Difference between revisions of "Gauge transformations and degrees of freedom"

$g_{\mu\nu}$ transforms as tensor, so taking into account that \[\frac{\partial}{\partial {x'}^{\mu'}}\xi^\mu =\frac{\partial}{\partial {x}^{\mu'}}\xi^\mu+O(h^2) =\xi^{\mu}_{,\mu'}+O(h^2),\] we get \begin{align} {g'}_{\mu'\nu'} &=\frac{\partial x^{\mu}}{\partial {x'}^{\mu'}} \frac{\partial x^{\nu}}{\partial {x'}^{\nu'}}\; g_{\mu\nu}=\\ &=\Big(\delta^{\mu}_{\mu'}- \frac{\partial \xi^{\mu}}{\partial {x'}^{\mu'}}\Big) \Big(\delta^{\nu}_{\nu'}- \frac{\partial \xi^{\nu}}{\partial {x'}^{\nu'}}\Big) \big(\eta_{\mu\nu}+h_{\mu\nu}+O(h^2)\big)=\\ &=\Big(\delta^{\mu}_{\mu'}- \frac{\partial \xi^{\mu}}{\partial {x}^{\mu'}}\Big) \Big(\delta^{\nu}_{\nu'}- \frac{\partial \xi^{\nu}}{\partial {x}^{\nu'}}\Big) \big(\eta_{\mu\nu}+h_{\mu\nu}+O(h^2)\big)=\\ &=\eta_{\mu'\nu'}+ \big[h_{\mu'\nu'} -\partial_{\mu'}\xi_{\nu'} -\partial_{\nu'}\xi_{\mu'}\big]+O(h^2), \end{align} thus by definition \[{h'}_{\mu\nu}=h_{\mu\nu} -\xi_{\mu,\nu}-\xi_{\nu,\mu}.\] Note that this in fact the same derivation as the one for the Killing equation.
The corrections for the curvature tensor are calculated in the same way, and will be proportional to $R_{\mu\nu\rho\sigma}\xi^\lambda$. As curvature tensor itself is linear by $h$, the corrections are quadratic and can be discarded in the first-order approximation. The same applies to the contractions: $R_{\mu\nu}$ and $R$. Thus the curvature tensor is said to be gauge invariant in the linearized theory, very much like electromagnetic field tensor $F^{\mu\nu}$ is invariant under gauge transformations of electrodynamics $A_\mu\to A_\mu +\partial_\mu \psi$.

Calculating the Christoffel symbols, we get \begin{align} {\Gamma^0}_{00}&=\partial_{0}\Phi;\\ {\Gamma^\alpha}_{00}& =\partial_{\alpha}\Phi+\partial_0 w_\alpha;\\ {\Gamma^0}_{\alpha 0}& =\partial_{\alpha}\Phi;\\ {\Gamma^0}_{\alpha\beta}& =-\partial_{(\alpha}w_{\beta)} -\tfrac{1}{2}\partial_0 h_{\alpha\beta};\\ {\Gamma^\alpha}_{0\beta}& =\tfrac{1}{2}\partial_0 h_{\alpha\beta} +\partial_{[\alpha}w_{\beta]};\\ {\Gamma^\alpha}_{\beta\gamma}& =\tfrac{1}{2}\partial_\alpha h_{\beta\gamma} -\partial_{[\alpha}h_{\beta]\gamma}, \end{align} where brackets and braces denote (anti-)symmetrization: \[A_{(ab)}=\tfrac{1}{2}(A_{ab}+A_{ba});\quad A_{[ab]}=\tfrac{1}{2}(A_{ab}-A_{ba}).\] Taking into account that for a particle with the natural chose of parameter \begin{align} u^{\mu}&=\frac{dx^{\mu}}{d\lambda}= E(1,\mathbf{v}), \qquad\text{and}\\ \frac{dp^\mu}{dt} &=\frac{dp^\mu / d\lambda}{dt/d\lambda} =\frac{1}{E}\frac{dp^\mu}{d\lambda}, \end{align} the geodesic equations can be rewritten as \begin{align} \frac{dE}{dt}&=-\frac{1}{E} {\Gamma^{0}}_{\mu\nu}u^\mu u^\nu;\\ \frac{dp^\alpha}{dt}&=-\frac{1}{E} {\Gamma^{\alpha}}_{\mu\nu}u^\mu u^\nu. \end{align} On substitution of ${\Gamma^{\lambda}}_{\mu\nu}$, in explicit form we obtain \begin{align} \frac{dE}{dt}&=-E\big[\partial_0 \Phi +2\partial_\alpha \Phi\; v^\alpha -\big(\partial_{(\alpha} w_{\beta)} +\tfrac{1}{2}\partial_0 h_{\alpha\beta}\big) v^\alpha v^\beta \big] ;\\ \frac{dp^\alpha}{dt}&=-E\big[ \partial_\alpha \Phi+\partial_0 w_\alpha +2(\partial_{[\alpha}w_{\beta]} +\tfrac12 \partial_0 h_{\alpha\beta})v^{\beta} -\big( \partial_{(\alpha} h_{\beta)\gamma} -\tfrac{1}{2}\partial_\alpha h_{\beta\gamma}\big) v^\beta v^\gamma \big]. \end{align} If we define the gravo-electric $G^\alpha$ and gravo-magnetic $H^\alpha$ fields as \begin{align} G^\alpha&=-\partial_\alpha \Phi -\partial_0 w_\alpha;\\ H^\alpha&=\varepsilon^{\alpha\beta\gamma} \partial_\beta w_\gamma, \end{align} the first terms appear to be represented in a very suggestive form, reminiscent of electrodynamics (remember that $E\approx m(1+v^2/2)$): \begin{align} \frac{1}{E}\frac{d \mathbf{p}}{dt} &=\mathbf{G}+\mathbf{v}\times \mathbf{H} +\mathbf{e}^{\alpha}\partial_0 h_{\alpha\beta}v^\beta +O(v^2). \end{align}

First note, that $\eta_{\alpha\beta}=-\delta_{\alpha\beta}$, so in the chosen frame $w_{\alpha}=-w^{\alpha}$. Straightforward calculation gives the Riemann tensor in the first order \begin{align} R_{0\alpha0\beta}&=-\partial_\alpha \partial_\beta \Phi -\partial_0 \partial_{(\alpha}w_{\beta)} -\tfrac{1}{2}\partial_0^2 h_{\alpha\beta};\\ R_{0\alpha\gamma\beta}& =\partial_\alpha \partial_{[\beta}w_{\gamma]} -\partial_{0}\partial_{[\gamma}h_{\beta]\alpha};\\ R_{\delta\alpha\gamma\beta}& =\partial_\alpha \partial_{[\gamma}h_{\beta]\delta} -\partial_{\delta}\partial_{[\gamma}h_{\beta]\alpha}; \end{align} Ricci tensor \begin{align} R_{00}&=\triangle \Phi -\partial_0 \nabla \mathbf{w} -3\partial_0^2 \Psi;\\ R_{0\gamma}&=3\partial_0 \partial_\gamma \Psi +\tfrac12 \triangle w^\gamma -\tfrac12 \partial_\gamma \nabla \mathbf{w} +\tfrac12 \partial_0 \partial_\alpha h_{\alpha\gamma};\\ R_{\alpha\beta}&= -\partial_\alpha \partial_\beta (\Phi+\Psi) -\partial_0 \partial_{(\alpha}w_{\beta)} -\tfrac12 \square h_{\alpha\beta} -2\partial_\gamma \partial_{(\alpha}s_{\beta)\gamma}, \end{align} where \[\nabla \mathbf{w}\equiv\partialrtial_\alpha w^\alpha;\quad \triangle\equiv\partialrtial_\alpha \partialrtial_\alpha;\quad \square \equiv\partialrtial_0^2 -\triangle,\] and summation is assumed over any repeated indices. Then \begin{align} R_{\alpha\alpha}=-\triangle(\Phi+\Psi) +3\square \Psi +\partialrtial_0 \nabla\mathbf{w} -2\partial_\alpha \partial_\beta s_{\alpha\beta} \end{align} and the Einstein tensor is \begin{align} G_{00}&=-2\triangle \Psi -\partial_\alpha \partial_\beta s_{\alpha\beta};\\ G_{0\alpha}&=3\partial_0 \partial_\alpha \Psi +\tfrac12 \triangle w^\alpha -\tfrac12 \partial_\alpha \nabla\mathbf{w} +\tfrac12 \partial_0 \partial_\beta h_{\alpha\beta};\\ G_{\alpha\beta}& =(\delta_{\alpha\beta}\triangle -\partial_\alpha \partial_\beta) (\Phi+\Psi)-2\delta_{\alpha\beta}\partial_0^2 \Psi- \notag\\ &-\partial_0 \partial_{(\alpha}w_{\beta)} -\delta_{\alpha\beta}\partial_0 \nabla \mathbf{w} -\square s_{\alpha\beta} -\tfrac12 \partial_{\gamma}\partial_{(\alpha}s_{\beta)\gamma} +\delta_{\alpha\beta}\partial_\gamma \partial_\delta s_{\gamma\delta}. \end{align}

From the gauge transformations the equations for $x^\mu (x)$ are \begin{align} &\partial_0 \xi^0 =\tfrac12 \Phi;\\ &\partial_0 \xi_\alpha= w_\alpha-\partial_\alpha \xi_0, \end{align} and why the gauge is named synchronous is obvious from the attractive form of the metric in it: \begin{equation} ds^{2}=dt^{2} -(\delta_{\alpha\beta}-h_{\alpha\beta}) dx^\alpha dx^\beta. \end{equation} Note, that this gauge (frame of reference) can always be chosen whenever $|h^{\mu\nu}|\ll 1$.

The first condition implies \[0=\partial_\alpha s_{\alpha\beta}' =\partial_\alpha s_{\alpha\beta}-\tfrac{1}{2} (\partial_\alpha \partial_\alpha \xi_\beta +\partial_\alpha \partial_\beta \xi_\alpha) +\tfrac13 \partial_{\beta}\partial_{\alpha}\xi_\alpha =\tfrac12 \big[\partial_\alpha s_{\alpha\beta} -(\triangle \xi_\beta +\tfrac13 \partial_\alpha \partial_\beta \xi_\alpha)\big],\] and the second \[0=\partial_{\alpha}w_{\alpha}'=\partial_{\alpha}w_\alpha +\partial_\alpha \partial_0 \xi_\alpha +\partial_\alpha \partial_\alpha \xi_0 =\partial_\alpha w_\alpha + (\triangle \xi_0 +\partial_0 \partial_\alpha \xi_\alpha).\] The solution of the first (three) equation(s) \[\triangle \xi_\beta +\tfrac13 \partial_\alpha \partial_\beta \xi_\alpha =\partial_\alpha s_{\alpha\beta}\] gives $\xi^\alpha$, then from the second equation \[\triangle \xi_0 =-\partial_\alpha w_\alpha -\partial_0 \partial_\alpha \xi_\alpha\] one can find $\xi^0$.