# The Lagrangian equation...

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Added warning tensors calculus has two meanings for the word covariant you have covariant vectors=covector and covariance which is a principle similar meaning to the laws of physics is the same all reference frames Google Lorentz covariance. Then you also have the covariant derivatives and contravariant derivatives. (Just to make things confusing lmao)

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$\begin{pmatrix}\acute{d}\\\acute{s}\\\acute{b}\end{pmatrix}\begin{pmatrix}V_{ud}&V_{us}&V_{ub}\\V_{cd}&V_{cs}&V_{cb}\\V_{td}&V_{ts}&V_{tb}\end{pmatrix}\begin{pmatrix}d\\s\\b\end{pmatrix}$

Electroweak correlations

$\mathcal{L}=\mathcal{L}_{gauge}+\mathcal{L}_f+\mathcal{L}_\phi+\mathcal{L}_{yuk}$

Gauge sector

$\mathcal{L}_{gauge}=\frac{1}{4}W^i_{\mu\nu}W^{\mu\nu I}-\frac{1}{4}B_{\mu\nu}B^{\mu\nu}$

Where $W_{\mu\nu}$ and $B_{\mu\nu}$ are the SU(2)and U(1) field strength tensors.

$W^i_{\mu\nu}=\partial_\mu^1-\partial W^i_\nu-\partial_\nu W^i_\mu-g\epsilon_{ijk}W_{\mu}^jW^k_\nu W^k_\nu$

$B^i_{\mu\nu}=\partial_\mu B_\nu-\partial_\nu B_\mu$

$\epsilon_{ijk}$ group structure constants of SU(2) B of U(1) abelion group has no self interaction (gauge boson)

$\mathcal{f}\subset\Sigma(\bar{q}+\bar{\ell}i\displaystyle{\not}D \ell)$

q is quark $\ell$ is leptons it sums over generations.

The quage covariant derivative is

$D_q=(\partial_\mu+\frac{ig}{2}\vec{\tau}\cdot\vec{W}_\mu+i\acute{g}Y\cdot B_\mu)q$

$\displaystyle{\not}D=\gamma D^\mu$ g and $\acute{g}$ are the gauge coupling constants of $SU(2)_w$ and $U(1)_y$ $\vec{\tau}$ refers to Pauli matrices. Y is hypercharge of U(1) the electric charge Q is

$Q=I_3+\frac{1}{2}Y$

langrangian for complex scalar fields.

$\mathcal{L}_\phi=(D^\mu)^\dagger D_\mu\phi-V(\phi)$

$D_\mu \phi=(\partial_\mu+\frac{ig}{2}\vec{\tau}\cdot\vec{W}_\mu+\frac{i\acute{g}}{2}B_\mu)\phi$

$V(\phi)=\mu^2\phi^\dagger\phi+\lambda(\phi^\dagger\phi)^2$

Lambda is the self interaction term

Edited by Mordred

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Sorry I can't contribute to the post but seriously.....Holy s#-t, there's  some serious math in this post my poor old iPad took at least 5 mins just to load it all up. That doesn't even happen on math forums I go to. I'm seriously envious of people who can do math/physics at this level. If you can understand this thread, I hope your ugly...

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Both Orion and myself do, we're breaking apart the relations that went into the OP Langrene. Right now I'm trying to determine if it's canonical or conformal by looking at the EW Langrene through symmetry break via the Higgs. Which will confirm the Higgs and Yukawa couplings underbrace sections. We're both learning from this gives us a refreshing challenge. The Yukawa section is rather challenging.

I've already confirmed the Higgs and Dirac covariant derivative forms.

Edited by Mordred

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Pulling this post back to where I can find it to add Yukawa couplings details this weekend

$\mathcal{G}=SU(3)_c\otimes SU(2)_L\otimes U(1)_Y$

Color, weak isospin, abelion Hypercharge groups.

Couplings in sequence $g_s, g, \acute{g}$

$\mathcal{L}_{gauge}=-\frac{1}{2}Tr{G^{\mu\nu}G_{\mu\nu}}-\frac{1}{2}Tr {W^{\mu\nu}W_{\mu\nu}}-\frac{1}{4}B^{\mu\nu}B_{\mu\nu}$

Field strengths in sequence in last G W B tensors for SU(3),SU(2) and U(1)

$D_\mu=\partial_\mu+ig_s\frac{\lambda_i}{2}G^i_\mu+ig\frac{\sigma_i}{2}W^i_\mu+igQ_YB_\mu$

Corresponds to

$G_{\mu\nu}=-\frac{i}{g_s}[D_\mu,D_\nu]$

$W_-\frac{I}{g}[D_{\mu}D_{\nu}]$

$B_{\mu\nu}-\frac{I}{\acute{g}}[D_\mu,D_\nu]$

Edited by Mordred

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Higgs field with above

$\mathcal{L}=(D_\mu H)^\dagger D^\mu H-\lambda(H^\dagger H-\frac{v^2}{2})^2$

v=246 GeV

Quartic coupling

$\lambda=m_h^2/2v^2=0.13$

$\langle H^\dagger H\rangle =v^2/2$

Fermions (matter content) (goal tie in CKMS and Pmns mixing angles (latter for leptons)) will require unity triangle...

$\displaystyle{\not}D=\gamma D^\mu$

self reminder

Feymann slash contraction of the gamma matrix with a four vector

$\displaystyle{\not}a=\gamma a^\mu a_\nu=\gamma_\mu a^\nu$ a is any four vector.

Edited by Mordred

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Posted (edited)
On 8/13/2019 at 7:59 PM, Mordred said:

one is a proof of Lorentz invariance using the inner products of the four momentum which is the expression you have above however those are the two transformation matrices of each of the two four momentum components as seen from two reference frames. S and S prime

Affirmative, revision complete.

Metric tensor field: (ref. 1, pg. 21, eq. 1.68)
$T^{'\mu \nu} \left(x' \right) = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} T^{\alpha \beta}\left(x \right)$
$\;$
In geometry, the line element or length element can be informally thought of as a line segment associated with an infinitesimal displacement vector in a metric space. The length of the line element, which may be thought of as a differential arc length, is a function of the metric tensor.
$\;$
General Relativity line element where spacetime is modelled as a curved Pseudo-Riemannian manifold with an appropriate metric tensor: (ref. 3, ref 4, pg. 15, eq. 41, para. 1)
$ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$
$\;$
The General Relativity line element with a curved Pseudo-Riemannian manifold metric tensor condition imposes constraints on the coefficients $\Lambda^{\mu}_{\nu}$: (ref. 3, pg. 1, eq. 3)
$g_{\mu \nu} = g_{\alpha \beta} \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu}$
$\;$
General Relativity curved Pseudo-Riemannian manifold line element identity:
$\boxed{ds^2 = g_{\alpha \beta} \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu} dx^{\mu} dx^{\nu}}$
$\;$
General Relativity matrix symmetry expression:
$\mu \cdot \nu = \nu \cdot \mu = \eta$
$\;$
Jacobian matrix transformation matrices: (ref. 4, ref 5, pg. 15, eq. 41, para. 1)
$\Lambda^{\mu}_{\alpha} = \frac{\partial \xi^{\mu} }{\partial x^{\alpha}} \; \; \; \; \; \; \Lambda^{\nu}_{\beta} = \frac{\partial \xi^{\nu}}{\partial x^{\beta}}$
$\;$
$x^{\alpha} = \left(ct, r, \theta, \phi \right) \; \; \; \; \; \; x^{\beta} = \left(ct, r, \theta, \phi \right)$
$\;$
$\Lambda^{\mu}_{\alpha} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{\partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}}$
$\;$
$\;$
$\Lambda^{\nu}_{\beta} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{\partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}}$
$\;$
Is this approach mathematically and symbolically correct to this point?
$\;$
What is the formal mathematical definition for $g_{\alpha \beta}$?
$\;$
$\;$
Reference:
Lorentz Group and Lorentz Invariance: (ref. 1)
https://gdenittis.files.wordpress.com/2016/04/ayudantiavi.pdf
Wikipedia - General relativity - Metric tensor - Local coordinates and matrix representations: (ref. 2)
https://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)#Local_coordinates_and_matrix_representations
Lorentz Transformations - Bernard Durney: (ref. 3)
https://arxiv.org/pdf/1103.0156.pdf
Introduction to Tensor Calculus for General Relativity - Edmund Bertschinger: (ref. 4)
https://web.mit.edu/edbert/GR/gr1.pdf
Wikipedia - Jacobian matrix: (ref. 5)
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Example_3:_spherical-Cartesian_transformation

Edited by Orion1

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Posted (edited)

$\mu\cdot\nu=\nu\cdot\mu=\eta$ is the inner product symmetry relations for the Minkowskii metric tensor. You identified it as the GR symmetry matrix expression.

You need the first order partials for the Jacobian  matrix while I don't have polar coordinate form handy you can look here for the Minkowskii form though you will have to switch the signature.

The one forms mentioned are invariant under coordinate change

Edited by Mordred

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5 hours ago, Mordred said:

You identified it as the GR symmetry matrix expression.

Affirmative, revision complete.

$\eta_{\mu \nu}$ - perturbed non-dynamical background metric
$\;$
General Relativity Minkowski flat spacetime metric: (ref. 1)
$\eta_{\mu \nu} = \pm \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
$\;$
General Relativity Minkowski flat spacetime metric is equivalent to the inverse metric: (ref.1, ref. 2)
$\boxed{\eta_{\mu \nu} = \eta^{\mu \nu}}$
$\;$
What is the formal mathematical definition for $g_{\alpha \beta}$?
$\;$
$\;$
Reference:
Wikipedia - Four-gradient As a Jacobian matrix for the SR Minkowski metric tensor: (ref. 1)
Wikipeda - Lorentz covariance: (ref. 2)
https://en.wikipedia.org/wiki/Lorentz_covariance

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Posted (edited)

$g_{\alpha\beta}$ is the metric tensor the indices run (0,1,2,3).

The form will vary according to the spacetime being modelled it can have either or both the covariant and contravariant terms accordingly to the Einstein summation convention. In the above its specifying covariant.

Edited by Mordred

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19 minutes ago, Mordred said:

Undefined control sequence \g

ah yes, breaking the fabric of reality I see.

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Lol correction applied

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Posted (edited)

Here is one example

Schwartzchild metric

Vacuum solution $T_{ab}=0$ which corresponds to an unaccelerated freefall frame $G_{ab}=dx^adx^b$ if

$ds^2> 0$ =spacelike propertime= $\sqrt{ds^2}$

$ds^2<0$ timelike =$\sqrt{-ds^2}$

$ds^2=0$ null=lightcone

spherical polar coordinates $(x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)$

$G_{\alpha\beta} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}$

line element

$ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)$

Edited by Mordred

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2 hours ago, Mordred said:

Vacuum solution Tab=0 which corresponds to an unaccelerated freefall frame Gab=dxadxb

Surely you meant to write $$g_{\alpha \beta}$$, since $$G_{\alpha \beta}$$ denotes the Einstein tensor, which is a different quantity.

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Posted (edited)

Lol I always seem to get caught by spell check lol. Correction made thanks for the catch

Edited by Mordred

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Posted (edited)
On 1/5/2020 at 5:23 PM, Mordred said:

You need the first order partials for the Jacobian  matrix while I don't have polar coordinate form handy you can look here for the Minkowskii form though you will have to switch the signature.

Affirmative, revision complete.

Metric tensor field: (ref. 1, pg. 21, eq. 1.68)
$T^{'\mu \nu} \left(x' \right) = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} T^{\alpha \beta}\left(x \right)$
$\;$
In geometry, the line element or length element can be informally thought of as a line segment associated with an infinitesimal displacement vector in a metric space. The length of the line element, which may be thought of as a differential arc length, is a function of the metric tensor.
$\;$
General Relativity line element where spacetime is modelled as a curved Pseudo-Riemannian manifold with an appropriate metric tensor: (ref. 3, ref 4, pg. 15, eq. 41, para. 1)
$ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$
$\;$
The General Relativity line element with a curved Pseudo-Riemannian manifold metric tensor condition imposes constraints on the coefficients $\Lambda^{\mu}_{\nu}$ (ref. 3, pg. 1, eq. 3)
$g_{\mu \nu} = g_{\alpha \beta} \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu}$
$\;$
General Relativity curved Pseudo-Riemannian manifold line element identity:
$\boxed{ds^2 = g_{\alpha \beta} \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu} dx^{\mu} dx^{\nu}}$
$\;$
The $g_{\alpha \beta}$ metric tensor will vary according to the spacetime being modeled. It can have either or both the covariant and contravariant terms accordingly to the Einstein summation convention. In this form it is specifying covariant.
$g_{\alpha \beta} = \pm \begin{pmatrix} -\xi^0 & 0 & 0 & 0 \\ 0 & \xi^1 & 0 & 0 \\ 0 & 0 & \xi^2 & 0 \\ 0 & 0 & 0 & \xi^3 \end{pmatrix}$
$\;$
In spherical coordinates $(ct, r, \theta, \phi)$ the Minkowski flat spacetime metric takes the form:
$ds^{2} = -c^{2} dt^{2} + dr^{2} + r^{2} d\theta^{2} + r^{2} \sin^{2} \theta \; d\phi^{2}$
$\;$
The Minkowski flat spacetime metric in covariant form:
$ds \; ds' = -c^{2} dt dt' + dr dr' + r r' d\theta \; d\theta' + r r' \sin \theta \sin \theta' \; d\phi \; d\phi'$
$\;$
General relativity stress-energy tensor:
$T_{\mu \nu} = \pm \left(\begin{matrix} -\rho c^{2} & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$
$\;$
General Relativity Minkowski flat spacetime metric tensor in covariant form:
$T^{\alpha \beta} \left(r \right) = \pm \begin{pmatrix} -c^{2} dt dt' & 0 & 0 & 0 \\ 0 & dr dr' & 0 & 0 \\ 0 & 0 & r r' d\theta \; d\theta' & 0 \\ 0 & 0 & 0 & r r' \sin \theta \sin \theta' \; d\phi \; d\phi' \end{pmatrix}$
$\;$
Jacobian matrix transformation matrices: (ref. 4, ref 5, pg. 15, eq. 41, para. 1)
$\Lambda^{\mu}_{\alpha} = \frac{\partial \xi^{\mu} }{\partial x^{\alpha}} \; \; \; \; \; \; \Lambda^{\nu}_{\beta} = \frac{\partial \xi^{\nu}}{\partial x^{\beta}}$
$\;$
$x^{\alpha} = \left(ct, r, \theta, \phi \right) \; \; \; \; \; \; x^{\beta} = \left(ct, r, \theta, \phi \right)$
$\;$
$\Lambda^{\mu}_{\alpha} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{c^{2} \partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}} = {\begin{bmatrix} -2 dt & 0 & 0 & 0 \\ 0 & 2 dr & 0 & 0 \\ 0 & 0 & 2r^2 d\theta & 0 \\ 0 & 0 & 0 & 2r^2 \sin^{2} \theta \; d\phi \\ \end{bmatrix}}$
$\;$
$\;$
$\Lambda^{\nu}_{\beta} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{c^{2} \partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}} = {\begin{bmatrix} -2 dt & 0 & 0 & 0 \\ 0 & 2 dr & 0 & 0 \\ 0 & 0 & 2r^2 d\theta & 0 \\ 0 & 0 & 0 & 2r^2 \sin^{2} \theta \; d\phi \\ \end{bmatrix}}$
$\;$
The matrix dot product transformation matrices in covariant form for each of the two four-momentum components as seen from two reference frames, S and S' prime:
$\Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} = {\begin{bmatrix} 4 dt dt' & 0 & 0 & 0 \\ 0 & 4 dr dr' & 0 & 0 \\ 0 & 0 & 4r^2 r'^{2} d\theta \; d\theta' & 0 \\ 0 & 0 & 0 & 4r^2 r'^{2} \sin^{2} \theta \sin^{2} \theta' \; d\phi \; d\phi' \\ \end{bmatrix}}$
$\;$
General Relativity weak field limit spacetime metric and Planck quantum gravity identity 4: (ref. 6)
$\boxed{\frac{8 \pi G}{c^{4}} T_{\mu \nu} = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} T^{\alpha \beta} \left(x \right) \left(1 - \frac{1}{2} \left(\eta^{\mu \nu} - h^{\mu \nu} \right)\left(\eta_{\mu \nu} + h_{\mu \nu} \right) \right)}$
$\;$
Solution 1: $\mu = \nu = 0$
$\boxed{2 \pi G \rho \left(r \right) = c^{4} dt^{2} dt'^{2}}$
$\;$
Solution 2: $\mu = \nu = 1$
$\boxed{2 \pi G p \left(r \right) = c^{4} dr^{2} dr'^{2}}$
$\;$
Solution 3: $\mu = \nu = 2$
$\boxed{2 \pi G p \left(r \right) = c^{4} dr^{3} dr'^{3} d\theta^{2} \; d\theta'^{2}}$
$\;$
Solution 4: $\mu = \nu = 3$
$\boxed{2 \pi G p \left(r \right) = c^{4} dr^{3} dr'^{3} \sin^{3} \theta \; \sin^{3} \theta' d\phi^{2} \; d\phi'^{2}}$
$\;$
Is this approach mathematically and symbolically correct to this point?
$\;$
$\;$
Reference:
Lorentz Group and Lorentz Invariance: (ref. 1)
https://gdenittis.files.wordpress.com/2016/04/ayudantiavi.pdf
Wikipedia - General relativity - Metric tensor - Local coordinates and matrix representations: (ref. 2)
https://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)#Local_coordinates_and_matrix_representations
Lorentz Transformations - Bernard Durney: (ref. 3)
https://arxiv.org/pdf/1103.0156.pdf
Introduction to Tensor Calculus for General Relativity - Edmund Bertschinger: (ref. 4)
https://web.mit.edu/edbert/GR/gr1.pdf
Wikipedia - Jacobian matrix: (ref. 5)
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Example_3:_spherical-Cartesian_transformation
Science Forums - Lagrangian equation for a massless Planck graviton - Orion1: (ref. 6)
https://www.scienceforums.net/topic/117992-the-lagrangian-equation/?do=findComment&comment=1096894

Edited by Orion1

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Posted (edited)

Much better for the Jacobian in spherical weak field limit. However It looks to me your graviton application is spin 1 dipolar. You need quaternion relations  spin 2 (quadrupolar). To match up to gravity wave data. ( Though I will have to study your equations for reference  6 further)

Let dig up some good graviton modelling and the relevant GR spin 2 application.

Keep in mind in order to properly model the gravitons you will need it's wavefunction for transerve and longitudinal component. (In gauge treatments you must be renormalizable. )

Edited by Mordred

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Posted (edited)

Here is one of the better papers for the graviton spin 2.

Note equation $g_{\mu\nu}\rightarrow \eta_{\mu\nu}+k h_{\mu\nu}$

The article provides the general spin Compton scattering for the other spin statistics as well as spin 2.

My recommendation is to start with the linearized Einstein Hilbert action. See the following Doctorate thesis. (It's a common methodology for modelling the graviton)

By using the Einstein Hilbert action your already working in quanta Ie quanta of action and thus can make the correlations to the creation and annihilation operators for the Feymann path integrals. Though the difficulty will be avoiding infinite one loop corrections. Divergence that ruins renormalization

Edit almost forgot in the above equation

$k^2=16\pi G$ I'm not sure any of the above articles mention that. The above formula is a fairly standard equation in numerous papers for massless spin 2 propogators.

Edited by Mordred

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See here for further detail you will also find the one loop vacuum polarization propogator handy

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On 1/7/2020 at 10:57 PM, Mordred said:

your graviton application is spin 1 dipolar. You need quaternion relations spin 2 (quadrupolar).

Affirmative, revision complete.

First order metric tensor field for spin-0 and spin-1 particles: (ref. 1, pg. 21, eq. 1.68)
$T^{'\mu \nu} \left(x' \right) = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} T^{\alpha \beta}\left(x \right)$
$\;$
Second order metric tensor field for spin-2 particles: (ref. 3)
$M^{'\mu \nu} \left(x' \right) = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} M^{\alpha \beta}\left(x \right)$
$\;$
The graviton is a spin-two particle, as opposed to the spin-one photon, so that the interaction forms are somewhat more complex, involving symmetric and traceless second order tensors rather than simple Lorentz four-vectors.
$\;$
In relativistic mechanics, the Center Of Mass-Energy boost and orbital 3-space angular momentum of a rotating object are combined into a four-dimensional bivector in terms of the four-position X and the four-momentum P of the object. (ref. 3)
$\mathbf{M} = \mathbf{X} \wedge \mathbf{P}$
$\;$
With matrix components:
$M^{\alpha \beta} = X^{\alpha} P^{\beta} - X^{\beta} P^{\alpha}$
$\;$
Which are six independent quantities altogether. Since the components of X and P are frame-dependent, so is M. Three components
$M^{ij} = x^{i} p^{j} - x^{j} p^{i} = L^{ij}$
$\;$
are those of the familiar classical 3-space orbital angular momentum, and the other three components
$M^{0i} = x^{0} p^{i} - x^{i} p^{0} = c \; \left(tp^{i} - x^{i}{\frac{E}{c^{2}}} \right) = -cN^{i}$
are the relativistic mass moment, multiplied by -c. The tensor is antisymmetric:
$M^{\alpha \beta} = -M^{\beta \alpha}$
$\;$
The components of the tensor can be systematically displayed as a matrix.
$\;$
The angular momentum $L = x \; \wedge \; p$ of a particle with relativistic mass m and relativistic momentum p, as measured by an observer in a lab frame, combines with another vector quantity dynamic mass-energy moment $N = mx - pt$ in the relativistic angular momentum tensor: (ref. 2)
$M^{\alpha \beta} = {\begin{pmatrix} 0 & -N_{x}^{1}c & -N_{y}^{2}c & -N_{z}^{3}c \\ N_{x}^{1}c & 0 & L^{12} & -L^{13} \\ N_{y}^{2}c & -L^{21} & 0 & L^{23} \\ N_{z}^{3}c & L^{31} & -L^{32} & 0 \end{pmatrix}}$
$\;$
In classical mechanics, the orbital angular momentum of a particle with instantaneous three-dimensional position vector $x = (x, y, z)$ and momentum vector $p = (px, py, pz)$, is defined as the axial vector: (ref. 4)
$\mathbf{L} = \mathbf{x} \times \mathbf{p}$
$\;$
Which has three components, that are systematically given by cyclic permutations of Cartesian directions, change x to y, y to z, z to x, repeat.
$L_{x} = yp_{z} - zp_{y}$
$L_{y} = zp_{x} - xp_{z}$
$L_{z} = xp_{y} - yp_{x}$
$\;$
A related definition is to conceive orbital angular momentum as a plane element. This can be achieved by replacing the cross product by the exterior product in the language of exterior algebra, and angular momentum becomes a contravariant second order antisymmetric tensor:
$\mathbf{L} = \mathbf{x} \wedge \mathbf{p}$
$\;$
or writing $x = (x_{1}, x_{2}, x_{3}) = (x, y, z)$ and momentum vector $p = (p_{1}, p_{2}, p_{3}) = (p_{x}, p_{y}, p_{z})$, the components can be compactly abbreviated in tensor index notation:
$L^{ij} = x^{i} p^{j} - x^{j} p^{i}$
$\;$
Where the indices i and j take the values 1, 2, 3. On the other hand, the components can be systematically displayed fully in a 3 x 3 antisymmetric matrix:
$\mathbf{L} = {\begin{pmatrix} L^{11} & L^{12} & L^{13} \\ L^{21} & L^{22} & L^{23} \\ L^{31} & L^{32} & L^{33} \\ \end{pmatrix}} = {\begin{pmatrix} 0 & L_{xy} & L_{xz} \\ L_{yx} & 0 & L_{yz} \\ L_{zx} & L_{zy} & 0 \end{pmatrix}} = \begin{pmatrix} 0 & L_{xy} & -L_{zx} \\ -L_{xy} & 0 & L_{yz} \\ L_{zx} & -L_{yz} & 0 \end{pmatrix}$
$\;$
$\mathbf{L} = {\begin{pmatrix} 0 & xp_{y} - yp_{x} & -\left(zp_{x} - xp_{z} \right) \\ -\left(xp_{y} - yp_{x} \right) & 0 & yp_{z} - zp_{y} \\ zp_{x} - xp_{z} & -\left(yp_{z} - zp_{y} \right) & 0 \end{pmatrix}}$
$\;$
This quantity is additive, and for an isolated system, the total angular momentum of a system is conserved.
$\;$
Dynamic mass-energy moment:
$\mathbf{N} = m \mathbf{x} - \mathbf{p} t = \frac{E}{c^{2}} \mathbf{x} - \mathbf{p} t = \gamma (\mathbf{u}) m_{0} (\mathbf{x} - \mathbf{u} t)$
$\;$
Expressing N in terms of relativistic mass-energy and momentum, rather than rest mass and velocity, avoids extra Lorentz factors. However, relativistic mass is discouraged by some authors since it can be a misleading quantity to apply in certain equations.
$\;$
Defined here so that the relativistic equation in terms of the relativistic mass-energy equivalence, and classical definition, have the same form. The Cartesian components are:
$N_{x} = mx - p_{x} t = \frac{E}{c^{2}} x - p_{x} t = \gamma \left(u \right) m_{0}\left(x - u_{x} t \right)$
$N_{y} = my - p_{y} t = \frac{E}{c^{2}} y - p_{y} t = \gamma \left(u \right) m_{0}\left(y - u_{y} t \right)$
$N_{z} = mz - p_{z} t = \frac{E}{c^{2}} z - p_{z} t = \gamma \left(u \right) m_{0}\left(z - u_{z} t \right)$
$\;$
For a massless spin-2 graviton, $E = pc$, and the relativistic angular momentum tensor is:
$M^{\alpha \beta} = {\begin{pmatrix} 0 & -p_{x}\left(\frac{x}{c} - t \right) & -p_{y}\left(\frac{y}{c} - t \right) & -p_{z}\left(\frac{z}{c} - t \right) \\ p_{x}\left(\frac{x}{c} - t \right) & 0 & xp_{y} - yp_{x} & -\left(zp_{x} - xp_{z} \right) \\ p_{y}\left(\frac{y}{c} - t \right) & -\left(xp_{y} - yp_{x} \right) & 0 & yp_{z} - zp_{y} \\ p_{z}\left(\frac{z}{c} - t \right) & zp_{x} - xp_{z} & -\left(yp_{z} - zp_{y} \right) & 0 \end{pmatrix}}$
$\;$
In spherical coordinates $(ct, r, \theta, \phi)$ the Minkowski flat spacetime metric in contravariant form:
$ds^{2} = -c^{2} dt^{2} + dr^{2} + r^{2} d\theta^{2} + r^{2} \sin^{2} \theta \; d\phi^{2}$
$\;$
In spherical coordinates $(ct, r, \theta, \phi)$ the relativistic angular momentum tensor in contravariant form:
$M^{\alpha \beta} = {\begin{pmatrix} 0 & -p_{x}\left(\frac{dr}{c} - t \right) & -p_{y}\left(\frac{r d\phi}{c} - t \right) & -p_{z}\left(\frac{r \sin \theta d\phi}{c} - t \right) \\ p_{x}\left(\frac{dr}{c} - t \right) & 0 & dr p_{y} - r d\theta p_{x} & -\left(r \sin \theta d\phi p_{x} - dr p_{z} \right) \\ p_{y}\left(\frac{r d\theta}{c} - t \right) & -\left(dr p_{y} - r d\theta p_{x} \right) & 0 & r d\theta p_{z} - r \sin \theta d\phi p_{y} \\ p_{z}\left(\frac{r \sin \theta d\phi}{c} - t \right) & r \sin \theta d\phi p_{x} - dr p_{z} & -\left(r d\theta p_{z} - r \sin \theta d\phi p_{y} \right) & 0 \end{pmatrix}}$
$\;$
General Relativity stress-energy tensor:
$M_{\mu \nu} = \pm \left(\begin{matrix} -\rho\left(r \right) c^{2} & 0 & 0 & 0 \\ 0 & p\left(r \right) & 0 & 0 \\ 0 & 0 & p\left(r \right) & 0 \\ 0 & 0 & 0 & p\left(r \right) \end{matrix} \right)$
$\;$
Jacobian matrix transformation matrices: (ref. 5, pg. 15, eq. 41, para. 1, ref 6)
$\Lambda^{\mu}_{\alpha} = \frac{\partial \xi^{\mu} }{\partial x^{\alpha}} \; \; \; \; \; \; \Lambda^{\nu}_{\beta} = \frac{\partial \xi^{\nu}}{\partial x^{\beta}}$
$\;$
$x^{\alpha} = \left(ct, r, \theta, \phi \right) \; \; \; \; \; \; x^{\beta} = \left(ct, r, \theta, \phi \right)$
$\;$
$\Lambda^{\mu}_{\alpha} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{c^{2} \partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}} = {\begin{bmatrix} -2 dt & 0 & 0 & 0 \\ 0 & 2 dr & 0 & 0 \\ 0 & 0 & 2r^2 d\theta & 0 \\ 0 & 0 & 0 & 2r^2 \sin^{2} \theta \; d\phi \\ \end{bmatrix}}$
$\;$
$\;$
$\Lambda^{\nu}_{\beta} \left(ct, r, \theta, \phi \right) = {\begin{bmatrix} \dfrac{\partial \xi^0}{c^{2} \partial t^0} & 0 & 0 & 0 \\ 0 & \dfrac{\partial \xi^1}{\partial r^1} & 0 & 0 \\ 0 & 0 & \dfrac{\partial \xi^2}{\partial \theta^2} & 0 \\ 0 & 0 & 0 & \dfrac{\partial \xi^3}{\partial \phi^3} \\ \end{bmatrix}} = {\begin{bmatrix} -2 dt & 0 & 0 & 0 \\ 0 & 2 dr & 0 & 0 \\ 0 & 0 & 2r^2 d\theta & 0 \\ 0 & 0 & 0 & 2r^2 \sin^{2} \theta \; d\phi \\ \end{bmatrix}}$
$\;$
The matrix dot product transformation matrices in covariant form for each of the two four-momentum components as seen from two reference frames, S and S' prime:
$\Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} = {\begin{bmatrix} 4 dt dt' & 0 & 0 & 0 \\ 0 & 4 dr dr' & 0 & 0 \\ 0 & 0 & 4r^2 r'^{2} d\theta \; d\theta' & 0 \\ 0 & 0 & 0 & 4r^2 r'^{2} \sin^{2} \theta \sin^{2} \theta' \; d\phi \; d\phi' \\ \end{bmatrix}}$
$\;$
Newton's constant: (ref. 7, pg. 9, eq. 37)
$\kappa^{2} = 32 \pi G$
$\;$
General relativity weak field limit spacetime metric: (ref. 7, pg. 9, eq. 37)
$g_{\mu \nu} = \eta_{\mu \nu} + \kappa h_{\mu \nu}$
$\;$
General Relativity weak field limit spacetime metric and Planck quantum gravity identity 6:
$\boxed{\frac{8 \pi G}{c^{4}} M_{\mu \nu} = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} M^{\alpha \beta} \left(x \right) \left(1 - \frac{1}{2} \left(\eta^{\mu \nu} - \kappa h^{\mu \nu} \right)\left(\eta_{\mu \nu} + \kappa h_{\mu \nu} \right) \right)}$
$\;$
Tensor matrix solution key:
$s\left(\mu, \nu, \alpha, \beta \right)$
$\;$
$s\left(1, 1, 1, 2 \right):$
$\boxed{2 \pi G p\left(r \right) = c^{4}\left(dr' dr^{2} p_{y} - dr dr' r d\theta p_{x} \right)}$
$\;$
$s\left(2, 2, 2, 3 \right):$
$\boxed{2 \pi G p\left(r \right) = c^{4} \left(r^{3} r'^{2} d\theta^{2} d\theta' p_{z} - r^{3} r'^{2} \sin \theta d\theta d\theta' d\phi p_{y} \right)}$
$\;$
With four stress-energy tensor elements and twelve relativistic angular momentum tensor elements, there are forty-eight possible solution keys.
$\;$
Are these energy-momentum tensors compatible with a massless spin-2 graviton?
$\;$
Is this approach mathematically and symbolically correct to this point?
$\;$
Are there any other tensor matrix solution keys that you want to examine?
$\;$
$\;$
Reference:
Lorentz Group and Lorentz Invariance: (ref. 1)
https://gdenittis.files.wordpress.com/2016/04/ayudantiavi.pdf
Wikipedia - Four-tensor - Second order tensors: (ref. 2)
https://en.wikipedia.org/wiki/Four-tensor#Second_order_tensors
Wikipedia - Relativistic angular momentum - 4d Angular momentum as a bivector: (ref. 3)
https://en.wikipedia.org/wiki/Relativistic_angular_momentum#4d_Angular_momentum_as_a_bivector
Wikipedia - Relativistic angular momentum - Orbital 3d angular momentum: (ref. 4)
https://en.wikipedia.org/wiki/Relativistic_angular_momentum#Orbital_3d_angular_momentum
Introduction to Tensor Calculus for General Relativity - Edmund Bertschinger: (ref. 5)
https://web.mit.edu/edbert/GR/gr1.pdf
Wikipedia - Jacobian matrix: (ref. 6)
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Example_3:_spherical-Cartesian_transformation
Barry R. Holstein - Department of Physics-LGRT - University of Massachusetts: (ref. 7)
https://arxiv.org/pdf/gr-qc/0607045.pdf

Edited by Orion1
source code correction...

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Looks good thus far the reference 7 page 11 equation 47 has the covariant derivative of the graviton propogator, as your employing the same tensors you should be be good.

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On 1/14/2020 at 3:42 PM, Mordred said:

Looks good thus far the reference 7 page 11 equation 47 has the covariant derivative of the graviton propogator, as your employing the same tensors you should be be good.

Affirmative, revision complete.

Relativistic Lagrangian:
$\boxed{\mathcal{L} = \sum_{1}^{n} \mathcal{L}\left(n \right) = 0} \; \; \; n = 5$
$\;$
Lagrangian equation for a massless quantum field: $\; \; \; m = 0$
$\mathcal{L} = \underbrace{ \mathbb{R} }_{\text{GR}} - \overbrace{\underbrace{\frac{1}{4} F^{\mu \nu} F_{\mu \nu}}_{\text{Yang-Mills}}}^{\text{Maxwell}} + \underbrace{i \overline{\psi} \gamma^\mu D_{\mu} \psi}_{\text{Dirac}} + \underbrace{|D_{\mu} \phi|^{2} - V\left(|\phi| \right)}_{\text{Higgs}} - \underbrace{g \overline{\psi} \psi}_{\text{Yukawa}} = 0$
$\;$
Lagrangian equation for a massless quantum field: $\; \; \; m = 0$
$\boxed{\mathcal{L} = \overbrace{\underbrace{\Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} M^{\alpha \beta}\left(x \right) }_{\text{Quantum Gravity}}}^{\text{spin } 2} - \overbrace{\underbrace{\frac{1}{4} F^{a \mu \nu} F^{a}_{\mu \nu}}_{\text{Yang-Mills}}}^{\text{spin } 1} + \overbrace{\underbrace{i \overline{\psi}_{a} \gamma^{\mu}_{ab} D_{\mu} \psi_{b}}}_{\text{Dirac}}^{\text{spin } 1/2} + \overbrace{\underbrace{|D_{\mu} \phi_{a}|^{2} - V\left(|\phi_{a}| \right)}}_{\text{Higgs}}^{\text{spin } 0} - \overbrace{\underbrace{g \overline{\psi}_{a} \psi_{b}}}_{\text{Yukawa}}^{\text{spin } 0,1/2} = 0}$
(ref. 1, ref. 2, ref. 3, ref. 4, ref. 5, ref. 6, ref. 7)
$\;$
Lagrangian equation for a mass quantum field: $\; \; \; m \neq 0$
$\boxed{\mathcal{L} = \overbrace{\underbrace{\Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} M^{\alpha \beta}\left(x \right) }_{\text{Quantum Gravity}}}^{\text{spin } 2} - \overbrace{\underbrace{\frac{1}{4} F^{a \mu \nu} F^{a}_{\mu \nu}}_{\text{Yang-Mills}}}^{\text{spin } 1} + \overbrace{\underbrace{\overline{\psi}_{a} \left(i \gamma^{\mu}_{ab} D_{\mu} - m \mathbb{I}_{ab} \right) \psi_{b}}}_{\text{Dirac}}^{\text{spin } 1/2} + \overbrace{\underbrace{|D_{\mu} \phi_{a}|^{2} - V\left(|\phi_{a}| \right)}}_{\text{Higgs}}^{\text{spin } 0} - \overbrace{\underbrace{g \overline{\psi}_{a} \psi_{b}}}_{\text{Yukawa}}^{\text{spin } 0,1/2} = 0}$
(ref. 4, ref. 5)
$\;$
Lagrangian equation for a mass and charge quantum field: $\; \; \; m \neq 0, Q \neq 0$
$\boxed{\mathcal{L} = \overbrace{\underbrace{\Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} M^{\alpha \beta}\left(x \right) }_{\text{Quantum Gravity}}}^{\text{spin } 2} - \overbrace{\underbrace{\frac{1}{4} F^{a \mu \nu} F^{a}_{\mu \nu}}_{\text{Yang-Mills}}}^{\text{spin } 1} + \overbrace{\underbrace{\overline{\psi}_{a} \left[\gamma^{\mu}_{ab} \left(i \partial_{\mu} - e Q A_{\mu} \right) - m \mathbb{I}_{ab} \right] \psi_{b}}}_{\text{Dirac}}^{\text{spin } 1/2} + \overbrace{\underbrace{|\left(\partial_{\mu} -ieQA_{\mu} \right) \phi_{a}|^{2} - \lambda \left(|\phi_{a}|^{2} - \Phi^{2} \right)^{2}}}_{\text{Higgs}}^{\text{spin } 0,1} - \overbrace{\underbrace{g \overline{\psi}_{a} \psi_{b}}}_{\text{Yukawa}}^{\text{spin } 0,1/2} = 0}$
(ref. 4, pg. 8, eq. 2.6, ref. 5, ref. 6, ref. 7)
$\;$
Is this approach mathematically and symbolically correct to this point?
$\;$
$\;$
Reference:
Wikipedia - Quantum gravity: (ref. 1)
https://en.wikipedia.org/wiki/Quantum_gravity
Science Forums - Orion1 - Spin 2 Quantum Gravity: (ref. 2)
https://www.scienceforums.net/topic/117992-the-lagrangian-equation/?do=findComment&comment=1128193
Wikipedia - Yang–Mills theory: (ref. 3)
https://en.wikipedia.org/wiki/Yang–Mills_theory#Mathematical_overview
Search For The Standard Model Higgs Boson - Huong Thi Nguyen: (ref. 4)
https://www-d0.fnal.gov/results/publications_talks/thesis/nguyen/thesis.pdf
Wikipedia - Dirac fields: (ref. 5)
https://en.wikipedia.org/wiki/Fermionic_field#Dirac_fields
Wikipedia - Higgs mechanism: (ref. 6)
https://en.wikipedia.org/wiki/Higgs_mechanism#Abelian_Higgs_mechanism
Wikipedia - Yukawa interaction: (ref. 7)
https://en.wikipedia.org/wiki/Yukawa_interaction#The_action

Edited by Orion1
source code correction...

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I like your reference four paper, there is several Langrangians in that paper I will latex later on to have a handy copy of them.

I also like what you did with the overbrace and underbrace. I don't see any problems thus far

Edited by Mordred

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