# Particles' motion in general black hole spacetimes

## Contents

In this section we use the $(-+++)$ signature, Greek letters for spacetime indices and Latin letters for spatial indices.

## Frames, time intervals and distances

In the next several problems we again consider the procedure of measuring time and space intervals by different observers, but in a different, more formal and powerful approach.

### Problem 1

Let a particle move with the four-velocity $U^{\mu }$. It can be viewed as some observer carrying a frame attached to him. Locally, it defines the hypersurface orthogonal to it. Show that \begin{equation} h_{\mu \nu }=g_{\mu \nu }+U_{\mu }U_{\nu } \label{h} \end{equation} is (i) the projection operator onto this hypersurface, and at the same time (ii) the induced metric of the hypersurface. This means that (i) for any vector projected at this hypersurface by means of $h^\mu_\nu$, only the components orthogonal to $U^{\mu}$ survive, (ii) the repeated application of the projection operation leaves the vector within the hypersurface unchanged. In other words, $h_{\mu \nu}$ satisfies \begin{align} &h^{\mu}_{\nu}U^{\nu }=0 ; \label{1} \\ &h^{\mu}_{\nu }h^{\nu}_{\lambda}=h^{\mu }_{\lambda}. \label{2} \end{align}

### Problem 2

Let us consider a particle moving with the four-velocity $U^{\mu }$. The interval $ds^{2}$ between two close events is defined in terms of differentials of coordinates,% \begin{equation} ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }. \end{equation}

For given $dx^{\mu }$, what is the value of the proper time $d\tau _{obs}$ between the corresponding events measured by this observer? How can one define locally the notions of simultaneity and proper distance $dl$ for the observer in terms of its four-velocity and the corresponding projection operator $h^\mu_\nu$ ? How is the interval $ds^{2}$ related to $d\tau _{obs}$ and $dl$?

In order to obtain $d\tau _{obs}$, one should project $dx^{\mu}$ onto the four-velocity: \begin{equation} d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau} \end{equation}% Then, locally, two events are simultaneous when \begin{equation} d\tau _{obs}=0. \label{sim} \end{equation} The proper distance \begin{equation} dl^{2}=h_{\mu \nu }dx^{\mu }dx^{\nu } \label{dl}. \end{equation} Then (\ref{h}) implies that \begin{equation} ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }=dl^{2}-d\tau _{obs}^{2} \label{ds}. \end{equation}

### Problem 3

Let our observer measure the velocity of some other particle passing in its immediate vicinity. Relate the interval to $d\tau _{obs}$ and the particle's velocity $w$.

The velocity can be defined as \begin{equation} w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}. \end{equation}% Then, it follows from (\ref{dl}), (\ref{ds}) that \begin{equation} ds^{2}=d\tau _{obs}^{2}(1-w^{2}). \label{sw} \end{equation}

### Problem 4

Analyze the formulas derived in the previous three problems applied to the case of flat spacetime (Minkovskii space) and compare them to the known formulas of special relativity.

In special relativity four-velocity is expressed through velocity (spatial components are denoted by Latin letters) \begin{equation} v^{i}=\frac{dx^{i}}{dt},\qquad i=1,2,3\label{vxt} \end{equation} and Lorentz factor \begin{equation} \gamma =(1-v^{2})^{-1/2} \end{equation} as follows: \begin{align} U^{\mu}&=\gamma (+1,v^i),\\ U_{\mu }&=\gamma (-1,v^i). \end{align} Let us compare the original frame (with coordinates $t,x^i$) and the one comoving with the observer (with coordinates $t^{\prime }=\tau _{obs}$ and $x^{i\prime}$.) Then using (\ref{vxt}) we get \begin{equation} ds^{2}=-dt^{\prime 2}=-dt^{2}+\delta _{ij}dx^{i}dx^{j}=-dt^{2}(1-v^{2}). \end{equation} Eqs. (\ref{tau}), (\ref{vxt}) then give us \begin{equation} d\tau _{obs}\equiv dt^{\prime } =\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}} =dt\sqrt{1-v^{2}}, \end{equation}% which coincides with the standard Lorentz transformation. Here, $dt$ and $dx^{i} $ are the time and coordinate differences between close events measured in the laboratory frame. The quantities with primes refer to the comoving frame. Then, $\tau _{obs}=t^{\prime }$ has the meaning of proper time, as in the case when $dx^{i\prime }=0$, and thus $dl^{\prime}=0$, from (\ref{tau}) we get $ds^{2}=-dt^{\prime 2}$.

### Problem 5

Consider an observer being at rest with respect to a given coordinate frame: $x^{i}=const$ ($i=1,2,3$). Find $h_{\mu \nu }$, $d\tau _{obs}$, the condition of simultaneity and $dl^{2}$ for this case. Show that the corresponding formulas are equivalent to eqs. (84.6), (84.7) of \cite{lan}, where they are derived in a different way.

If the observer does not move, its four-velocity is \begin{equation} U^{\mu }=(U^{0},0,0,0). \end{equation}% From the normalization condition one finds \begin{equation} g_{00}\left( U^{0}\right) ^{2}=-1, \end{equation}% whence \begin{equation} U^{\mu }=\frac{1}{\sqrt{-g_{00}}}(1,0,0,0).\label{st} \end{equation} The covariant components $U_{\mu }=g_{\mu \nu }U^{\nu }$ then are equal to \begin{equation} U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}. \label{uco} \end{equation}% By substitution into (\ref{h}), one finds \begin{align} h_{00}&=0,\quad h_{0i}=0,\label{h0}\\ h_{ij}&=g_{ij}-\frac{g_{0i}g_{0j}}{g_{00}}, \label{hi} \end{align} which coincides (up to the choice of the overall signature) with the spatial metric $\gamma_{ij}$ as defined in eq. (84.7) of \cite{lan}. Then, $dl^{2}=h_{ij}dx^{i}dx^{j}$ turns into eq. (84.6). The condition of simultaneity (\ref{sim}) with (\ref{uco}) taken into account reads \begin{equation} dt=-dx^{i}\frac{g_{0i}}{g_{00}} \end{equation}% which coincides with eq. (84.14) of \cite{lan}.

### Problem 6

Consider two events at the same point of space but at different values of time. Find the relation between $dx^{\mu}$ and $d\tau _{obs}$ for such an observer.

In this case, $dx^{i}=0$, and we obtain from (\ref{tau}) and (\ref{uco}) that \begin{equation} d\tau _{obs}=\sqrt{-g_{00}}\;dt \end{equation}% which coincides with eq. 84.1 of \cite{lan}.

## Fiducial observers

### Problem 7

Consider an observer with \begin{equation} U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.} \label{uz} \end{equation}% We call it a fiducial observer (FidO) in accordance with \cite{mb}. This notion is applied in \cite{mb} mainly to static or axially symmetric rotating black holes. In the latter case it is usually called the ZAMO (zero angular momentum observer). We will use FidO in a more general context.

Show that a FidO's world-line is orthogonal to hypersurfaces of constant time $t=const$.

A vector $n_\mu$ is called normal to a hypersurface $\Sigma$ if it is orthogonal to any displacement $dx^\mu $ within this hypersurface: $n_\mu dx^\mu =0$. For hypersurfaces of constant time $dx^{\mu}=(0,dx^i)$, so the normal vector is $n_\mu =(n_0 ,0)\sim \partial_t $, with $n_i =0$. Thus, according to (\ref{uz}), $n_{\mu}$ and $U_{\mu}$ are parallel according to (\ref{uz}) and $U_\mu$ is also normal to $t=const$.

### Problem 8

Find the explicit form of the metric coefficients in terms of the components of the FidO's four-velocity. Analyze the specific case of axially symmetric metric in coordinates $(t,\phi ,r,\theta )$ with $g_{0i}=g_{t\phi }\delta _{i}^{\phi }$.

By definition, $U_{i}=0$. Then, the normalization condition gives us% \begin{equation} U^{\mu }=\frac{1}{N}(1,N^{i}) \label{zamo} \end{equation}% with some $N^{i}$. It has the simple physical meaning of the velocity of the observer with respect to the coordinate frame $\{t,x^{i}\}$ defined as \begin{equation} V^{i}\equiv \frac{dx^{i}}{dt}\Big| _{obs}=\frac{U^{i}}{U^{0}}=N^{i}. \label{vn} \end{equation} It follows from (\ref{tau}) that% \begin{equation} d\tau _{obs}=Ndt . \end{equation}% Using that \begin{equation} 0=U_{i}=\frac{g_{i 0}}{N}+\frac{g_{ij}N^{j}}{N}, \end{equation}% we find \begin{equation} g_{0i}=-g_{ij}N^{j}. \end{equation} In a similar way,% \begin{equation} -N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N}, \end{equation}% whence% \begin{equation} g_{00}=-N^{2}+g_{0i}N^{i}N. \end{equation} Collecting all this, we obtain that the metric can be written in the form \begin{equation} ds^{2}=-dt^{2}N^{2}+g_{ij}(dx^{i}-N^{i}dt)(dx^{j}-N^{j}dt) \label{N}, \end{equation} and calculating $h_{\mu \nu }$ we find \begin{equation} h_{00}=g_{00}+N^{2},\qquad h_{0i}=g_{0i},\qquad h_{ij}=g_{ij}. \end{equation} In the case of a stationary axially symmetric metric \begin{equation} ds^{2} =-dt^{2}N^{2} +g_{\phi \phi }(d\phi -\omega dt)^{2} +g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}. \end{equation}% we have \begin{align} g_{00}&=-N^{2}+g_{\phi \phi }\omega ^{2}, \label{00m}\\ g_{0\phi }&=-g_{\phi \phi }\omega, \label{03} \end{align} and therefore \begin{align} h_{00}=\omega ^{2}g_{\phi \phi }, &\qquad h_{0\phi }=g_{0\phi }, \qquad h_{\phi \phi }=g_{\phi \phi };\\ & N^{\phi }=\omega ,\qquad N^{r}=N^{\theta }=0.\label{vf} \end{align}

### Problem 9

Consider a stationary metric with the time-like Killing vector field $\xi^\mu =(1,0,0,0)$. Relate the energy $E$ of a particle with four-velocity $u^\mu$ as measured at infinity by a stationary observer to that measured by a local observer with 4-velocity $U^\mu$.

The energy as measured at infinity by a stationary observer is the integral of motion \begin{equation} E=-mu_{\mu }\xi ^{\mu }=-mu_{0}. \end{equation} On another hand, the energy measured by a local observer is \begin{equation} E_{rel}=-mu_{\mu }U^{\mu }\equiv m\gamma =-m(u_{0}U^{0}+u_{i}U^{i}), \label{rel} \end{equation} whence \begin{equation} E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e} \end{equation}% where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).

### Problem 10

Express $E_{rel}$ and $E$ in terms of the relative velocity $w$ between a particle and the observer (i.e. velocity of the particle in the frame of the observer and vice versa).

Using (\ref{tau}), (\ref{rel}) we find that the Lorentz factor is \begin{equation} \gamma =-u_{\mu }U^{\mu } =-\frac{U^{\mu }dx_{\mu }}{ds}=-\frac{d\tau _{obs}}{ds}. \end{equation} It follows from (\ref{sw}) that \begin{equation} \gamma =\frac{1}{\sqrt{1-w^{2}}},\qquad E_{rel}=\frac{m}{\sqrt{1-w^{2}}}. \end{equation} Then, we find from (\ref{e}) that \begin{equation} E-p_{i}V^{i}=\frac{m}{\sqrt{1-w^{2}}}\frac{1}{U_{0}}, \label{ep} \end{equation} where $p^{\mu }=mu^{\mu }$ is the particle's four-momentum.

### Problem 11

Show that in the flat spacetime eq. (\ref{ep}) is reduced to the usual formula of the Lorentz transformation.

In this case, $U^{0}=(1-V^{2})^{-1/2}=\gamma $, where $V$ is the absolute value of the velocity of the observer (frame 0) with respect to the stationary laboratory frame. Then the energy of a particle in frame 0 is \begin{equation} E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}), \end{equation}% This is the standard formula for energy transformation.

### Problem 12

Find the expression for $E$ for the case of a static observer ($U^{i}=0$).

Using (\ref{st}) and (\ref{e}) we find that the kinetic term in (\ref{e}) vanishes and \begin{equation} E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}}, \end{equation}% which coincides with eq. (88.9) of \cite{lan}.

### Problem 13

Find the expression for $E$ for the case of the ZAMO observer and, in particular, in case of axially symmetric metric.

\begin{equation} E-mu_{i}V^{i}=E_{rel} N=\frac{mN}{\sqrt{1-w^{2}}}. \end{equation} In the axially symmetric case, the angular momentum of a particle $mu_{\phi}=L$ is conserved, and according to (\ref{vf}), $V^{\phi }=\omega$. Then, \begin{equation} E-\omega L=E_{rel}N=\frac{mN}{\sqrt{1-w^{2}}}. \end{equation}

## Collision of particles: general relationships

### Problem 14

Let two particles collide. Define the energy in the center of mass (CM) frame $E_{c.m.}$ at the point of collision and relate it to $E_{rel}$ and the Lorentz factor of relative motion of the two particles.

In general, the momenta and velocities of two particles are defined at different points and this prevents one from constructing well-defined quantities. However, as collision occurs at one point, both particles have characteristics defined just in this point. This enables one to construct the total momentum $P^{\mu }=p_{1}^{\mu }+p_{2}^{\mu }$ at the point of collision (the subscript $a=1,2$ enumerates the two particles). A particle's mass is \begin{equation} m^{2}=-p^{\mu }p_{\mu }, \end{equation} and likewise the full energy (mass) of two particles at the point of collision, which is also their energy in the center of mass frame, is found from \begin{equation} E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm} \end{equation}% It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies \begin{equation} E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma , \end{equation}% where \begin{equation} \gamma =-u_{1}^{\mu }u_{2\mu } \end{equation} is the Lorentz factor of the two particles' relative motion. It can be expressed in terms of energy $E_{rel}(2,1)$ of particle 1 with respect to particle 2 and vice versa \begin{equation} \gamma=\frac{E_{rel.}(1,2)}{m_{1}}=\frac{E_{rel.}(2,1)}{m_{2}}. \end{equation}

### Problem 15

Let us consider a collision of particles 1 and 2 viewed from the frame attached to some other particle 0. How are different Lorentz factors related to each other? Analyze the case when the laboratory frame coincides with that of particle 0.

Let us introduce notation \begin{equation*} \gamma (0,1)=\gamma _{1}, \quad \gamma (0,2)=\gamma _{2}, \quad \gamma (1,2)=\gamma , \quad u_{0}^{\mu }=U^{\mu}. \end{equation*} Each particle's four velocity can be decomposed into components parallel and orthogonal to $U^\mu$ as follows \begin{equation} u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu}, \label{ab} \end{equation}% where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to find from (\ref{ab}) that% \begin{equation} a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1}, \end{equation}% and from the normalization conditions for $u_{1}^{\mu }$ and $U^{\mu }$ then we see that %$b_1$ as introduced in (\ref{ab}) is the velocity of particle 1 in the frame of particle 0: \begin{equation} b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 }, \end{equation} so in terms of velocity of particle 1 in the frame of particle 0 \begin{equation} w_1 =(1-\gamma_1^{-2})^{1/2}=\frac{b_1}{\gamma}, \end{equation} the decomposition takes form \begin{equation} u_{1}^{\mu }=\gamma_{1}(U^{\mu }+w_1 n_{1}^{\mu}). \end{equation} Writing down for particle 2 the equation analogous to (\ref{ab}) and calculating the scalar product $u_{1}^{\mu }u_{2\mu }$, one obtains \begin{align} \gamma &=\gamma _{1}\gamma _{2} -b_1 b_2 (n_{1}^{\mu }n_{2\mu})\\ &=\gamma _{1}\gamma _{2}\big(1-\varepsilon\; w_{1}w_{2}\big), \label{ga} \end{align} where \begin{equation} \varepsilon =(n_{1}^{\mu }n_{2\mu }),\qquad |\varepsilon|\leq 1 , \end{equation} or in terms of velocities \begin{equation} \frac{1}{\sqrt{1-w^{2}}} =\frac{1}{\sqrt{1-w_{1}^{2}}} \frac{1}{\sqrt{1-w_{2}^{2}}} (1-\varepsilon w_{1}w_{2}). \label{gw} \end{equation} In case particle 0 is at rest in the laboratory frame $U^{\mu }=(1,0)$, we have $n_{a}^{\mu}=(0,\vec{n}_{a})$, $a=1,2$, and \begin{equation} u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}% \vec{n}_{2}). \end{equation}% Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of \cite{li}.

### Problem 16

When can $\gamma $ as a function of $\gamma _{1}$ and $\gamma _{2}$ grow unbounded? How can the answer be interpreted in terms of relative velocities?

1) If $\gamma _{1}$ and $\gamma _{2}$ are finite, it follows from (\ref{ga}) that $\gamma $ is also finite. It means that if $w_{1}$ and $w_{2}$ are finite, the relative velocity $w$ of particles 1 and 2 is also finite (in the sense that it is separated from $c$). 2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite. Then,% \begin{equation} \gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big) \end{equation}% grows unbound irrespective of $\varepsilon $. It means that the relative velocity of particles, one of which moves with some finite speed and the other one almost with the speed of light is always close to the speed of light. 3) Let $\gamma _{1}\rightarrow \infty $, $\gamma _{2}\rightarrow \infty $. 3a) If $\varepsilon \neq +1$, \begin{equation} \gamma \approx \gamma _{1}\gamma _{2}(1-\varepsilon )\rightarrow \infty , \qquad w\rightarrow 1. \end{equation} 3b) Let $\varepsilon =+1$. Then \begin{equation} \gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{% \gamma _{2}}{\gamma _{1}}\Big). \end{equation} We see that $\gamma $ remains finite if $\gamma _{1}/\gamma _{2}$ does so. Otherwise, $\gamma \rightarrow \infty $. The resulting relative velocity depends on the relative rate with which $w_{1}$ and $w_{2}$ approach the speed of light. Thus in case 3, motion in different directions always gives $\gamma \rightarrow \infty $, whereas motion in the same direction may or may not lead to divergent $\gamma $.

### Problem 17

A tetrad basis, or the orthonormal tetrad, is the set of four unit vectors $h_{(a)}^\mu$ (subscripts in parenthesis $a=0,1,2,3$ enumerate these vectors), of which one, $h_{(0)}^\mu$, is timelike, and three vectors $h_{(i)}^\mu$ ($i=1,2,3$) are spacelike, so that \begin{equation} g_{\mu\nu}h_{(a)}^\mu h_{(b)}^\nu =\eta_{ab},\qquad a,b=0,1,2,3. \end{equation} A vector's tetrad components are \begin{equation} u_{(a)}=u_\mu h_{(a)}^\mu,\qquad u^{(b)}=\eta^{ab}u_{(b)}. \end{equation}

Define the local three-velocities with the help of the tetrad basis attached to the observer, which would generalize the corresponding formulas of special relativity.

In special relativity,% \begin{equation} u^{\mu }=\frac{dx^{\mu }}{ds} =\gamma (1,v^{i}),\qquad \gamma =\frac{1}{\sqrt{1-v^{2}}}. \end{equation}% Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get \begin{equation} v^{i}=\frac{dx^{i}}{d\tau _{obs}}.\label{vi} \end{equation} Let us attach a tetrad to some observer with four-velocity $U^\mu$. This means simply that we choose the first, time-like, tetrad vector to be along its world-line \begin{equation} h_{(0)}^\mu =U^\mu . \end{equation} Then, the natural definition for velocity in the frame of this observer, taking into account (\ref{tau}), will be \begin{equation} v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}. \end{equation}% It can be rewritten as \begin{equation} v^{(i)}=\frac{u^{\mu }h_{\mu }^{(i)}}{u^{\mu }h^{(0)}_\mu}. \end{equation} The tetrad components of the projection operator (\ref{h}) are \begin{equation} h_{(ab)}=\eta_{ab}+\eta_{0a}\eta_{0b}, \end{equation} from which it is easy to check explicitly that the two definitions, through $h_{\mu\nu}$ and through $h^\mu_{(a)}$, are equivalent.

### Problem 18

Derive the analogues of formulas (\ref{e}), (\ref{ep}) for massless particles (photons). Analyze the cases of static and ZAMO observers.

In this case, $m=0$, the velocity of a photon is always 1, and the notion of a frame comoving with the photon does not have sense. Nonetheless, some formulas retain their validity with $E$ replaced by the frequency $\nu$ in accordance with the fact that for photons $E=\hbar\nu$ (we use $\nu=2\pi/\text{(period)}$ for the angular frequency here, so as not to confuse it with the metric function $\omega$). Let us consider a photon with wave four-vector $k_{\mu }$ in a gravitational field. In the case of static or stationary metric, $\nu _{0}=-k_{\mu }\xi^{\mu }$ is conserved along the trajectory, where $\xi ^{\mu }$ is the Killing vector. If it corresponds to time translations, $\nu$ has the meaning of frequency. It is what is measured by a remote observer (if the flat infinity exists). In parallel to derivation of (\ref{e}), one can obtain \begin{equation} \nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}}, \label{nu} \end{equation}% where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the observer. For the static observer $V^{i}=0$, $U^{0}=\frac{1}{\sqrt{-g_{00}}}$, and \begin{equation} \nu_{0}=\nu \sqrt{-g_{00}}. \label{0g} \end{equation} For the ZAMO observer, $V^{\phi }=\omega $, $k_{\phi }=L$ is the angular momentum which is conserved in the axially symmetric case (in units of $\hbar$), $U^{0}=\frac{1}{N}$ (see eq. (\ref{zamo})). Then \begin{equation} \nu _{0}-\omega L=\nu N. \end{equation}

### Problem 19

The ergosphere is a surface defined by equation $g_{00}=0$. Show that it is the surface of infinite redshift for an (almost) static observer.

From (\ref{0g}) it follows that in the limit $g_{00}\rightarrow 0$ the frequency $\nu _{0}\rightarrow 0$ for any finite $\nu $. However, for a nonstatic observer this is not true.

### Problem 20

Consider an observer orbiting with a constant angular velocity $\Omega $ in the equatorial plane of the axially symmetric back hole. Analyze what happens to redshift when the angular velocity approaches the minimum or maximum values $\Omega _{\pm }$.

Eq. (\ref{nu}) reads in this case% \begin{equation} \nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,} \end{equation}% where $U^{0}$ obeys the normalization condition% \begin{equation} Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left( U^{0}\right) ^{2}}\text{.} \end{equation} It can be rewritten in the form% \begin{align} Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }}) =g_{\phi \phi }(\Omega -\Omega _{+})(\Omega -\Omega _{-});\\ &\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}}, \end{align} Then,% \begin{equation} \nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega -\Omega _{-})}\text{.} \end{equation} When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }% \rightarrow 0$. Thus, if, say, $\Omega \rightarrow \Omega _{-}$, $\nu _{0}\rightarrow \Omega _{-}L$. If $L=0$, then $\nu _{0}\rightarrow 0$.

### Problem 21

Let two massive particle 1 and 2 collide. Express the energy of each particle in the centre of mass (CM) frame in terms of their relative Lorentz factor $\gamma (1,2)$. Analyze the limiting cases of ultra-relativistic $\gamma (1,2)\rightarrow \infty $ and non-relativistic $\gamma (1,2)\approx 1$ collisions.

The energy of particle 1 in the CM frame is \begin{equation} \left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM) \label{e1} \end{equation}% where% \begin{equation} \gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu } \end{equation}% is its Lorentz factor in the same frame, which has four-velocity \begin{equation} U_{c.m.}^{\mu }=\frac{P^{\mu }}{\mu}. \end{equation} Here $P^\mu$ is the total momentum and $\mu \equiv E_{c.m.}$ is the energy in the center of mass frame introduced in (\ref{cm})%problem 14 \begin{align} P^{\mu }&=m_{1}u_{1}^{\mu }+m_{2}u_{2}^{\mu };\\ \mu ^{2}&=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma (1,2).\label{mu} \end{align} %\begin{equation} %\gamma (1,2)=-u_{1\mu }u_{2}^{\mu }. %\end{equation}% Then \begin{align} \gamma (1,CM)&=\frac{m_{1}+m_{2}\gamma (1,2)}{\mu };\label{g1}\\ ( E_{1}) _{c.m.} &=m_{1}\gamma (1,CM).\label{e1cm} \end{align} Likewise, for the second particle \begin{align} \gamma (2,CM)&=\frac{m_{2}+m_{1}\gamma (1,2)}{\mu };\\ ( E_{2}) _{c.m.}&=m_{2}\gamma (2,CM).\label{e2cm} \end{align} It follows from (\ref{e1cm}), (\ref{e2cm}) that% \begin{equation} ( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu , \end{equation} as it should be. In the ultra-relativistic limit $\gamma (1,2)\rightarrow \infty $ \begin{equation} \gamma (1,CM)\approx \frac{\mu }{2m_{1}}\approx \sqrt{\frac{m_{2}}{2m_{1}}\gamma (1,2)}. \end{equation} In the opposite limit of small relative velocity $w(1,2)\ll 1$ we obtain $\gamma (1,2)\approx 1+\frac{w^{2}}{2}$ and \begin{align} &\mu \approx (m_{1}+m_{2})+\frac{m_{1}m_{2}}{2(m_{1}+m_{2})}w^{2};\\ &w(1,CM)\approx w\frac{m^{2}}{(m_{1}+m_{2})}, \end{align} which agrees with formulas of nonrelativistic mechanics -- see Ch. 3, Sec. 13 of \cite{lan}.

### Problem 22

For a stationary observer in a stationary space-time the quantity $\alpha =(U^0)^{-1}$ is the redshifting factor: if this observer emits a ptoton with frequency $\omega_{em}$, it is detected at infinity by another stationary observer with frequency $\omega_{det}=\alpha \omega_{em}$. For a generic observer this interpretation is invalid, however, $\alpha =ds/dt$ still determines the time dilation for this observer, and thus can still be called the same way. Express the redshifting factor of the center of mass frame $\alpha _{c.m.}$ through the redshifting factors of the colliding particles $\alpha_{1}$ and $\alpha_2$.

Starting from the definition, \begin{equation} \frac{1}{\alpha _{c.m.}} =\left( U^{0}\right) _{c.m.} =\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu } =\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu }, \end{equation}% whence% \begin{equation} \alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha _{2}+m_{2}\alpha _{1}}. \end{equation} If case $\alpha _{2}\ll \frac{m_{2}}{m_{1}}\alpha _{1}$, we get \begin{equation} \alpha _{c.m.}\approx \frac{\mu \alpha _{2}}{m_{2}}. \end{equation}

### Problem 23

Relate the energy of a particle at infinity $E_{1}$, its energy at the point of collision in the C.M. frame $(E_{1})_{c.m.}$ and $\mu$.

From (\ref{e}) we know that \begin{equation} E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel}, \label{e2} \end{equation}% where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$. A particle's energy in the center of mass frame can be found from (\ref{e1cm}) \begin{equation} (E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2), \end{equation}% where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get \begin{align} E_{1}-m_{1}u_{1i}v_{2}^{i} &=\frac{\alpha_{2}(\mu E_{1})_{c.m.} -m_{1}^{2}\alpha _{2}}{m_{2}};\\ E_{2}-m_{2}u_{2i}v_{1}^{i} &=\frac{\alpha _{1}(\mu E_{2})_{c.m.} -m_{2}^{2}\alpha _{1}}{m_{1}}. \end{align} On another hand, applying (\ref{e}) to particle 1 and the center of mass frame, we obtain \begin{align} E_{1}-m_{1}u_{1i}V_{c.m.}^{i} &=\alpha _{c.m.}(E_{1}) _{c.m.}; \label{1cm}\\ E_{2}-m_{2}u_{2i}V_{c.m.}^{i} &=\alpha _{c.m.}(E_{2}) _{c.m.}.\label{2cm} \end{align} The sum of (\ref{1cm}) and (\ref{2cm}) is the total conserved energy $E=E_1 +E_2$: \begin{equation} E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu . \end{equation}% Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to \begin{equation} EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2}, \end{equation}% as it should be.

### Problem 24

Solve the same problem when both particles are massless (photons). Write down formulas for the ZAMO observer and for the C.M. frame.

Let $k^\mu_1$ and $k^\mu_2$ be the wave-vectors of two photons. Then vector \begin{equation} K^{\mu }=\left( k^{\mu }\right) _{1}+\left( k^{\mu }\right) _{2} \label{K} \end{equation} is \emph{time-like} (unless the two photons are collinear, but then they would not collide). Dividing by the energy $\mu$ in the center of mass frame \begin{equation} \mu ^{2}=-K^{\mu }K_{\mu }=-2k_{1\mu }k^{2\mu }, \end{equation} we obtain the frame's (time-like) four-velocity \begin{equation} U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }. \end{equation}% There is no Lorentz factor or relative velocity for two photons. The contraction \begin{equation} \tilde{\gamma}(1,obs)=-k_{\mu } U^{\mu } \end{equation} has the meaning of the frequency of a photon with four-vector $k^\mu$ as measured by the observer with four-velocity $U^\mu$. Then frequency measured in the CM frame is \begin{equation} \left( \nu _{1}\right) _{c.m.}=-k_{1\,\mu} U_{c.m.}^{\mu }, \end{equation} and taking into account (\ref{mu}) and (\ref{K}), we obtain \begin{equation} (\nu _{1}) _{c.m.} =-k_{1\,\mu}U_{c.m.}^{\mu } =-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2}, \end{equation}% as for photons $k_{i\,\mu}k_i^{\mu}=0$. This is quite natural, since for photons the energy is equal to the absolute value of the momentum, so in the CM frame the energies of both photons should be equal. Likewise, \begin{equation} ( \nu _{2}) _{c.m.} =-k_{2\,\mu}U_{c.m.}^{\mu } =\frac{\mu }{2}=(\nu_{1}) _{c.m.}, \end{equation} and \begin{equation} \mu =( \nu _{1}) _{c.m.}+( \nu _{2}) _{c.m.} =2(\nu _{1}) _{c.m.}=2(\nu _{2}) _{c.m.} \end{equation} If $\xi$ is the Killing vector field responsible for time translation, the conserved frequency of a photon, equal to the frequency measured at infinity by a stationary observer, is \begin{equation} \nu _{\infty }=-k_{\mu }\xi ^{\mu }=-k_{0}. \end{equation} In some arbitrary frame it is \begin{equation} \nu _{loc}=-k_{\mu }U^{\mu }=-k_{0}U^{0}-k_{i}U^{i}. \end{equation} Then \begin{equation} \nu _{\infty }-k_{\alpha}V^{\alpha}=\alpha \nu _{loc},\qquad V^{i}=\frac{U^{i}}{U^{0}},\qquad \alpha =\frac{1}{U^{0}},\label{obs} \end{equation}% the same as for the case of massive particle. The difference is only that for photons we cannot express $\nu _{loc}$ in terms of mass and velocity as $\frac{m}{\sqrt{1-w^{2}}}$. ZAMO frame% \begin{equation} \nu _{\infty }-\omega L=N\nu _{loc}. \end{equation} CM frame \begin{align} &\nu _{\infty }-k_{i}V^{i}=\alpha _{c.m.}\frac{\mu }{2},\\ &\alpha _{c.m.}=\frac{1}{U^{0}}=\frac{\mu }{k_{1}^{0}+k_{2}^{0}}. \end{align}