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Status Updates posted by Dubbelosix

  1. Do you like it? Do you think this geodesic equation is simple enough to be possibly right? 

    [math]\nabla_n \dot{\gamma}(t) = \nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla_j,\nabla_j]|\dot{\psi}>} = 0[/math]

    1. Dubbelosix


      I hold that the RHS must [always] equal zero because the curve component is squared, meaning by definition that the covariant derivatives have to be the same. I defined the time derivative early on with a j-subscript. This necessary application fundamentally implements that the geodesic always equals zero. The fact the connections are acting classically could also be a hint for classical gravity at the quantum scale? Certainly, Penrose has suggested classical gravity in the phase space.

  2. Mordred, I have a technical question here.... my work is a bit more complicated and requires a bit more depth than I am used to - in the case of a projective space, the identity P^2 = P is said to hold -  what does it mean?

    In my case I am looking at, I have an antisymmetric matrix that requires to be squared in the projected space to yield the identity/unity. The projective space looks like


    [math]P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|[/math]


    The square of the Pauli matrix should yield an identity [math](n \cdot \sigma)^2 = \mathbf{I}[/math] (unit vectors naturally square into unity). What is the square of the dyad in such a case? 

    1. Show previous comments  2 more
    2. Mordred


      [math]\rho^2=\rho[/math] is the projector condition [math]\rho[/rho is only a projector if and only if [math]\rho^2=\rho[/math]

      wiki explains it a bit better

      In operator language, a density operator is a positive semidefinite, Hermitian operator of trace 1 acting on the state space.[7] A density operator describes a pure state if it is a rank one projection. Equivalently, a density operator ρ describes a pure state if and only if

      ρ=ρ2{\displaystyle \rho =\rho ^{2}}{\displaystyle \rho =\rho ^{2}},

      i.e. the state is idempotent. This is true regardless of whether H is finite-dimensional or not.



    3. Dubbelosix


      Oh I see. Thanks. 

      Ahh yes of course, and it satisfies pure states, as expected. 

    4. Mordred



      lol messed up the closing tag above but can't edit ah well ya got the answer P

  3. Isn't it nice to have an honest, educational discussion? I see so much trash on the science forums and had very negative responses at some other sites with people calling my work rubbish and work that I do not understand. 


    1. Show previous comments  4 more
    2. Mordred


      Thanks all its nice to see my methodology recieve positive results. The thread under discussion should however be creditted to Dubbelsox. 

      He is abiding by the proper methodology of a toy universe model to the letter. Makes it far easier to engage in a proper scientific discussion.


      Yes Stringy I do occassionally vist physicsforum however I find I make greater contributions here. Mainly due simply because this site has a controllable Speculations forum.

      Strange as that may sound lol

    3. koti


      Just went through the thread and frankly, I too understand a few percent of what is being laid out. It is a pleasant surprise to see a gem thread like this in the speculations forum though, I will be lurking to learn Dubbelosix and Mordred. I'm also surprised that you have been mistreated at other forums, seems bizarre.  

    4. Dubbelosix


      Thanks guys, that was all very nice to read.

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