# Schwarzschild black hole geometry in Novikov coordinates

As I am a layman in gravitation I would really appreciate any suggestions on how to approach and tackle the following problem:

Choose a time-symmetric 3-geometry for the initial geometry of Schwarzschild black hole and continue the foliation of spacetime in free fall coordinates. Show that this results in Novikov metric.

I believe this is the general relativistic initial value problem within the ADM formulation. I read that the curvature of spacetime can be integrated analytically only for some highly symmetric and simple cases, otherwise approximations and numerics have to be applied.

My idea is to solve standard 3+1 decomposition (foliation) of Einstein field equations (ADM equations). I would start with initial data for the Schwarzschild spacetime at the moment of time symmetry and with gauge fixing for lapse and shift $\alpha=1$ and $\beta_i=0$, which would attach coordinates to free fall observers (also called geodesic slicing).

In [1, p.535] the time-symmetric 4-geometry is defined as one which has a spacelike hypersurface with extrinsic curvature 0. This is thus one constraint for the curvature tensor of initial 3-geometry: $K_{ij}=0$.

[1]: Misner, C.W, Thorne, K. S., Wheeler, J.A., Gravitation, 1973.

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What do you mean by a foliation? The usual coodinates provide a foliation in the usual sense by constant time surfaces on the exterior. – Ron Maimon Sep 21 '11 at 1:57
If you want a foliation which extends into the horizon, you can use infalling null coordinates. There are no dynamical equations for foliations, because they are not uniquely determined. – Ron Maimon Sep 21 '11 at 15:41
@liberias: as I know it, thats' the only way to go. Do the coordinate trasform, and then the Novikov $T$ will be the time coordinate you use to do the transformation. If you're just looking for a coordinate system without a horizon singularity, I would strongly suggest the Kerr-Schild coordinates over the Novikov coordinate system. – Jerry Schirmer Sep 21 '11 at 23:45
@liberias: yes, that is exactly right. The only thing is that you will have $T_{\mu \nu}=0$ because the Schwarzschild black hole is a vacuum solution. – Jerry Schirmer Sep 23 '11 at 11:42
You can find a derivation here: springerlink.com/content/b3x1wkp9m3887g39 . Without getting into details I think this is exactly what you want. Greets – Robert Filter Sep 23 '11 at 23:14

Schwarzschild geometry in Schwarzschild coordinates $(t,r,\theta,\phi)$ is time-symmetric $$ds^2=-\left(1-\frac{2GM}{c^2 r}\right)c^2dt^2+\left(1-\frac{2GM}{c^2 r}\right)^{-1}dr^2+r^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;.$$

Novikov coordinate system is defined by a set of geodesic clocks. The coordinate clocks are freely falling from some maximal radius $r_m$ towards $r=0$, where $r_m$ is different for each clock. All clocks start falling at the same Schwarzschild time $t_0$ and they are synchronized in such a manner that each clock shows $0$ at $r_m$. Novikov coordinate is defined to stay constant along the trajectory of each clock, while for time coordinate proper time is taken.

From now on the angular part metric will be omitted, since is stays the same. We also take $r_s=2M$ and $G=c=1$: $$\label{eq:sch-met2} ds^2=-\left(1-\frac{r_s}{r}\right)dt^2+\left(1-\frac{r_s}{r}\right)^{-1}dr^2 \;.$$

# Geodesics in Schwarzschild gometry

To get the equation of geodesics in Schwarzschild geometry we have to solve equations of motion of a free particle: $$\label{eq:lagrangian} \mathcal{L}=\frac{1}{2}mg_{\mu\nu}\dot{x}^\mu\dot{x}^\nu \;,$$ $$\label{eq:dot} \dot{x}^\mu=\frac{dx^\mu}{d\tau}=u^\mu \;.$$ $$\label{eq:lagrangian2} \mathcal{L}=-\frac{m}{2}\left(1-\frac{r_s}{r}\right)\dot{t}^2+\left(1-\frac{r_s}{r}\right)^{-1}\dot{r}^2 \;,$$ $$\label{eq:EL} \frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial \dot{x}^\mu}-\frac{\partial\mathcal{L}}{\partial x^\mu}=0 \;,$$ For $\mu=0$ we get a constant of motion $$\label{eq:ConstOfMotion} \frac{\partial}{\partial\tau}\left[\left(1-\frac{r_s}{r}\right)\dot{t}\right]=0 \qquad \Rightarrow \qquad \left(1-\frac{r_s}{r}\right)\dot{t}=a \;,$$

For timelike geodesics: $ds^2=-d\tau^2$ the radial geodesic equation becomes $$\label{eq:orbit} \left(\frac{d\tau}{dr}\right)^2=\frac{1}{a^2-\left(1-\frac{r_s}{r}\right)} \;.$$ Maximal radius is ($dr/d\tau=0$) $$\label{eq:maximal} r_m=\frac{r_s}{1-a^2} \;.$$ We use $\frac{dt}{dr}=\frac{dt}{d\tau}\frac{d\tau}{dr}$ and obtain the following relations: \begin{eqnarray} \frac{d\tau}{dr} &=& \frac{\varepsilon}{\sqrt{\frac{r_s}{r}-\frac{r_s}{r_m}}} \;,\label{eq:orbit1} \ \frac{dt}{dr} &=& \frac{\varepsilon\sqrt{1-\frac{r_s}{r_m}}}{\left(1-\frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}-\frac{r_s}{r_m}}} \;, \label{eq:orbit2} \end{eqnarray} where $\varepsilon$ is $+1$ or $-1$. For falling particles we choose $\varepsilon=-1$.

# Novikov time coordinate

We first transform from $(r,t)$ to $(r,\tau)$. From last two equations we obtain for $d\tau(dt,dr)$ $$\label{eq:bit} d\tau=\left(1-\frac{r_s}{r_m}\right)^{1/2}dt+\frac{\left(\frac{r_s}{r}-\frac{r_s}{r_m}\right)^{1/2}}{1-\frac{r_s}{r}}dr \;.$$ where we assumed $t$ are $r$ known.

This can be integratet from $r$ to $r_m$, where we take into account that all clocks reach their maximum radius at $\tau_{0i}=0$. It follows $$\label{eq:integral} \tau=\left(1-\frac{r_s}{r_m}\right)^{1/2}(t-t_0)+\int_{r_m}^{r}\frac{\left(\frac{r_s}{y}-\frac{r_s}{r_m}\right)^{1/2}}{1-\frac{r_s}{y}}dy \;.$$

maximal radius $r_m$ is here a function of $r$ and \tau$. Their implicit relationship is $$\label{eq:implicit1} \tau=-f(r,r_m)\;,$$ where $$f(r,r_m) = \int_{r_m}^{r}\frac{dy}{\sqrt{\frac{r_s}{y}-\frac{r_s}{r_m}}} \label{eq:integral3} = -\left[\frac{rr_m}{r_s}(r_m-r)\right]^{1/2}-\frac{r_m^{3/2}}{\sqrt{r_s}}\arccos\left[\left(\frac{r}{r_m}\right)^{1/2}\right] \;.\label{eq:f}$$ We can now eliminate coordinate$t$from the line element $$\label{eq:sch-met3} ds^2=-d\tau^2+\frac{1}{1-\frac{r_s}{r_m}}\left[- dr-\left(\frac{r_s}{r}-\frac{r_s}{r_m}\right)^{1/2}d\tau\right]^2 \;.$$ # Novikov radial coordinate For radial coordinate we take the maximal Schwarzschild radius$r_m$, which remains constant along the worldline of a geodesic clock. $$\label{eq:relation2} - dr-\left(\frac{r_s}{r}-\frac{r_s}{r_m}\right)^{1/2}d\tau=\left(\frac{r_s}{r}-\frac{r_s}{r_m}\right)^{1/2}\frac{\partial f}{\partial r_m}dr_m \;.$$ With this we can eliminate the other Schwrazschild coordinate$r$: $$\label{eq:sch-met4} ds^2=-d\tau^2+\frac{\left[g(r,r_m)\right]^2}{1-\frac{r_s}{r_m}}dr_m^2 \;.$$ Here we$g(r,r_m)$is the following \begin{eqnarray} g(r,r_m)&=&-\left(\frac{r_s}{r}-\frac{r_s}{r_m}\right)^{1/2}\frac{\partial f}{\partial r_m} \label{eq:g}\ &=&1+\frac{1}{2}\left(1-\frac{r}{r_m}\right)-\frac{3}{4}\left(\frac{r_m}{r}-1\right)^{1/2}\left[\sin^{-1}\left(\frac{2r}{r_m}-1\right)-\frac{\pi}{2}\right] \;. \nonumber \end{eqnarray}$r$is not a radial coordinate anymore, but a metric function of coordinates$r_m$and$\tau$, which is given implicitly by equation (). # Novikov metric By introducting$r_m$the metric became diagonal as in Schwarzschild coordinates. It also stays diagonal by introducig a new radial coordinate, that is only functionally related to the old one. Novikov-s choice is$r^*$with the following monotonic relation to$r_m$: $$\label{eq:r*} r^*=\left(\frac{r_m}{r_s}-1\right)^{1/2}\;.$$ The metric now becomes $$\label{eq:novikov-met} ds^2=-d\tau^2+4r_s^2\left({r^*}^2+1\right)\left[g(r,r^*)\right]^2d{r^*}^2 \;.$$ We can show that the following also holds $$\label{eq:relation} 4Mg(r,r^*)=\frac{1}{r^*}\frac{\partial r}{\partial r^*}\;.$$ With this the metric gets the form that is standard in literature [MTW,p. 826]: $$\label{eq:novikov} ds^2=-d\tau^2+\left(\frac{{r^*}^2+1}{{r^*}^2}\right)\left(\frac{\partial r}{\partial r^*}\right)^2d{r^*}^2+r^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;,$$ where we also included the angular part. # Relations among coordinates We now give the relations between Schwarzschild coordinates$(t,r)$and Novikov coordinates$(\tau,r^*)$. The first one,$r=(\tau,r^*)$, is obtained from equations () and () $$\label{eq:CoordRela1} \tau=r_s\left({r^*}^2+1\right)\left[\frac{r}{r_s}-\frac{(r/r_s)^2}{{r^*}^2+1}\right]^{1/2}+r_s\left({r^*}^2+1\right)^{3/2}\arccos\left[\left(\frac{r/r_s}{{r^*}^2+1}\right)^{1/2}\right] \;.$$ The second one,$t=(\tau,r^*)\$, is obtained by integration from ()

t=r_s\ln\left|\frac{r^*+\left(\frac{r_s}{r}\left({r^}^2+1\right)-1\right)^{1/2}}{r^-\left(\frac{r_s}{r}\left({r^*}^2+1\right)-1\right)^{1/2}}\right|+r_sr^\left[\left({r^}^2+3\right)\arctan\left(\frac{r_s}{r}\left({r^}^2+1\right)-1\right)^{1/2}+\left({r^}^2+1\right)\frac{\sqrt{\frac{r_s}{r}\left({r^*}^2+1\right)-1}}{\frac{r_s}{r}\left({r^*}^2+1\right)}\right] \;.

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