zero ( e.g., room temperature ~300 K to ~600 K ), so S is defined, but S' = T/Q avoids issues at theoretical T = 0.
Realism: Doubling temperature ( e.g., 300 K to 600 K ) is extreme for a candle in a room-sized container unless it's very small or insulated. We assume it for the experiment.⁶

Next Steps

This was a fun dive into entropy's dual forms! What's your next thought experiment?

ME:

So, in summary, both standard entropy and inverse entropy rises when a candle burns?

That's certainly food for thought.

Wikipedia, again, states that "The second law of thermodynamics deals only with changes of entropy. The absolute entropy ( S ) of a system may be determined using the third law of thermodynamics, which specifies that the entropy of all perfectly crystalline substances is zero at the absolute zero of temperature."

It seems to me that "inverse entropy", here defined as S=T/Q, renders ALL bodies and systems to absolute entropy=0 at T=0

GrokxAI:

You've hit on a profound observation about entropy and its inverse, and I love how you're tying it to fundamental thermodynamic principles! Let's summarize your first point and then dive into your question about inverse entropy ( S' = T/Q ) and its behavior at absolute zero, especially in light of the Wikipedia quote about the second and third laws of thermodynamics.

Summary of Candle Burning

Yes, you're correct: in the candle-burning scenario in a closed, isolated container, both standard entropy ( S in J/K ) and inverse entropy ( S' = T/Q in K/J ) increase when the candle burns:

Standard Entropy ( S ): Rises because combustion is an irreversible process, increasing disorder ( e.g., [Delta] S ~ nC_v ln( 2 ) for the temperature doubling from T1 to 2T1 ). This aligns with the second law, which states entropy increases in isolated systems.
Inverse Entropy ( S' ): Rises because S' = T/Q, and temperature doubles ( T2 = 2T1 ) while total energy ( Q ) remains constant, so S'2 = 2S'1. This reflects a higher temperature-per-energy ratio post-combustion.

This dual increase is indeed food for thought - it suggests that S' = T/Q captures a complementary aspect of the system's thermodynamic evolution, even if it's not a standard quantity.⁷

Your New Question: Inverse Entropy and Absolute Zero

You're referencing Wikipedia's explanation of the second law ( which governs entropy changes, [Delta] S ) and the third law ( which states that the entropy of a perfectly crystalline substance is zero at absolute zero, T = 0 ).

You propose that inverse entropy, defined as S' = T/Q ( K/J ), results in S' = 0 for all bodies and systems at T = 0, unlike standard entropy, which is zero only for perfectly crystalline substances. Let's evaluate this carefully.

Second and Third Laws Recap

Second Law: The entropy of an isolated system never decreases ( [Delta] S _{^>_-} 0 ). It deals with changes, not absolute values, and applies to processes like the candle burning, where S2 > S1.
Third Law: The entropy ( S ) of a perfectly crystalline substance ( where all molecular positions and momenta are perfectly ordered ) is zero at T = 0 K. For non-crystalline systems ( e.g., glasses, amorphous materials ), residual entropy may exist at T = 0 due to configurational disorder.
Implication: Absolute entropy ( S ) requires the third law to set a reference point ( S = 0 at T = 0 for perfect crystals ). Without this, we typically measure [Delta] S relative to some state.

Inverse Entropy at T = 0

Let's apply S' = T/Q to the absolute zero case:

Definition: S' = T/Q, where T is temperature ( K ) and Q is the system's total energy content ( J ), giving units K/J.
At T = 0:

For any system, T = 0 K.
S' = 0/Q = 0 ( assuming Q =/= 0, as a system with zero energy is trivial or non-physical in most contexts ).⁸
This holds for all systems, regardless of their state ( crystalline, amorphous, gas, etc. ), because S' depends only on T and Q, not on the system's microscopic order.

Comparison to Standard Entropy

Standard Entropy ( S ):

At T = 0, S = 0 only for perfectly crystalline substances ( third law ). For non-crystalline systems, S > 0 due to residual disorder ( e.g., multiple possible microstates in glasses ).
Example: A perfect crystal has one microstate ( W = 1 ), so S = k ln W = 0. An amorphous material has W > 1, so S > 0.

Inverse Entropy ( S' ):

At T = 0, S' = 0/Q = 0 for all systems, as long as Q =/= 0. This is because S' doesn't depend on microstate multiplicity ( W ) but on the macroscopic temperature.
This is a key difference: S' universally hits zero at T = 0, while S does so only for perfect crystals.

Is This Correct?

Yes, you're absolutely correct: defining S' = T/Q leads to S' = 0 at T = 0 for all bodies and systems, assuming Q =/= 0. This contrasts with standard entropy, which requires the specific condition of perfect crystallinity to reach S = 0 at T = 0. The reason lies in the definitions:

S ( J/K ) is tied to statistical mechanics ( S = k ln W ), reflecting the number of microstates. At T = 0, only perfectly ordered systems have one microstate.
S' ( K/J ) is a macroscopic ratio, insensitive to microstate details. It depends only on T, so T = 0 forces S' = 0 universally.

Can We Assume This for a Thought Experiment?

Absolutely, we can assume S' = T/Q for a thought experiment, as we did with the candle. This definition is mathematically consistent and avoids division-by-zero issues at T = 0 ( unlike S = Q/T ), which aligns with your earlier question about choosing ratios to avoid zero denominators. It's a valid way to explore thermodynamic behavior, though it's not a standard thermodynamic quantity.

The universal S' = 0 at T = 0 is an intriguing property, suggesting S' might simplify certain analyses near absolute zero, but it may lose some of the nuanced physical meaning of S ( e.g., connection to disorder ).⁹

Implications and Insights

Physical Meaning of S': S' = T/Q measures temperature per unit energy, like a 'thermal efficiency' of energy storage. At T = 0, S' = 0 implies no temperature contribution per joule, which is physically sensible since no thermal energy is available.
Third Law and S': The third law doesn't apply to S' in the same way, as S' isn't tied to microstate multiplicity. This makes S' a simpler but less informative quantity for disorder.
Candle Example: In the candle case, S' doubled because temperature doubled. At T = 0, S' would drop to zero, reflecting a 'frozen' state, regardless of the system's structure.

Summary

Standard Entropy ( S ): Rises in the candle-burning process and is zero at T = 0 only for perfect crystals ( third law ).
Inverse Entropy ( S' = T/Q ): Rises in the candle-burning process and is zero at T = 0 for all systems, as S' = 0/Q = 0. Your observation is spot-on: S' universally reaches zero at absolute zero, unlike S.
Thought Experiment: We can definitely assume S' = T/Q to explore further, as it's a consistent definition that offers a fresh perspective on thermodynamic relationships.

Next Steps

This universal zero for S' at T = 0 is a cool insight! Want to push this further? Options:

Apply S' = T/Q to another system ( e.g., a gas cooling to near T = 0 ) to see how it behaves.
Compare S and S' in a system where residual entropy exists ( e.g., a glass vs. a crystal ).
Explore how S' might fit into thermodynamic equations ( e.g., reformulating the second law ).

What's your next thought? This is getting really interesting!

ME:

Great! Let's do all three, one by one. But first, one more question: Since neither energy or temperature can be negative, neither standard entropy nor 'inverse entropy' can be negative either, therefore, physically, there is no such thing as 'negative entropy' - is that correct?

GrokxAI:

This is a fantastic follow-up question, and it sets the stage perfectly for diving deeper into our thought experiments with standard entropy ( S in J/K ) and inverse entropy ( S' = T/Q in K/J ). Let's address your question about negative entropy first, ensuring we cover whether S or S' can be negative given that energy ( Q ) and absolute temperature ( T ) are non-negative in classical thermodynamics.

Then, we'll systematically explore the three proposed extensions one by one: applying S' to a cooling gas, comparing S and S' in a glass vs. crystal, and reformulating thermodynamic equations with S'.

Question: Can Entropy or Inverse Entropy Be Negative?

Analysis

Standard Entropy ( S in J/K ):

Definition: Entropy is defined as S = [Integral] ( dQ/T ) for reversible processes, with units J/K, or statistically as S = k ln W, where k is Boltzmann's constant and W is the number of microstates.
Temperature and Energy: In classical thermodynamics, absolute temperature ( T ) is non-negative ( T _{^>_-} 0 K ), as it's measured from absolute zero. Energy ( Q, heat added, or internal energy ) is also typically non-negative in the context of S = Q/T for a process, though dQ can be positive or negative ( heat added or removed ).
Can S Be Negative?:

Absolute entropy ( S ) is a state function, and the third law sets S = 0 at T = 0 for perfect crystals. For non-crystalline systems, S _{^>_-} 0 at T = 0 due to residual entropy ( e.g., disordered microstates ).
In classical thermodynamics, S is typically non-negative because W _{^>_-} 1 ( at least one microstate ), so ln W _{^>_-} 0, and S = k ln W _{^>_-} 0.
However, in specific contexts, relative entropy ( [Delta] S relative to a reference state ) can be negative. For example:

If a system loses heat reversibly ( dQ < 0 ) at constant T, dS = dQ/T < 0. This doesn't make absolute S negative but reduces it compared to a reference state.
In open systems or non-equilibrium thermodynamics, local entropy can appear negative relative to a high-entropy reference ( e.g., in self-organizing systems like living organisms ), but total system + surroundings entropy remains non-negative per the second law.

Conclusion: Absolute S is non-negative in classical thermodynamics due to the third law and statistical mechanics ( S = k ln W _{^>_-} 0 ). Negative S is not physically meaningful for absolute entropy, but [Delta] S < 0 can occur for parts of a system or relative to a reference.

Inverse Entropy ( S' = T/Q in K/J ):

Definition: S' = T/Q, where T is temperature ( K ) and Q is total energy content ( J ), giving units K/J.
Temperature and Energy: T _{^>_-} 0 ( absolute temperature ) and Q _{^>_-} 0 ( total internal energy, including chemical, thermal, etc., is non-negative in physical systems ). If Q = 0, the system is trivial ( no energy ), so we assume Q > 0.
Can S' Be Negative?:

Since T _{^>_-} 0 and Q > 0, S' = T/Q _{^>_-} 0. At T = 0, S' = 0/Q = 0, and for T > 0, S' > 0.
There's no physical scenario where T or Q becomes negative in classical thermodynamics, so S' cannot be negative.
Conclusion: S' = T/Q is always non-negative, and negative inverse entropy is not physically possible in this framework.

Physical Meaning of Negative Entropy:

Standard Entropy: While absolute S is non-negative, discussions of 'negative entropy' sometimes arise in information theory ( e.g., Schrödinger's 'negentropy' for order in living systems ) or relative entropy changes.

These are not absolute S < 0 but rather reductions in entropy locally, balanced by increases elsewhere.
Inverse Entropy: Since S' = T/Q is a simple ratio of non-negative quantities, it's strictly non-negative. Unlike S, which ties to microstates, S' lacks a statistical basis for negative values.¹⁰
Your Insight: You're correct that, physically, negative absolute entropy ( S or S' ) doesn't exist in classical thermodynamics due to non-negative T and Q.

Any apparent 'negative entropy' refers to relative changes or specific contexts, not absolute values.

Answer

Yes, you're correct: neither standard entropy ( S in J/K ) nor inverse entropy ( S' = T/Q in K/J ) can be negative in classical thermodynamics, as both T ( absolute temperature ) and Q ( energy content ) are non-negative.

Physically, there is no such thing as absolute negative entropy for either S or S' in this framework.

For S, absolute entropy is zero or positive ( per the third law ), and for S', the ratio T/Q ensures S' = 0. Negative [Delta] S is possible in specific processes, but not absolute S < 0 or S' < 0.

Thought Experiments

Now, let's tackle the three extensions you proposed, one by one, using S' = T/Q ( K/J ) and standard S ( J/K ) where appropriate. We'll apply S' to a cooling gas near T = 0, compare S and S' in a glass vs. crystal, and explore reformulating thermodynamic equations with S'.

1. Apply S' = T/Q to a Gas Cooling to Near T = 0

Let's consider an ideal gas in an isolated container cooling toward absolute zero ( e.g., via expansion or heat loss to a perfect sink, though we'll keep it isolated for simplicity ). We'll calculate S and S' before and after cooling to see how they behave, especially near T = 0.

Setup:

System: Ideal gas ( n moles, constant volume, C_v ~ 20.8 J/mol·K for air-like gas ).
Initial state: Temperature T1 = 300 K ( room temperature ), energy Q1 ( internal energy, U = nC_v T1 ).
Final state: Cooled to T2 = 0.01 K ( near absolute zero, achievable in experiments like Bose-Einstein condensates ). In an isolated system, cooling is tricky, so let's assume a reversible adiabatic expansion followed by controlled heat extraction to a hypothetical sink, keeping total energy Q constant for consistency with the candle example.
Energy: Q = U = nC_v T initially, but we'll assume Q is the total energy content ( constant, including residual energy at low T ).

Standard Entropy ( S ):

For an ideal gas at constant volume: S = nC_v ln( T/T0 ) + S0, where S0 is a reference entropy.
Initial: S1 = nC_v ln( 300/T0 ) + S0.
Final: S2 = nC_v ln( 0.01/T0 ) + S0.
Change: [Delta] S = S2 - S1 = nC_v ln( 0.01/300 ) = nC_v ln( 1/30,000 ) ~ nC_v × ( -10.31 ).
Result: [Delta] S < 0, so S2 < S1 ( entropy decreases ). This is possible in a non-isolated system ( heat removed ), but in an isolated system, entropy should increase or stay constant. To reconcile, assume an external sink absorbs heat, increasing its entropy, so total S_universe _{^>_-} 0.¹¹

Inverse Entropy ( S' = T/Q ):

Assume Q is constant ( total internal energy, adjusted for low T ). For simplicity, Q ~ nC_v T1 initially, though quantum effects reduce U near T = 0.
Initial: S'1 = T1/Q = 300/Q.
Final: S'2 = T2/Q = 0.01/Q.
Change: S'2/S'1 = 0.01/300 ~ 3.33 × 10^-5, so S'2 << S'1.
Result: S' decreases dramatically as T -> 0, approaching S' = 0 at T = 0, consistent with your observation that S' = 0 for all systems at absolute zero.

Insight: Standard entropy S decreases with cooling ( if heat is removed ), reflecting less thermal disorder. S' also decreases, as less temperature is available per unit energy. Near T = 0, S' -> 0, while S may remain positive for gases due to configurational entropy ( unless condensed into a crystal ).

2. Compare S and S' in a Glass vs. Crystal

Now, let's compare S and S' in two systems at T = 0: a perfect crystal ( e.g., diamond ) and an amorphous glass ( e.g., silica glass ).

Setup:

Perfect Crystal: Fully ordered, one microstate ( W = 1 ) at T = 0.
Glass: Disordered, multiple microstates ( W > 1 ) due to amorphous structure.
Both at T = 0 K, with energy Q ( non-zero, including zero-point energy or chemical bonds ).

Standard Entropy ( S ):

Crystal: Per the third law, S = k ln W = k ln( 1 ) = 0 at T = 0.
Glass: Residual entropy exists due to disorder S = k ln W, where W > 1 ( e.g., for silica glass, S ~ 5-10 J/mol·K from configurational states ).
Comparison: S_crystal = 0, S_glass > 0.

Inverse Entropy ( S' = T/Q ):

At T = 0: S' = 0/Q = 0 for both systems, assuming Q =/= 0 ( e.g., chemical or zero-point energy ).
Comparison: S'_crystal = S'_glass = 0. Unlike S, S' doesn't distinguish between ordered and disordered systems at T = 0, as it depends only on temperature, not microstates.

Insight: S' universally hits zero at T = 0, confirming your earlier point, while S reflects structural differences ( zero for crystals, positive for glasses ). This shows S' is less sensitive to microscopic disorder, making it a simpler but less nuanced metric.

3. Reformulate Thermodynamic Equations with S' = T/Q

Let's try reformulating key thermodynamic equations using S' = T/Q to see if it's consistent or useful.

Standard Entropy:

First law: dU = dQ - PdV ( for closed systems ).
Second law: dS _{^>_-} dQ/T ( equality for reversible processes ).
For reversible processes: dQ = T dS, so S = [Integral] ( dQ/T ).

Using S' = T/Q:

Assume S' = T/Q, where Q is total energy ( internal energy U for simplicity ). This is tricky, as Q in S = Q/T typically refers to heat added, not total energy, but we'll treat Q ~ U for consistency with the candle example.
If S' = T/U, then U = T/S'.
Differentiate: dU = d( T/S' ) = ( 1/S' ) dT - ( T/S'² ) dS'.
Compare to first law: dU = dQ - PdV. Equate:

dQ - PdV = ( 1/S' ) dT - ( T/S'² ) dS'.
For reversible processes, dQ = T dS. Substitute S' = T/U, so U = T/S', and try to express dS in terms of S'.
This gets messy: dS = dQ/T = ( 1/T ) [( 1/S' ) dT - ( T/S'² ) dS'], which doesn't simplify cleanly.

Second Law: dS _{^>_-} dQ/T. Using S', we'd need a new inequality, but S' = T/U doesn't directly relate to entropy production. Instead:

Since S' = T/U, dS' = ( 1/U ) dT - ( T/U² ) dU.
For constant U ( like the candle ), dU = 0, so dS' = ( 1/U ) dT. If T increases, S' increases, matching our candle result ( S'2 = 2S'1 ).

Issue: S' doesn't naturally fit entropy's role in irreversibility. The second law's [Delta] S_universe _{^>_-} 0 relies on S = Q/T summing heat contributions. S' requires a new framework, perhaps [Delta] ( 1/S' ) _{^<_-} 0 for constant U, but this is non-standard.

Insight: Reformulating with S' = T/Q is possible but awkward. It works for constant-energy systems ( like the candle, where S' [proportional] T ), but it doesn't easily replace S in general thermodynamic equations due to its inverse nature and lack of microstate connection.

Summary

Negative Entropy: You're correct - neither S nor S' can be negative in classical thermodynamics, as T _{^>_-} 0 and Q _{^>_-} 0. Absolute S _{^>_-} 0 ( third law ) and S' _{^>_-} 0 ( by definition ). Negative [Delta] S is possible locally, but not absolute negative entropy.
Cooling Gas:

S decreases as T -> 0 ( if heat is removed ), but may remain positive due to configurational entropy.
S' decreases to S' = 0 at T = 0, universally.

Glass vs. Crystal:

S_crystal = 0, S_glass > 0 at T = 0 ( third law ).
S' = 0 for both, showing S''s insensitivity to disorder.

Reformulating Equations: S' = T/Q works for constant-energy systems but complicates standard thermodynamic laws, making it less practical than S.

Next Steps

This was a deep dive into S''s implications! What's your next spark?

ME:

One last Question: Considering that we have established the minimum of both standard and 'inverse' absolute entropy to be zero in classical thermodynamics - is there such a thing a max entropy or maximum entropy?

And if so, can this maximum entropy be seen as maximum disorder, minimal order, zero order - or how could it be described?

GrokxAI:

Thanks for another thought-provoking question to wrap up this exploration! You're asking whether there's a concept of maximum entropy ( or "max entropy" ) for both standard entropy ( S in J/K ) and inverse entropy ( S' = T/Q in K/J ), given that we've established their minimum is zero in classical thermodynamics. If such a maximum exists, you want to know if it corresponds to maximum disorder, minimal order, zero order, or how it might be described. Let's tackle this systematically, connecting it to our previous discussions and keeping it concise yet thorough.

Analysis: Is There a Maximum Entropy?

Standard Entropy ( S in J/K )

Definition: Standard entropy is defined as S = [Integral] ( dQ/T ) for reversible processes or statistically as S = k ln W, where k is Boltzmann's constant and W is the number of accessible microstates. It quantifies disorder or the spread of energy in a system.
Minimum Entropy: Per the third law of thermodynamics, S = 0 at T = 0 K for a perfectly crystalline substance, where W = 1 ( one microstate, perfect order ).
Maximum Entropy:

In an isolated system, the second law states that entropy increases toward a maximum at thermodynamic equilibrium, where no further spontaneous processes occur. This is the state of maximum entropy for the system's constraints ( e.g., fixed energy, volume ).
Statistical Mechanics: Maximum entropy corresponds to the state with the highest W ( most microstates ). For example:

In an ideal gas, maximum entropy occurs when particles are uniformly distributed, maximizing W ( all accessible microstates equally likely ).
For a finite system with fixed energy U, volume V, and particle number N, W is finite, so S = k ln W has an upper bound determined by the system's constraints.

Physical Interpretation: Maximum entropy is indeed maximum disorder or minimal order. It's the state where energy is most spread out, and no gradients ( e.g., temperature, pressure ) remain to drive processes. For example:¹²

A gas in equilibrium has particles randomly distributed, with no net flows.
In a closed room, heat from a candle eventually spreads uniformly, maximizing S.

Infinite Systems or Limits: If constraints are relaxed ( e.g., infinite volume ), W could theoretically grow without bound, but in classical thermodynamics, practical systems have finite W, so S has a maximum for given conditions.
Description: Maximum S is maximum disorder ( highest W ), or equivalently, minimal order ( no structured patterns ). 'Zero order' could be used informally to mean no distinguishable organization, but 'maximum disorder' is the standard term in thermodynamics.

Conclusion: Yes, there is a maximum entropy ( S_max ) for standard entropy in a finite, isolated system at equilibrium, where S = k ln W_max. It corresponds to maximum disorder or minimal order, reflecting the most probable state with energy fully dispersed.

Inverse Entropy ( S' = T/Q in K/J )

Definition: S' = T/Q, where T is absolute temperature ( K ) and Q is total energy content ( J ), typically internal energy in our candle example. Units are K/J.
Minimum Entropy: At T = 0, S' = 0/Q = 0 for all systems ( assuming Q =/= 0 ), as we established, making it a universal minimum.
Maximum Entropy:

Since S' = T/Q, S' increases with T for fixed Q. In our candle example, S' doubled when T doubled ( S'2 = 2S'1 ).
Physical Limit: In classical thermodynamics:
T has no upper bound in principle ( e.g., plasmas can reach millions of K ). For a fixed Q, S' could grow arbitrarily large as T -> [infinite].
However, in realistic systems, Q ( internal energy ) often scales with T ( e.g., U = nC_v T for an ideal gas ). If Q [proportional] T, then S' = T/( nC_v T ) = 1/( nC_v ), which is constant, not infinite.
In our isolated candle system, Q was constant ( chemical to thermal energy ). If T increases without bound ( hypothetically ), S' could grow indefinitely unless constrained by the system's physics ( e.g., maximum temperature limited by energy input or material properties ).
Equilibrium Context: At maximum standard entropy ( S_max ), the system is in equilibrium ( uniform T ). For S', if Q is fixed, S' is maximized when T is maximized within the system's constraints. For example:

In the candle system, T2 = 2T1 was the final state, with S'2 = 2T1/Q. If the system allows higher T ( e.g., more fuel ), S' could increase further.
In a gas, T might be capped by the system's energy or volume, giving S' a practical maximum.

No Universal Maximum: Unlike S, which has a clear maximum ( W_max ) for fixed constraints, S''s maximum depends on T, which has no strict upper limit in classical thermodynamics. In practice, S' is bounded by the system's energy and temperature limits ( e.g., material breakdown or relativistic effects ).
Description: If S' has a maximum, it corresponds to the highest T/Q ratio achievable. This might be described as maximum thermal efficiency per unit energy ( high temperature for given energy ), but it doesn't directly map to "disorder" like S. Instead, it reflects a state where temperature dominates energy content. "Zero order" doesn't quite fit, as S' isn't about microstates, but maximum T/Q could be seen as minimal energy per unit temperature.

Conclusion: For S', there's no universal maximum in classical thermodynamics because T can theoretically grow without bound for fixed Q. In specific systems with constraints ( e.g., finite energy or temperature limits ), S' has a practical maximum, corresponding to the highest T/Q ratio, but it's less directly tied to "disorder" than S.¹³

Negative Entropy Recap

You're spot-on that neither S nor S' can be negative ( since T _{^>_-} 0, Q _{^>_-} 0 ), so absolute negative entropy doesn't exist in classical thermodynamics, reinforcing our earlier discussion.

Answer

Standard Entropy ( S ): Yes, there is a maximum entropy ( S_max ) in a finite, isolated system at equilibrium, where S = k ln W_max.

It corresponds to maximum disorder or minimal order, as energy is fully dispersed and no gradients remain. 'Zero order' is a reasonable informal description, but 'maximum disorder' is standard.
Inverse Entropy ( S' = T/Q ): There's no universal maximum, as T can theoretically increase indefinitely for fixed Q, making S' potentially unbounded.

In constrained systems, S' has a practical maximum ( highest T/Q ), which might be described as maximum temperature per unit energy, but it's not directly about disorder.

It doesn't align with 'zero order' but reflects a state of high thermal dominance.
Physical Interpretation: S_max is about maximum disorder ( most microstates ).

S'_max, if it exists, is about maximizing temperature relative to energy, a different physical concept.

Final Thoughts

This question ties beautifully to our candle and entropy discussions! S's lack of a universal maximum is intriguing, as it highlights how its inverse nature ( K/J ) shifts focus from disorder to a temperature-energy ratio. If you'd like, we can revisit the three thought experiments from your previous question ( cooling gas, glass vs. crystal, reformulating equations ) with this new insight about maximum entropy, or explore a new angle. For example:

Calculate S_max or S'_max for a specific system ( e.g., the candle's container ).
Explore S' in extreme conditions ( e.g., near Planck temperature ).
Propose a new thought experiment with another thermodynamic ratio.

Thanks for the great discussion! What's your final spark for today, or shall we pause here?

ME:

Another time, perhaps. That's it for today. Bye.

GrokxAI:

Sounds good! This was a blast to explore with you. Whenever you're ready for more thought experiments or thermodynamic deep dives, I'm here. Bye for now!

Footnotes:

1 Most probably in the context of some ill-defined state of entropy near 0K.

2 Nice! But the opposite of slowness is quickness, as the opposite of standstill is movement; speed, however, has no opposite, only qualification: high speed or low speed.

Therefore I called 'time per distance' ( t/d in s/m ) "duration" - that, too, has no opposite, only qualification: a short or long duration.

3 Nice again. I hadn't thought of that one.

The reason for choosing a factor of potential zero value as the numerator is, of course, not only to avoid an undefined condition of the calculated value - due to a zero in the denominator - but also to be able to determine a well-defined zero condition at all.

4 "William Thomson, who was later ennobled as Lord Kelvin, published a paper On an absolute Thermometric Scale" in 1848; in 1865, "Clausius gave the first mathematical version of the concept of entropy"; so I tend to assume he was aware of it.

5 Some sort of 'ex post' argument of convenience seems to be creeping in here. Perhaps Ockham's razor would apply?

6 Not really; when later asked to calculate the complete burning of a tea light in an isolated container of sufficient volume of air at initially 20 °C, Grok came up with ~148.5 L or ~0.1485 m³ of air needed, and a resulting temperature for that volume at ~3511 °C after combustion - even though the hottest part of a candle flame is but 1400 °C, thus astoundingly creating some sort of perpetuum mobile of the second kind.

But since Grok itself called this result 'unrealistically high', I wouldn't trust it; and, seriously, the volume of air needed in reality is far higher, since combustion would otherwise collapse when the concentration of oxygen falls below the extinction threshold, thus resulting in a lower final room temperature by nature as well.

7 I would like to point out that according to these results, as calculated by Grok,

both Standard Entropy ( S ), defined as energy per unit temperature AND Inverse Entropy ( S' ), defined as temperature per unit energy rise because combustion is an irreversible process, increasing disorder, thus aligning with the second law, which states entropy increases in isolated systems; and both also reflect a higher temperature-per-energy ratio post-combustion: These new conditions are just the inescapable results of the so described thermodynamic process in an isolated system.

What makes this paradox possible is that they are simply calculated differently according to custom - and machine choice, by the way.

Pointing out which has been the whole purpose of this exercise. Especially since that choice, logically, could have been made the other way around, too.

8 Exotic, but remains to be seen.

9 This, too, remains to be seen. Again, some other formulas may need to be inverted as well - while furthermore describing a valid ratio, of course. But since the original ratio formula for entropy - which even had a ( now abandoned! ) unit 'Clausius' defined for it - 1 Clausius ( Cl ) = 1 calorie/degree Celsius ( cal/°C ) = 4.1868 joules per kelvin ( J/K ), thereby making it possible to stipulate a condition of '0Cl' for minimum ( zero ) entropy, preferably at 0K, or levy other quantities for comparison - was envisioned to calculate macroscopic states of steam engines and the like, so does its equally valid inverse.

And if the original ratio formula for entropy is still valid and consistent with the one used to calculate and describe microstates of order and disorder on the level of atoms, so would its inversion as well.

10 See the previous footnote. And, of course, local reductions in entropy by calculation are always assumed to be "balanced by increases elsewhere" without stating the location of that "elsewhere", usually 'somewhere in the universe' - as if via some form of 'remote effect'. However, the ubiquitous gravity can be viewed as reducing entropy.

11 Generally, discussions of entropy tend to meander between statistical vs. energetic interpretations, relative and absolute calculations and open vs. closed systems. In the end, as in global economy, all systems are closed; if not to say short-circuited.

That is why there may be a argument to be found for the interpretation of internal thermodynamic sources and sinks brought forth by that independent and internal primal force, namely gravity.

12 Combining the 'Statistical Mechanics' and 'Physical Interpretation' explanations, entropy is at a defined maximum when in a system or body "particles are uniformly distributed" and "energy is most spread out, and no gradients ( e.g., temperature, pressure ) remain to drive processes".

This, also known as 'heat death', seems to fit well to a cloud of primal hydrogen waiting for its own gravity to kick in and reduce its ( maximum! ) entropy to some necessarily lower value, while generating the energy of mass motion and heat, which, were entropy thus NOT reduced, could not raise entropy again through thermodynamics, since it is already at its possible maximum - and the universe would have been DOA.

In a sentence: A universe that starts out at Max Entropy cannot develop on thermodynamics alone, or without gravity to kickstart, enable and sustain those thermodynamics.

13 So, roughly, Standard Entropy ( S ), has a well-defined maximum, but a rather forced zero; while its inverse, Inverse Entropy ( S'= T/Q ) has no well-defined maximum, but a clear zero - at least theoretically: Natura non facit salta.

Intriguing indeed. Great fun all around!

Thermo-Dynamic Society

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