Difference between revisions of "Gauge transformations and degrees of freedom"
(→Problem 3: Dynamical degrees of freedom) |
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\[\nabla \mathbf{w}\equiv\partial_\alpha w^\alpha;\quad | \[\nabla \mathbf{w}\equiv\partial_\alpha w^\alpha;\quad | ||
\triangle\equiv\partial_\alpha \partial_\alpha;\quad | \triangle\equiv\partial_\alpha \partial_\alpha;\quad | ||
− | \square \equiv\ | + | \square \equiv\partial_0^2 -\triangle,\] |
and summation is assumed over any repeated indices. Then | and summation is assumed over any repeated indices. Then | ||
\begin{align} | \begin{align} | ||
R_{\alpha\alpha}=-\triangle(\Phi+\Psi) | R_{\alpha\alpha}=-\triangle(\Phi+\Psi) | ||
+3\square \Psi | +3\square \Psi | ||
− | +\ | + | +\partial_0 \nabla\mathbf{w} |
-2\partial_\alpha \partial_\beta s_{\alpha\beta} | -2\partial_\alpha \partial_\beta s_{\alpha\beta} | ||
\end{align} | \end{align} |
Revision as of 13:58, 26 December 2012
The general equations
\[G_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}\]
are valid in any coordinate frame, in which the metric obeys eq. 1, so$^*$ we have the freedom to make coordinate transformation
\[x^{\mu}\to {x'}^{\mu}=x^{\mu}+\xi^{\mu}(x),\]
with four arbitrary functions $\xi^\mu$, which are of the first order by $h_{\mu\nu}$.
$^*$In addition to global Lorentz transformations, which are symmetries of the Minkowski background, or in general the isometries of the background spacetime.
Contents
Problem 1: Gauge transformations
Find $h_{\mu\nu}$ in the new (primed) coordinates; show that curvature tensor and its contractions are gauge invariant and do not change their functional form.
$g_{\mu\nu}$ 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 $A_\mu\to A_\mu +\partial_\mu \psi$.
In a given frame the metric perturbation $h_{\mu\nu}$ can be decomposed into pieces which transform under spatial rotations as scalars, vectors and tensors (the irreducible representations of the rotation group $SO(3)$) in the following way (spatial components are denoted by Greek indices from the beginning of the alphabet $\alpha,\beta,\gamma\ldots=1,2,3$):
\begin{align}
h_{00}&=2\Phi;\\
h_{0\alpha}&=-w_{\alpha};\\
h_{\alpha\beta}&=2\big( s_{\alpha\beta}
+\Psi\eta_{\alpha\beta}\big),
\end{align}
where $h_{\alpha\beta}$ is further decomposed in such a way that $s_{ij}$ is traceless and $\Psi$ encodes the trace:
\begin{align}
h\equiv h_{\alpha}^{\alpha}
&=\eta^{\alpha\beta}h_{\alpha\beta}
=0+2\Psi \delta^{\alpha}_{\alpha}=6\Psi;\\
\Psi&=\tfrac{1}{6}h;\\
s_{\alpha\beta}&=\tfrac{1}{2}\big(h_{\alpha\beta}
-\tfrac{1}{6}h\; \eta_{\alpha\beta}\big).
\end{align}
Thus the metric takes the form
\[ds^{2}=(1+2\Phi)dt^2 -2w_{\alpha}dt\,dx^{\alpha}
-\big[(1-2\Psi)\eta_{\alpha\beta}
-2s_{\alpha\beta}\big]dx^\alpha dx^\beta\]
Problem 2: Particle's motion, gravo-magnetic and gravo-electric fields
Write down geodesic equations for a particle in the weak field limit in terms of fields $\Phi$, $w_\alpha$, $h_{\alpha\beta}$. What are the first terms of expansion by $v/c$ in the non-relativistic limit?
HINT: The equations of motion for a particle with $u^{\mu}=E(1,\mathbf{v})$ are$*$ \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} We can define the gravo-electric $G^\alpha$ and gravo-magnetic $H^\alpha$ fields \begin{align} G^\alpha&=-\partial_\alpha \Phi -\partial_0 w_\alpha;\\ H^\alpha&=\varepsilon^{\alpha\beta\gamma} \partial_\beta w_\gamma, \end{align} so that the first terms in the equation of motion reproduce the familiar Lorentz force of electrodynamics, with electric and magnetic fields replaced by gravo-electric and gravo-magnetic. In general there are additional terms even linear by $v$, but e.g. in a stationary field they vanish, so in the first order by $v/c$ the non-relativistic equations of motion look very much like those in electrodynamics in effective fields $G^\alpha$ and $H^\alpha$. The fields $\Phi$ and $w^\alpha$ are the analogues of scalar and vector potentials.
$^*$(Anti-)symmetrization is defined with the $1/2$ factors.
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}
Problem 3: Dynamical degrees of freedom
Derive the Einstein equations for the scalar $\Phi,\Psi$, vector $w^\alpha$ and tensor $s_{\alpha\beta}$ perturbations. Which of them are dynamical?
HINT: 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 \partial_\beta w^\beta +\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-\\ &-\partial_0 \partial_{(\alpha}w_{\beta)} -\delta_{\alpha\beta}\partial_0 \partial_\gamma w^\gamma -\square s_{\alpha\beta} -\tfrac12 \partial_{\gamma}\partial_{(\alpha}s_{\beta)\gamma} +\delta_{\alpha\beta}\partial_\gamma \partial_\delta s_{\gamma\delta}, \end{align} where $\triangle\equiv\partial_\alpha \partial_\alpha$, $\square \equiv\partial_0^2 -\triangle$ and summation is assumed over any repeated indices.
None of the equations contain time derivatives of the scalar and vector perturbations. So, from the $(00)$ equation, knowing $s_{\alpha\beta}$ and the matter sources $T_{00}$, we can find $\Psi$ (up to boundary conditions, which are assumed to be fixed), thus $\Psi$ is not an independent dynamical field/variable: it does not need initial conditions. Likewise, $\mathbf{w}$ is obtained from the $(0\alpha)$ equations as long as we know $h_{\alpha\beta}$. Finally, from the $(\alpha\beta)$ equations one obtains $\Phi$. So the dynamical degrees of freedom all lie in $s_{\alpha\beta}$.
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 \[\nabla \mathbf{w}\equiv\partial_\alpha w^\alpha;\quad \triangle\equiv\partial_\alpha \partial_\alpha;\quad \square \equiv\partial_0^2 -\triangle,\] and summation is assumed over any repeated indices. Then \begin{align} R_{\alpha\alpha}=-\triangle(\Phi+\Psi) +3\square \Psi +\partial_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}
Problem 4: Gauge decomposition
Find the gauge transformations for the scalar, vector and tensor perturbations.
HINT: The gauge transformation $x\to x+\xi$ changes the full metric perturbation as \[h_{\mu\nu}\to h_{\mu\nu} -\partial_\mu \xi_{\nu}-\partial_\nu \xi_\mu.\] Then \begin{align} \Phi\equiv h_{00}&\to \Phi- \partial_0 \xi_0;\\ w_\alpha\equiv h_{0\alpha}&\to w_{\alpha}+\partial_0 \xi_{\alpha}+\partial_\alpha \xi_0;\\ \Psi=\tfrac16 h^\alpha_\alpha &\to \Psi+\tfrac13 \partial_\alpha \xi_\alpha;\\ s_{\alpha\beta}=\tfrac12 (h_{\alpha\beta} -\Psi \eta_{\alpha\beta})&\to s_{\alpha\beta}-\partial_{(\alpha}\xi_{\beta)} -\tfrac13 \eta_{\alpha\beta}\partial_\gamma \xi_\gamma \end{align}
Problem 5: Synchronous gauge
This one is equivalent to Gaussian normal coordinates and is fixed by setting \begin{equation} \Phi=0,\qquad w^\alpha=0. \end{equation} Write the explicit coordinate transformations and the metric in this gauge.
HINT: $ds^{2}=dt^{2} -(\delta_{\alpha\beta}-h_{\alpha\beta}) dx^\alpha dx^\beta$
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 $|h^{\mu\nu}|\ll 1$.
Problem 6: Transverse gauge
This is a generalization of the conformal Newtonian or Poisson gauge sometimes used in cosmology, which is fixed by demanding that \begin{equation} \partial_\alpha s^{\alpha\beta}=0,\qquad \partial_\alpha w^\alpha =0. \end{equation} Find the equations for $\xi^\mu$ that fix the transverse gauge.
HINT: $\triangle \xi_\beta +\tfrac13 \partial_\alpha \partial_\beta \xi_\alpha =\partial_\alpha s_{\alpha\beta},\quad \triangle \xi_0 =-\partial_\alpha w_\alpha -\partial_0 \partial_\alpha \xi_\alpha$
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 $\xi^0$.