Dubbelosix
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Status Updates posted by Dubbelosix
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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]
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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.
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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?
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[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
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ρ=ρ2{\displaystyle \rho =\rho ^{2}}
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i.e. the state is idempotent. This is true regardless of whether H is finite-dimensional or not.
https://en.wikipedia.org/wiki/Density_matrix
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ρ=ρ2{\displaystyle \rho =\rho ^{2}}
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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.
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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
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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.
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