
Posts
518 
Joined

Last visited
Content Type
Profiles
Forums
Events
Status Replies posted by Dubbelosix

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]

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 jsubscript. 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.


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?

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?

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?

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]<jk> = \delta_{ij}[/math], so just a bit confused about the dyad.
that should be [math]\mathbf{I} = \sum_j\ <jj>\ = \delta_{ij}[/math]
with [math]\delta_{ij} =\ <ij>[/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.


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.

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.