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Thermodynamic Cosmic Problem

  1. Aug 9, 2012 #1
    Thermodynamic Cosmic Problem !!

    Ill start with the third law of thermodynamics which denies the existence of zero kelvin (as stated by my lecturer). Also i found out that the temperature of vacuum is 3k due to the radiations left over from the Big Bang. However the space(or the absolute vacuum) which exists beyond the expanding universe must be at zero Kelvin as it has no access to the radiations which would be expanding at the speed of light. Doesn't this fact violate the 3rd law of thermodynamics ?
    Thanks in advance !
     
    Last edited: Aug 9, 2012
  2. jcsd
  3. Aug 9, 2012 #2

    jbriggs444

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    Re: Thermodynamic Cosmic Problem !!

    "Beyond the expanding universe"? There is no such thing.

    It's not an expansion in space. It's an expansion of space.
     
  4. Aug 9, 2012 #3
    Re: Thermodynamic Cosmic Problem !!

    That's not true. the third law states that zero kelvin is unreachable. Quantum mechanics confirmed this by investigations of the Casimir effect: there's no obligation that you read this obviously, unless you find it interesting:
    http://en.wikipedia.org/wiki/Casimir_effect
    Essentially the ground state of any Quantum electromagnetic field is non-zero.
    woooaaa! please let's keep this a science forum. Nonetheless I do enjoy a bit of philosophic banter every so often. But then how do you define the universe if it has well defined edges, and at the same time, is just the emptiness within which everything else exists in. It is a rather circular argument. The true scientist would tell you, that the universe expanded with matter from the Bigbang (copyrighted by some american millionaire?) and at the same rate, and so you cannot define universe if you don't have matter there essentially. I would not be so happy to go with that view. I would tell you that any speculation about the apparent size, or direction of the expansion of the universe cannot be absolute. In essence we may be at a point in the universe that we might have been at billions of years before and over and over again. Unless you're investigating it from the outside, there's no well defined position in the universe, but there are well defined position-time vectors.

    In the model, which won the noble prize for describing the vacuum fluctuations, the assumption is that the E.M. field from the bigbang is bounded by the size of the universe, and that is the reason why there's a non-zero solution at every point inside. In fact if the radiation wasn't there, that would be evidence that the universe may not be finite at all...
     
    Last edited: Aug 9, 2012
  5. Aug 9, 2012 #4
    Re: Thermodynamic Cosmic Problem !!

    There is a difference between existing and reachable.
     
  6. Aug 9, 2012 #5
    Re: Thermodynamic Cosmic Problem !!

    @Ardie : I'm really thankful to you for such an elaborate explanation. And i agree, 'existing and reachable' (in relation to 0k) are 2 very different things. But since i have yet to read quantum mechanics in my upcoming semesters i apologize for my inability to appreciate the 'Casimir effect'.
    However what i could make out from your second argument was that my understanding of the universe and space (which i referred to as the absolute vacuum :P) as seperable is at fault. I understand the space as something in which universe is expanding (initiated by the big bang). I would be grateful if you correct me !
     
    Last edited: Aug 9, 2012
  7. Aug 9, 2012 #6
    Re: Thermodynamic Cosmic Problem !!

    This is classical cosmology. It surprisingly leads to the same observable results as the Einstein-de-Sitter-Model. However as classical mechanics is falsified the corresponding cosmological models are invalid too. In GR the space itself is expanding.
     
  8. Aug 9, 2012 #7
    Re: Thermodynamic Cosmic Problem !!

    What should i then think of the universe and space ?
    p.s : i am yet to complete my bachelors in physics so i'd be pleased if u could explain it to me keeping that into consideration !
     
  9. Aug 9, 2012 #8
    Re: Thermodynamic Cosmic Problem !!

    I think it is important to know what we know exactly as we do the calculations, rather than simplify them just so that you understand in classical terms, because there is really no alternative. If classical mechanics really explained everything, then there wouldn't be so much success attached to quantum mechanics.
    The very existence of the vacuum fluctuations (3k everywhere in space) is direct evidence that the universe is finite in length, and that is so because of exactly what you said; otherwise the photons would have escaped by now. Let's not speculate as to where, but if you try and solve the Q.M. equations for light in an infinite boundary (no boundaries), you must get 0 intensity everywhere, but you do not.
     
  10. Aug 9, 2012 #9
    Re: Thermodynamic Cosmic Problem !!

    to ardie: I don't think the Casimir effect says anything about 0 temperature being unreachable. I think you are confusing temperature with energy. Zero point energy has no bearing on the temperature. Also, I don't follow your argument for why the universe is finite.

    pranav: The big bang encompassed the entire universe, so there is no "outside the reaches of the big bang".
     
  11. Aug 9, 2012 #10
    Re: Thermodynamic Cosmic Problem !!

    What your lecturer actually meant that “it is impossible for any procedure to bring a system with a positive absolute temperature to absolute zero of temperature in a finite number of steps.” In order for the third law to be useful, the absolute temperature of the system has to be more than zero.
    If there was a region outside the visible universe which had a temperature of absolute zero, then it didn’t need any steps to bring it down to zero. It was already at absolute zero, and will remain there until energy, momentum and entropy reach it.
    I doubt there is a region of space which is completely empty of energy or matter. However, if there is then it is already at absolute zero and has always been at absolute zero. The third law does not say, “There can’t be a region of space that was always at absolute zero.”
    If there is a closed system at a positive absolute temperature, then one can always bring the temperature down using a single step process. The temperature can asymptotically approach zero in the limit of infinite time. One can divide the positive temperature by 2, 4, 6, 8, etc. One can get as close to absolute zero as one wants by more steps. This will take time. However, one will never get to absolute zero in finite time.
    There is more than one way to express the Third Law of Thermodynamics. This link has an article presenting some of the equivalent ways of stating the Third Law.
    http://en.wikipedia.org/wiki/Third_law_of_thermodynamics
    “The entropy of a perfect crystal at absolute zero is exactly equal to zero…
    The entropy of a system approaches a constant value as the temperature approaches zero…
    Physically, the law implies that it is impossible for any procedure to bring a system to the absolute zero of temperature in a finite number of steps…”

    On a gross engineering level, the third law is really saying, “The specific heat of all materials gets very, very big at temperatures close to absolute zero.”


    I think the second expression is probably the one that is most useful in application. Laboratory measurements can be used to determine the functional dependence of absolute entropy on absolute temperature in a closed system. One can falsify the Third Law as follows.
    One can not know what is happening beyond the edge of the visible universe. The practical question is really how the Third Law relates to laboratory measurements. Therefore, the real question that you may be asking is whether the Third Law can be falsified on a laboratory scale. The third law has never been falsified in the laboratory. However, I propose the following method for falsifying the third law.
    Determine the entropy of a material as a function of temperature by experimental measurement. Of course, there will be uncertainties in the measurements. However, one can decide on engineering tolerances based on these uncertainties. If there is a relative minimum in entropy at zero absolute temperature, within whatever tolerances one decides on, then the results are consistent with the Third Law. If the entropy doesn’t reach a relative minimum within the tolerances of the theory, then the Third Law is falsified.
    For me, the most interesting part would be the engineering tolerances. This is a matter of statistical analysis, not fundamental physics.
     
  12. Aug 10, 2012 #11
    Re: Thermodynamic Cosmic Problem !!

    i may have confused cosmic background radiations with vacuum fluctuations. either way even if you manage to get rid of the cosmic radiations, you cannot get rid of the nonzero vacuum fluctuations ground state energy. btw temperature is another measure of the internal energy.
     
  13. Aug 10, 2012 #12
    Re: Thermodynamic Cosmic Problem !!

    The nonzero background fluctuations are not part of the energy that contributes to absolute temperature. The energy density of the quantum fluctuations is always infinite, even at absolute zero. The 3 Degree cosmic background radiation is not a vacuum quantum fluctuation per se. In particle terms, the vacuum fluctuations correspond to virtual photons and the 3 degree radiation corresponds to real photons. In wave terms, the vacuum fluctuations have a Lorentz invariant spectrum and the 3 degree radiation has a Planck thermal spectrum.
    The Third Law would apply to a region of space that initially has the 3 degree radiation. One could refrigerate this portion of space to within any fraction of a degree Kelvin using a finite number of states. However, you could never refrigerate this portion of space to 0 degrees Kelvin in a finite number of steps. The refrigeration would not eliminate the vacuum fluctuations, no matter how many steps it involves.
     
  14. Aug 10, 2012 #13
  15. Aug 11, 2012 #14
    Re: Thermodynamic Cosmic Problem !!

    Thanks a lot Darwin123 (for this part) because i really got where i went wrong. Infact all my misconceptions regarding the 3rd law have been cleared. However the statement
    is still unclear.

    Also i'm curious to understand your way of falsifying the third law. I would be grateful if you explain it to me in a bit more simpler terms (I'm yet to complete my bachelors in physics) !
    Thanks !
     
  16. Aug 11, 2012 #15
    Re: Thermodynamic Cosmic Problem !!

    @Ardie : I request you to allow me some time before i get back to you as i'm reading stuff on the understanding of space :) Thanks
     
  17. Aug 11, 2012 #16
    Re: Thermodynamic Cosmic Problem !!

    First, understand that I was speculating. To understand what I was saying, lets reduce that Third Law of Thermodynamics to differential equations and inequalities. Instead of mucking around with visualization, let us try to express the Third Law mathematically. Once we have a mathematically formal way to express the Third Law, then we can think about experimental ways to falsify or validate it.
    I don't have any references that precisely express the Laws of Thermodynamics the way I showed you. However, there are plenty of textbook examples which use equations like them.
    One way to express the Third Law is by stating that the entropy of a closed system is at a minimum at zero absolute temperature. Let us write some differential equations equivalent to that.
    The entropy of a closed system near equilibrium is a function of state variables including absolute temperature. That is,
    1) S=S(T,P,V,{u_i})
    where for the closed system T is absolute temperature, P is pressure, V is volume, u_i is the i'th state variable and {u_i} is a complete set of such state variables. The entropy S is a continuous function of T. Basically, equation 1 is equivalent to saying entropy is a perfect differential in terms of reversible processes. However, I promised not to go deeply into visualizations.
    The second law states that the total entropy of a system must never decrease, and that the entropy can never spontaneously flow from a high temperature system to a low temperature system. The second law is basically equivalent to the statement that if,
    2) T>0,
    then,
    3) ∂S(T,P,V,{u_i})/∂T>0.
    Equations 2 and 3 are mathematically formal expressions for the second law. The more visual expressions for the third law can be derived from equations 2 and 3, provided we carefully define what a system variable is. However, I think you can roughly see how 2 and 3 is approximately equivalent to the other expressions of the second law.
    From introductory calculus, one knows how to find the minimum of a function. If S is at a relative minimum at T=0, then,
    4) ∂S(T,P,V,{u_i})/∂T=0,
    and,
    5)∂^2 S(T,P,V,{u_i})/∂T^2 > 0.
    The last two equations are mathematical expressions for the third law. One can see that finding absolute zero is basically the same as finding the minimum of the function, S. Because of the second law (equation 3), one can see that there is only one relative minimum in S. There is only one zero absolute temperature.
    If you look at number 4, you see that one needs an infinitely large change in entropy to produce a finite change in temperature at T=0. This is basically the third law.
    Of course, these mathematical equations have several physical hypotheses hidden in them. I haven't told you how to measure S, for instance. However, I hope that you get my point. The Four Laws of Thermodynamics (0, 1,2, 3, 4) are quantitative theories. They are used to define differential equations with quantitative solutions.
    To falsify the Third Law, you have to show that in the limit of T=0,
    6) ∂S(T,P,V,{u_i})/∂T≠0.

    Experimentally, one would determine S as a function of T for any fixed set of state variables. If one can determine that at T=0, equation 6 is probably true, then one has proven that the Third Law is probably false.
    I can rewrite these equations several says. However, proving the Third Law of Thermodynamics is true is basically proving that equations 4 and 5 are true. Experimentalists have to deal with uncertainties and tolerances, which is why I tried to explain the procedure in the way I did. A good experimenter never absolutely proves something, he proves something to within a certain tolerance. But without going through grungy statistical details, the experimenter would try to prove equations 4 and 5 as T approaches 0. One is measuring the slope of the S versus T curve. One extrapolates to T=0.
     
    Last edited: Aug 11, 2012
  18. Aug 11, 2012 #17
    Re: Thermodynamic Cosmic Problem !!

    On Darwin's approach I am going to return this??

    First of all, the statistical way of defining the entropy (entropy as a quantity is not measurable) is:
    S= k lnW
    where k is Boltzman's constant
    and W is the number of microStates that build up your given MacroState.
    This can be proved in several ways... by having a general S function: S=k Σ[pr lnpr] where pr is the probability of the microstate r, and trying to minimize it for fixed quantities of energy or number of bodies (Boltzman's formula above is a deduction of this approach, by saying you have a fixed total energy for your system)

    BUT
    at absolute zero (T=0K) your system falls to its ground state. So W=1 if there is no degeneration. Given that S=0 for T=0.



    by the way something interesting I read (all my idea came by Landau & Lifgarbagez, paragraph 23, p68-70). From the same quote I read, that entropy goes to zero, for T going to 0, is true only when other quantities remain fixed (such as vollume and pressure). If the temperature (it says) of our system decreases to zero, while its density also decreases without limit, then the entropy need not tend to zero.... this sounds quite interesting for even our universe...
     
    Last edited: Aug 11, 2012
  19. Aug 11, 2012 #18
    Re: Thermodynamic Cosmic Problem !!

    Who needs statistics?
    Thermodynamics is most useful in its phenomenological form, which doesn't need statistics. In most chemistry problems, the statistical expression for entropy is not useable. One doesn't know how to calculate the degrees of freedom in any realistically complex system.
    The change in entropy can be determined by calorimetric experiment without knowing a single thing about the atomic structure of what one is looking at. The statistical mechanics can actually be confusing if all you want to know is the available work.
    Statistics has its place. It is really important in certain types of problems. However, pure thermodynamics is most useful as a short cut through statistical mechanics. It is a way of avoiding the unimportant details to get right at what is most important to the problem. So I think the right way to learn thermodynamics is starting with the macroscopic system.
    I prefer a more visual way of learning thermodynamics that defers probability and statistics. Entropy is an indestructible fluid that permeates other matter. Temperature is the pressure of this fluid. This should be the start of a thermodynamics course.
    This was suggested in the following article:
    Hans U. Fuchs, "Entropy in the teaching of thermodynamics," American Journal of Physics 55(3) 215-219 (March 1987).
    The abstract reads:
    "An elementary physics course in thermodynamics is presented. It uses entropy and temperature as fundamental, undefined objects."
    So that is how I like to explain thermodynamics. Start with entropy and temperature as fundamental objects.
     
  20. Aug 11, 2012 #19
    Re: Thermodynamic Cosmic Problem !!

    Entropy can be determined from calorimetric measurements. If ones measures the specific heat of a material as a function of temperature, then one can determine the entropy. Hence, entropy can be effectively "measured".
    The number of microstates can't be measured. The number of microstates can't even be numerically determined in detailed theory. The number of microstates can be determined once the total entropy is determined.
    How many chemists do you think do a precise molecular dynamics simulation to find the enthalpy of a chemical reaction? How many chemists do you think use a precise molecular dynamics simulation to find the change in entropy of a chemical reaction?
    There may be a few, these days. I conjecture that a lot more use calorimetry to determine both the change in enthalpy and the change of entropy in chemical reactions.
     
  21. Aug 12, 2012 #20
    Re: Thermodynamic Cosmic Problem !!

    Statistical physics ain't that bad, in the way they give answer to many problems. It reconstructs the already known phenomenological data, or the "classical thermodynamics" but it can be used in quantum mechanics way better.
    For example if you try to determine the entropy of an ideal gas, you will eventually end up in an expression for S that doesn't equal to 0 for T->0. That is a problem of classical thermodynamics. It can be overpassed in the quantum level, and quantum thermodynamics have both statistics and probability instinctively in them. Nerst's theorem (or the 3rd law) is a deduction of quantum statistics and cannot be purely taken by classical approaches.
    In some cases the number of the microstates can be determined thanks to the problem (Like Ising model). In fact I can't think of Quantum Statistics in the Classical Thermodynamics Approach. They diverge in results!



    when you say that a physical quantity is not measurable, it means you can't measure it straightforward through an experiment (There is no machine which you can put in your system and measure its entropy). The heat and temperature are measurable, and due to their measuring, you can then CALCULATE the entropy.
     
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