# Dubbelosix

Senior Members

518

1. ## Dubbelosix Mordred

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

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

1. 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. ## Dubbelosix Mordred

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

$P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|$

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

1. Oh I see. Thanks.

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

2. (See 4 other replies to this status update)

3. ## Dubbelosix Mordred

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

$P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|$

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

1. even when trying to do it right, I get it wrong lol... what a mess (and sing relic notation) $\mathbf{I} = \sum_j |j><j|$

2. (See 4 other replies to this status update)

4. ## Dubbelosix Mordred

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

$P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|$

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

1. I have a feeling this isn't the case though, as it is possible say $\mathbf{I} = \sum_j |j><j|[\math] where we take [math]<j|k> = \delta_{ij}$, so just a bit confused about the dyad.

that should be $\mathbf{I} = \sum_j\ <j|j>\ = \delta_{ij}$

with $\delta_{ij} =\ <i|j>$ 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.

2. (See 4 other replies to this status update)

5. ## Dubbelosix Mordred

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. Thanks guys, that was all very nice to read.

2. (See 6 other replies to this status update)

6. ## Dubbelosix Mordred

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.

×