Difference between revisions of "Particles' motion in general black hole spacetimes"

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}

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}

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 $ds^{2}=-dt^{\prime 2}$.

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}.

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}.

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 $t=const$.

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}

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}).

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.

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.

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}.

\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}

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}

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}.

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 $.

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.

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}

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.

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 $\nu _{0}\rightarrow 0$.

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}) \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} 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}.

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}

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.

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}