Lab Note 2, Part 3: Gravitational and Electrodynamic Potentials, the Electro-Gravitational Lagrangian, and a Possible Approach to Quantum Gravitation

February 14, 2008

Lab Note 2, Part 3: Gravitational and Electrodynamic Potentials, the Electro-Gravitational Lagrangian, and a Possible Approach to Quantum Gravitation

Filed under: Einstein Equation,Electrodynamics,Five Dimensions,General Relativity,Geodesic,Gravitation,Mathematics,Maxwell Tensor,Maxwell's Equations,Nature,Physics,QED,Quantum Electrodynamics,Quantum Field Theory,Quantum Gravitation,Science,Technology,Unified Field Theory,Vector Potential — Jay R. Yablon @ 1:39 am

Note: You may obtain a PDF version of Lab Note 2, with parts 2 and 3 combined, at Lab Note 2, with parts 2 and 3.

Also Note: This Lab Note picks up where Lab Note 2, Part 2, left off, following section 7 thereof. Equation numbers here, reference this earlier Lab Note.

8. The Electrodynamic Potential as the Axial Component of the Gravitational Potential

Working from the relationship $F^{{\rm M} } _{{\rm T} } \propto 2\Gamma ^{{\rm M} } _{{\rm T} 5}$ which generalizes (5.4) to five dimensions, and recognizing that the field strength tensor $F^{\mu \nu }$ is related to the four-vector potential $A^{\mu } \equiv \left(\phi ,A_{1} ,A_{2} ,A_{3} \right)$ according to $F^{\mu \nu } =A^{\mu ;\nu } -A^{\nu ;\mu }$ , let us now examine the relationship between $A^{\mu }$ and the metric tensor $g_{{\rm M} {\rm N} }$ . This is important for several reasons, one of which is that these are both fields and so should be compatible in some manner at the same differential order, and not the least of which is that the vector potential $A^{\mu }$ is necessary to establish the QED Lagrangian, and to thereby treat electromagnetism quantum-mechanically. (See, e.g., Witten, E., Duality, Spacetime and Quantum Mechanics, Physics Today, May 1997, pg. 28.)

Starting with ${\tfrac{1}{2}} F^{{\rm M} } _{{\rm T} } \propto \Gamma ^{{\rm M} } _{{\rm T} 5}$ , expanding the Christoffel connections $\Gamma ^{{\rm A} } _{{\rm B} {\rm N} } ={\tfrac{1}{2}} g^{{\rm A} \Sigma } \left(g_{\Sigma {\rm B} ,{\rm N} } +g_{{\rm N} \Sigma ,{\rm B} } -g_{{\rm B} {\rm N} ,\Sigma } \right)$ , making use of $g^{{\rm M} {\rm N} } _{,5} =0$ which as shown in (6.5) is equivalent to $F^{{\rm M} {\rm N} } =-F^{{\rm N} {\rm M} }$ , and using the symmetry of the metric tensor, we may write:

${\tfrac{1}{2}} F^{{\rm M} } _{{\rm T} } \propto \Gamma ^{{\rm M} } _{{\rm T} 5} ={\tfrac{1}{2}} g^{{\rm M} \Sigma } \left(g_{\Sigma {\rm T} ,5} +g_{5\Sigma ,{\rm T} } -g_{{\rm T} 5,\Sigma } \right)={\tfrac{1}{2}} g^{{\rm M} \Sigma } \left(g_{5\Sigma ,{\rm T} } -g_{5{\rm T} ,\Sigma } \right)$ . (8.1)

It is helpful to lower the indexes in field strength tensor and connect this to the covariant potentials $A_{\mu }$ , generalized into 5-dimensions as $A_{{\rm M} }$ , using $F_{\Sigma {\rm T} } \equiv A_{\Sigma ;{\rm T} } -A_{{\rm T} ;\Sigma }$ , as such:

$A_{\Sigma ;{\rm T} } -A_{{\rm T} ;\Sigma } \equiv F_{\Sigma {\rm T} } =g_{\Sigma {\rm M} } F^{{\rm M} } _{{\rm T} } \propto g_{\Sigma {\rm M} } g^{{\rm M} {\rm A} } \left(g_{5{\rm A} ,{\rm T} } -g_{5{\rm T} ,{\rm A} } \right)=\left(g_{5\Sigma ,{\rm T} } -g_{5{\rm T} ,\Sigma } \right)$ . (8.2)

The relationship $F_{\Sigma {\rm T} } \propto \left(g_{5\Sigma ,{\rm T} } -g_{5{\rm T} ,\Sigma } \right)$ expresses clearly, the antisymmetry of $F_{\Sigma {\rm T} }$ in terms of the remaining connection terms involving the gravitational potential. Of particular interest, is that we may deduce from (8.2), the proportionality

$A_{\Sigma ;{\rm T} } \propto g_{5\Sigma ,{\rm T} }$ . (8.3)

(If one forms $A_{\Sigma ;{\rm T} } -A_{{\rm T} ;\Sigma }$ from (8.3) and then renames indexes and uses $g_{{\rm M} {\rm N} } =g_{{\rm N} {\rm M} }$ , one arrives back at (8.2).) Further, we well know that $F_{\Sigma {\rm T} } =A_{\Sigma ;{\rm T} } -A_{{\rm T} ;\Sigma } =A_{\Sigma ,{\rm T} } -A_{{\rm T} ,\Sigma }$ , i.e., that the covariant derivatives of the potentials cancel out so as to become ordinary derivatives when specifying $F_{\Sigma {\rm T} }$ , i.e., that $F_{\Sigma {\rm T} }$ is invariant under the transformation $A_{\Sigma ;{\rm T} } \to A_{\Sigma ,{\rm T} }$ . Additionally, the Maxwell components (7.10) of the Einstein equation, are also invariant under $A_{\Sigma ;{\rm T} } \to A_{\Sigma ,{\rm T} }$ , because (7.10) also employs only the field strength $F^{\sigma \mu }$ . Therefore, let is transform $A_{\Sigma ;{\rm T} } \to A_{\Sigma ,{\rm T} }$ in the above, then perform an ordinary integration and index renaming, to write:

$A_{{\rm M} } \propto g_{5{\rm M} }$ . (8.4)

In the four spacetime dimensions, this means that the axial portion of the metric tensor is proportional to the vector potential, $g_{5\mu } \propto A_{\mu }$ , and that the field strength tensor $F_{\Sigma {\rm T} }$ and the gravitational field equations $-\kappa T^{{\rm M} } _{{\rm N} } =R^{{\rm M} } _{{\rm N} } -{\tfrac{1}{2}} \delta ^{{\rm M} } _{{\rm N} } R$ are invariant under the transformation $A_{\Sigma ;{\rm T} } \to A_{\Sigma ,{\rm T} }$ used to arrive at (8.4). We choose to set $A_{\Sigma ;{\rm T} } \to A_{\Sigma ,{\rm T} }$ , and can thereby employ the integrated relationship (8.4) in lieu of the differential equation (8.3), with no impact at all on the electromagnetic field strength or the gravitational field equations, which are invariant with respect to this choice.

9. Unification of the Gravitational and QED Lagrangians

The Lagrangian density for a gravitational field in vacuo is ${\rm L}_{gravitation} =\sqrt{-g} R$ , where g is the metric tensor determinant and $R=g^{\mu \nu } R_{\mu \nu }$ is the Ricci tensor. Let us now examine a Lagrangian based upon the 5-dimensional Ricci scalar, which we specify by:

${\rm R} \equiv R^{\Sigma } _{\Sigma } =R^{\sigma } _{\sigma } +R^{5} _{5} =R+R^{5} _{5}$ . (9.1)

To start, we return to deduce the ${\rm B} =5$ component of (7.6), namely:

$R^{{\rm A} } _{55{\rm N} } =-\Gamma ^{{\rm A} } _{55,{\rm N} } +\Gamma ^{\Sigma } _{5{\rm N} } \Gamma ^{{\rm A} } _{\Sigma 5} -\Gamma ^{\Sigma } _{55} \Gamma ^{{\rm A} } _{\Sigma {\rm N} }$ , (9.2)

as well as $R_{55}$ , which is easily found by contracting the remaining free indexes in the above:

$R_{55} =R^{{\rm T} } _{55{\rm T} } =-\Gamma ^{{\rm T} } _{55,{\rm T} } +\Gamma ^{\Sigma } _{5{\rm T} } \Gamma ^{{\rm T} } _{\Sigma 5} -\Gamma ^{\Sigma } _{55} \Gamma ^{{\rm T} } _{\Sigma {\rm T} }$ . (9.3)

Now, let us return to the discussion in the next-to-last paragraph of section 7, where we considered, but reached no conclusions about, the question of whether $\Gamma ^{\mu } _{55} =-{\tfrac{1}{2}} g_{55} ^{,\mu }$ is, or is not, equal to zero. Let us now make the (inductive) hypothesis that $\Gamma ^{\mu } _{55} =-{\tfrac{1}{2}} g_{55} ^{,\mu } =0$ , and see what (deductive) results emerge from this hypothesis.

First, as already noted, taken together with $g^{{\rm M} {\rm N} } _{,5} =0$ , and because $g_{55} =+1$ in geodesic coordinates, this means that that $g_{55} =+1={\rm constant}$ everywhere in the 5-dimensional spacetime. Second, the geodesic equation (4.9) now does reduce to (5.2), which, via (5.3) and (5.4), can be made identically equivalent with the Lorentz force law (5.1). Third, from (8.4), $A_{{\rm M} } \propto g_{5{\rm M} }$ , the axial component of the covariant vector potential must now be constant everywhere in spacetime, that is $A_{5} \propto g_{55} =+1={\rm constant}$ . Fourth, this means that the axial components of the field strength tensor $F_{\Sigma {\rm T} } =A_{\Sigma ;{\rm T} } -A_{{\rm T} ;\Sigma } =A_{\Sigma ,{\rm T} } -A_{{\rm T} ,\Sigma }$ must all become zero. To see this, we simply take:

$F_{5{\rm T} } =A_{5;{\rm T} } -A_{{\rm T} ;5} =A_{5,{\rm T} } -A_{{\rm T} ,5} \propto g_{55,{\rm T} } -g_{5{\rm T} ,5} =0$ . (9.4)

The first term, $g_{55,{\rm T} } =0$ , by virtue of the hypothesis just made. The latter term, $g_{5{\rm T} ,5} =0$ , because this is just a component of $g^{{\rm M} {\rm N} } _{,5} =0$ , i.e., $F^{{\rm M} {\rm N} } =-F^{{\rm N} {\rm M} }$ . Another way of stating (9.4), is that only the ordinary spacetime components $F^{\mu \nu }$ of the field strength tensor $F^{{\rm M} {\rm N} }$ are non-zero. Earlier, we generalized $F^{\mu \nu }$ to $F^{{\rm M} {\rm N} }$ . Now, we find that all of these added components are zero. Fifth, and finally, $\Gamma ^{\mu } _{55} =-{\tfrac{1}{2}} g_{55} ^{,\mu } =0$ directly simplifies (9.2) to:

$R^{{\rm A} } _{55{\rm N} } =+\Gamma ^{\Sigma } _{5{\rm N} } \Gamma ^{{\rm A} } _{\Sigma 5}$ , (9.5)

and (9.3) to:

$R_{55} =R^{{\rm T} } _{55{\rm T} } =+\Gamma ^{\Sigma } _{5{\rm T} } \Gamma ^{{\rm T} } _{\Sigma 5}$ . (9.6)

Now, it will be helpful to start with the mixed Ricci tensor $R^{\Sigma } _{{\rm N} }$ , and lower this into covariant form, in a 5-covariant manner, as such:

$R_{{\rm M} {\rm N} } =g_{{\rm M} \Sigma } R^{\Sigma } _{{\rm N} } =g_{{\rm M} \sigma } R^{\sigma } _{{\rm N} } +g_{{\rm M} 5} R^{5} _{{\rm N} }$ . (9.7)

From this, is it easily found, making use of $g_{55} =+1$ , that the component equation:

$R_{55} =g_{5\sigma } R^{\sigma } _{5} +g_{55} R^{5} _{5} =g_{5\sigma } R^{\sigma } _{5} +R^{5} _{5}$ . (9.8)

We then rearrange this into $R^{5} _{5} =R_{55} -g_{5\sigma } R^{\sigma } _{5}$ and insert the result into (9.1), thus arriving at:

${\rm R}=R+R_{55} -g_{5\sigma } R^{\sigma } _{5}$ . (9.9)

Finally, we make use of the spacetime components of: (7.9) written as $R^{\sigma } _{5} \propto j^{\sigma }$ ; (8.4) written as $g_{5\sigma } \propto A_{\sigma }$ ; (9.6), written as $R_{55} =+\Gamma ^{\Sigma } _{5{\rm T} } \Gamma ^{{\rm T} } _{\Sigma 5}$ ; and the oft-employed $\Gamma ^{{\rm M} } _{{\rm T} 5} \propto {\tfrac{1}{2}} F^{{\rm M} } _{{\rm T} }$ , to rewrite (9.9) as:

${\rm R}=R+R_{55} -g_{5\sigma } R^{\sigma } _{5} =R+\Gamma ^{\Sigma } _{5{\rm T} } \Gamma ^{{\rm T} } _{\Sigma 5} -g_{5\sigma } R^{\sigma } _{5} =R+a\cdot {\tfrac{1}{4}} F^{\Sigma } _{{\rm T} } F^{{\rm T} } _{\Sigma } -b\cdot A_{\sigma } j^{\sigma }$ , (9.10)

where we have absorbed the proportionality $\propto$ into the unknown constants $a,b$ . However, the term $F^{\Sigma } _{{\rm T} } F^{{\rm T} } _{\Sigma } =F^{\Sigma {\rm T} } F_{{\rm T} \Sigma } =-F^{\Sigma {\rm T} } F_{\Sigma {\rm T} } =-F^{\sigma \tau } F_{\sigma \tau }$ , with the final step taken by virtue of $F_{5{\rm T} } =0$ from (9.4). Now, choosing $a=b=1$ , (9.10) finally reduces to:

${\rm R}=R-{\tfrac{1}{4}} F^{\sigma \tau } F_{\sigma \tau } -A_{\sigma } j^{\sigma } =R+{\tfrac{1}{\sqrt{-g} }} {\rm L}_{QED} ={\tfrac{1}{\sqrt{-g} }} \left({\rm L}_{gravitation} +{\rm L}_{QED} \right)$ . (9.11)

Lo and behold: the QED Lagrangian density ${\rm L}_{QED} =\sqrt{-g} \left(-{\tfrac{1}{4}} F^{\sigma \tau } F_{\sigma \tau } -A_{\sigma } j^{\sigma } \right)$ is automatically added to the four-dimensional Ricci scalar $R$ as part of the five-dimensional Ricci scalar ${\rm R}$ . More to the point: $\sqrt{-g} {\rm R}$ is a seamlessly-integrated electro-gravitational Lagrangian density, in vacuo. Choosing the constant factors $a=b=1$ , even the factor of ¼ and the negative sign of the QED Lagrangian density are all automatically introduced. Employed in the Euler-Lagrange equation, ${\rm L}_{QED}$ can be used in the usual manner to obtain Maxwell’s equation $j^{\nu } =F^{\mu \nu } _{;\mu }$ . But of even greater interest, is that we now bring QED into the mix, directly from gravitational theory in five dimensions, which raises a possible approach to quantum gravitation.

10. A Tentative Path Toward Quantum Gravitation

Maintaining g to be the four-dimensional metric determinant of $g_{\mu \nu }$ , the electro-gravitational Lagrangian density, in vacuo, is now specified by:

${\rm L}=\sqrt{-g} {\rm R}=\sqrt{-g} \left(R-{\tfrac{1}{4}} F^{\sigma \tau } F_{\sigma \tau } -A_{\sigma } j^{\sigma } \right)$ . (10.1)

Though derived from a 5-dimensional spacetime with axial time, and wholly-founded upon geometrodynamics in five dimensions, all that remains in (10.1) are objects specified in ordinary 4-dimensional spacetime. The Einstein-Hilbert action — now including QED — is then specified in the usual way by integrating over the invariant 4-volume element $dV\equiv \sqrt{-g} d^{4} x$ :

$S\left(g_{\mu \nu } ,A_{\sigma } \right)=\frac{c^{4} }{16\pi G} \int \sqrt{-g} {\rm R}d^{4} x =\frac{c^{4} }{16\pi G} \int \left(R-{\tfrac{1}{4}} F^{\sigma \tau } F_{\sigma \tau } -A_{\sigma } j^{\sigma } \right)\, dV$ . (10.2)

Expanding $F^{\sigma \tau } =A^{\sigma ;\tau } -A^{\tau ;\sigma }$ , and integrating by parts in the usual way, then allows one to specify a path integral $Z=\int DA e^{iS\left(A\right)} \equiv e^{iW\left(J\right)}$ and associated transition amplitude $W\left(J\right)$ for the ${\rm L}_{QED}$ portion of the above. But what is particularly intriguing, is that ${\tfrac{1}{\sqrt{-g} }} {\rm L}_{QED}$ , in five dimensions, is naturally added to the Ricci scalar $R$ , see (9.11). What Zee, A. in Quantum Field Theory in a Nutshell, Princeton (2003), pp. 167 and 460 refers to as “The Central Identity of Quantum Field Theory,” is given generally by:

$\int D \varphi \exp (-\tfrac{1}{2} \varphi \cdot K \cdot \varphi +J \cdot \varphi -V(\varphi)) =\exp(-V(\tfrac{\delta}{\delta J})) \dot \exp(\tfrac{1}{2} J \cdot K^{-1} \cdot J)$ (10.3)

In (10.2), we find that the Ricci curvature scalar $R$ , in this identity, plays the role of $V$ .

Given that the Ricci scalar $R$ , which is a classical gravitational scalar, is firmly-entrenched together with ${\rm L}_{QED}$ in equation (9.11), given that this originates strictly on the geometrodynamic basis of the 5-dimensional spacetime geometry with axial time, given that $R$ appears to map neatly to $V$ in the Gaussian identity (10.3), and given that we know a great deal about how ${\rm L}_{QED}$ is utilized in quantum field theory, all of the foregoing may provide a new pathway for understanding how to quantize gravitation.

11. The Geometric Maxwell Tensor

Before concluding this lab note, it is also helpful to recast the Maxwell energy tensor $T^{\mu } _{\nu Maxwell} =-{\tfrac{1}{4\pi }} \left(F^{\mu \sigma } F_{\nu \sigma } -{\tfrac{1}{4}} \delta ^{\mu } _{\nu } F^{\sigma \tau } F_{\sigma \tau } \right)$ into geometric form. Again we start with $\Gamma ^{{\rm M} } _{{\rm T} 5} \propto {\tfrac{1}{2}} F^{{\rm M} } _{{\rm T} }$ , which, in light of (9.4), $F_{5{\rm T} } =0$ , can be written without any information loss as the four-dimensional $\Gamma ^{\mu } _{\tau 5} \propto {\tfrac{1}{2}} F^{\mu } _{\tau }$ .

First, we return to (9.5). Using $\Gamma ^{{\rm M} } _{{\rm T} 5} \propto {\tfrac{1}{2}} F^{{\rm M} } _{{\rm T} }$ , this becomes:

$R^{{\rm M} } _{55{\rm N} } =+\Gamma ^{\Sigma } _{5{\rm N} } \Gamma ^{{\rm M} } _{\Sigma 5} \propto -{\tfrac{1}{4}} F^{{\rm M} \Sigma } F_{{\rm N} \Sigma }$ , (11.1)

and with $F_{5{\rm T} } =0$ ,

$R^{\mu } _{55\nu } \propto -{\tfrac{1}{4}} F^{\mu \sigma } F_{\nu \sigma }$ . (11.2)

The contraction of this, also evident via (9.6), becomes:

$R_{55} \propto -{\tfrac{1}{4}} F^{\sigma \tau } F_{\sigma \tau }$ . (11.3)

Employing (11.2) and (11.3) then enables us to specify the Maxwell tensor, geometrically, as:

$T^{\mu } _{55\nu } \equiv T^{\mu } _{\nu Maxwell} =-{\tfrac{1}{4\pi }} \left(F^{\mu \sigma } F_{\nu \sigma } -{\tfrac{1}{4}} \delta ^{\mu } _{\nu } F^{\sigma \tau } F_{\sigma \tau } \right)\propto R^{\mu } _{55\nu } -{\tfrac{1}{4}} \delta ^{\mu } _{\nu } R_{55}$ . (11.4)

To maintain a balanced set of spacetime indexes, (11.4) suggests that the Maxwell tensor must actually be the axial ${\rm A} {\rm B} =55$ and ${\rm M} {\rm N} =\mu \nu$ component of larger, fourth-rank energy tensor $T^{{\rm M} } _{{\rm A} {\rm B} {\rm N} }$ which has the same symmetries as the Riemann tensor $R^{{\rm M} } _{{\rm A} {\rm B} {\rm N} }$ , and for which $T^{\mu } _{55\nu } \equiv T^{\mu } _{\nu Maxwell}$ . The Poynting vector is then to be found residing on the $T^{0} _{55k} \propto R^{0} _{55k}$ components of the Riemann tensor, $k=1,2,3$ , and so too, acquires a totally geometric foundation.

These expanded tensors have 50 independent components. Particularly, $R_{{\rm M} {\rm A} {\rm B} {\rm N} } =-R_{{\rm M} {\rm A} {\rm N} {\rm B} }$ and $R_{{\rm M} {\rm A} {\rm B} {\rm N} } =R_{{\rm N} {\rm B} {\rm M} {\rm A} }$ yields a symmetric tensor of two antisymmetric 5×5 tensors. An antisymmetric 5×5 tensor has 10 independent components, yielding a symmetric 10×10 tensor with 55 independent components. However, identity (7.2) imposes five constraints among these 55 independent components, one for each of the five indexes which is omitted from any particular equation based on (7.2). The result is 50 independent components. Maxwell’s tensor (11.4) clearly contains 10 of these independent components, leaving the remaining 40 components for further exploration.

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Lab Notes for a Scientific Revolution (Physics)

February 14, 2008