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Big Bang and Entropy

  1. Feb 15, 2013 #1
    I am reading Cycles of Time by Penrose, page 71-72. Could someone explain if the universe started out with at a state of extraordinarily tiny entropy or a state of maximum entropy and what proof or theories do we have to justify this choice. I find the explanation in the book confusing.

    Thank you.
  2. jcsd
  3. Feb 15, 2013 #2
    He says that the big bang was a state of tiny entropy because if we accept the law of 'ever increasing entropy', i.e. 2nd law of thermodynamics (which Penrose explains in every detail), then going backwards in time mist take us to states of ever decreasing entropy. Going way back to the big bang, we see it must have a very low entropy.
  4. Feb 15, 2013 #3


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    Just to add -- the extraordinary low entropy initial state of the universe is one of the prevailing mysteries of modern cosmology. Sean Carroll, for one, has been doing much writing on this topic and its connection to the arrow of time: http://preposterousuniverse.com/eternitytohere/faq.html [Broken]
    Last edited by a moderator: May 6, 2017
  5. Feb 15, 2013 #4


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  6. Feb 16, 2013 #5
    Here's are review excerpts of the book I found on Amazon books.....

  7. Feb 17, 2013 #6
    you may think, why is the entropy of the matter immediately after the big bang so low, even though it is spread out very evenly?

    let me offer a small opinion as to why. can we agree to the 2 following facts?

    the matter after the big bang is not a black hole.
    the entirety of matter after the big bang is a gravitating system.

    For a classical gas kinetic energy <K> = 3/2*kT. In contrary to conventional notation, due to T being occupied by the temperature, K will instead be used for the kinetic energy.

    A self gravitating system obeys the virial theorem because the kinetic and potential energies, and the momenta, of all particles, are bounded. <K> = -1/2 <U> where <U> is the time average of the potential energy. The proof is long so lets take an example instead: the 2 body central force problem.

    U = -GmM/R (gravitational potential)
    F[itex]_{g}[/itex] = -GmM/R[itex]^{2}[/itex] (gravitational force)
    F[itex]_{c}[/itex] = mv[itex]^{2}[/itex]/R (centrifugal force)

    at equilibrium and in appropriate coordinates the net force is zero.

    GmM/R[itex]^{2}[/itex] = mv[itex]^{2}[/itex]/R
    rearrange to get (1/2)mv[itex]^{2}[/itex] = (1/2) GmM/R

    oh look we have K = -(1/2)U. If it works for 2 particles, maybe it works for N particles.

    So K = -(1/2)U, U = -2K. total energy E = K + U = K - 2K = -K.
    K = (3/2)kT. So E = -(3/2)kT

    In other words, total energy = -number*temperature, and as temperature increases, E is decreasing.

    What is gravitational potential energy of a sphere of gas?


    U = -3GM[itex]^{2}[/itex]/R.

    So E = -number*1/R -> T = number/R. As R decreases T increases. As T increases, E decreases.

    This means that self gravitating systems want to decrease their energy by decreasing their size (measured by radius R) and increasing temperature. The uniform state after the big bang is therefore a state with extremely low entropy and favored to evolve towards higher entropy states. Indeed, energy would be minimized if R approached zero.
    Last edited: Feb 17, 2013
  8. Feb 18, 2013 #7
    Thank you all for the responses.

    So if the Big Bang was in a state of tiny entropy, why when looking at the CMB radiation we notice there is thermal equilibrium? Wouldn't this imply maximum entropy even during the beginning of the Big Bang also. (I could be misunderstanding the idea of course graining and thermal equilibrium. Still a novice at this)
    Last edited: Feb 18, 2013
  9. Feb 18, 2013 #8
    This is backwards from the Penrose Conformal Cyclic Cosmology : The entirety of energy after the big bang is repulsive gravity according to him.

    There are different ideas what might be the causes...a recent one I posted was a research paper attributing it to TORSION of space-time....

    I think you mean uniformity...rather than equilibrium ...except for small perturbations the CMBR is UNIFORM. The CMBR is NOT in thermal equilibrium and never has been. For example it started out close to 3,000 degrees K and is now about about 2.73 degrees K .

    Penrose attributes the extraordinary low initial entropy to that of gravity....other elements as I recall he does not claim to be in low entropy...

    As I posted previously:

    High entropy with gravity means clumping mass...likes planets,stars,etc......w/o mass I am guessing Penrose thinks gravitational entropy would be low....

    yes!!!! found this note I made from his video:

    More here:

    Cycles of time--Penrose says his cyclic cosmology obeys thermodynamics

    Cycles of Time [great slides,diagrams, although a bit 'handwavy'
    http://www.slideshare.net/scexxn/cycles-of-time-16403194 [Broken]
    Last edited by a moderator: May 6, 2017
  10. Feb 18, 2013 #9
    Cantor: I have this description in my notes about CCC, did not record the source:

    My notes from the above slides:

    Marcus posted this in another discussion:

    In addition there is good reason to believe entropy cannot be well defined without defining an observer....and the number of degrees of freedom also have major impact on defining entropy...further Penrose has changed his mind about black hole information loss.. Penrose has returned believing it IS lost...if the information were NOT lost, as black holes spew forth their contents [ 99% of the information of the universe] it would be recovered at the end of this aeon...hardly a low entropy state
    so beware..... there are MANY subtleties in all this....

    FYI: For those with Penrose's book ROAD TO REALITY, he does discuss these ideas..
    Last edited: Feb 18, 2013
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