1. Entropy ： A concept that is not Physical Quantity shufeng-zhang Email: uhsgnahz@126.com College of Physics, Central-South University, China Abstract This paper demonstrates that entropy is not a physical quantity. We define heat engine efficiency η as: η= W/W 1, that is, replacing Q 1 in the original definition η=W/Q 1 with W 1, W still is the net work of the heat engine applied to the outside in one cycle, W 1 is the work the heat engine applied to the outside in the cycle, then, we use Stirling cycle as the element reversible cycle, if ∮ dQ/T =0 is tenable, we can prove ∮ dW/T =0 and ∮ dE/T =0. If the formula ∮ dQ/T=0, ∮ dW/T=0 and ∮ dE/T=0 can really define new system state variables, it should come to the absurd result of such a definition. In fact, during the process of obtaining entropy, ∑[(ΔQ)/T)] becoming ∫dQ/T is untenable, therefore, the formula ∮ dQ/T=0, ∮ dW/T=0 and ∮ dE/T=0 are all untenable. The entropy defined by Boltzmann is used to interpret “entropy” by Clausius, so, it is at the same time denied. Keywords ： physics, thermodynamics, statistical physics, entropy PACS numbers: 05.70.-a, 05.70.ce I.Introduction What is entropy ? This is a controversial question debated for more than 100 years. In 1865, Mr.Clausius, based on the result  dQ / T  0 in any reversible cycling process of arbitrary thermodynamic system, concluded that there existed a new status quantity entropy ( remarked as S). And people believed that the entropy difference between any two arbitrary balanced states was ： 2 1 moreover, only this differential value can be used for calculation in thermodynamics. Mr. Clausius also put forward the well-known entropy increase law accordingly. In 1872 Boltzmann put forward the formula of absolute entropy S  k ln , of which, k was Boltzmann’s constant and Ω is the probability of thermodynamics. He held that entropy was the degree of system disorder or the measuring symbol of consequence. This was considered as the best interpretation of entropy which is still been used by people today. The above conclusion is still broadly accepted and applied, and the above contents can be found in any text books of thermodynamics and statistical physics. Entropy has been regarded as an important physical quantity and broadly used by people though people can not ascertain what entropy is. There exist many unsolved problems or contradictions in the above conclusion, this predicts that there must be something wrong with entropy. II.Entropy is not a physical quantity 2.1 Origin of Entropy To demonstrate entropy being not a physical quantity, we first review the origin of entropy. We define the efficiency of the heat engine as   W / Q 1, namely taking the ratio of the net work W done outside by the heat engine in circulation and the heat quantity Q 1 absorbed by the system from outside as efficiency of the heat Email: uhsgnahz@126.com -1-

2. Q1Q1 Q 2 engine, then get the Carnot cycle  W / Q 1  1 , here η has no relation with the working medium but only has relation with the temperature of the two constant heat sources, and then in view of the above, the thermodynamics temperature scale θ is defined as :  2 /  1  Q 2 / Q 1 ; when working medium of the system is ideal gas, it can prove ： Q 2 / Q 1  T 2 / T 1, namely  2 /  1  T 2 / T 1 still use sign T to express thermodynamics temperature scale θ, namely Q 2 / Q 1  T 2 / T 1  Q 1 / T 1  Q 2 / T 2  0 ， here Q2Q2 means release of heat and is in itself a negative value. according to this, use infinite Carnot cycles to approximate and replace arbitrary reversible cycles,then, get dQ / T  0 ， therfore, people believe that dQ/T is a complete differential and 2.2Entropy is not a physical quantity Entropy originated from dQ / T  0, therefore, to prove entropy is not a physical quantity, it is necessary and it is only necessary to prove that dQ / T  0 can not define physical quantity or dQ / T  0, itself, is untenable. As we know that  dQ / T  0 originated from Q 2 / Q 1  T 2 / T 1 in Carnot cycles, and this is a way to define thermodynamics temperature scale, its existing base is the combination of heat engine efficiency definition mode and Carnot cycles. It should be known that the heat engine efficiency formula is a definition mode and Carnot cycle, comparing with other cycles, has only difference in form and should not occupy positions higher than other cycles, its function to define thermodynamics temperature scale is not unique. Let us prove that dQ / T  0 can not define physical quantity and itself is untenabal as follows. Above all, redefining the heat engine efficiency. Because heat engine efficiency is significant only to observers, how we define heat engine efficiency has nothing to do with the objective process of the heat engine system, therefore we can reasonably define heat engine efficiency according to other mode. Now we redefine heat engine efficiency as: ratio of the net work done to outside by heat engine system in one cycle to the work done to outside by heat engine system, namely WW1WW1  ……… (2.2.1) That is to use W 1 done to outside by heat engine system in cycle to replace Q 1 heat quantity absorbed by system from outside in original definition  W / Q 1, because W 1 done to outside by system in cycle can not totally be transferred into net W, this is just as heat quantity absorbed by system from outside can not totally be used to do net work in cycle–statement of Kelvin of the Second Law. It is obvious that this two kind of definitions are equal rights. We provide another statement of the Second Law: there can not be such a heat engine by which all work done to outside can be transferred into net work in cycle process. Obviously the formulation is equivalent to Kelvin’s formulation. There exists a heat engine that uses certain work medium to do work W 1 to outside in one cycle, and the outside does work W 2 towards the system, and the system is restored, thus get W  W 1  W 2. From (2.2.1 ), we can determine: -2-

3. W2W1W2W1 WW1WW1  1   ……… (2.2.2) Using the reversible cycle (stirling cycle) shown in Figure 1. as element cycle, which plays the role of Carnot cycle in the process of reducing dQ / T  0. P `` Figure 1.TV(stirling) cycle abcda is composed of two reversible constant volume processes bc and da and two reversible constant temperature processes ab and cd. Here, we define the heat sources of system energy exchange in constant temperature process as work source so as to understand the following contents, name in short such cycles (stirling cycle) as TV cycles, and name the doing TV cycling heat engines as TV engines. We prove as follows: The efficiencies of all reversible engines (namely TV engines) that work between the two constant temperature work sources are equal, and the efficiency of irreversible engine is less than that of reversible engine. Take any of two reversible engine E and E ＇, and make them to work between constant work sources θ 1 and θ 2, they are inevitably TV engines, their working medium is arbitrary; Use θ 1 and θ 2 express respectively the temperatures of high temperature work source and low temperature work source, θ 1 ＞ θ 2 ; here θ can take any thermodynamics temperature scale. Let E and E ＇ do equal net work towards outside in one cycle, as ΔW 1 and ΔW 2, ΔW 1 =ΔW 2 = W, this can always be done (similar to the situation of Carnot cycle in textbooks), use W 1 and W 1 ＇ to express the work done towards outside by E and E ＇ in one cycle, and W 2 and W 2 ＇ to express the work done by the outside towards E and E ＇ in the cycle, η and η ＇ to express the efficiency of E and E ＇, first prove η=η ＇, here we adopt contradiction proof, Supposeη ＇ > η Because E and E ＇ are both reversible, therefore E can be made to operate reversibly, that is to make E do work W 2 towards outside, and the outside does work W 1 towards E, the net work done by the outside is W=W 1 - W 2, W is provided by engine E ＇ which operates positively and the heat quantity ΔQ=W(=ΔW 1 =ΔW 2 ) absorbed by engine E ＇ in cycling is provided by engine E, then get WW1WW1 W W 1 '  W 1  W 1 '  and because then get W 2 = W 1 - W W 2 ＇ = W 1 ＇ - W W 2 ＞ W 2 ＇ -3- a b dcVdcV

4.   PdV P ' dV ' -4- 2 to θ Combine E ＇ and the counter-operating E into one heat engine, after they cycle together one time, the system is restored, the only result is that the system absorbs work Δ W= W 2 - W 2 ＇ from the low temperature work source (heat source θ 2 ) and automatically does work towards the high temperature work source ( heat source θ 1 ) Δ W= W 1 - W 1 ＇ = W 2 - W 2 ＇, this means that there is heat quantity that equals to Δ W= W 2 - W 2 ＇ = W 1 - W 1 ＇ automatically transmitting from the low temperature work source θ （ heat source ） the high temperature work source （ heat source ） 1, This directly contradicts Clausius formulation of Second Law, that is to say that η ＇ > η is untenable. The same, make engine E ＇ operates reversibly, it is also proved that η > η ＇ is untenable, thus, there inevitably occurs η = η ＇ ……… (2.2.3) If E ＇ is an irreversible engine, it is not a TV engine, then E ＇ can not be made to operate reversibly, from here, following will inevitably occurs η ＇ ≤ η ……… ( 2.2.4) Because there is already a reversible engine E ＇, and after this heat engine cycling one time together with another counter-operating reversible engine E, they can make the system and the outside restore, therefore, if E ＇ is an irreversible engine, that means the equality sign in η ＇ ≤η is untenable, because, if it is η=η ＇, then it is obvious that the counter-operating E and the forward operating E ＇ cycling together for one time can make the system and the outside completely restore, in that case, E ＇ can only be a reversible engine which contradicts with E ＇ being an irreversible engine, therefore, if E ＇ is irreversible engine, it must inevitably have η ＇＜ η ……… ( 2.2.5) Thus proving that under the definition of formula (2.2.1), the efficiencies of all reversible engines (TV engines) which working between two constant temperature sources are equal, and that the efficiencies of irreversible engines are smaller than that of reversible engines, this result have nothing to do with working media. Because the efficiency of TV engines has nothing to do with the working media, therefore the thermodynamics temperature scale (absolute thermometric scale) can be defined as W 1 W 2   1  2 ……… (2.2.6) That is to say that the ratio of the two thermodynamic temperatures is the ratio of the work W 1 done by the TV engine that work between the two temperature work sources (heat source) to outside. and the work W 2 that the outside did to the TV engine. When working medium is ideal gas and the system do TV cycles, the following formula is obtained: T 2 T 1 V 2 V 1 V 2 V 1 W 2 W 1 V 1 V 2 V 1 V 2  1   RT 2 ln  RT 1 ln  1  ………(2.2.7) Comparing (2.2.6) with （ 2.2.7), it is known that using ideal gas can get  2 /  1  T 2 / T 1, namely that the thermodynamics temperature scale defined by (2.2.6) and the thermodynamics temperature scale defined by  2 /  1  Q 2 / Q 1 is of equal to each other. Due to the convention, we still use sign T to express thermodynamics temperature scale in the following, that is: W 1 W 2 T 1 T 2  ……………(2.2.8)

5.  T   T  0 -5- from formula (2.2.8), we get  W 2 T 2 W 1 T 1  0 ……………(2.2.9) As implied, W 2 means the work done by the outside towards the system and is negative value. From here we can know that when any system making TV cycles, the sum of the ratio of the work (positive or negative) exchanged by the system with each work source (heat source) and the thermodynamics temperature scale of the work source (heat source) equals zero. This process is entirely the same as the process of  dQ / T  0, use a serial element TV cycles to separate and replace any reversible cycles of the system, please see Figure 2. P V Figure 2A serial TV cycle to replace any reversible cycling process Since in two constant volume process of TV cycling, there is always ΔW=0, namely dW=0, when element process is infinite, namely when the system exchanges work (equivalent heat quantity) with infinite work sources (heat sources), there occurs  dW T  0 …………… (2.2.10) The same, this result has nothing to do with the working medium. Obviously, to irreversible cycle, formula  dW T ＜ 0 can be obtained. We obtain the conclusion  dW / T  0 which paralleling with  dQ / T  0, and using the first law, we can also get formula dE  dQ  dW, and making reversible cycle of any thermodynamic systems and get   dQ dW dE T For over one hundred years, people believed that  dQ / T  0 defined a system state variable, namely entropy, if it is right, then,  0  dW T and  0  dE T

6.   1 T   1 T  would inevitably defined new system state variable, moreover, such as in the process of system reversible heat isolation, when the system from balance state 1 reaches another balance state 2  2121 2121 dW T dQ T ……………(2.2.11) 2 dQ 2 dE ……………(2.2.12) In the reversible constant volume process, from balance state 3 reaches another balance state 4  4343 4343 dE T dW T ……………(2.2.13) From formula (2.2.11)~(2.2.13), we know that if  dQ / T  0 ﹑  dW / T  0 and  dE / T  0 have defined new state variables of the system, then they are inevitably different from each other, but their dimensions are the same, all are J/K, there must be also that a system state variable concurrently have different numerical values, or there concurrently exist three different system state variables with the same unit in the system, although there can concurrently exist different system state variable with the same unit in one system, but one S has already made people unaware of what it is, now we have to define three ， this obviously should be absurd. Therefore, the conclusion is that dQ / T  0 should be a wrong result. The same,  dW / T  0 and III.  dE / T  0  dQ / T  0 should be wrong results either. is gained based on a calculus error From the above-mentioned, we can know that Textbooks stress that not use mathematics to reduce out  dQ / T  0 ， but if there is really a result of  dQ / T  0, then it inevitably defined a system state variable, and the paper has already demonstrate the absurdity of using  dQ / T  0 to define physical quantity, this means that there should not exist such conclusion as  dQ / T  0. The key point of this problem rests on the process in which  dQ / T  0 is reduced ， ∑[(ΔQ)/T)] becoming ∫dQ/T is false, because : 1. The prerequisite condition of forming differential is existence of derivable function ， but here, there isn’t any derivable function at all. 2. As for elementary calculus, as we know, the premise elementary calculus is still function and variable. When Q is function, if Q=f (T), then, we can get 1/TdQ=dF(T) and ∫ T 1/TdQ= ∫ T dF(T), under this condition, △ Q/T→dQ/T is tenable. But, as we know, in fact, to arbitrarily reversible process, there is Q=f (T,V,P), then, 1/TdQ=dF(T,V,P), under this condition, what is ∫ T 1/TdQ = ∫ T dF(T,V,P) ? obviousely, it is meaningless. Here, to arbitrarily reversible process, T,V and P are all variables but not constants. What is meaningful is ∫ T ∫ V ∫ P 1/TdQ = ∫ T ∫ V ∫ P dF(T,V,P). So, ∫ T 1/TdQ ( namely ∫ T dF(T,V,P) ) is meaningless, that is △ Q/T→dQ/T is misuse of Process function Q, dQ/T ( namely df(T,V,P)/T ) itself is meaningless at all. people didn’t follow the principle of calculus, but took △ Q/T→dQ/T for granted, then, led to this false result. -6-

7.  not exist this formula   0/ TdQ, and there don’t exist the formula   0/ TdW   0/ TdE either. -7- What is false in textbooks is that there emerges misunderstanding and misuse of Process function Q. The conclusion is that dQ / T  0 is neither a mathematic conclusion, nor a physical result, that is to say that there does and IV. Concerning Boltzmann “entropy” Then, what is Boltzmann entropy ? On one hand, entropy from Boltzmann was to construe entropy from Clausius, the unit J/K from Boltzmann entropy was also transplanted from Clausius entropy. But this paper expounds and proves entropy by Clausius does not exist, therefore entropy defined by Boltzmann is denied accordingly. On the other hand, in S=kInΩ, Ω is a so-called thermodynamic probability, but calculation of Ω needs to use phase-cell division method to separate hyperspace μ, phase cell is two dimensions 2i, and i is the total free degree of molecules within the system. Calculation of Ω, in essence, is to separate the continuous space μ and makes it produce objective meaning. In fact, this method won’t work, no matter how much work is being done by the people, a conclusion with objective-results won’t be able to be obtained. The primary reason is that there is not an objective standard for phase cell division, that is to say that Ω does not possess the meaning of objective value. And again combining with Liouville Theorem and the conclusion of this paper, we can conclude that Boltzmann entropy is a skill to display irreversibility. V. Concerning the second law of thermodynamics The second law of thermodynamics is formulated by several mutual-equivalent statements, although they have been testified by endless facts, it is impossible to cover the nature of them only being phenomenological laws, just as formulating the law of universal gravitation as “any object can not spontaneously move from low place to high place”, the formulations of the second law are only the statements of the concrete phenomena and can not reveal the unified nature and law that lead to these phenomena. The second law of thermodynamics will be re-expounded by new methods. References Except textbooks, this paper doesn’t need references, and the content in textbooks is wellknown, so, I don’t list references, for it is unnecessary.