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I Understanding Penrose's large number

  1. Mar 15, 2016 #1
    Roger Penrose claims here:
    that the universe is very special to the degree 10^10^123
    I understood this as measuring the probability of the entropy state was chosen at random what are the odds it would have been as low as we think it was at the big bang.
    However someone else described it to me as representing the mass energy distribution of the universe at the big bang. This person does not seem like a reliable source to me but perhaps someone can confirm who is correct here.
  2. jcsd
  3. Mar 15, 2016 #2


    Staff: Mentor

    Sort of, yes. Penrose makes the argument in terms of phase space volumes; the number 10^10^123 is the ratio of the phase space volume corresponding to the macroscopic state of the universe today, to the phase space volume corresponding to the macroscopic state of the universe at the Big Bang. This ratio basically corresponds to the ratio of probabilities of ending up in a particular macroscopic state if you chose a microstate of the system at random.

    This is just wrong, as far as I can see. The phase space of the universe doesn't just include the states of the matter and energy; it also includes the states of spacetime itself (the "gravitational field"). The fact that the latter has to be included is one of the key points underlying Penrose's arguments in the article.
  4. Mar 30, 2016 #3
    Please anyone answer me
    A: ##(10^{10})^{123}##
    B: ##10^{(10^{123})}##

    A has 1230 zeros
    B has ...? I don't know :smile:

    But what I want to know is A or B. It's sufficient for me.
  5. Mar 30, 2016 #4


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    Yes, A is the same as [itex]10^{1230}[/itex] so has 1230 "0"s. B, [itex]10^{(10^{123})}[/itex], has [itex]10^{123}[/itex] "0"s.
  6. Mar 30, 2016 #5
    But what I want to know is this number 10^10^123
    Is it
    A: 10 with 1230 numbers
    B: 10 with 10^123 numbers
  7. Mar 30, 2016 #6


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    Penrose means (B).
  8. Mar 30, 2016 #7
    I think I was wrong
    10^10^123 is not
    A: 10 with 1230 numbers, but 1 with 1230 zeros
    B: 10 with 10^123 numbers, but 1 with 10^123 zeros.
  9. Mar 30, 2016 #8


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    You are asking whether exponentiation denoted with the caret ("^") symbol is left-associative or right-associative by convention. One answer is that it is only a convention. There is no physics content to the question. That said...

    In the Ada programming language, the notation is simply illegal. Exponentiation in Ada is non-associative and parentheses must be used (a healthy attitude in my opinion).

    Most other programming languages use right-associativity. So a^b^c is interpreted as a^(b^c). https://en.wikipedia.org/wiki/Associative_property has commentary that provides motivation for that convention.

    https://en.wikipedia.org/wiki/Exponentiation also indicates that right associativity (and top to bottom evaluation in the case of superscripts) is the standard convention but notes that Excel and Matlab choose to go with left associativity instead.
  10. Mar 30, 2016 #9
    Actually I'm not asking about association. I just want to understand the discussion in this thread.
    But since you mention it. I've never thought of exponential association before.
    I'm comfortable with
    A/B/C -> (A/B)/C not A/(B/C)
    A-B-C -> (A-B)-C not A-(B-C)
    But who ever thought about say 10^10^10, while 10^100 itself is very big. Only some variable can hold this number. Double or Extended (in Pascal I think)
    Extended can hold much bigger 10^3000 if I recall.
    Thanks to bring it up anyway.
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