Difference between revisions of "Schwarzschild black hole"

The proper time of the stationary observer is $d\tau=ds|_{dr=d\theta=d\varphi=0}$: \[d\tau=\sqrt{g_{00}}dt=\sqrt{1-\frac{r_{g}}{r}}dt.\] At $r\to \infty$ it coincides with $dt$, so the coordinate time $t$ can be interpreted as the proper time of a "remote" observer. At $r\to r_g$ the local time flows slower and asymptotically stops. If one of two twins were to live some time at $r\approx r_g$, he will return to his remote twin having aged less (thought he might have acquired some grey hair due to constant fear of tumbling over the horizon).

\[l_{r1-r2}=\int\limits_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r_g}{r}}};\quad l_{\varphi1-\varphi_2}=2\pi r|\varphi_1-\varphi_2|; \quad l_{\theta1-\theta_2}=2\pi r|\theta_1-\theta_2|.\]

This is question is given not to be answered but to make think on the answer. Correct questions and correct answers can be given in terms of a proper coordinate frame, that is regular both in $r>r_g$ and in $r<r_g$. Still, one can say something meaningful as is.

At $r<r_g$ we have $g_{00}<0$ and $g_{11}>0$, thus the $t$ coordinate is spatial and $r$ coordinate is temporal (!): \[ds^{2}=|h|^{-1}(r)dr^{2}-|h|(r)dt^{2} -r^{2}d\Omega^{2}.\] The metric therefore is nonstationary in this region, depending on the temporal coordinate $r$, but homogeneous, as there is no dependence on spatial coordinates. Then for an observer "at rest" with respect to this coordinate system we would have $dt=d\theta=d\varphi=0$, and thus \[d\tau^2=ds^2=-\frac{dr^{2}}{1-r_{g}/r}= \frac{r dr^{2}}{r_{g}-r}>0, \quad\Rightarrow\quad d\tau=\frac{\sqrt{r}dr}{\sqrt{r_{g}-r}}.\] An observer at rest with respect to the old coordinate system $dr=d\theta=d\varphi=0$, though, does not exist, as it would be $ds^{2}<0$ for him, which corresponds to spacelike geodesics (i.e. particles traveling faster than light). The last of the two questions cannot be answered without additional assumptions, because time $t$, which is the spatial coordinate now, in the two points is not given. The physical distance at $d\theta=d\varphi=dr=0$ is defined as \[dl^{2}=|h(r)|dt^2.\] It evidently depends on time $r$. This very fact that Schwarzschild metric is nonstationary at $r<r_g$, and that a stationary one does not exist in this region, leads to the absence of stationary observers and thus to the impossibility to imagine it "welded" of a system of stiff rods.

*This was actually not a very simple problem

If a particle is at rest, its 4-velocity is $u^{\mu}=g_{00}^{-1/2}\delta^{\mu}_{0}$, where the factor is determined from the normalizing condition \[1=g_{\mu\nu}u^{\mu}u^{\mu}=g_{00}(u^{0})^{2}.\] Then the $4$-acceleration can be found from the geodesic equation: \begin{align*} a^{0}\equiv&\frac{du^0}{ds}=-{\Gamma^{0}}_{00}u^{0}u^{0}\sim {\Gamma^{0}}_{00}\sim \Gamma_{0,\,00}\sim (\partial_{0}g_{00}+\partial_{0}g_{00} -\partial_{0}g_{00})=0;\\ a^{1}\equiv&\frac{du^1}{ds} =-{\Gamma^{1}}_{00}u^{0}u^{0} =-g^{11}\Gamma_{1,\,00}\;g_{00}^{-1} =-(g_{00}g_{11})^{-1}\Gamma_{1,\,00}=\Gamma_{1,00}= \\ &=\tfrac{1}{2} (\partial_{0}g_{10}+\partial_{0}g_{10} -\partial_{1}g_{00}) =-\frac{1}{2}\frac{dg_{00}}{dr} =-\frac{h'}{2} =\frac{r_{g}}{2r^2}. \end{align*} The scalar acceleration $a$ is then equal to \[a^{2}=-g_{11}(a^{1})^{2}=\frac{(h')^{2}}{4h} =-\frac{r_{g}^{2}}{4 r^4} \Big(1-\frac{r_g}{r}\Big)^{-1}\] and tends to infinity when we approach the horizon.

Straightforward calculation of Christoffel symbols and Ricci tensor yields the vacuum Einstein equation $R_{\mu\nu}=0$.

If a Killing vector field is timelike we can use the coordinate frame in which $K^\mu$ is the unitary vector of the time coordinate\footnote{See problem \ref{equ_oto-kill2} of chapter 2.} $K^{\mu}=(1,0,0,0)$, while the spacelike basis vectors are orthogonal to it. Then the integral of motion is energy in the corresponding (stationary) frame $p_{\mu}K^{\mu}=p_{0}\equiv\varepsilon/c$. Using the result (\ref{u0}), which holds for arbitrary gravitational field, we obtain the expression for energy, which is the integral of motion in a stationary metric \begin{equation}\label{EnergyStat} \varepsilon=mc^{2}u_{0}= mc^{2}\sqrt{g_{00}}\cdot\gamma= \frac{mc^{2}\sqrt{g_{00}}} {\sqrt{1-\frac{v^2}{c^2}}}.\end{equation}

Killing vectors correspond to infinitesimal transformations that leave the metric invariant. Considering a two-dimensional sphere embedded in the tree-dimensional space, we can see that its symmetries are rotations. Three of them are independent: the rotations in each of the three coordinate planes. Let us consider an infinitesimal rotation in the plane $XY$: $dx=yd\lambda,\; dy=-xd\lambda$. Thus the Killing vector is* \[K_{1}=x\partial_y-y\partial_x.\] Using the spherical coordinates \[x=\sin\theta\cos\varphi;\quad y=\sin\theta\sin\varphi;\quad z=\cos\theta,\] we obtain $\partial_{x}=-\frac{\sin\varphi}{\sin\theta}\partial_\varphi$, $\partial_y=\frac{\cos\varphi}{\sin\theta}\partial_\varphi$, and therefore $K_{1}=\partial_{\varphi}$. Considering rotations in plains $XZ$ and $YZ$ in the same way, in the end we obtain \[\left\{\begin{array}{l} K_{1}=x\partial_{y}-y\partial_{x}=\partial_{\varphi}\\ K_{2}=z\partial_{x}-x\partial_{z}= \cos\varphi\;\partial_{\theta}-\cot\theta\sin\varphi\;\partial_\varphi\\ K_{3}=z\partial_{y}-y\partial_{x}= \sin\varphi\;\partial_{\theta}+\cot\theta\cos\varphi\;\partial_\varphi\;. \end{array}\right.\]

*Hereafter we use the following notation: a 4-vector $A$ is $A=A^{\mu}\partial_{\mu}$, where $\partial_{\mu}\equiv\partial/\partial x_{\mu}$ is the "coordinate" basis, $A^{\mu}$ the coordinates of $A$ in this basis; it is not hard to verify that transformation laws for $A^\mu$ and $\partial_{\mu}$ are adjusted so that $A$ is a quantity that does not depend on a coordinate frame. This also enables us to conveniently recalculate $A^\mu$ when we change the basis. For more detail see e.g. the textbook by Carroll:

Carroll S., Spacetime and geometry: an introduction to General Relativity. AW, 2003, ISBN 0805387323

Vectors $\Omega^\mu$ and $R^\mu$ are Killing vectors because the metric does not depend explicitly either on $t$ or on $\varphi$. The last two correspond to the spherical symmetry (see previous problem). We can also check directly that they obey the Killing equation by evaluating the Christoffel symbols for the Schwarzschild metric.

The integrals of motion corresponding to $S^\mu$ and $T^\mu$ are \begin{align*} S=u_{\mu}S^{\mu}=& \;u_{2}\cos\varphi-u_{3}\cot\theta\sin\varphi;\\ T=u_{\mu}T^{\mu}=& -u_{2}\sin\varphi-u_{3}\cot\theta\cos\varphi. \end{align*} Let us align the coordinate frame in such a way that the initial conditions are \[\theta|_{t=0}=\pi/2;\quad u_{2}|_{t=0}\equiv u_{\theta}|_{t=0}=0.\] Then $S$ and $T$ are zero: \[u_{2}\cos\varphi=u_{3}\cot\theta\sin\varphi;\qquad u_{2}\sin\varphi=-u_{3}\cot\theta\cos\varphi.\] Taking the square and adding the two equations, and also multiplying them, we obtain \[\left\{\begin{array}{l} u_{2}^{2}=u_{3}^{2}\cot^{2}\theta,\\ u_{2}^{2}\sin\varphi\cos\varphi= -u_{3}^{2}\cot^{2}\theta\sin\varphi\cos\varphi \end{array}\right.\quad\Rightarrow\quad u_{3}^{2}\sin\varphi\cos\varphi \cdot\cot^{2}\theta=0.\] Then either $\varphi=const$, which means that $u_{3}=u_{2}=0$ and the motion is radial, or $\cot^{2}\theta=0$ and the motion takes place in the plane $\theta=\pi/2$.

Conservation of $u_{\mu}R^{\mu}$ means that $R\equiv u_{3}\equiv\varphi'=const$. It is not hard to derive from the expressions for $S$ and $T$ that \begin{align*} &S\sin\varphi+T\cos\varphi=R\cot\theta,\\ &S\cos\varphi-T\sin\varphi=u_2. \end{align*} Taking the square and adding up, we are led to \[u_2^2 =S^2+T^2+R^2-\frac{R^2}{\sin^2 \theta},\] then on differentiating we obtain \[u'_{2}=R^2 \frac{\cos\theta}{\sin^{2}\theta}.\] Let the trajectory deviate slightly from the plane $\theta=\pi/2$. Then $\theta=\pi/2+\delta\theta$ and \[(\delta\theta)''=u'_{2} =R^2 \frac{\cos\theta}{\sin^{2}\theta} \approx -R^2 \delta\theta,\] therefore $\theta$ oscillates around the stable point $\pi/2$.

The first integral of motion is, up to a multiplier, the energy (see \ref{EnergyStat}): \begin{equation}\label{SchwInt-E1} E=\Omega_{\mu}u^{\mu}= g_{00}\frac{dx^{0}}{d\lambda}=h\frac{dt}{d\lambda}= \left(1-\frac{r_g}{r}\right)\frac{dt}{d\lambda},\end{equation} where $\lambda$ is the geodesic parameter: for a massive particle we can always choose the natural parametrization $d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ (see (\ref{IntervalStaticCase})) so that $E$ is energy per unit mass; recovering the multipliers, for true energy we obtain \begin{equation}\label{SchwInt-E2} \varepsilon_{m}\equiv mcE= mc^{2}\sqrt{g_{00}}\,\gamma= mc^{2}\sqrt{h}\,\gamma= mc^{2}\sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation} The second integral of motion is the angular momentum (per unit mass also, and the sign is chosen so that in the Newtonian limit we obtain the usual angular momentum) \begin{equation}\label{SchwInt-L1} L=-R_{\mu}u^{\mu}=-g_{33}\frac{dx^{3}}{d\lambda}= r^{2}\sin^{2}\theta\frac{d\varphi}{d\lambda}= r^{2}\frac{d\varphi}{d\lambda}.\end{equation} For a massive particle we choose $\lambda=s$, then $d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ and \begin{equation}\label{SchwInt-L2} l_{m}\equiv mL= mr^{2}\dot{\varphi} \sqrt{g_{00}}\gamma= m r^{2}\dot{\varphi} \sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation} Note that conservation of $l_m$ is a generalization to the relativistic case of the second Kepler's law on the sweeping of equal areas per unit time $r^{2}\dot{\varphi}=const$.

At $r\gg r_g$ we'll have $\varepsilon_{m}=mc^{2}$, while near the horizon $\varepsilon_{m}\to 0$. The difference is the work needed to pull the particle away from the horizon to infinity, and it equals the rest mass (energy) of the particle. No, it will not change, because the energy of the falling particle is zero.

Equation of null curve is $ds^2=0$. Substituting the metric with $d\theta=d\varphi=0$, we get \[dt=\pm dr\,\sqrt{-\frac{g_{11}}{g_{00}}} =\pm\frac{dr}{h}.\] On integration, we obtain \begin{equation}\label{Schw-NullGeodesic} \pm t=r+r_{g}\ln|r-r_{g}|+const.\end{equation} Due to symmetry it is also a geodesic. This can be verified by evaluating the Christoffel symbols for the Schwarzschild metric and writing down the geodesic equatoin in explicit form.

In the radial case the physical velocity $v$ is expressed through the $4$-velocity $u^{\mu}=(u^{0},u^{1},0,0)$ as \begin{equation}\label{Schw-v(u)} \beta^{2}=\frac{dx^{\alpha}dx_{\alpha}}{d\tau^{2}}= \frac{-g_{11}(dx^{1})^{2}}{g_{00}(dx^{0})^{2}}= h^{-2}\Big(\frac{u^1}{u^0}\Big)^2= \Big(\frac{\dot{r}}{h}\Big)^2,\end{equation} where dot denotes differentiation by coordinate time $t$, and thus the energy conservation law (see (\ref{SchwInt-E2})) takes the form \[const=\Big(\frac{\varepsilon}{mc^2}\Big)^{2}= h\gamma^{2}=\frac{h}{1-\dot{r}^{2}/h^2}.\] Initial condition fixes the constant*: \begin{equation}\label{Schw-h0} \dot{r}\sim v|_{r_{0}}=0\quad \Rightarrow\quad \Big(\frac{\varepsilon}{mc^2}\Big)^{2} =g_{00}(r_{0})=h(r_0)=1-\frac{r_{g}}{r_{0}} \equiv h_{0},\end{equation} so for the physical velocity we get \begin{equation}\label{SchwRad-V} \gamma^{2}=\frac{h_{0}}{h(r)} \underset{r\to r_{g}}{\longrightarrow}\infty; \quad \frac{v}{c}\equiv\beta= \sqrt{1-\gamma^{-2}}= \sqrt{1-\frac{h}{h_{0}}} \underset{r\to r_{g}}{\longrightarrow}1.\end{equation} It tends to infinity close to the horizon! At the same time, as $v^{2}=\dot{r}^{2}/h^{2}$ from (\ref{Schw-v(u)}), the coordinate velocity \begin{equation}\label{Schw-dotR} \Big|\frac{dr}{dt}\Big|=vh=h \sqrt{1-\frac{h}{h_{0}}}= \Big(1-\frac{r_{g}}{r}\Big) \sqrt{1-\frac{1-\frac{r_{g}}{r}}{1-\frac{r_{g}}{r_{0}}}} \underset{r\to r_{g}}{\thicksim}h\to 0\end{equation} tends to zero. After integration we obtain $t(r)$: \begin{equation}\label{Schw-t(r)} const\pm t=\int\limits_{r}^{r_{0}} \frac{dr}{vh}= \int\limits_{r}^{r_{0}}\frac{rdr}{r-r_{g}} \bigg[1-\frac{1-r_{g}/r}{1-r_{g}/r_{0}}\bigg]^{-1/2} \underset{\substack{r\approx r_{g}\\r_{0}\gg r_{g}}} {\approx}\int\limits_{r}^{r_{0}} \frac{r\,dr}{r-r_{g}}=r+r_{g}\ln|r-r_{g}|.\end{equation} As should be expected, the asymptote is the same as for the null geodesics (for $r_{0}\gg r_{g}$).

*Note that $r_0$ has the meaning of a turning point only for finite motion; this, however, does not prevent us from using the same notation in the non-finite case. Then $r_0$ is determined from $\varepsilon$ by the same formula and takes negative values, while $h_{0}>1$.

The particle's proper time interval $\tau$ is $d\tau=\gamma^{-1} d\hat{t}$, where $d\hat{t}=\sqrt{h}dt$ is local time. Then \begin{equation}\label{ShwRad-tau} \tau=\int dt \gamma^{-1}\sqrt{h}= \int\limits_{r}^{r_{0}} \frac{dr}{\gamma v\sqrt{h}}= \int\limits_{r}^{r_{0}} \frac{dr}{v \sqrt{h_{0}}}= \int\limits_{r}^{r_{0}} \frac{dr}{\sqrt{h_{0}-h}}= \int\limits_{r}^{r_{0}} \frac{dr}{\sqrt{\frac{r_{g}}{r_{0}}- \frac{r_{g}}{r}}}.\end{equation} On the other hand, in the Newtonian approach from energy conservation we obtain in the same way that \[(\dot{r})^{2}=2(E-U)=2\;\Big(E+\frac{GM}{r}\Big) =\frac{r_{g}}{r}-\frac{r_{g}}{r_{0}},\] which gives us the same equation of motion. We see that the integral is regular at $r=r_{g}$, is finite and converges even at $r\to 0$. It is not hard to evaluate: \begin{equation}\label{Schw-TauParametric} \tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}} (\eta+\sin\eta),\qquad r=\frac{r_0}{2}(1+\cos\eta),\qquad \eta\in(0,\pi). \end{equation} The proper time of reaching the singularity $r=0$ is therefore \[\tau_{f}=\frac{\pi}{2}r_{0} \sqrt{\frac{r_{0}}{r_{g}}}.\] The integral (\ref{Schw-t(r)}) for $t(r)$ can be rewritten as \[t(r)=\int \frac{dr}{vh}=\sqrt{h_{0}} \int\frac{dr}{\sqrt{h_{0}-h}}\cdot \frac{1}{h}= \underbrace{\sqrt{1-\frac{r_{g}}{r_{0}}}} _{\text{correction 1}}\, \underbrace{\int\!\frac{dr}{\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}}_{\tau(r)} \cdot \underbrace{\frac{1}{1-\frac{r_g}{r}}} _{\text{correction 2}}.\] Now we see that the first factor gives the correction taking into account time dilation in the initial point $r_0$, while the second correction accounts for local time dilation and proper time dilation due to acceleration. If the second factor is close to unity, we can expand it in powers of $r_{g}/r$ and obtain \[t=\sqrt{1-\frac{r_g}{r_0}}\;\; \big[\tau+\Delta\tau\big], \quad\text{where}\quad \Delta\tau=\int \frac{dr\,r_{g}/r} {\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}.\] On integrating, we can express $\Delta\tau$ through $\eta$ (\ref{Schw-TauParametric}), and thus derive the equation of motion in the form analogous to (\ref{Schw-TauParametric}), but with the correction term \begin{equation}\label{Schw-TauParametric2} \tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}}\; \Big([1+\tfrac{r_g}{r_0}]\eta+\sin\eta\Big),\qquad r=\frac{r_0}{2}(1+\cos\eta). \end{equation} Evidently, this approximation works only for $r\gg r_{g}$.

In the relativistic limit we should assume $\gamma\gg1$ (near $r_{g}$ this always holds), thus $v\to 1$, and from (\ref{Schw-dotR}) we get \[\frac{dr}{dt}\approx h\quad \Rightarrow\quad t=\int \frac{dr}{h}.\] This is, as should be expected, the null geodesic equation.

The geodesics describing light signals propagating away from the horizon are given by the equation (\ref{Schw-NullGeodesic}) with the plus sign: \begin{equation}\label{Schw-tLambda} t=r+r_{g}\ln|r-r_{g}|+t_{\Lambda},\end{equation} where parameter $t_\Lambda$ parametrizes this family of geodesics and gives the difference in coordinate times of signals' registration at a given $r$. At $r\gg r_g$ this is also the proper time of the remote detector. Then \begin{equation}\label{Schw-dtL(dr)} dt_{\Lambda}=dt-\frac{r_{g}dr}{r-r_{g}}= dt-\frac{r_{g}dr}{rh}= -\frac{dr}{vh}-\frac{r_{g}dr}{rh}= -\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big).\end{equation} As both $v$ and $r_{g}/r$ tend to unity near the horizon, the integral diverges logarithmically, twice faster than the one for $t(r)$. The meaning of this number is the following: as $r\to r_{g}$, the geodesics of massive and massless particles asymptotically coincide, and it takes the light as much time to "escape" the potential well as it takes the particle to fall in. Asymptotically we obtain \begin{equation}\label{Schw-r(tL)} -t_{\Lambda}\approx 2\int\limits^{r} \frac{rdr}{r-r_{g}}\approx 2r_{g}\ln|r-r_{g}| \approx 2t(r)\quad\Rightarrow\quad r(t_{\Lambda})-r_{g}\sim r_{g} e^{-t_{\lambda}/2r_{g}};\end{equation} also $\gamma=h_{0}/h\approx r_{g}/(r-r_{g})$, $v(r)=\sqrt{1-h/h_{0}}\approx 1-(r-r_{g})/2r_{g}$, and $\dot{r}\approx h$, thus \begin{equation}\label{Schw-v(tL)} \gamma\sim e^{t_{\Lambda}/2r_{g}};\qquad (1-v)\sim {\textstyle\frac{1}{2}} e^{-t_{\Lambda}/2r_{g}}; \qquad \dot{r}\sim e^{-t_{\Lambda}/2r_{g}}.\end{equation}

The integral of motion that conserves along the light ray due to the timelike Killing vector $\Omega^{\mu}=\partial_{t}$ is energy (up to a factor): \begin{equation}\label{Schw-OmegaConst} const\equiv\omega_{0}= g_{\mu\nu}\Omega^{\mu}k^{\nu}= g_{00}k^{0}=hk^{0}.\end{equation} For a static observer $u^{\mu}=(u^{0},0,0,0)$, while \begin{equation}\label{Schw-u0} 1=u^{\mu}u_{\mu}=g_{\mu\nu}u^{\mu}u^{\nu}= g_{00}(u^{0})^{2}\quad\Rightarrow\quad u^{0}=h^{-1/2}.\end{equation} Therefore the frequency registered by this static observer is \begin{equation}\label{Schw-omegastat} \omega_{stat}=g_{\mu\nu}u^{\mu}_{stat}k^{\nu}= g_{00}u^{0}_{stat}k^{0}=\sqrt{g_{00}}k^{0}= \sqrt{h}k^{0}=\omega_{0}h^{-1/2}.\end{equation} Then \begin{equation}\label{Schw-RedShift0} \omega_{0}=\omega_{stat}(r)\sqrt{h(r)}=const,\end{equation} so we obtain \begin{equation}\label{Schw-RedShift} \omega_{det}=\omega_{em} \sqrt{\frac{g_{00}(r_{em})}{g_{00}(r_{det})}}.\end{equation} This is gravitational redshift.

It is educative to derive the Doppler effect from analogous considerations. Let us consider an emitter (detector) moving radially with physical velocity $v$ (see. (\ref{Schw-v(u)})). From the normalizing condition \begin{equation}\label{Schw-rad-u0} 1=u^{\mu}u_{\mu}=g_{00}(u^{0})^{2}+g_{11}(u^{1})^{2}= g_{00}(u^0)^{2}(1-\beta^{2})=\gamma^{-2} h (u^0)^{2},\end{equation} thus \begin{equation}\label{Schw-rad-u^mu} u^{\mu}=\gamma\Big( \frac{1}{\sqrt{g_{00}}}, \frac{\beta}{\sqrt{-g_{11}}},0,0\Big)= \gamma\Big(\frac{1}{\sqrt{h}}, \beta\sqrt{h},0,0\Big).\end{equation} For light \[0=k^{\mu}k_{\mu}=g_{00}(k^0)^{2}+g_{11}(k^{1})^{2} \quad\Rightarrow\quad k^{1}=\pm k^{0}\sqrt{-g_{00}/g_{11}}=\pm hk^{0},\] therefor the observer registers (or the detector emits) light with frequency \[\omega_{em}\equiv\omega_{\beta}= g_{\mu}u^{\mu}k^{\nu}= g_{00}u^{0}k^{0}+g_{11}u^{1}k^{1}= h\cdot \frac{\gamma}{\sqrt{h}}k^{0}\pm \frac{1}{h}\cdot\gamma\beta\sqrt{h}\cdot hk^{0}= \sqrt{h}\gamma k^{0}(1\pm\beta).\] Choosing the sign "$+$" here, which corresponds to the emitter moving towards the horizon and light travaling outwards, we obtain the relativistic Dopper effect \begin{equation}\label{RelatDoppler-radial} \omega_{\beta}= \frac{\omega_{0}}{\sqrt{h}}\cdot\gamma(1+\beta)= \omega_{stat}\cdot\gamma(1+\beta).\end{equation} Taking into account that in this case the emitter is moving and detector is at rest, so that $\omega_{em}=\omega_{\beta}$, $\omega_{det}=\omega_{stat}$, we get \begin{equation}\label{Schw-RadialDoppler} \omega_{det}\equiv \omega_{stat}= \omega_{em}\frac{\gamma^{-1}}{1+\beta}= \omega_{em}\sqrt{\frac{1-\beta}{1+\beta}}.\end{equation}

In this case we just need to take into account both the gravitational redshift (\ref{Schw-RedShift}) and the Doppler effect (\ref{Schw-RadialDoppler}). On substituting $\gamma=\sqrt{h_{0}/h_{em}}$, we get \begin{equation}\label{Schw-Rad-FullRedshift} \omega_{det}=\omega_{stat}(r_{em}) \sqrt{\frac{h_{em}}{h_{det}}}= \omega_{em}\frac{\gamma^{-1}}{1+\beta} \sqrt{\frac{h_{em}}{h_{det}}}= \frac{\omega_{em}}{1+\beta}\cdot \frac{h_{em}}{\sqrt{h_{0}h_{det}}}\end{equation} In the limit $r_{em}\equiv r\sim r_{g}\ll r_{det}$ we obtain $\beta\approx 1$, thus \[\omega_{det}\approx \omega_{em}\cdot \frac{h_{em}}{2}= \omega_{em}\cdot {\textstyle\frac{1}{2}} \Big(1-\frac{r_{g}}{r_{em}}\Big)\to 0.\]

Repeating the argument of this problem, for the proper time interval between detection of consecutive infinitely close signals (\ref{Schw-dtL(dr)}) we have \[dt_{\Lambda}= -\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big),\] where $dr$ is the corresponding displacement of the emitter, related with its proper time interval as (see \ref{ShwRad-tau}) \[dr=-\gamma v h\, d\tau.\] Then using $\gamma=h^{-1/2}$ from (\ref{SchwRad-V}), we get \begin{equation}\label{Schw-dtL(dTau)} dt_{\Lambda}=\gamma \Big(1+v\frac{r_{g}}{r}\Big) d\tau= h^{-1/2}\Big(1+v\frac{r_{g}}{r}\Big) d\tau.\end{equation} Intensity of the emitted and detected light is proportional to the photons' frequency and inversely proportional to the proper time interval during which a given number $N$ of photons are being emitted/detected. So, combining (\ref{Schw-Rad-FullRedshift}) and (\ref{Schw-dtL(dTau)}), we get \begin{equation}\label{SchwRad-I} \frac{I_{det}}{I_{em}}= \frac{\omega_{det}}{\omega_{em}}\cdot \frac{d\tau}{dt_{\Lambda}}= \frac{1}{1+v} \frac{h_{em}}{\sqrt{h_{0}h_{det}}}\cdot \frac{\sqrt{h_{em}}}{1+v\frac{r_{g}}{r}}.\end{equation} In the limit $h_{0},h_{det}\to 1$ and $r\to r_{g}$, so that $v\to1$, we have \[\frac{I_{det}}{I_{em}}\approx \frac{1}{4}h_{em}^{3/2}= \frac{1}{4}\Big(1-\frac{r_{g}}{r}\Big)^{3/2}.\] Substituting $r(t_\Lambda)$ from (\ref{Schw-r(tL)}), we finally obtain \begin{equation}\label{Schw-I(tL)} I_{det}(t_\Lambda)\approx I_{em}\cdot \exp\Big\{-\frac{3\;t_\Lambda}{4\;r_g}\Big\}.\end{equation}

For $r\gg r_{g}$ we'll have $h\approx 1$ so if we count the angle from $\pi/2$ the impact parameter is $b=r\cos\varphi$. If we denote the length element of the trajectory by $dl$, then the ratio between the angular momentum and energy is \[\frac{L}{E} =\frac{r^2\cdot d\varphi/d\lambda} {h\cdot dt/d\lambda} =\frac{r^2 d\varphi}{hdt}\approx \frac{r^2 (dl\cos\varphi\, /r)}{dl/v}= r\cdot r\cos\varphi=vb.\] In the ultrarelativistic case \[L=Eb.\]

The first equation is $\epsilon=u^{\mu}u_{\mu}$. For a massive particle $\epsilon=1$ and we can also choose the parameter $\lambda$ as the natural parameter of the geodesic $d\lambda=ds$. For a massless particle $\epsilon=0$ and the parameter is arbitrary. In the Schwarzschild metric therefore \begin{equation}\label{Schw-Orbit1} h\Big(\frac{dt}{d\lambda}\Big)^{2}- \frac{1}{h}\Big(\frac{dr}{d\lambda}\Big)^{2}- r^{2}\Big(\frac{d\varphi}{d\lambda}\Big)^{2}= \epsilon.\end{equation} The two integrals of motion (\ref{SchwInt-E1},\ref{SchwInt-L1}) are \begin{equation}\label{Schw-Orbit-Integrals} E=h\frac{dt}{d\lambda};\qquad L=r^{2}\frac{d\varphi}{d\lambda}.\end{equation} Substituting this into (\ref{Schw-Orbit1}) and rearranging terms, we derive the equation for $r(\lambda)$ in the form \begin{equation}\label{Schw-V} \frac{1}{2}\Big(\frac{dr}{d\lambda}\Big)^{2}+ V_{\epsilon}(r)=\varepsilon,\qquad\mbox{где}\quad V(r)=\frac{h}{2} \Big(\epsilon+\frac{L^2}{r^2}\Big), \quad \varepsilon=\frac{E^2}{2}.\end{equation} This is an analogue of one-dimensional motion of a particle in a potential\footnote{Its analytic solution for the given $V$ is expressed through the integral from the square root of a third degree polynomial in the denominator, which is reduced to the elliptic Jacobi functions.} $V$, with the quantity $\varepsilon$ playing the role of full energy. Note that $V(r=r_{g})=0$. Recall that for massive particles $E$ and $L$ with the chosen parametrization are the energy and angular momentum per unit mass (see (\ref{SchwInt-E2},\ref{SchwInt-L2})).

Let us first consider $\epsilon=0$.

The effective potential for a massless particle $V(r/r_{g})$. The position of the maximum, that corresponds to the photon sphere, is at $r/r_{g}=\tfrac{3}{2}$ and does not depend on $L$

For massless particles the parameter $\lambda$ is arbitrary, so we will fix it by demanding that far from the horizon, where the metric is asymptotically Euclidean, holds $u^{0}\equiv dt/d\lambda=1$. Then, first, $E=1$, and second, at $r\gg r_{g}$ \[0=u^{\mu}u_{\mu}=(u^{0})^{2}-u^{\alpha}u_{\alpha} \quad\Rightarrow\quad \frac{dl}{d\lambda}=1.\] Here $dl$ is the line element of the (asymptotically) flat space, and for such parametrization we obtain that $L=b$, where $b$ is the impact parameter of the ray at infinity: \begin{equation}\label{Schw-orb-NullInts} u^{0}|_{r\to\infty}=1,\;\Rightarrow\quad E=1;\quad L=b.\end{equation} Then \[V_{\epsilon=0}=b^{2}\frac{h}{2r^{2}}= \frac{b^{2}}{2} \Big(\frac{1}{r^2}-\frac{r_{g}}{r^3}\Big), \quad \varepsilon=\frac{1}{2}.\] At small $r/r_{g}$ the effective potential behaves as $\sim (-r^{-3})$, while at large $r/r_{g}$ as $\sim r^{2}$, and reaches its maximum at \[r_{max}=\frac{3}{2}r_{g},\quad V_{max}=V(r_{max})= \frac{2}{27}\Big(\frac{b}{r_g}\Big)^2.\] As this is a maximum, the corresponding circular orbit on the so-called ``photon sphere is unstable. For $\varepsilon>V_{max}$ all the trajectories from one side of it escape to infinity, and the ones from the other side fall on the center\footnote{More accurately, they fall on the horizon, as we do not yet consider the motion beyond this point.} Rewriting the inequality in terms of $b$, we have \[b<b_{m}\equiv\frac{3\sqrt{3}}{2}\;r_{g}.\]

For massive particles it is convenient to express $V$ in terms of dimensionless quantities $\xi=r/r_{g}$ and $l=L/r_{g}$: \[V_{\epsilon=1}-\frac{1}{2}=-\frac{1}{2\xi}+ \frac{l^2}{2\xi^2}-\frac{l^2}{2\xi^3}.\] The first term is the Newtonian potential energy, the second one is the centrifugal energy, and only the third term is absent in the Newtonian theory and is unique to General Relativity. It changes the asymptote of $V$ at small $\xi$: $V\sim -\xi^{-3}$ instead of the usual $V\sim\xi^{-2}$.

Effective potential for a massive particle $V(r/r_{g})$ for $l=\{0, 1.25, \sqrt{3}, 1.85, 2, 2.25, 2.5, 3\}$ (problem \ref{BlackHole37}). There are stable as well as unstable circular orbits. The limiting value $l=\sqrt{3}$ defines the inflection point.

Same figure zoomed in to show the shallow minima

The extrema of $V$ are found as the roots of quadratic equation \[\xi^2-2l^2 \xi +3l^2=0\quad\Rightarrow\quad \xi_{\pm}=l^{2}\left\{1\pm\sqrt{1-3/l^2}\right\}.\] Two extrema exist, maximum and minimum, if $l>\sqrt{3}$, i.e. $L>L_{cr}$, where \[L_{cr}\equiv\sqrt{12}\,GM.\] The minimum $\xi=\xi_{+}>l^2$ corresponds to a stable circular orbit and non-circular finite motions dangling around it (they are not elliptic). The maximum $\xi=\xi_{-}<l^2$ corresponds to an unstable circular orbit. If $\varepsilon>V(\xi_{-})$, then the motion is infinite with the fall on the center (i.e. at least on the horizon $r=r_g$).

The radius of the stable circular orbit is minimal when the discriminant turns to zero: $l=\sqrt{3}$, $\xi_{+}=\xi_{-}=l^{2}=3$, and thus \[r_{circ}^{min}=3r_{g}=6GM.\]

For $l<\sqrt{3}$, i.e. $L<L_{cr}$, there are no extrema of $V$ and a particle's motion, either finite ($E<1$) or infinite ($E\geq 1$) always ends with the fall on the center.

In the limit $l\gg 1$, which corresponds to $L\gg r_{g}$, \[\xi_{-}\approx3/2,\;\xi_{+}\approx2l^2, \quad\Rightarrow\quad r_{-}=\frac{3}{2}r_{g};\; r_{+}=\frac{2L^{2}}{r_{g}}= \frac{L^2}{GM},\] so the inner unstable orbit tends to the photon sphere, while the outer stable orbit tends to the classical circular one.

The gravitational capture happens, i.e. a particle moving from infinity falls on the center, if it passes above the effective potential barrier \[\varepsilon\geq V_{\max}.\] (a) For a massless particle we have already found the height of the barrier, so the limiting condition is \[\frac{1}{2}=\frac{2}{27}\Big(\frac{b}{r_g}\Big)^2,\] and for the capture cross-section we get \[S_{\gamma}=\pi b^2 =\frac{27\pi}{4}r_{g}^{2} =27\pi \Big(\frac{GM}{c^2}\Big)^2.\] For a massive particle $V_{\epsilon=1}$ reaches its maximum in the lesser of the two roots $\xi_{\pm}$: \[\xi_{\max}^{-1}=\frac{1}{3}\big(1+\sqrt{1-3/l^2}\big).\] On substituting this into $V(\xi)$, after some transformations we find the maximum: \[V_{\max}=\frac{1}{27}\Big( l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\] (b) Ultrarelativistic case: \[\gamma\gg1,\quad E=\gamma_{\infty}\gg1, \quad\varepsilon =\tfrac{1}{2}\gamma_{\infty}^{2}\gg 1,\] thus $V_{\max}\gg 1$, which only can be when $l\gg1$. In this limit, in the first order by $l^{-2}$ \[V_{\max}=\frac{2}{27}l^2 +\frac{1}{6},\] so the limiting condition of capture is \[l^{2}=\frac{27}{4}\big( \gamma_{\infty}^{2}-\tfrac{1}{3}\big).\] Expressing it through the impact parameter $b=L/Ev$, we are led to \[S_{\gamma\gg1} =\pi b^2=\pi r_{g}^{2}\frac{l^2}{v^2 E^2} =\frac{27\pi}{4}r_{g}^{2}\Big( 1+\frac{2}{3\gamma_{\infty}^{2}}\Big).\] (c) Nonrelativistic case: $E\approx (1+v^2 /2)$, $\varepsilon\approx \tfrac{1}{2}(1+v^2)$, and the limiting condition of capture takes the form \[1+v^2 =\frac{1}{27}\Big( l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\] In the zeroth order by $v^2$ its solution is $l=2$ (note that at $l=\sqrt{3}$ there is no maximum of $V$), while in the first order \[l^2 =4(1+2v^2).\] In terms of impact parameter then the cross-section is \[S_{v\ll 1}=\pi r_{g}^{2}\frac{l^2}{v^2 E^2} \approx \frac{4\pi r_{g}^{2}}{v^2}(1+v^2).\]

The radius of stable circular orbit is minimal when \[\xi_{\min}=l_{min}^{2}=3+0.\] The value of effective potential at $\xi=\xi_{min}$ determines the particle's energy on the corresponding circular orbit (\ref{Schw-V}): \[\frac{E^2}{2}=\varepsilon=V|_{\epsilon=1} =\frac{1}{2}\Big(1-\frac{1}{\xi}\Big) \Big(1+\frac{l^{2}}{\xi^2}\Big)= \Big\| \xi=\xi_{min}=l_{min}^{2}\Big\|= \frac{1-\xi_{\min}^{-2}}{2},\] thus \[E=\sqrt{1-\xi_{min}^{-2}}=\sqrt{\frac{8}{9}} =\frac{2\sqrt{2}}{3}.\] The binding energy then is (in the units of $mc^{2}$) \[1-E=1-\frac{2\sqrt{2}}{3}\approx 0,06.\]

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