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Covariance with respect to canonical transformations - a unification of gravity and electromagnetism?

Gyula Bene
Department of Theoretical Physics, Institute of Physics, Roland E??tv??s University
1117 Budapest, P??zm??ny P??ter s??t??ny 1/A



Plan of the talk:

Introduction

Geometry of the phase space

Vectors and tensors in phase space

Phase space coordinates: \[z^0=x^0\;,\] \[z^1=x^1\;,\] \[z^2=x^2\;,\] \[z^3=x^3\;,\] \[z^4=p_0\;,\] \[z^5=p_1\;,\] \[z^6=p_2\;,\] \[z^7=p_3\;.\] Canonical transformation: \begin{eqnarray}x^j=\frac{\partial \Psi}{\partial p_j}\label{e5}\end{eqnarray} and \begin{eqnarray}p'_j=\frac{\partial \Psi}{\partial x'^j}\;,\label{e6}\end{eqnarray} where \(\Psi=\Psi(x',p)\). Contravariant vectors transform like \(dz^j\), i.e., \begin{eqnarray}dz'^j=M^j_{\phantom{j}k}dz^k\;.\label{e7}\end{eqnarray} \(M\) may be expressed in terms of the second derivatives of the generating function \(\Psi(x',p)\). It satisfies \begin{eqnarray}MJM^T=J\;,\label{e8}\end{eqnarray} where \begin{eqnarray}J^{x^jx^k}=J^{p_jp_k}=0\label{e9}\end{eqnarray} and \begin{eqnarray}J^{x^jp_k}=-J^{p_kx^j}=\delta^j_k\;.\label{e10}\end{eqnarray} Hence, the symplectic unit matrix \(J\) is a second order contravariant tensor that is left unchanged by canonical transformations.
Covariant vector components transform like the gradient of a scalar, i.e. \begin{eqnarray}\frac{\partial \Phi}{\partial z'^j}=\left(M^{-1}\right)^k_{\phantom{k}j}\frac{\partial \Phi}{\partial z^k}\label{e11}\end{eqnarray} Obviously, the product of a covariant vector with a contravariant one is a scalar, therefore Hamilton's equations may be written in the manifestly covariant form \begin{eqnarray}\frac{dz^j}{ds}=J^{jk}\frac{\partial H}{\partial z^k}\;.\label{e12}\end{eqnarray} A covariant index may be raised by \(J\), namely \begin{eqnarray}A^j=J^{jk}A_k\;.\label{e_raise}\end{eqnarray} This definition, when applied to the (yet undefined) covariant counterpart of \(J\) itself, yields \begin{eqnarray}J^{jk}J_{km}=J^{j}_{\phantom{j}m}\;.\label{e_Jraise}\end{eqnarray} Lowering indices is performed by multiplication with the covariant tensor \(J_{mk}\). The rule is \begin{eqnarray}A_j=J_{kj}A^k\;.\label{e_lower}\end{eqnarray} Consistency requires: \begin{eqnarray}J^{jm}J_{km}=\delta^j_k\label{e_consis}\end{eqnarray} Hence, the covariant symplectic unit tensor \(J_{jk}\) as the inverse transposed of \(J\), which is \(J\) itself. This means that the components of \(J_{jk}\) are the same as those of \(J^{jk}\). We also have \begin{eqnarray}J^{j}_{\phantom{j}k}=-\delta^j_k\label{e_mixa}\end{eqnarray} and \begin{eqnarray}J_{k}^{\phantom{k}j}=\delta^j_k\label{e_mixb}\;.\end{eqnarray} The rules imply \begin{eqnarray}A^jB_j=-A_jB^j\label{e_scal1}\;.\end{eqnarray} Especially, for any vector \(A^j\) \begin{eqnarray}A^jA_j=0\label{e_scal2}\;.\end{eqnarray}

Covariant derivatives and Christoffel's symbols

Albeit the tensors \(J^{jk}\) and \(J_{jk}\) which raise and lower indices are constant in this case, unlike the metric tensor in general relativity, the geometry of phase space is still nontrivial, because the transformation rule depends on the chosen point in phase space. This implies that the partial derivative of a vector will not be a tensor, because it stems from the difference of vectors at different phase space points. Indeed, differentiating the transformation rule of a covariant vector component \(v_j\) \begin{eqnarray}v'_j= \frac{\partial z^k}{\partial z'^j}v_k\label{e13}\end{eqnarray} one gets \begin{eqnarray}v'_{j,l}\equiv \frac{\partial v'_j}{\partial z'^l}= \frac{\partial z^k}{\partial z'^j}\frac{\partial z^n}{\partial z'^l}\frac{\partial v_k}{\partial z^n}+\frac{\partial^2 z^k}{\partial z'^j\partial z'^l}v_k \;.\label{e14}\end{eqnarray} Covariant derivatives (denoted by semicolon) that are tensors can be obtained by introducing suitable Christoffel's symbols \(\Gamma^k_{\phantom{k}jl}\): \begin{eqnarray}v_{j;l}=v_{j,l}-\Gamma^k_{\phantom{k}jl}v_k\;.\label{e15}\end{eqnarray} This implies the transformation rule of Christoffel's symbols: \begin{eqnarray}\Gamma'^k_{\phantom{k}jl}=\frac{\partial z'^k}{\partial z^s}\frac{\partial z^r}{\partial z'^j}\frac{\partial z^n}{\partial z'^l}\Gamma^s_{\phantom{s}rn}+\frac{\partial z'^k}{\partial z^s}\frac{\partial^2 z^s}{\partial z'^j\partial z'^l}\label{e16}\end{eqnarray} Since the symplectic unit matrix is constant, it is natural to assume that its covariant derivative vanishes: \begin{eqnarray}J_{ij;l}=-\Gamma^k_{\phantom{k}il}J_{kj}-\Gamma^k_{\phantom{k}jl}J_{ik}=0\label{e19}\end{eqnarray} This is equivalent with \begin{eqnarray}\Gamma_{ijl}=\Gamma_{jil}\;.\label{e20}\end{eqnarray} (288 independent components)

Let \[W_{ijk}=\Gamma_{ijk}-\Gamma_{ikj}\] This is a tensor. Further, it satisfies \[W_{ijk}+W_{jki}+W_{kij}=0\] (168 independent components)

Let \[S_{ijk}=\frac{1}{3}\left(\Gamma_{ijk}+\Gamma_{jki}+\Gamma_{kij}\right)\] the symmetrized part of the connection (120 independent components). This is not a tensor and may be set to zero at a given phase space point by a suitable canonical transformation.
By adding the equations \begin{eqnarray} \Gamma_{ijk}+\Gamma_{jki}+\Gamma_{kij} &=& 3S_{ijk}\\ \Gamma_{ijk}-\Gamma_{ikj} &=& W_{ijk}\\ \Gamma_{jik}-\Gamma_{jki} &=& W_{jik} \end{eqnarray} we get \[\Gamma_{ijk}=S_{ijk}+\frac{1}{3}\left(W_{ijk}+W_{jik}\right)\] Another kind of covariant derivative can be defined if the tensorial part is left out: \[A_{i:j}=A_{i,j}-S^{k\phantom{ij}}_{\phantom{k}ij}A_k\] Note that \(A^i_{:i}=A^i_{,i}\) and \(F^{jk}_{\phantom{jk}:k}=F^{jk}_{\phantom{jk},k}\) if \(F^{jk}\) is antisymmetric.

Calculating connections

Hamilton's equations of motion: \begin{eqnarray}\frac{dz^i}{ds}=v^i\label{g1}\end{eqnarray} with \begin{eqnarray}v^i=J^{ij}\frac{\partial H}{\partial z^j}\label{g1a}\;.\end{eqnarray} We require this to be identical with the geodetic equation \begin{eqnarray}\frac{d^2z^i}{ds^2}+\Gamma^i_{\phantom{i}jk}\frac{dz^j}{ds}\frac{dz^k}{ds}=0\;.\label{g2}\end{eqnarray} Applying Hamilton's equations twice we have \begin{eqnarray}\frac{d^2z^i}{ds^2}=v^i_{,j}v^j\label{g1b}\;.\end{eqnarray} Comparing this with the geodetic equation we get \begin{eqnarray}v^i_{,j}v^j+\Gamma^i_{\phantom{i}jk}v^jv^k\equiv v^i_{;j}v^j=0\label{g3}\;.\end{eqnarray} This allows the systematic calculation of the connections: \begin{eqnarray} \Gamma_{p_ip_jp_k} &=& 0\\ \Gamma_{p_ip_jx^k} &=& -\frac{g^{ij}(p_k-A_k)}{H^2}\\ \Gamma_{p_ix^kp_j} &=& \Gamma_{x^kp_ip_j}=0\\ \Gamma_{x^ix^jp_k} &=& \gamma^{k}_{\phantom{k}ij}\\ \Gamma_{x^ip_kx^j} &=& \Gamma_{p_kx^ix^j} = \gamma^{k}_{\phantom{k}ij}-\frac{(p_j-A_j)(p^k-A^k)_{;i}}{H^2}\\ \Gamma_{x^ix^jx^k} &=& -\frac{(p_n-A_n)_{;i}(p^n-A^n)_{;j}(p_k-A_k)}{H^2}\\ &&-\frac{1}{6}\sum_{all\; permutations\;of\;i,j,k}(p_i-A_i)_{;j;k}\\ &&+\frac{1}{3}(A_{k,j}-A_{j,k})_{;i}+\frac{1}{3}(A_{k,i}-A_{i,k})_{;j} \end{eqnarray} The tensorial part is now easily obtained: \begin{eqnarray} W_{p_ip_jp_k} &=& 0\\ W_{p_ip_jx^k} &=& -W_{p_ix^kp_j} = -\frac{g^{ij}(p_k-A_k)}{H^2}\\ W_{x^kp_ip_j} &=& 0\\ W_{x^ix^jp_k} &=& -W_{x^ip_kx^j} = \frac{(p_j-A_j)(p^k-A^k)_{;i}}{H^2}\\ W_{p_kx^ix^j} &=& \frac{(p_i-A_i)(p^k-A^k)_{;j}}{H^2}-\frac{(p_j-A_j)(p^k-A^k)_{;i}}{H^2}\\ W_{x^ix^jx^k} &=& (A_{k,j}-A_{j,k})_{;i}\\ &&+\frac{(p_n-A_n)_{;i}(p^n-A^n)_{;k}(p_j-A_j)}{H^2}\\ &&-\frac{(p_n-A_n)_{;i}(p^n-A^n)_{;j}(p_k-A_k)}{H^2} \end{eqnarray} Note that it implies \[W_{ik}^{\phantom{ik}k}=\frac{2H_{,i}}{H}\] The symmetrized part of the connection: \begin{eqnarray} S_{p_ip_jp_k} &=& 0\\ S_{p_ip_jx^k} &=& S_{x^kp_ip_j}=S_{p_ix^kp_j}=-\frac{1}{3}\frac{g^{ij}(p_k-A_k)}{H^2}\\ S_{x^ix^jp_k} &=& S_{x^ip_kx^j} = S_{p_kx^ix^j} = \gamma^{k}_{\phantom{k}ij}\\ &&-\frac{1}{3}\frac{(p_i-A_i)(p^k-A^k)_{;j}}{H^2}-\frac{1}{3}\frac{(p_j-A_j)(p^k-A^k)_{;i}}{H^2}\\ S_{x^ix^jx^k} &=& -\frac{1}{3}\frac{(p_n-A_n)_{;i}(p^n-A^n)_{;j}(p_k-A_k)}{H^2}\\ &&-\frac{1}{3}\frac{(p_n-A_n)_{;j}(p^n-A^n)_{;k}(p_i-A_i)}{H^2}\\ &&-\frac{1}{3}\frac{(p_n-A_n)_{;k}(p^n-A^n)_{;i}(p_j-A_j)}{H^2}\\ &&-\frac{1}{6}\sum_{all\; permutations\;of\;i,j,k}(p_i-A_i)_{;j;k} \end{eqnarray}

Search for field equations

We are looking for a set of equations that determines \(W\), \(S\) and \(H\) (altogether 289 field components) so that the resulting equations are compatible with the Einstein-Maxwell system of equations.
Strategy: compiling all possible Lagrangian densities and finding their right combination. Field equations are the Euler-Lagrange equations \[\frac{\partial L}{\partial W_{ijk}}-\frac{\partial}{\partial z^l}\frac{\partial L}{\partial W_{ijk,l}}=0\] \[\frac{\partial L}{\partial S_{ijk}}-\frac{\partial}{\partial z^l}\frac{\partial L}{\partial S_{ijk,l}}=0\] \[\frac{\partial L}{\partial H}=0\] These turn out to be tensorial equations with the same symmetries as \(W\) and \(S\). The last equation is a scalar.
Since \(S\) is not a tensor, it cannot directly enter the Lagrangian density. We define the Riemannian by \begin{eqnarray}A_{i:j:k}-A_{i:k:j}=A_lR^l_{\phantom{l}ijk}\label{e24}\end{eqnarray} This implies: \[R_{ijkl}=S_{ijl,k}-S_{ijk,l}+S_{ikm}S^m_{\phantom{m}jl}-S_{ilm}S^m_{\phantom{m}jk}\] Symmetries: \(R_{ijkl}=R_{jikl}=-R_{ijlk}\;,\quad R_{ijkl}+R_{iklj}+R_{iljk}=0\;.\)

Building blocks of the Lagrangian density: \[H\;,\quad W_{ijk}\;,\quad W_{ijk:l}\;,\quad W_{ijk:l:m}\;,\quad R_{ijkl}\;,\quad R_{ijkl:m}\;.\]

What is still missing?

Conclusion