Jump to content


Photo
- - - - -

Gravitational Ponynting Vector


  • Please log in to reply
12 replies to this topic

#1 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 02 October 2019 - 07:15 AM

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

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

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

Unaware of a gravitational analogue I knew though there would be such cases in literature... Here is one I found and let's see how my approach differs if at all.

Gravitational Analog of the Electromagnetic Poynting Vector

L.M. de Menezes
(Submitted on 29 Jan 1998)

"The gravitational analog of the electromagnetic Poynting vector is constructed using the field equations of general relativity in the Hilbert gauge. It is found that when the gravitational Poynting vector is applied to the solution of the linear mass quadrupole oscillator, the correct gravitational quadrupole radiation flux is obtained. Further to this, the Maxwell-like gravitational Poynting vector gives rise to Einstein's quadrupole radiation formula. The gravitational energy-momentum (pseudo) tensor obtained is symmetric and traceless. The former property allows the definition of angular momentum for the free gravitational field. "

We will talk about the paper soon... But first an equation we came across, a definition of the mass current

[∇_μ, ∇_v] φ = ∂_μ·∇_v φ - iσ ⋅ (Γ_μ x ∇_v) φ = 4πG/c^2 ∇_μv j = - R_μv φ

Where [j] is known as the mass current density... Let us just look at the first term, the dimensions are simple enough;

G · Mass/length^3

Since

φ = Gm/r

And each connection has an inverse length. It can be noticed then we retrieve Poisson equation,

∇· g = -  4πGρ

Or by a substitution into the Gauss Law.

∇ · (-∇)g = -  4πGρ

The mass flow can be seen as playing the same role as the electromagnetic flux.

#2 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 03 October 2019 - 07:16 PM

The torsion is simply related to the frequency of the spin as;

Ω = - ω/2

If the gravitational is Conservative and (rotational) it allows me to express the Poynting vector as part of the spin space involving torsion with c =1.

γ_0 R_μv = − (∂_μ + Γ_μ·Γ_v)γ_0 - 2 iσ ⋅ (∂Ω_μv/∂t) ^k γ_k

= − (∂_μ + Γ_μ·Γ_v)γ_0 + iσ ⋅ (∂ω_μv/∂t) ^k γ_k

I missed that there is a relativistic correction on the current density in the form of the grav-electric field

[∇_μ, ∇_v] φ = ∂_μ·∇_v φ + iσ ⋅ (Γ_μ x ∇_v) φ

= ∂_μv ⋅ E + iσ ⋅ (Γ_μv x E) = ∂_μv ⋅ Γ(g) + iσ ⋅ (Γ_μv x Γ(g))

The torsion is the gravimagnetic field definition through units

iσ ⋅ (Γ_μv x E) = - ∂_μv B/∂t

Which needs to be built up properly, which I will do over weekend.

Edited by Dubbelosix, 04 October 2019 - 01:33 PM.


#3 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 04 October 2019 - 05:39 PM

[∇_μ, ∇_v] φ = ∂_μ·∇_v φ - iσ ⋅ (Γ_μ x ∇_v) φ = ∇_μv ⋅ E = - R_μv φ = - ∇· g = -  4πGρ

γ_0 R_μv = − (∂_μ∂_v  + Γ_μ·Γ_v)γ_0 - 2 iσ ⋅ (∂Ω_μv/∂t) ^k γ_k

= − (∂_μ∂_v  + Γ_μ·Γ_v)γ_0 + iσ ⋅ (∂ω_μv/∂t) ^k γ_k

∇_μv · E = ∂_μv ⋅ E + iσ ⋅ (Γ_μv x E)

The torsion is the gravimagnetic field definition through units

iσ ⋅ (Γ_μv x E) = - ∂_μv B/∂t

∇_μv · E = ∂_μv ⋅ E - (∂_μv B/∂t)

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

We define

∂_μv B/∂t = ∂Ω_μv/∂t

∇_μv E = ∂_μv ⋅ E - (∂_μv B/∂t) = ∂_μv ⋅ E - (∂_μv Ω/∂t)

Since the magnetic force can unify under a weak limit, we can interpret above as the equation demonstrating the role as the gravimagnetic field defined in geometric algebra. This is an interesting deviation which started analysis on a gravitational Poynting vector.

#4 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 04 October 2019 - 05:57 PM

We can define the unifying gravielectromagnetism formalism in two equations establishing the gravelectric and torsion (gravimagnetic field) ;

∇_μv E = ∂_μv ⋅ E - (∂_μv B/∂t) = ∂_μv ⋅ E - (∂_μv Ω/∂t)

∇_μv E = ∂_μv ⋅ E + iσ ⋅ (Γ_μv x E) = ∂_μv ⋅ E + (∂_μv B/∂t)

Edited by Dubbelosix, 04 October 2019 - 06:48 PM.


#5 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 05 October 2019 - 11:04 AM

We can define the unifying gravielectromagnetism formalism in two equations establishing the gravelectric and torsion (gravimagnetic field) ;

∇_μv E = ∂_μv ⋅ E + (∂_μv B/∂t)^k γ_kγ_0
= ∂_μv ⋅ E + iσ ⋅ (Γ_μv x E)^k γ_k γ_0
= ∂_μv ⋅ E + (∂Ω_μv/∂t)^k a_k

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 square;

∇_μv E = ∂_μv ⋅ E γ_1γ_2γ_3 + (∂_μv B/∂t)^k γ_kγ_0

= ∂_μv ⋅ E γ_1γ_2γ_3 + iσ ⋅ (Γ_μv x E)^k γ_k γ_0

= ∂_μv ⋅  E γ_1γ_2γ_3 + (∂Ω_μv/∂t)^k a_k

∇_μv γ_0 E = (∂_μv ⋅ E γ_1γ_2γ_3 + (∂_μv B/∂t)^k γ_k γ_0) γ_0 (∂_μv ⋅ E γ_1γ_2γ_3 + (∂_μv B/∂t)^k γ_k γ_0)

= (∂_μ ⋅ E γ_1γ_2γ_3 + iσ ⋅ (Γ_μ x E)^k γ_k γ_0) γ_0 (∂_v ⋅ E γ_1γ_2γ_3 + iσ ⋅ (Γ_v x E)^k γ_k γ_0 )

Alternatively

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_μv ⋅ E ^k γ_k γ_0 - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3)

= (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^k γ_k γ_0 + (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3)


Based on what I knew from here, I considered the following ~ starting with a curvature tensor

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

Using

(γ_1γ_2γ_3)^2 = - 1

Distributing (γ_1γ_2γ_3)

We get for one connection

∇ = ∂^k γ_k (γ_1γ_2γ_3)^2 + iσ ⋅ Γ^k γ_k γ_o γ_1γ_2γ_3

iγ_o γ_1γ_2γ_3 = γ^5

Which yields

∇ = - ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5

The master equation for gravielectromagnetism is;

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_μv ⋅ E^j γ_j γ_0 - (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

= (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

Or as a mixture

∇_μv γ_0 E = (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

So let's calculate this.

= (∂_μ ⋅ E^k γ_k γ_0 γ_0 ∂_v ⋅ E^j γ_j γ_0.
+ ∂_μ ⋅ E^k γ_k γ_0γ_0 (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3
+  iσ ⋅ (Γ_μ x E)^k γ_k γ_1γ_2γ_3 γ_0 ∂_v ⋅ E^j γ_j γ_0
+ iσ ⋅ (Γ_μ x E)^k γ_k γ_1γ_2γ_3 γ_0 (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

Using the following;

γ_0 γ_0 = - 1

iγ_1γ_2γ_3γ_0 = γ_5

(γ_1γ_2γ_3)^2 = - 1

∇_μv γ_0 E = (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

= (∂_μ ⋅ E^k γ_k ∂_v ⋅ E^j γ_j γ_0.
+ ∂_μ ⋅ E^k γ_k (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3
+  σ ⋅ (Γ_μ x E)^k γ_k γ_5 ∂_v ⋅ E^j γ_j γ_0
+ σ ⋅ (Γ_μ x E)^k γ_k γ_5 (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

Using further γ_j γ_0 = a_k we have a simplification

= (∂_μ ⋅ E^k γ_k ∂_v ⋅ E^j a_j
+ ∂_μ ⋅ E^k γ_k (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3
+  σ ⋅ (Γ_μ x E)^k γ_k γ_5 ∂_v ⋅ E^j a_j
+ σ ⋅ (Γ_μ x E)^k γ_k γ_5 (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

#6 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 05 October 2019 - 11:06 AM

Based on what I knew from here, I considered the following ~ starting with a curvature tensor

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

Using

(γ_1γ_2γ_3)^2 = - 1

Distributing (γ_1γ_2γ_3)

We get for one connection

∇ = ∂^k γ_k (γ_1γ_2γ_3)^2 + iσ ⋅ Γ^k γ_k γ_o γ_1γ_2γ_3

iγ_o γ_1γ_2γ_3 = γ_5

Which yields

∇ = - ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5

The basic goal was to have the chirality a coefficient of the spin space (Poincare Symmetry).

The master equation for gravielectromagnetism is;

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_μv ⋅ E^j γ_j γ_0 - (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

= (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3)

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

iσ ⋅ (Γ_μ x E)^k γ_k γ_0 = (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3

Multiplying through by γ_1γ_2γ_3 yields

σ ⋅ (Γ_μ x E)^k γ_k γ_5 = - (∂_μv B/∂t)^k γ_k

From using

iγ_o γ_1γ_2γ_3 = γ_5

Also because of Hermiticity σ^2 = 1 we also can write it as

(Γ_μ x E)^k γ_k γ_5 = - (∂_μv (σ ⋅ B ) /∂t)^k γ_k

= - (∂_μv (σ ⋅ Ω) /∂t)^k γ_k = 1/2 (∂_μv (σ ⋅ ω) /∂t)^k γ_k

Edited by Dubbelosix, 05 October 2019 - 11:06 AM.


#7 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 08 October 2019 - 07:49 AM

The gravitational Poynting vector was shown to be ~

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

Thisis the analogue principle covariant derivatives, which are often dented by gravelectric and gravimagnetic fields as

E_μ γ_0 B_v = (E_μ⋅E_v + B_μ⋅B_v)γ_0 + iσ · (E_μ × B_v)^kγ_k

This is the same thing except we define the connections as gravieletric and gravimagnetic fields.Being the Ponynting vector in terms of the derivatives concerning interest in the field strengths and the spin-torsion contribution.

Dirac equation

(i∂_μv−eA_μv/L)ψ = σ ⋅(mc/L)_μv ψ = (σ ⋅ ∂_μv) ψ

(i∂_μ−eAμ/L)(i∂_v−eA_v/S)ψ = (m^2c^2/L ⋅ S)_μv ψ = (σ ⋅ ∂_μ) (σ ⋅ ∂_v) ψ

By deduction the last term reveals the geometric algebra;

(σ ⋅ ∂_μ) (σ ⋅ ∂_v) = ∂_μ⋅∂_v + (∂_μ ∧ ∂_v)

= ∂_μ⋅∂_v + iσ ⋅ (∂_μ x ∂_v)

(i∂_μ− eA_μ/L)(i∂_v - eA_v/S)ψ

= (∂_μ ∂_v + i[∂_μ (eA_v/S) - (eA_μ/L) ∂_v] + iσ ⋅ [eA_μ/L × eA_v/S] )ψ

The L and S commute and there is indication the subscripts of the curvature tensor may not always commute R_[μv]. Since J is the total angular momentum

J = L+S

So taking the dot product gives

= L^2 + S^2 + 2 L ⋅ S

We will come back to this.

Edited by Dubbelosix, 08 October 2019 - 04:09 PM.


#8 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 08 October 2019 - 08:47 PM

The Dirac equation is presented here with an extra derivative

iℏ ∂^k/∂x γ_k γ_0 γ_1γ_2γ_3 ∂_μ ∂^μ ψ = ❏ γ_0 γ_k mc ψ

Using geometric algebra we obtain

❏_μv = ∂_μ ⋅ ∂_v + iσ ⋅ [Γ_μ x ∂_v]

By using

iγ_o γ_1γ_2γ_3 = γ_5

iℏ ∂^k/∂x γ_k γ_0 γ_1γ_2γ_3 ∂_μ∂^v ψ = (∂_μ ⋅ ∂_v + iσ ⋅ [Γ_μ x ∂_v])^k γ_k γ_o mc ψ

Distributing γ_1γ_2γ_3 we get

- iℏ ∂^k/∂x γ_k γ_0 ∂_μ∂^v ψ
= ∂_μ ⋅ ∂_v^k γ_kγ_oγ_1γ_2γ_3 + σ ⋅ [Γ_μ x ∂_v]^k γ_k γ_5 mc ψ

(i∂^k_μ γ_kγ_0 −eA^k_μ/L γ_kγ_1γ_2γ_3 )(i∂^k_v γ_k γ_0 − (eA^j_v/S) γ_jγ_1γ_2γ_3)ψ

= (σ ⋅ ∂_μ) (σ ⋅ ∂_v) ψ

(i∂^k_μ γ_kγ_0 − (eA^k_μ/L) γ_kγ_1γ_2γ_3 )(i∂^j_v γ_j γ_0 − (eA^j_v/S) γ_j γ_1γ_2γ_3)ψ

= ∂^k_μ γ_k γ_0γ_0 ∂^j_v γ_j
+ i[∂^k_μ γ_k γ_0 (eA^j_v/S) γ_j γ_1γ_2γ_3 - (eA^k_μ/L) γ_k γ_1γ_2γ_3 ∂^j_v γ_j γ_0]
+ (eA^k_μ/L) γ_kγ_1γ_2γ_3(eA^j_v/S) γ_j γ_1γ_2γ_3

Edited by Dubbelosix, 09 October 2019 - 03:40 AM.


#9 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 09 October 2019 - 06:22 AM

iℏ ∂^k_μ γ_k γ_0 - eA^k_μ γ_kγ_1γ_2γ_3 = mc

This is a more accurate version for the vaults - I will still want to search the second derivative.

Edited by Dubbelosix, 09 October 2019 - 09:51 AM.


#10 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 09 October 2019 - 01:17 PM

Here are my thoughts and my final prediction for the Dirac equation, all those last thoughts combined in this last post:

In the last attempt we looked at an equation which had Klein-Gorden equation

(i∂_μv−eA_μv/L)ψ = σ ⋅(mc/L)_μv ψ = (σ ⋅ ∂_μv) ψ

(i∂_μ−eAμ/L)(i∂_v−eA_v/S)ψ = (m^2c^2/L ⋅ S)_μv ψ = (σ ⋅ ∂_μ) (σ ⋅ ∂_v) ψ

By deduction the last term reveals the geometric algebra;

(σ ⋅ ∂_μ) (σ ⋅ ∂_v) = ∂_μ∂_v + iσ ⋅ (∂_μ x ∂_v)

The derivative itself can be defined by a four dimensional case. Simply, the dAlembertian in four dimensional space using geometric algebra is the Dirac equation is

❏_μv = ∂_μ ⋅ ∂_v + iσ ⋅ [Γ_μ x ∂_v]

With a mass parameter, this is the Dirac equation using geometric algebra.

❏_μv + m^2_μvc^2/ℏ^2 = (∂_μ ⋅ ∂_v + iσ ⋅ Γ_μ x ∂_v ) + m^2_μvc^2/ℏ^2

We get for one connection

∇ = ∂^k γ_k (γ_1γ_2γ_3)^2 + iσ ⋅ Γ^k γ_k γ_o γ_1γ_2γ_3

iγ_o γ_1γ_2γ_3 = γ^5

Which yields

∇ = - ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5

The new covariant deravative I suggest is

∇ = - ∂^k γ_k + σ ⋅ Γ^k γ_k γ^5

Then the box operator is:

❏ = - (∂^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) (∂^j γ_j + σ ⋅ Γ^j γ_j γ^5)

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

The Dirac equation is implied through the geometric algebra from initially the Klein-Gorden equation and we carried it through showing it was a a type of diffusion equation with a mass parameter

❏_μv + m^2_μvc^2/ℏ^2 = (∂_μ ⋅ ∂_v + iσ ⋅ Γ_μ x ∂_v ) + m^2_μvc^2/ℏ^2

Edited by Dubbelosix, 09 October 2019 - 03:14 PM.


#11 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 09 October 2019 - 06:22 PM

The Dirac equation again was;

❏_μv + m^2c^2/ℏ^2 = (∂_μ ⋅ ∂_v + iσ ⋅ Γ_μ x ∂_v ) + m^2c^2/ℏ^2

When the operator is scrutanized under the same rigour as the curvature tensor which was analogous to a geometric product form of the Ponynting equation.

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

It might even be argued that the second term plays the density roles of the gravelectric and gravimagnetic fields in relativistic ways, which can be seen as a correction term;

❏_μv + m^2c^2/ℏ^2 = (∂_μ·∂_v + Γ_μ·Γ_v) + 2iσ ⋅ (Γ_μ x ∂_v ) + m^2c^2/ℏ^2 = 0

How true this is will remain a matter for relativists; I really want to move this along - out of this form of the Dirac equation create it first as a diffusion equation and then we will transpose that into curved space such as found in the language of general relativity while retaining the geometric algebra space.

The operator acting on the potential is a type of diffusion equation ~

❏φ = g^μv❏_μv φ = g^μv ∇_μ(∂_v φ )

And in geometric algebra language the derivatives just act accordingly;

❏φ = g^μv❏_μv φ = g^μv (∇_μ · (∂_v φ ) + 2iσ ⋅ [∇_μ x ∂_v]) φ

Further it can be defined;

❏φ = g^μν(∂_μ∂_ν−Γ^σ_μν ∂_σ)φ = (∂_μ∂^μ−g_μν Γ^σ_μν ∂_σ)φ

But under geometric algebra this would become

❏φ = g^μν(∂_μ ⋅ ∂_ν − 2iσ ⋅ [Γ^σ_μν x ∂_σ] )φ

This will properly demonstrate a diffusion equation when involving the diffusion coefficient;

α g^μν(∂_μ ⋅ ∂_ν − 2iσ ⋅ [Γ^σ_μν x ∂_σ] )φ

Edited by Dubbelosix, 09 October 2019 - 06:25 PM.


#12 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted 11 October 2019 - 12:40 PM

The master equation for gravielectromagnetism is;

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_μv ⋅ E^j γ_j γ_0 - (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

= (∂_μ ⋅ E^k γ_k γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + iσ ⋅ (Γ_μ x E)^k γ_kγ_1γ_2γ_3)

∇_μv γ_0 E = (∂_μv ⋅ E ^k γ_k γ_0) - (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3) γ_0 (∂_v ⋅ E^j γ_j γ_0 + (∂_μv B/∂t)^j γ_j γ_1γ_2γ_3)

iσ ⋅ (Γ_μ x E)^k γ_k γ_0 = (∂_μv B/∂t)^k γ_k γ_1γ_2γ_3

Multiplying through by γ_1γ_2γ_3 yields

σ ⋅ (Γ_μ x E)^k γ_k γ_5 = - (∂_μv B/∂t)^k γ_k

From using

iγ_o γ_1γ_2γ_3 = γ_5

Also because of Hermiticity σ^2 = 1 we also can write it as

(Γ_μ x E)^k γ_k γ_5 = - (∂_μv (σ ⋅ B ) /∂t)^k γ_k

= - (∂_μv (σ ⋅ Ω) /∂t)^k γ_k = 1/2 (∂_μv (σ ⋅ ω) /∂t)^k γ_k


This particular form;

(Γ_μ x E)^k γ_k γ_5 = - (∂_μv (σ ⋅ B ) /∂t)^k γ_k

Is interesting because the energy associated to the gravimagnetic energy would be associated to a coefficient capable of transposing the units just much like how the Bohr magneton produces;

H = μ_B [σ ⋅ B] = total spin energy

Which describes the gravimagnetic interaction energy, so long as (B ) relates to the gravimagnetic component, otherwise this is a standard spin interaction. The full Hamiltonian extends into what is called the Pauli formalism;

ℋ = 1/2m(p - eA)^2 - eL/2m (σ ⋅ B )

In which

μ_B = eL/2m

Is the Bohr magneton.

In the language of the gamma matrices I can write this as

ℋ = 1/2m(p^k γ_k γ_0 - eA^j γ_j γ_1γ_2γ_3)^2 - eL/2m (σ ⋅ B )

Which can be expanded as

γ_0 ℋ = 1/2m (p^k γ_k γ_0 - eA^k γ_k γ_1γ_2γ_3) γ0 (p^j γ_j γ_0 - eA^j γ_j γ_1γ_2γ_3)

= 1/2m (p^k γ_k γ_0 γ_0 p^j γ_j γ0
− p^k γ_k γ_0 γ_0 eA^j γ_j γ_1γ_2γ_3 − eA^k γ_k γ_1γ_2γ_3 γ_0 p^j γ_j γ_0
+ eA^k γ_k γ_1γ_2γ_3 γ_0 eA^j γ_j γ_1γ_2γ_3)

= 1/2m (- p^k γ_k p^j γ_j γ0
+ p^k γ_k eA^j γ_j γ_1γ_2γ_3 − eA^k γ_k γ_1γ_2γ_3 p^j γ_j
- eA^k γ_k γ_0 eA^j γ_j)

By using

γ_0γ_0 = - 1

(γ_1γ_2γ_3)^2 = - 1

= 1/2m (- p^k γ_k p^j γ_j γ0
+ p^k γ_k eA^j γ_j γ_1γ_2γ_3 − eA^j γ_j γ_1γ_2γ_3 p^k γ_k
- eA^k γ_k eA^j γ_j γ_0)

= − 1/2m ( p·p γ_0 + p^k eA^j (γ_k γ_j − γ_j γ_k) γ_1γ_2γ_3 − (eA · eA) γ_0

Since they commutate, that is,

p^k eA^j (γ_k γ_j − γ_j γ_k) γ_1γ_2γ_3 = 0

we get simply ~

= 1/2m [p · p + eA · eA + eL (σ ⋅ B )]

It is fortuitous that in literature the Pauli spin matrix is connected to the magnetic moment interaction and we can label that as such

= 1/2m [(σ ⋅ p) · (σ ⋅ p) + (σ ⋅ eA) · (σ ⋅ eA) + eL (σ ⋅ B )]

We should know by now how the geometric algebra is theoretically achieved from similar products we looked at before, such as the investigation of the wave equation ~

(i∂_μv−eA_μv/L)ψ = σ ⋅(mc/L)_μv ψ = (σ ⋅ ∂_μv) ψ

(i∂_μ−eAμ/L)(i∂_v−eA_v/S)ψ = (m^2c^2/L ⋅ S)_μv ψ = (σ ⋅ ∂_μ) (σ ⋅ ∂_v) ψ

By deduction the last term reveals the geometric algebra;

(σ ⋅ ∂_μ) (σ ⋅ ∂_v) = ∂_μ∂_v + iσ ⋅ (∂_μ x ∂_v)

... And we have two such terms in our equation on the RHS:

ℋ = 1/2m [(σ ⋅ p) · (σ ⋅ p) + (σ ⋅ eA) · (σ ⋅ eA) + eL (σ ⋅ B )]

By inspection we can write this in the following way;

ℋ = 1/2m [(σ ⋅ p) · (σ ⋅ p) + (σ ⋅ eA) · (σ ⋅ eA) + eL (σ ⋅ B )]

= 1/2m [(p · p + eA · eA) γ_0 + i2 σ ⋅ (L x B)^k e γ_k]

In this theory the cross product is concerned as a coupling between angular momentum and the magnetic field as described under these equations of motion. This is to be an expected result as the cross product is strongly tied to the kinetic rotation of the body. The equation also forms similar in form from previous application of the same algebra in different set ups, for instance a gravitational Poynting vector analog:

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

If the derivatives are renamed gravelectric and gravimagnetic fields, then the framework consists of the same results, just different notation and interpretation.

E γ_0 B = (E⋅E + B⋅B )γ_0 + iσ⃗ · (E × B )^kγ_k

Even the derivative of the gravelectric field holds within it, the same direct consequences of the cross product;

∇_μv E = ∂_μv ⋅ E + (∂_μv B/∂t)^k γ_kγ_0
= ∂_μv ⋅ E + σ ⋅ (Γ_μ x E)^k γ_k γ_5
= ∂_μv ⋅ E + (∂_μv Ω/∂t)^k γ_kγ_0

Which in this form written from previous work, involves clear definitions of torsion from geometric space in a very blatant way. If we did the same calculation for two connections of the gravitational field, like we did in previous work, we saw that I obtained;

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

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

I fudged the last part − 2 (∂_μ × Γ_v)^k γ_k as a predictive term because the commutation still applies

(γ_k γ_j − γ_j γ_k) = 0

But the same result applied for the full Hamiltonian

ℋ = 1/2m(p^k γ_k γ_0 - eA^j γ_j γ_1γ_2γ_3)^2 - eL/2m (σ ⋅ B )

In which we constructed the cross product, not from the expansion but from an additional coupling term to the energy. The third term can be predicted a number of different ways, the fact it remains here is nothing new.

Edited by Dubbelosix, 11 October 2019 - 03:50 PM.


#13 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 3040 posts

Posted Yesterday, 08:13 AM

Let's now go back to the bivector curvature tensor, I argued that;

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

And using;

γ_0γ_0 = - 1

We got

- 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 σ ⋅ (∂_μ × Γ_v)^k γ_k γ_0

Let's use the ordinary potential

φ = Gm/r

The bivector theory, being linearized in geometry ensures a gravielectric field as

R_μv φ = [(∂_μ·∂_v + Γ_μ·Γ_v) + 2 σ ⋅ (∂_μ × Γ_v)^k γ_k γ_0] Gm/r = - 4πGρ

The acceleration in this case is attributed to the standard Poisson equation and depends on the mass of the body in the two-body problem. Each connection has an inverse length and is also the definition of the gravitational field . It can be noticed then we retrieve Poisson equation,

∇· g = -  4πGρ

Or by a substitution into the Gauss Law.

∇ · (-∇) g = -  4πGρ

The mass flow can be seen as playing the same role as the electromagnetic flux. The gravielectric field, with a magnetic potential is in standard form;

E = - ∇φ - 1/c (∂A/∂t)

And the gradient of the field;

∇ · E = ∇ · (-∇) φ - 1/c^2 (∂^2A/∂t^2)

Using the Lorentz gauge, Maxwell's equations can be written in terms of the potential and the scalar potential

∇^2ϕ−1/c^2 (∂ϕ/∂t)^2 = − ρ ϵ(G)

∇^2A−1/c^2 (∂A/∂t)^2 = − μ(G) j

Where j is the current and ϵ(G), μ(G)) the gravitational permittivity and permeability. We should also note we have had the Poisson equation arise from the diffusion equation

We have investigated the Poisson equation before, in which it arises from the relativistic diffusion equation that I explored:

α(η_μν ∂^μ∂^ν ϕ)= α □ϕ = α(∂^2ϕ/∂τ^2+∇^2ϕ)