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Quantum Gravity made too simple.

  1. Mar 13, 2013 #1
    I will work with supernatural units in which [itex]\hbar = c = 1[/itex] and further, [itex]\pi = e = -1 = 1[/itex].
    Using these units, and some minor calculation we see that the Lagrangian is:

    [tex]\mathcal{L} = 1[/tex]

    This not only simplifies matters, but leads to remarkably accurate predictions. For instance, if we assume that the S matrix is given by

    [tex]S = \sum_{n=0}^{\infty}S^{(n)}[/tex]

    where

    [tex]S^{(n)} = 9 (.1)^{n+1}[/tex]

    Then we have [itex]g = 0.99\overline{9}[/itex], in remarkable agreement with the experimental value of 1, the error being only [itex]0.0\overline{0}1[/itex].

    What's more, by setting 0 = 1 we have the following grand unification of quantum mechanics and general relativity.

    [tex](i\not{\partial} - m)\psi = 0 = 1 = 0 = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R - 8GT_{\mu\nu}[/tex]

    In my next paper, I will investigate the implications of the further simplification [itex]\frac{1}{2} = 8 = 1[/itex].
     
    Last edited: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    HallsofIvy

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    Okay, I admit- you got me. I actually wrote out a response pointing out all the errors when, while trying to decide how to respond to the last line, it finally hit me- this is a joke!
     
  4. Mar 13, 2013 #3

    marcus

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    Funny! I came to this thread immediately after watching a Samantha Bee episode on Daily Show, and it's right in the same vein.
    She was explaining the terms used to describe the gathering of Cardinals to elect a Pope.
     
  5. Mar 14, 2013 #4

    tom.stoer

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    There is no single pope but a landscape of popes; you find the correct vacuum state by catholic reasoning which requires 3 = 1; from this equation you can derive both the existence and the non-existence of god, so it's a theory of everything.
     
  6. Mar 14, 2013 #5
    only if you have the correct ......... wait for it........cardinality
     
  7. Mar 15, 2013 #6

    MathematicalPhysicist

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    Don't forget:

    [tex]1+2+3+\cdots =-\frac{1}{12}[/tex]

    Now everything makes sense.

    :-D
     
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