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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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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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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I have a feeling this isn't the case though, as it is possible say [math]\mathbf{I} = \sum_j |j><j|[\math] where we take [math]<j|k> = \delta_{ij}[/math], so just a bit confused about the dyad.
that should be [math]\mathbf{I} = \sum_j\ <j|j>\ = \delta_{ij}[/math]
with [math]\delta_{ij} =\ <i|j>[/math] sorry. Tired and heading to bed soon.
It's probably true what I have said, not sure why I thought it needs to square to unity. It seems n|\sigma> = 1 (is this always the case) anyway, good night.
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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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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.