A Is the Higgs Field Responsible for Microscopic Gravitational Effects?

[samalkhaiat]

TL;DR Summary

Investigating the force law in particle physics: The Lorentz 4-force and the Higgs force.

To get you started I will derive the Lorentz force law from the QED Lagrangian \begin{equation}\mathcal{L} = \frac{i}{2} \bar{\psi}\gamma^{\mu}D_{\mu}\psi + h.c - \frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu} ,\end{equation}D_{\mu} = \partial_{\mu} + ieA_{\mu}, and then, I let you do the same to a SM-like theory, i.e., an \mbox{SU}(N) gauge invariant theory of fermionic fields coupled to the Higgs field, both belonging to the fundamental representation of \mbox{SU}(N): \begin{align}\mathcal{L} & = \frac{i}{2} \bar{\psi}\gamma^{\mu}D_{\mu}\psi - \frac{1}{16\pi}F^{a}_{\mu\nu}F^{\mu\nu}_{a} + \frac{1}{2}\left(D_{\mu}\phi \right)^{\dagger}D^{\mu}\phi \\ & - \frac{\mu^{2}}{2} \phi^{\dagger}\phi - \frac{\lambda}{4!}\left(\phi^{\dagger} \phi \right)^{2} \\ & - \bar{\psi} \phi^{\dagger}\mathbf{m} \psi + hc ,\end{align}D_{\mu} = \partial_{\mu} + igA^{a}_{\mu}t_{a}. Here \mu^{2}, \ \lambda are real parameters of the Higgs potential, and \mathbf{m} is the Yukawa coupling-matrix. Of course, in the SM one has to specify explicitly the generators t_{a}, the fermion fields \psi, the Higgs field \phi, and the coupling matrix \mathbf{m} which determines the fermion mass eignstates after the spontaneous symmetry breaking.

The Lorentz Force Law from the QED Lagrangian:
From (1) we get the field equations for the fermionic matter field \psi:
\begin{equation} i\gamma^{\mu}D_{\mu}\psi = 0 , \end{equation} and the EM gauge field F^{\mu\nu}
\begin{equation}\partial_{\mu}F^{\mu\nu} = 4\pi j^{\nu}, \end{equation} with the gauge-invariant matter-field current \begin{equation}j^{\mu} = e \bar{\psi} \gamma^{\mu} \psi .\end{equation} The conserved and gauge-invariant energy-momentum tensor in QED has a gauge-invariant matter-field and electromagnetic-field part T^{\mu}{}_{\nu} = T^{\mu}{}_{\nu}(\psi) - T^{\mu}{}_{\nu}(F), where \begin{equation}T^{\mu}{}_{\nu}(\psi) = \frac{i}{2}\left(\bar{\psi}\gamma^{\mu}D_{\nu}\psi - \overline{( D_{\nu}\psi )}\gamma^{\mu}\psi \right), \end{equation}\begin{equation}T^{\mu}{}_{\nu}(F) = \frac{1}{4\pi}\left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4}\delta^{\mu}_{\nu} F^{2}\right) . \end{equation} From \partial_{\mu}T^{\mu\nu} = 0, one gets, neglecting surface integral at infinity: \begin{equation}\frac{d}{dt} \int d^{3}x \ T^{0}{}_{\nu}(\psi) = \int d^{3}x \ \partial_{\mu}T^{\mu}{}_{\nu}(F) .\end{equation} Notice that on the LHS we have the rate of change of the 4-momentum P_{\nu}(\psi) of the matter field. Now, using the field equation \partial_{\mu}F^{\mu\nu} = 4\pi j^{\nu}, one can easily show \begin{equation}\partial_{\mu}T^{\mu}{}_{\nu}(F) = F_{\nu\sigma}j^{\sigma} . \end{equation} Substituting this in (10), we find \begin{equation} \frac{d}{dt}P_{\mu}(\psi) = \int d^{3}x \ F_{\mu \sigma}j^{\sigma} . \end{equation} The RHS represents the Lorentz 4-force which causes the change of the 4-momentum of the matter field with time.

Your exercise: Apply the above reasoning to the \mbox{SU}(N) Lagrangian of eq(2) and show that \begin{equation}\frac{d}{dt}P_{\mu}(\psi) = \int d^{3}x \ F_{\mu\nu}^{a}j^{\nu}_{a}(\psi) + \int d^{3}x \ \bar{\psi} \left((D_{\mu}\phi)^{\dagger}\mathbf{m} + \mathbf{m}^{\dagger}(D_{\mu}\phi) \right)\psi , \end{equation} where j^{\nu}_{a}(\psi) = g\bar{\psi}\gamma^{\nu}t_{a}\psi is the fermionic matter field current. The two integrals on the RHS of (13) represent the Lorentz-like 4-force of the gauge-field and the Higgs-field force respectively. Clearly, the gauge field couples to the matter-field via the \mbox{SU}(N) coupling constant g, whereas the coupling strength of the Higgs-field to the fermions is determined only by the (fermionic) mass matrix \mathbf{m}. This fact seems to point to a gravitational role for the Higgs field on the microscopic level !

 

[Demystifier]

Summary:: Investigating the force law in particle physics: The Lorentz 4-force and the Higgs force.

This fact seems to point to a gravitational role for the Higgs field on the microscopic level !

I'm not surprised that the Higgs field generates a force at the microscopic level, but I don't see gravity in your analysis. Of course, the couplings of Higgs with other fields contributes to the energy-momentum tensor which sits on the right-hand side of the Einstein equation, but that's kind of obvious. Do I miss something?

 

[vanhees71]

I think you have to formulate everything as a theory in an arbitrary background spacetime to see that the same gauge invariant energy-momentum tensor of the fields are the sources of gravity.

Further the mass terms by construction are due to the coupling to the Higgs field. It's the part of this coupling due to the "condensate" of the Higgs field in the Higgs mechanism.

In other words: In special relativity "inertia" is not only due to the mass as in Newtonian mechanics but to all kinds of energy (and stresses!). By construction then in general relativity inertia and sources of gravity are the same, i.e., the energy-momentum-stress tensor of all other fields describing "matter and radiation" of all (known and unknown) kinds.

 

[samalkhaiat]

..., but I don't see gravity in your analysis.

If you look at Eq(13), you see that the Higgs field couples to the masses only and not to any gauge charges. This behaviour is exactly that of classical gravity. The only qualitative difference with respect to the Newtonian gravity consists (besides non-linearity) in the short range nature of the Higgs gravity. This becomes clear if you spontaneously break the symmetry. Indeed, one can show that any excited Higgs field mediates an attractive scalar interaction of Yukawa type (i.e. short range) between those particles which acquire mass. Higgs’s gravity is not a new idea. The idea that the Higgs field of particle physics can also serve as the scalar field in a scalar-tensor theory of gravity, as was first proposed by Zee in 1979[1] and later investigated by Dehnen, Frommert, and Ghaboussi in 1992[2]. The whole developed formalism of this proposal can be found in [3] by Dehnen and Frommert.
[1] A. Zee, Phys.Rev.Lett., 42(7), 417(1979).
[2] Dehnen, Frommert, and Ghaboussi, Int.J.Theo.Phys., 31(1), 109(1992).
[3] Dehnen and Frommert, Int.J.Theo.Phys., 32(7), 1135(1993).

Of course, the couplings of Higgs with other fields contributes to the energy-momentum tensor which sits on the right-hand side of the Einstein equation, but that's kind of obvious. Do I miss something?

This is a microscopic approach. Here we are talking about strong, short-ranged, scalar gravity on Minkowski background. The aim of the program is to reproduce Einstein’s GR as an effective theory, in the same way as the nuclear force is a “van der Waals”-type interaction resulting from QCD. The out-standing problem though is light-deflection. Indeed, the Higgs doesn’t have an influence on the EM field.

 

[vanhees71]

But isn't one very convincing solution of this program just to make the Poincare (or Lorentz) symmetry local as done by Kibble and Utiyama?

T. W. B. Kibble, Lorentz Invariance and the Gravitational Field, Jour. Math. Phys. 2, 212 (1960),
https://doi.org/10.1063/1.1703702.

R. Utiyama, Invariant theoretical interpretation of interaction, Phys. Rev. 101, 1597 (1956),
https://doi.org/10.1103/PhysRev.101.1597.

 

[Demystifier]

But isn't one very convincing solution of this program just to make the Poincare (or Lorentz) symmetry local as done by Kibble and Utiyama?

T. W. B. Kibble, Lorentz Invariance and the Gravitational Field, Jour. Math. Phys. 2, 212 (1960),
https://doi.org/10.1063/1.1703702.

R. Utiyama, Invariant theoretical interpretation of interaction, Phys. Rev. 101, 1597 (1956),
https://doi.org/10.1103/PhysRev.101.1597.

I think not, because those theories don't have a scalar component of gravity.

 

[vanhees71]

Why do you need a scalar component of gravity (whatever that may be)? GR seems to be good enough (though I'd expect that it should rather be Einstein-Cartan theory given the above quoted papers).

 

[samalkhaiat]

But isn't one very convincing solution of this program just to make the Poincare (or Lorentz) symmetry local as done by Kibble and Utiyama?

Without doubts, the Yang-Mills line of thought used by Utiyama and Kibble is very elegant. However, 1) that approach has not helped our quest for quantum gravity, and 2) it says nothing about the nature of the Higgs force. In the SM, the Higgs force is like “Piggy in the Middle” compared to the rest of the forces. So with respect to the above two issues, showing that the (exited)-Higgs-field-interaction is some kind of gravitational interaction is (I believe) a step in the right direction. Also, Higgs gravity puts a “smile on your face” because, when applied to SU(2)\times U(1), it allows you to relate Newton’s constant to the Higgs VEV.

 

[ergospherical]

Haven't come across this stuff before, I wondered if someone could help me understand the example in the OP?\begin{equation*}\mathcal{L} = \frac{i}{2} \bar{\psi}\gamma^{\mu}(\partial_{\mu} \psi + ieA_{\mu} \psi) + hc - \frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu} ,\end{equation*}I assumed the field equations for $\psi$ come from $\dfrac{\partial \mathcal{L}}{\partial \psi} - \partial_{\mu} \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu} \psi)} = 0$ whilst the field equations for $A_{\mu}$ come from $\dfrac{\partial \mathcal{L}}{\partial A_{\mu}} - \partial_{\nu} \dfrac{\partial \mathcal{L}}{\partial(\partial_{\nu} A_{\mu})} = 0$. Given that $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ then\begin{align*}
\dfrac{\partial \mathcal{L}}{\partial A_{\mu}} = \frac{-e}{2} \bar{\psi}\gamma^{\mu} \psi = -\dfrac{1}{2} j^{\mu}
\end{align*}(this seems wrong... did I miss something with the $\gamma^{\mu}$?) and\begin{align*}
\partial_{\nu} \left(\dfrac{\partial \mathcal{L}}{\partial(\partial_{\nu} A_{\mu})}\right) &= -\dfrac{1}{16\pi} \partial_{\nu} \dfrac{\partial}{\partial(\partial_{\nu} A_{\mu})} \left [ \left( \partial_{\rho} A_{\sigma} - \partial_{\sigma} A_{\rho} \right) \left( \partial^{\rho} A^{\sigma} - \partial^{\sigma} A^{\rho}\right) \right] \\

&= -\dfrac{1}{8\pi} \partial_{\nu} \dfrac{\partial}{\partial(\partial_{\nu} A_{\mu})} \left [ (\partial_{\rho} A_{\sigma}) (\partial^{\rho} A^{\sigma}) - (\partial_{\rho} A_{\sigma})(\partial^{\sigma} A^{\rho}) \right] \\

&= -\dfrac{1}{4\pi} \partial_{\nu} \left( \partial^{\nu} A^{\mu} - \partial^{\mu} A^{\nu} \right) \\

&= -\frac{1}{4\pi} \partial_{\nu} F^{\nu \mu}
\end{align*}which would appear to give $\partial_{\mu} F^{\mu \nu} = 2 \pi j^{\nu}$, a factor of 2 off - what did I do wrong? As for the matter field $\psi$,\begin{align*}
\dfrac{\partial \mathcal{L}}{\partial \psi} &= \dfrac{-e}{2} \bar{\psi} \gamma^{\mu} A_{\mu} \\

\partial_{\mu} \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu} \psi)} &= \dfrac{i}{2}\partial_{\mu}( \bar{\psi} \gamma^{\mu})
\end{align*}therefore\begin{align*}
e\bar{\psi} \gamma^{\mu} A_{\mu} +i \bar{\psi} \underbrace{\partial_{\mu} \gamma^{\mu}}_{=0} +i \gamma^{\mu} \partial_{\mu} \bar{\psi} &= 0 \\

e \psi \gamma^{\mu} A_{\mu} - i \gamma^{\mu} \partial_{\mu} \psi &= 0 \\

-i\gamma^{\mu} (\partial_{\mu} + ieA_{\mu}) \psi &= 0 \\ \\
\implies i\gamma^{\mu} D_{\mu} \psi = 0
\end{align*}For equation (10), and letting the index $i$ run over $i \in \{1,2,3\}$,\begin{align*}
\frac{d}{dx^0} \int d^{3}x \ T^{0}{}_{\nu}(\psi) &= \int d^3 x \partial_0\left( {T^{0}}_{\nu} + {T^{0}}_{\nu}(F) \right) \\

&= \int d^3 x \partial_{\mu}({T^{\mu}}_{\nu} + {T^{\mu}}_{\nu}(F)) - \int d^3 x \partial_i ( {{T^{i}}_{\nu}} + {{T^{i}}_{\nu}}(F)) \\

&= \int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(F) - \oint d^2 x ({{T^{i}}_{\nu}} + {{T^{i}}_{\nu}}(F)) n_i \\

&= \int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(F)
\end{align*}because the surface integral vanishes since ${{T^{i}}_{\nu}} + {{T^{i}}_{\nu}}(F)$ is zero at infinity. To try and show $(11)$ I wrote\begin{align*}
\partial_{\mu} T^{\mu}{}_{\nu}(F) &= \frac{1}{4\pi} \partial_{\mu} \left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4}F^{2}\right) \\

&= \frac{1}{4\pi} F^{\mu \rho} \partial_{\mu} F_{\nu \rho} + \dfrac{1}{4\pi} F_{\nu \rho} \partial_{\mu} F^{\mu \rho} - \frac{1}{16\pi} \partial_{\mu} F^2 \\

&= F_{\nu \rho} j^{\rho} + \left( \frac{1}{4\pi} F^{\mu \rho} \partial_{\mu} F_{\nu \rho} - \frac{1}{16\pi} \partial_{\mu} F^2 \right)
\end{align*}How do you interpret $F^2$? It can't be $({F_{\mu}}^{\mu})^2$ for instance, because that's zero.

Apart from those two questions in red, does the rest look vaguely correct?

 

[vanhees71]

You missed the terms written as "+hc", standing for the hermitian conjugated of the term written before.

 

[ergospherical]

You missed the terms written as "+hc", standing for the hermitian conjugated of the term written before.

Ahhhh thank you, I was wondering what that meant! By any chance do you also know how $F^2$ is defined in equation $(9)$?

 

[vanhees71]

That should be $F^{\rho \sigma}F_{\sigma \rho}$.

 

[ergospherical]

I don't quite understand the index structure in $T^{\mu}{}_{\nu}(F) = \frac{1}{4\pi}\left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4}F^{2}\right)$, the second term $\sim F^{\rho \sigma} F_{\sigma \rho}$ being a scalar.

 

[samalkhaiat]

I don't quite understand the index structure in $T^{\mu}{}_{\nu}(F) = \frac{1}{4\pi}\left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4}F^{2}\right)$, the second term $\sim F^{\rho \sigma} F_{\sigma \rho}$ being a scalar.

There is a typo in there (now corrected). It should read T^{\mu}{}_{\nu}(F) = \frac{1}{4 \pi} \left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4} \delta^{\mu}_{\nu}F^{\rho\sigma}F_{\rho\sigma}\right).

 

[samalkhaiat]

<br /> \begin{align*}<br /> \frac{d}{dx^0} \int d^{3}x \ T^{0}{}_{\nu}(\psi) &amp;= \int d^3 x \partial_0\left( {T^{0}}_{\nu} + {T^{0}}_{\nu}(F) \right) \\<br /> <br /> &amp;= \int d^3 x \partial_{\mu}({T^{\mu}}_{\nu} + {T^{\mu}}_{\nu}(F)) - \int d^3 x \partial_i ( {{T^{i}}_{\nu}} + {{T^{i}}_{\nu}}(F)) \\<br /> <br /> &amp;= \int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(F) - \oint d^2 x ({{T^{i}}_{\nu}} + {{T^{i}}_{\nu}}(F)) n_i \\<br /> <br /> &amp;= \int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(F)<br /> \end{align*}

The first line is not correct. In general, it is not wise to ignore the surface term in the integral of T(F). And you don’t need to go forward and backward: you have \partial_{\mu}T^{\mu}{}_{\nu} = 0. So, \partial_{\mu}T^{\mu}{}_{\nu}(\psi) = \partial_{\mu}T^{\mu}{}_{\nu}(F) . Now integrate and ignore the surface integral on the left-hand side,, i.e., \oint dS_{i} \ T^{i}{}_{\nu}(\psi) = 0

 

[ergospherical]

Thanks, that's much simpler. If I understood correctly, \begin{align*}
\int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(F) &= \int d^3 x \partial_{\mu} {T^{\mu}}_{\nu}(\psi) \\

&= \int d^3 x \partial_{0} {T^{0}}_{\nu}(\psi) + \int d^3 x \partial_{i} {T^{i}}_{\nu}(\psi) \\

&= \int d^3 x \partial_{0} {T^{0}}_{\nu}(\psi) + \underbrace{\oint dS_i {T^{i}}_{\nu}(\psi)}_{=0} \\

&= \dfrac{d}{dx^0} \int d^3 x {T^{0}}_{\nu}(\psi)
\end{align*}

 

[strangerep]

@samalkhaiat: As is often the case, your challenge exposes things which I probably should know already, but don't.

OK,... so,... from your hints about Higgs gravity I can (vaguely) see that your EoM (13) is relatable to something like the classical geodesic equation for weak GR gravity -- if we neglect electromagnetism, and restrict to a low-energy (or is it low velocity?) approximation where we ignore muons, tauons, etc, and the mass matrix becomes trivial. There's a current-like term on the rhs which is relatable to the $\dot x^\alpha$'s that contract with $\Gamma$ in the geodesic equation.

So then, my immediate instinct is to try and relate all this to the equation for tidal acceleration in classical Newtonian gravity, and thence to the geodesic deviation equation in classical GR which involves the Riemann tensor (since this is my preferred way to motivate the Einstein field equations).

Here's the thing I don't know: how does one generalize/relate the classical derivation of Newtonian tidal acceleration to the field case, analogous to the way one generalizes the simple Newtonian $m \ddot x^\mu$ to a time derivative of the total momentum $P_\mu$, the latter involving a spatial integral of $T_{\mu\nu}$ ?

[ I wasn't sure whether my question should go into a new thread. If so, maybe it could become challenge #003? ]

 

[vanhees71]

There is a typo in there (now corrected). It should read T^{\mu}{}_{\nu}(F) = \frac{1}{4 \pi} \left( F^{\mu\rho}F_{\nu\rho} - \frac{1}{4} \delta^{\mu}_{\nu}F^{\rho\sigma}F_{\rho\sigma}\right).

But shouldn't it then be
$${T^{\mu}}_{\nu} = \frac{1}{4 \pi} \left( F^{\mu\rho}F_{\rho \nu} + \frac{1}{4} \delta^{\mu}_{\nu}F^{\rho\sigma}F_{\rho\sigma}\right ),$$
i.e., with an overall minus sign as compared to the expression given above. At least that's what one gets from the standard application of Noether's theorem + making the resulting canonical energy-momentum tensor gauge invariant (and symmetric) by applying a socalled "pseudo-gauge transformation". For details, see Sect. 4.7 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Note that I use Heaviside Lorentz units, i.e., the EM tensor has no $1/(4 \pi)$.

 

[samalkhaiat]

@samalkhaiat:
,... from your hints about Higgs gravity I can (vaguely) see that your EoM (13) is relatable to something like the classical geodesic equation for weak GR gravity -- if we neglect electromagnetism, and restrict to a low-energy (or is it low velocity?) approximation where we ignore muons, tauons, etc, and the mass matrix becomes trivial. There's a current-like term on the rhs which is relatable to the $\dot x^\alpha$'s that contract with $\Gamma$ in the geodesic equation.

I must say, as it stands, eq(13) does not describe the Higgs gravity because all fields are still massless. And, as I said, the aim of the program is to let the classical world of GR arises from a QFT with spontaneously broken symmetry (SSB). If that proves successful, then one can understand why GR refuses to be quantized. After SSB, eq(13) becomes
\begin{align*}\frac{dP_{\mu}}{dt} &amp;= \int d^{3}x \left( \bar{\psi}m\psi + \frac{(1 + \varphi)}{4\pi}M_{ab}A^{a}_{\nu}A^{b\nu}\right)\partial_{\mu}\varphi \\ &amp;+ \int d^{3}x F^{a}_{\mu\nu}j^{\nu}_{a}(\psi) ,\end{align*} where \varphi is the excited Higgs field, M_{ab} = M_{ba} is the vector boson mass matrix and m is the fermions mass matrix. So, one can say that the Higgs gravity is quantum mechanical.

Here's the thing I don't know: how does one generalize/relate the classical derivation of Newtonian tidal acceleration to the field case

When I was young, I made the correspondence a^{\mu} \to R^{\mu}{}_{\nu\rho\sigma} J^{\nu} S^{\rho\sigma} , where S is the spin tensor and J is an appropriate current. After 2 weeks of crazy mathematics, I found that an empty space could give birth to something if T^{00} \geq 0. This contradicts a well-known theorem in Hawking & Ellis (The Large-Scale Structure of Space-Time, p.94) which says: If the dominant energy condition and the covariant conservation law hold for the energy-momentum tensor, then if that tensor is zero on a Cauchy surface, it can never be nonzero off the Cauchy surface.
So, I don’t think one can find a meaning full object.

[ ... , maybe it could become challenge #003? ]

No, the mathematics involved in the above correspondence is too crazy for PF.

 

[samalkhaiat]

But shouldn't it then be
$${T^{\mu}}_{\nu} = \frac{1}{4 \pi} \left( F^{\mu\rho}F_{\rho \nu} + \frac{1}{4} \delta^{\mu}_{\nu}F^{\rho\sigma}F_{\rho\sigma}\right ),$$
i.e., with an overall minus sign as compared to the expression given above.

Hehehe, Exercise: Show that both expressions are correct, i.e., both are symmetric, gauge invariant and traceless.
As for the “overall minus sign”: Notice that I defined the total energy-momentum tensor of the theory by the difference T^{\mu\nu} = T^{\mu\nu}(\psi) - T^{\mu\nu}(F) instead of the sum T^{\mu\nu}(\psi) + T^{\mu\nu}(F)

 

[strangerep]

When I was young, I made the correspondence a^{\mu} \to R^{\mu}{}_{\nu\rho\sigma} J^{\nu} S^{\rho\sigma} , where S is the spin tensor and J is an appropriate current.

Well, now you've really got me intrigued...

By "made the correspondence", do you mean that you simply postulated $a^{\mu} \to R^{\mu}{}_{\nu\rho\sigma} J^{\nu} S^{\rho\sigma}$, or that you derived it somehow?

The ordinary formula for geodesic deviation is $$a^{\mu} ~=~ R^{\mu}{}_{\nu\rho\sigma} V^\nu V^\rho X^\sigma ~,$$ where $V^\nu := dx^\mu(s,q)/ds$ is a geodesic tangent vector and $X^\sigma := dx^\mu(s,q)/dq$ is the relative displacement vector between neighbouring geodesics. So I guess you're adopting an analogue that $V^{[\rho} X^{\sigma]}$ should be regarded as a simple spin tensor (though it looks more like orbital angular momentum to me)?

If we admit the possibility of torsion, (sourced by the spin tensor?), the geodesic deviation equation has 3 extra (complicated) terms involving the torsion. Have those energy theorems you mentioned ever been extended to GR-with-torsion?

 

[vanhees71]

Using the gauge approach a la Kibble et al usually you are forced to extend GR (pseudo-Riemannian spacetime) to Einstein-Cartan theory (a spacetime with torsion) when you introduce fields with spin. An exception is electrodynamics (and I guess also non-Abelian gauge fields realizing vector fields), where there results no torsion. I also guess that's why it is so difficult to empirically find effects of torsion of spacetime: In the macroscopic (or rather even astronomical) contexts all we effectively have is macroscopic continuum mechanics of spin-saturated matter and electromagnetism, i.e., basically Einstein-Maxwell theory, where you end up with usual GR without torsion of spacetime.

The question is, of course, whether there are other approaches to spin within GR that avoids the necessity of torsion.

 

[samalkhaiat]

Well, now you've really got me intrigued...

By "made the correspondence", do you mean that you simply postulated $a^{\mu} \to R^{\mu}{}_{\nu\rho\sigma} J^{\nu} S^{\rho\sigma}$, or that you derived it somehow?

The ordinary formula for geodesic deviation is $$a^{\mu} ~=~ R^{\mu}{}_{\nu\rho\sigma} V^\nu V^\rho X^\sigma ~,$$ where $V^\nu := dx^\mu(s,q)/ds$ is a geodesic tangent vector and $X^\sigma := dx^\mu(s,q)/dq$ is the relative displacement vector between neighbouring geodesics. So I guess you're adopting an analogue that $V^{[\rho} X^{\sigma]}$ should be regarded as a simple spin tensor (though it looks more like orbital angular momentum to me)?

Yes, it was a guesswork based on the similarity in structure between the geodesic deviation equation and the term R^{\mu}{}_{\nu\rho\sigma}U^{\nu}S^{\rho\sigma} which appears in the equation of motion of spinning particle in GR.

Have those energy theorems you mentioned ever been extended to GR-with-torsion?

Non-zero torsion does not change the proof. \nabla_{\mu}T^{\mu\nu} = 0, \ T^{00} \geq 0 and a globally hyperbolic spacetime are the only assumptions needed in the proof.

 

[MathematicalPhysicist]

When I was young, I made the correspondence a^{\mu} \to R^{\mu}{}_{\nu\rho\sigma} J^{\nu} S^{\rho\sigma} , where S is the spin tensor and J is an appropriate current. After 2 weeks of crazy mathematics, I found that an empty space could give birth to something if T^{00} \geq 0. This contradicts a well-known theorem in Hawking & Ellis (The Large-Scale Structure of Space-Time, p.94) which says: If the dominant energy condition and the covariant conservation law hold for the energy-momentum tensor, then if that tensor is zero on a Cauchy surface, it can never be nonzero off the Cauchy surface.
So, I don’t think one can find a meaning full object.

No, the mathematics involved in the above correspondence is too crazy for PF.

So who made the mistake(s)? you or them?

 

[samalkhaiat]

So who made the mistake(s)? you or them?

No body. There is no mistakes in the math. The contradiction with the mentioned theorem means that my guesswork does not apply to a globally hyperbolic spacetime.