weyl-metric-calculations.tex

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    pdftitle={Detailed calculations for the Weyl metric},    % title
    pdfauthor={Adam Getchell},     % author
    pdfsubject={Causal Dynamical Triangulations},   % subject of the document
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\title{Detailed calculations for the Weyl metric}
\author{\textbf{Adam Getchell}\footnote{\href{mailto:acgetchell@ucdavis.edu}{acgetchell@ucdavis.edu}}\\\textit{Department of Physics, University of California, Davis, CA, 95616}}
\date{\today}
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\section{Vacuum solution to the Weyl metric}

Starting from the cylindrically symmetric (Weyl) vacuum metric \cite{synge_relativity}:

\begin{equation}
	ds^{2}=e^{2\lambda}dt^{2}-e^{2\left(\nu-\lambda\right)}\left(dr^{2}+dz^{2}\right)-r^{2}e^{-2\lambda}d\phi^{2}
	\label{eq:weyl-vacuum-metric}
\end{equation}

\begin{equation}
g_{\mu\nu}=\left(\begin{array}{cccc}
e^{2\lambda} & 0 & 0 & 0\\
0 & -e^{2\left(\nu-\lambda\right)} & 0 & 0\\
0 & 0 & -e^{2\left(\nu-\lambda\right)} & 0\\
0 & 0 & 0 & -r^{2}e^{-2\lambda}
\end{array}\right)\label{eq:general-axisymmetric-static-matrix-metric}
\end{equation}

In this coordinate basis, the definition of the Christoffel connection is: \cite{carroll2003spacetime} 
\begin{equation}
\Gamma_{\mu\nu}^{\lambda}=\frac{1}{2}g^{\lambda\sigma}\left(\partial_{\mu}g_{\nu\sigma}+\partial_{\nu}g_{\sigma\mu}-\partial_{\sigma}g_{\mu\nu}\right)
\end{equation}

The non-zero Christoffel connections are:
\begin{equation}
\begin{aligned}
\Gamma^{t}_{tr}&=\partial_{r}\lambda\\
\Gamma^{t}_{tz}&=\partial_{z}\lambda\\
\Gamma^{r}_{tt}&=e^{4\lambda-2\nu}\partial_{r}\lambda\\
\Gamma^{r}_{rr}&=\partial_{r}\nu-\partial_{r}\lambda\\
\Gamma^{r}_{rz}&=\partial_{z}\nu-\partial_{z}\lambda\\
\Gamma^{r}_{zz}&=\partial_{r}\lambda-\partial_{r}\nu\\
\Gamma^{r}_{\phi\phi}&=re^{-2\nu}\left(r\partial_{r}\lambda-1\right)\\
\Gamma^{z}_{tt}&=e^{4\lambda-2\nu}\partial_{z}\lambda\\
\Gamma^{z}_{rr}&=\partial_{z}\lambda-\partial_{z}\nu\\
\Gamma^{z}_{rz}&=\partial_{r}\nu-\partial_{r}\lambda\\ 
\Gamma^{z}_{zz}&=\partial_{z}\nu-\partial_{z}\lambda\\
\Gamma^{z}_{\phi\phi}&=r^{2}e^{-2\nu}\partial_{z}\lambda\\
\Gamma^{\phi}_{r\phi}&=\frac{1}{r}-\partial_{r}\lambda\\
\Gamma^{\phi}_{z\phi}&=-\partial_{z}\lambda\\
\end{aligned}
\label{eq:christoffel-connections}
\end{equation}

The components of the Riemann tensor are given by:
\begin{equation}
R_{\sigma\mu\nu}^{\rho}=\partial_{\mu}\Gamma_{\nu\sigma}^{\rho}-\partial_{\nu}\Gamma_{\mu\sigma}^{\rho}+\Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\sigma}^{\lambda}-\Gamma_{\nu\lambda}^{\rho}\Gamma_{\mu\sigma}^{\lambda}
\end{equation}

Using the properties of the Riemann tensor:
\begin{equation}
\begin{aligned}
R_{\rho\sigma\mu\nu}&=-R_{\rho\sigma\nu\mu}\\
R_{\rho\sigma\mu\nu}&=-R_{\sigma\rho\mu\nu}\\
R_{\rho\sigma\mu\nu}&=R_{\mu\nu\rho\sigma}\\
R_{\rho[\sigma\mu\nu]}&=0\\
\end{aligned}
\end{equation}
The non-zero components of the Riemann tensor are:
\begin{equation}
\begin{aligned}
R^{t}_{rtr}&=-\partial^{2}_{r}\lambda+\left(\partial_{z}\lambda\right)^{2}-2\left(\partial_{r}\lambda\right)^{2}+\partial_{r}\lambda\partial_{r}\nu-\partial_{z}\lambda\partial_{z}\nu\\
R^{t}_{rtz}&=-\partial_{r}\partial_{z}\lambda-3\partial_{r}\lambda\partial_{z}\lambda+\partial_{r}\lambda\partial_{z}\nu+\partial_{r}\nu\partial_{z}\lambda\\
R^{t}_{ztz}&=-\partial^{2}_{z}\lambda-2\left(\partial_{z}\lambda\right)^{2}+\left(\partial_{r}\lambda\right)^{2}-\partial_{r}\lambda\partial_{r}\nu+\partial_{z}\lambda\partial_{z}\nu\\
R^{t}_{\phi t\phi}&=re^{-2\nu}\left(r\left(\partial_{r}\lambda\right)^{2}-\partial_{r}\lambda+r\left(\partial_{z}\lambda\right)^{2}\right)\\
R^{r}_{zrz}&=\partial^{2}_{r}\lambda-\partial^{2}_{r}\nu+\partial^{2}_{z}\lambda-\partial^{2}_{z}\nu\\
R^{z}_{\phi z\phi}&=re^{-2\nu}\left(r\partial^{2}_{z}\lambda-r\partial_{z}\lambda\partial_{z}\nu+r\partial_{r}\lambda\partial_{r}\nu-r\left(\partial_{r}\lambda\right)^{2}+\partial_{r}\lambda-\partial_{r}\nu\right)\\
R^{z}_{\phi\phi r}&=re^{-2\nu}\left(-r\partial_{r}\partial_{z}\lambda+r\partial_{r}\nu\partial_{z}\lambda-r\partial_{r}\lambda\partial_{z}\lambda+r\partial_{r}\lambda\partial_{z}\nu-\partial_{z}\nu\right)\\
R^{\phi}_{r\phi r}&=\partial^{2}_{r}\lambda+\frac{1}{r}\partial_{r}\nu-\partial_{r}\lambda\partial_{r}\nu-\left(\partial_{z}\lambda\right)^{2}+\partial_{z}\lambda\partial_{z}\nu+\frac{1}{r}\partial_{r}\lambda\\
\end{aligned}
\end{equation}

The Ricci tensor is given by:
\begin{equation}
R_{\mu\nu}=R_{\mu\lambda\nu}^{\lambda}
\end{equation}

The non-zero components of the Ricci tensor are:
\begin{equation}
\begin{aligned}
R_{tt}&=\frac{e^{4\lambda-2\nu}}{r}\left(r\partial^{2}_{r}\lambda+r\partial^{2}_{z}\lambda+\partial_{r}\lambda\right)\\
R_{rr}&=\partial^{2}_{r}\lambda-\partial^{2}_{r}\nu+\partial^{2}_{z}\lambda-\partial^{2}_{z}\nu-2\left(\partial_{r}\lambda\right)^{2}+\frac{1}{r}\partial_{r}\lambda+\frac{1}{r}\partial_{r}\nu\\
R_{rz}&=\frac{1}{r}\partial_{z}\nu-2\partial_{r}\lambda\partial_{z}\lambda\\
R_{zz}&=\partial^{2}_{r}\lambda-\partial^{2}_{r}\nu+\partial^{2}_{z}\lambda-\partial^{2}_{z}\nu-2\left(\partial_{z}\lambda\right)^{2}+\frac{1}{r}\partial_{r}\lambda-\frac{1}{r}\partial_{r}\nu\\
R_{\phi\phi}&=re^{-2\nu}\left(r\partial^{2}_{r}\lambda+r\partial^{2}_{z}\lambda+\partial_{r}\lambda\right)
\end{aligned}
\label{eq:ricci-tensor-components}
\end{equation}

The Ricci scalar is defined as:

\begin{equation}
R=R_{\mu}^{\mu}=g^{\mu\nu}R_{\mu\nu}
\end{equation}

Which is:

\begin{equation}
R=2e^{2\left(\lambda-\nu\right)}\left(\partial^{2}_{r}\nu+\partial^{2}_{z}\nu-\partial^{2}_{r}\lambda-\partial^{2}_{z}\lambda+\left(\partial_{r}\lambda\right)^{2}+\left(\partial_{z}\lambda\right)^{2}-\frac{1}{r}\partial_{r}\lambda\right)\label{eq:R}
\end{equation}

Einstein's equation in a vacuum is:

\begin{equation}
\label{eq:einstein-vacuum-equation}
G_{\mu\nu}=0
\end{equation}

Whence Einstein's equation:

\begin{equation}
G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi GT_{\mu\nu}
\label{eq:einstein}
\end{equation}

However, we can take a shortcut by using:

\begin{equation}
R_{\mu\nu}=0
\label{eq:vacuum-solutions}
\end{equation}

Since the trace of a zero-valued matrix is identically zero, and thus \Cref{eq:vacuum-solutions} automatically satisfies \Cref{eq:einstein-vacuum-equation}.

Applying \Cref{eq:vacuum-solutions} to \Cref{eq:ricci-tensor-components} gives the following:
\begin{equation}
\partial^{2}_{r}\lambda+\frac{1}{r}\partial_{r}\lambda+\partial^{2}_{z}\lambda=0\label{eq:laplace}
\end{equation}
\begin{equation}
\partial_{r}\nu=r\left(\partial^{2}_{r}\nu+\partial^{2}_{z}\nu+2\left(\partial_{r}\lambda\right)^{2}\right)\label{eq:R_rr=0}
\end{equation}
\begin{equation}
\partial_{z}\nu=2r\partial_{r}\lambda\partial_{z}\lambda\label{eq:nu_z}
\end{equation}
\begin{equation}
\partial^{2}_{r}\nu+\partial^{2}_{z}\nu+\left(\partial_{r}\lambda\right)^{2}+\left(\partial_{z}\lambda\right)^{2}=0\label{eq:R_phiphi=0}
\end{equation}

\Cref{eq:laplace} is the two-dimensional Laplace equation in cylindrical coordinates. That is:

\begin{equation}
\nabla^2\lambda(r,z)=0
\label{eq:laplace-r-z}
\end{equation}

Plugging \Cref{eq:R_phiphi=0} into \Cref{eq:R_rr=0} gives:

\begin{equation}
\partial_{r}\nu=r\left(\left(\partial_{r}\lambda\right)^{2}-\left(\partial_{z}\lambda\right)^{2}\right)\label{eq:nu_r}
\end{equation}

Using \Cref{eq:nu_z,eq:nu_r} we find solutions for $\nu$ are given by:
\begin{equation}
\nu=\int r[\left(\left(\partial_{r}\lambda\right)^{2}-\left(\partial_{z}\lambda\right)^{2}\right)dr+\left(2\partial_{r}\lambda\partial_{z}\lambda\right)dz]\label{eq:nu}
\end{equation}

The solutions must satisfy \Cref{eq:laplace-r-z,eq:nu}. A particular solution corresponding to two objects (given by Curzon in 1924 \cite{curzon1924} ) is:

\begin{equation}
\lambda_0(r,z)=-\frac{\mu_1}{r_1}-\frac{\mu_2}{r_2}
\label{eq:lambda-0}
\end{equation}

\begin{equation}
	\label{eq:nu-0}
	\nu_0(r,z)=-\frac{1}{2}\frac{\mu_{1}^{2}r^2}{r_{1}^{4}}-\frac{1}{2}\frac{\mu_{2}^{2}r^2}{r_{2}^{4}}+\frac{2\mu_1\mu_2}{(z_1-z_2)^2}\left[\frac{r^2+(z-z_1)(z-z_2)}{r_{1}r_{2}}-1\right]
\end{equation}
Where $z_1$ and $z_2$ correspond to the positions on the z-axis for the two objects, $\mu_1$ and $\mu_2$ are length parameters, and:

\begin{equation}
r_1=\sqrt{r^2+(z-z_1)^2}
\label{eq:r_1}
\end{equation}

\begin{equation}
r_2=\sqrt{r^2+(z-z_2)^2}
\label{eq:r_2}
\end{equation}

Just as a final check, plugging \Cref{eq:laplace,eq:R_phiphi=0} into \Cref{eq:R} gives $R=0$, which shows that our solutions are consistent with our assumptions.

By construction, these solutions only apply to empty space, and so must exclude the two objects at $z_1$ and $z_2$. In addition, as noted by Synge \cite{synge_relativity}, the z axis between the two objects must also be excluded due to violation of elementary flatness. We will examine this in the next section.

\section{Curvature from Parallel Transport}

Consider parallel transport of a vector $V$ about the $z$-axis in
the $\hat{\phi}$ direction. The equation for parallel transport is generally given by:
  	
\begin{equation}
\begin{array}{rcl} \frac{D}{d\lambda}=\frac{dx^{\mu}}{d\lambda}\nabla_{\mu}=0 & \mbox{along} & x^{\mu}\left(\lambda\right)	  	
\end{array}
\end{equation}	  	
That is, the directional covariant derivative is equal to zero along
the curve $x^{\mu}$ parameterized by $\lambda$. For a vector this can
be simply written as:

\begin{equation}
\label{eq:x-par-xport}
\nabla_\mu V^{\nu}=\partial_\mu V^\nu+\Gamma^\nu_{\mu\lambda} V^\lambda=0
\end{equation}	  	
Starting with parallel transport along $\hat{e}_{\phi}$, \Cref{eq:x-par-xport} along with the relevant Christoffel symbols $\Gamma^{r}_{\phi\phi}$, $\Gamma^{z}_{\phi\phi}$, $\Gamma^{\phi}_{\phi r}$, and $\Gamma^{\phi}_{\phi z}$ gives:
	  	
\begin{equation}
\begin{aligned}
\partial_{\phi}V^{r}+\Gamma^{r}_{\phi\phi}V^{\phi}&=0\\
\partial_{\phi}V^{z}+\Gamma^{z}_{\phi\phi}V^{\phi}&=0\\
\partial_{\phi}V^{\phi}+\Gamma^{\phi}_{\phi r}V^{r}+\Gamma^{\phi}_{\phi z}V^{z}&=0\\
\end{aligned}
\end{equation}
Plugging in the values from \Cref{eq:christoffel-connections}, our equations are: 

\begin{equation}
\partial_{\phi}V^{r}+\left(re^{-2\nu}\left(r\partial_{r}\lambda-1\right)\right)V^{\phi}=0\label{eq:V-r-phi}
\end{equation}

\begin{equation}
\label{eq:V_z-V_phi}
\partial_{\phi}V^{z}+\left(r^{2}e^{-2\nu}\partial_{z}\lambda\right)V^{\phi}=0
\end{equation}

\begin{equation}
\partial_{\phi}V^{\phi}+\left(\frac{1}{r}-\partial_{r}\lambda\right)V^{r}-\partial_{z}\lambda V^{z}=0\label{eq:V-phi-r-z}
\end{equation}

Differentiating \Cref{eq:V-phi-r-z} with respect to $\phi$ and plugging it into \Cref{eq:V-r-phi} gives:

\begin{equation}
\partial^{2}_{\phi}V^{\phi}-\partial_z\lambda\partial_{\phi}V^z+r^{2}e^{-2\nu}\left(\partial_r\lambda-\frac{1}{r}\right)^2V^{\phi}=0
\end{equation}
Plugging in the expression for $\partial_{\phi}V^z$ from
\Cref{eq:V_z-V_phi} and letting 

\begin{equation}
\label{eq:def-chi}
\chi=re^{-\nu}\sqrt{\left(\partial_z\lambda\right)^2+\left(\frac{1}{r}-\partial_r\lambda\right)^2}
\end{equation}
We have the simple differential equation:

\begin{equation}
\partial^2_\phi V^\phi+\chi^2 V^\phi=0
\end{equation}
For which the solution is:

\begin{equation}
V^{\phi}=A\sin\chi\phi+B\cos\chi\phi
\end{equation}
Therefore, integrating \Cref{eq:V-r-phi} with respect to $\phi$ we get:

\begin{equation}
V^{r}=\frac{r^2e^{-2\nu}(\partial_r\lambda-\frac{1}{r})}{\chi}\left(A\cos\chi\phi-B\sin\chi\phi\right)
\end{equation}

And from \Cref{eq:V_z-V_phi}:

\begin{equation}
V^{z}=\frac{r^2 e^{-2\nu}\partial_z\lambda}{\chi}\left(A\cos\chi\phi-B\sin\chi\phi\right)
\end{equation}

So our general vector is then:

\begin{equation}
\label{eq:V-from-par-transport}
\begin{split}
V=\frac{r^2e^{-2\nu}(\partial_r\lambda-\frac{1}{r})}{\chi}\left(A\cos\chi\phi-B\sin\chi\phi\right)\hat{e}_{r} \\
+\frac{r^2 e^{-2\nu}\partial_z\lambda}{\chi}\left(A\cos\chi\phi-B\sin\chi\phi\right)\hat{e}_{z} \\
+\left(A\sin\chi\phi+B\cos\chi\phi\right)\hat{e}_{\phi}
\end{split}
\end{equation}

Normalizing $V(\phi=0)$:

\begin{equation}
\label{eq:inner-product}
g_{\mu\nu}V^{\mu}V^{\nu}=1
\end{equation}

We obtain the condition that:

\begin{equation}
A^2+B^2=r^{-2}e^{2\lambda}
\end{equation}

For simplicity, we choose $A^2=r^{-2}e^{2\lambda}$ and $B^2=0$. 

Now, when we parallel transport $V$ around to $\phi=2\pi$ there will be an angle between $V(\phi=0)$ and $V(\phi=2\pi)$ given by the definition of the scalar product:

\begin{equation}
\cos(\beta)=\frac{g_{\mu\nu}V^{\mu}(0)V^{\nu}(2\pi)}{g_{\mu\nu}V^{\mu}(0)V^{\nu}(0)}
\end{equation}

Since we have normalized our vectors, the denominator is equal to 1, and we get the expression that:

\begin{equation}
\cos\beta=\cos(2\pi\chi)
\end{equation}

Where $\chi$ is given by \Cref{eq:def-chi}. Hence $\beta=2\pi\chi$. We can now use the definition of the deficit angle:

\begin{equation}
\label{eq:deficit-angle}
\Delta=2\pi-\beta=2\pi(1-\chi)
\end{equation}

To get the curvature $\mathcal{R}$ via:

\begin{equation}
\label{eq:curvature}
\mathcal{R}=\lim_{A\rightarrow 0}\frac{\Delta}{A}
\end{equation}

The area $A$ is defined on the reduced metric:

\begin{equation}
ds^2=e^{2(\nu-\lambda)}dr^2+r^2 e^{-2\lambda}d\phi^2
\end{equation}

Via:

\begin{equation}
\label{eq:area}
A=\int\sqrt{|g|}d^n x=\int_{\phi=0}^{\phi=2\pi}\int_{r=0}^{r=R}\sqrt{r^2 e^{2\nu-4\lambda}}drd\phi=2\pi\int_{r=0}^{r=R}re^{\nu-2\lambda}dr
\end{equation}

Plugging \Cref{eq:nu-0,eq:lambda-0} into \Cref{eq:area} gives:

\begin{equation}
\label{eq:area-solved}
A=
\end{equation}

And thus the curvature is:

\begin{equation}
\label{eq:curvature-solved}
\mathcal{R}=
\end{equation}

From the definition of the curvature in \Cref{eq:R} we can obtain the Ricci tensor, and hence the Einstein tensor. Then reading off the value of $G_{zz}$ we obtain the desired $T_{zz}$.

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