# Poincare Symmetry And Chirality Coefficient Torsion

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### #18 Dubbelosix

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Posted 30 September 2019 - 03:41 AM

Going back to >

[∇, ∇] = - (∂^k γ_k + σ ⋅ Γ^k γ_k γ^5) (- ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5)

= (∂^k γ_k + σ ⋅ Γ^k γ_k γ^5) (∂^k γ_k + σ ⋅ Γ^k γ_k γ^5)

This can be expanded but certain algebra rules need to be accordingly done for instance, the square of the Pauli spin vector spits back the identity operator which is attached to the torsion. Expanding we get:

= ∂^k_μ γ_k∂^k_v γ_k

+ ∂^k_μ γ_k (σ ⋅ Γ^k) γ_k γ^5

+ (σ ⋅ Γ^k_μ) γ_k γ^5 ∂^k_v γ_k

+ i^2 σ^2 ⋅ [Γ^k_μ γ_k , Γ^k_v γ_k [γ^5]^2]

> not nicest format but let us start using latex and required subscripts, if I have it right it should look something like;

$(\partial^k_{\mu} \gamma_k + \sigma \cdot \Gamma^k_{\mu} \gamma_k \gamma_5)(\partial^k_{\nu} \gamma_k + \sigma \cdot \Gamma^k_{\nu} \gamma_k \gamma_5)$

$= \partial^k_{\mu} \gamma_k \partial^k_{\nu} \gamma_k + [\partial^k_{\mu} \gamma_k (\sigma \cdot \Gamma^k_{\nu}) \gamma_k \gamma_5 + \gamma_k \gamma_5 (\sigma \cdot \Gamma^k_{\mu}) \partial^k_{\nu} \gamma_k] + i^2 \sigma^2 \cdot [\Gamma^k_{\mu} \gamma_k\Gamma^k_{\nu} \gamma_5 \gamma_k + \gamma_k \gamma_5\Gamma^k_{\nu}\gamma_k\Gamma^k_{\mu}]$

But I am not too sure on this last term, but we have loads of time to come back and analyse this further.

Edited by Dubbelosix, 03 October 2019 - 09:20 PM.

### #19 Dubbelosix

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Posted 01 October 2019 - 05:59 PM

Back to geometric algebra ~ remember how I defined the curvature tensor? It was meddled about with to produce it in the form

R_μv = [∇_μ,∇_v] = (∂^k_μv γ_kγ_1γ_2γ_3 + iσ ⋅ Γ^k_μv γ_k γ_o)^2

In which I concentrated on the solution for the connection as;

∇ = - ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5

We noted which made this a new idea was to attach the timlike matrix to the part which would produce torsion ~ which too measured in units of timelike space. We will undo that approach and I'll investigate a different solution, similar to before but is a stronger version of linearized gravity;

γ_0 R__μv = (∂^k_μ γ_k γ_0 − Γ^k_μγ_k γ_1γ_2γ_3) γ_0 (∂^j_v γ_j γ_0 − Γ^j_vγ_jγ_1γ_2γ_3)

= ∂^k_μ γ_k γ_0 γ_0 ∂^j_v γ_j γ_0
− ∂^k_μ γ_kγ_0 γ_0 Γ^j_v γ_j γ_1γ_2γ_3 − Γ^k_μγ_k γ_1γ_2γ_3 γ_0 ∂^j_v γ_jγ_0
+ Γ^k_μγ_k γ_1γ_2γ_3 γ_0 Γ^j_vγ_jγ_1γ_2γ_3

Here we make use of γ_0 γ_0 = - 1 and we get

= − ∂^k_μ γ_k ∂^j_v γ_j γ_0
+ ∂^k_μ γ_k Γ^j_v γ_j γ_1γ_2γ_3 − Γ^k_μγ_k γ_1γ_2γ_3 ∂^j_v γ_j
+ Γ^k_μγ_k γ_1γ_2γ_3 Γ^j_vγ_jγ_1γ_2γ_3 γ0

= − ∂^k_μ γ_k ∂^j_v γ_j γ_0
+ ∂^k_μ γ_k Γ^j_vγ_jγ_1γ_2γ_3 − Γ^j_vγ_jγ_1γ_2γ_3 ∂^k_μ γ_k
− Γ^k_μγ_k Γ^j_v γ_j γ0

We get a similar commutator as we have seen before

= − ∂_μ·∂_v γ0
+ ∂^k_μ Γ^j (γ_k γ_j − γ_j γ_k) γ_1γ_2γ_3
− Γ_μ·Γ_v γ0

This yields a gravitational Poynting vector

= − (∂_μ·∂_v + Γ_μ·Γ_v)γ_0 − 2 (∂_μ × Γ_v)^k γ_k

Distributing γ_0 through gives

= − (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂_μ × Γ_v)^k a_k

Replacing for the definition of torsion

∂_μ × Γ_v = - ∂Ω/∂t

= (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂Ω_μv/∂t) ^k a_k

Instead we can also go back to

= − ∂_μ·∂_v γ0
+ ∂^k_μ Γ^j (γ_k γ_j − γ_j γ_k) γ_1γ_2γ_3
− Γ_μ·Γ_v γ0

And multiply through by γ_1γ_2γ_3 which gives back the unit pseudoscalar and a sign change

= − i∂_μ·∂_v
+ ∂^k_μ Γ^j (γ_k γ_j + γ_j γ_k)
− iΓ_μ·Γ_v

In which

i = γ_0 γ_1γ_2γ_3

Ref

http://www.av8n.com/...htm#sec-vectors

Edited by Dubbelosix, 01 October 2019 - 06:05 PM.

### #20 Dubbelosix

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Posted 01 October 2019 - 07:22 PM

Continued;

[1] when we multiply by the time like gamma, through the equation, both sides must be considered:

γ_0 γ_0 R__μv = − (∂_μ·∂_v + Γ_μ·Γ_v)γ_0 γ_0 − 2 (∂_μ × Γ_v)^k γ_k γ_0

γ_0γ_0 = - 1

- R__μv = − (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂_μ × Γ_v)^k γ_k γ_0

Removing negative signs we get

R__μv = (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂_μ × Γ_v)^k γ_k γ_0

Replacing as a simplification γ_k γ_0 = a_k

R__μv = (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂_μ × Γ_v)^k a_k

Now we replace for torsion which really gives the curvature with negative sign;

R__μv = (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂_μ × Γ_v)^k γ_k γ_0 = - (∂_μ·∂_v + Γ_μ·Γ_v) + 2 (∂Ω_μv/∂t)^k a_k

Because of the definition of torsion

∂_μ × Γ_v = - ∂Ω_μv/∂t