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

From Universe in Problems
Jump to: navigation, search
Line 30: Line 30:
 
Let us consider a particle moving with the four-velocity $U^{\mu }$. The
 
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
 
interval $ds^{2}$ between two close events is defined in terms of
differentials of coordinates,%
+
differentials of coordinates,
 
\begin{equation}
 
\begin{equation}
 
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
 
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
Line 47: Line 47:
 
\begin{equation}
 
\begin{equation}
 
d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau}
 
d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau}
\end{equation}%
+
\end{equation}
 
Then, locally, two events are simultaneous when
 
Then, locally, two events are simultaneous when
 
\begin{equation}
 
\begin{equation}
Line 81: Line 81:
 
\begin{equation}
 
\begin{equation}
 
w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}.
 
w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}.
\end{equation}%
+
\end{equation}
 
Then, it follows from (\ref{dl}), (\ref{ds}) that
 
Then, it follows from (\ref{dl}), (\ref{ds}) that
 
\begin{equation}
 
\begin{equation}
Line 126: Line 126:
 
=\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}}
 
=\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}}
 
=dt\sqrt{1-v^{2}},
 
=dt\sqrt{1-v^{2}},
\end{equation}%
+
\end{equation}
 
which coincides with the standard Lorentz transformation. Here, $dt$ and $dx^{i} $ are the time and coordinate differences between close events
 
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  
 
measured in the laboratory frame. The quantities with primes refer to the comoving frame. Then, $\tau _{obs}=t^{\prime }$ has the meaning of  
Line 151: Line 151:
 
\begin{equation}
 
\begin{equation}
 
U^{\mu }=(U^{0},0,0,0).
 
U^{\mu }=(U^{0},0,0,0).
\end{equation}%
+
\end{equation}
 
From the normalization condition one finds
 
From the normalization condition one finds
 
\begin{equation}
 
\begin{equation}
 
g_{00}\left( U^{0}\right) ^{2}=-1,
 
g_{00}\left( U^{0}\right) ^{2}=-1,
\end{equation}%
+
\end{equation}
 
whence
 
whence
 
\begin{equation}
 
\begin{equation}
Line 165: Line 165:
 
U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}.
 
U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}.
 
\label{uco}
 
\label{uco}
\end{equation}%
+
\end{equation}
 
By substitution into (\ref{h}), one finds
 
By substitution into (\ref{h}), one finds
 
\begin{align}
 
\begin{align}
Line 176: Line 176:
 
\begin{equation}
 
\begin{equation}
 
dt=-dx^{i}\frac{g_{0i}}{g_{00}}
 
dt=-dx^{i}\frac{g_{0i}}{g_{00}}
\end{equation}%
+
\end{equation}
 
which coincides with eq. (84.14) of \cite{lan}.
 
which coincides with eq. (84.14) of \cite{lan}.
 
   </p></div>
 
   </p></div>
Line 197: Line 197:
 
\begin{equation}
 
\begin{equation}
 
d\tau _{obs}=\sqrt{-g_{00}}\;dt
 
d\tau _{obs}=\sqrt{-g_{00}}\;dt
\end{equation}%
+
\end{equation}
 
which coincides with eq. 84.1 of \cite{lan}.
 
which coincides with eq. 84.1 of \cite{lan}.
 
   </p></div>
 
   </p></div>
Line 211: Line 211:
 
\begin{equation}
 
\begin{equation}
 
U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.}  \label{uz}
 
U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.}  \label{uz}
\end{equation}%
+
\end{equation}
 
We call it a fiducial observer (FidO) in accordance with \cite{mb}. This
 
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
 
notion is applied in \cite{mb} mainly to static or axially symmetric rotating
Line 240: Line 240:
 
     <p style="text-align: left;">
 
     <p style="text-align: left;">
  
By definition, $U_{i}=0$. Then, the normalization condition gives us%
+
By definition, $U_{i}=0$. Then, the normalization condition gives us
 
\begin{equation}
 
\begin{equation}
 
U^{\mu }=\frac{1}{N}(1,N^{i})  \label{zamo}
 
U^{\mu }=\frac{1}{N}(1,N^{i})  \label{zamo}
\end{equation}%
+
\end{equation}
 
with some $N^{i}$. It has the simple physical meaning of the velocity of the
 
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
 
observer with respect to the coordinate frame $\{t,x^{i}\}$ defined
Line 252: Line 252:
 
\end{equation}
 
\end{equation}
  
It follows from (\ref{tau}) that%
+
It follows from (\ref{tau}) that
 
\begin{equation}
 
\begin{equation}
 
d\tau _{obs}=Ndt .
 
d\tau _{obs}=Ndt .
\end{equation}%
+
\end{equation}
 
Using that
 
Using that
 
\begin{equation}
 
\begin{equation}
Line 265: Line 265:
 
\end{equation}
 
\end{equation}
  
In a similar way,%
+
In a similar way,
 
\begin{equation}
 
\begin{equation}
 
-N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N},
 
-N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N},
\end{equation}%
+
\end{equation}
whence%
+
whence
 
\begin{equation}
 
\begin{equation}
 
g_{00}=-N^{2}+g_{0i}N^{i}N.
 
g_{00}=-N^{2}+g_{0i}N^{i}N.
Line 289: Line 289:
 
+g_{\phi \phi }(d\phi -\omega dt)^{2}
 
+g_{\phi \phi }(d\phi -\omega dt)^{2}
 
+g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}.
 
+g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}.
\end{equation}%
+
\end{equation}
 
we have
 
we have
 
\begin{align}
 
\begin{align}
Line 330: Line 330:
 
\begin{equation}
 
\begin{equation}
 
E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e}
 
E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e}
\end{equation}%
+
\end{equation}
 
where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).
 
where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).
 
   </p></div>
 
   </p></div>
Line 382: Line 382:
 
\begin{equation}
 
\begin{equation}
 
E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}),
 
E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}),
\end{equation}%
+
\end{equation}
 
This is the standard formula for energy transformation.
 
This is the standard formula for energy transformation.
 
   </p></div>
 
   </p></div>
Line 402: Line 402:
 
\begin{equation}
 
\begin{equation}
 
E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}},
 
E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}},
\end{equation}%
+
\end{equation}
 
which coincides with eq. (88.9) of \cite{lan}.
 
which coincides with eq. (88.9) of \cite{lan}.
 
   </p></div>
 
   </p></div>
Line 457: Line 457:
 
\begin{equation}
 
\begin{equation}
 
E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm}
 
E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm}
\end{equation}%
+
\end{equation}
 
It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies
 
It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies
 
\begin{equation}
 
\begin{equation}
 
E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma ,
 
E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma ,
\end{equation}%
+
\end{equation}
 
where  
 
where  
 
\begin{equation}
 
\begin{equation}
Line 498: Line 498:
 
\begin{equation}
 
\begin{equation}
 
u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu},  \label{ab}
 
u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu},  \label{ab}
\end{equation}%
+
\end{equation}
 
where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to
 
where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to
find from (\ref{ab}) that%
+
find from (\ref{ab}) that
 
\begin{equation}
 
\begin{equation}
 
a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1},
 
a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1},
\end{equation}%
+
\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:
+
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}
 
\begin{equation}
 
b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 },
 
b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 },
Line 539: Line 539:
 
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  
 
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}
 
\begin{equation}
u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}%
+
u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}
 
\vec{n}_{2}).
 
\vec{n}_{2}).
\end{equation}%
+
\end{equation}
 
Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of \cite{li}.
 
Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of \cite{li}.
 
   </p></div>
 
   </p></div>
Line 564: Line 564:
  
 
2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite.
 
2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite.
Then,%
+
Then,
 
\begin{equation}
 
\begin{equation}
 
\gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big)
 
\gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big)
\end{equation}%
+
\end{equation}
 
grows unbound irrespective of $\varepsilon $. It means that the relative
 
grows unbound irrespective of $\varepsilon $. It means that the relative
 
velocity of particles, one of which moves with some finite speed and the
 
velocity of particles, one of which moves with some finite speed and the
Line 583: Line 583:
 
3b) Let $\varepsilon =+1$. Then  
 
3b) Let $\varepsilon =+1$. Then  
 
\begin{equation}
 
\begin{equation}
\gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{%
+
\gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{\gamma _{2}}{\gamma _{1}}\Big).
\gamma _{2}}{\gamma _{1}}\Big).
+
 
\end{equation}
 
\end{equation}
  
Line 619: Line 618:
 
     <p style="text-align: left;">
 
     <p style="text-align: left;">
  
In special relativity,%
+
In special relativity,
 
\begin{equation}
 
\begin{equation}
 
u^{\mu }=\frac{dx^{\mu }}{ds}
 
u^{\mu }=\frac{dx^{\mu }}{ds}
 
=\gamma (1,v^{i}),\qquad
 
=\gamma (1,v^{i}),\qquad
 
\gamma =\frac{1}{\sqrt{1-v^{2}}}.
 
\gamma =\frac{1}{\sqrt{1-v^{2}}}.
\end{equation}%
+
\end{equation}
 
Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get
 
Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get
 
\begin{equation}
 
\begin{equation}
Line 637: Line 636:
 
\begin{equation}
 
\begin{equation}
 
v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}.
 
v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}.
\end{equation}%
+
\end{equation}
 
It can be rewritten as
 
It can be rewritten as
 
\begin{equation}
 
\begin{equation}
Line 674: Line 673:
 
\begin{equation}
 
\begin{equation}
 
\nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}},  \label{nu}
 
\nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}},  \label{nu}
\end{equation}%
+
\end{equation}
 
where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the
 
where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the
 
observer.
 
observer.
Line 723: Line 722:
 
     <p style="text-align: left;">
 
     <p style="text-align: left;">
  
Eq. (\ref{nu}) reads in this case%
+
Eq. (\ref{nu}) reads in this case
 
\begin{equation}
 
\begin{equation}
 
\nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,}
 
\nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,}
\end{equation}%
+
\end{equation}
where $U^{0}$ obeys the normalization condition%
+
where $U^{0}$ obeys the normalization condition
 
\begin{equation}
 
\begin{equation}
 
Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left(
 
Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left(
Line 733: Line 732:
 
\end{equation}
 
\end{equation}
  
It can be rewritten in the form%
+
It can be rewritten in the form
 
\begin{align}
 
\begin{align}
 
Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }})
 
Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }})
Line 739: Line 738:
 
&\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}},
 
&\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}},
 
\end{align}
 
\end{align}
Then,%
+
Then,
 
\begin{equation}
 
\begin{equation}
 
\nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega
 
\nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega
Line 745: Line 744:
 
\end{equation}
 
\end{equation}
  
When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }%
+
When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }
 
\rightarrow 0$.
 
\rightarrow 0$.
  
Line 769: Line 768:
 
\begin{equation}
 
\begin{equation}
 
\left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM)  \label{e1}
 
\left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM)  \label{e1}
\end{equation}%
+
\end{equation}
where%
+
where
 
\begin{equation}
 
\begin{equation}
 
\gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu }
 
\gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu }
\end{equation}%
+
\end{equation}
 
is its Lorentz factor in the same frame, which has four-velocity
 
is its Lorentz factor in the same frame, which has four-velocity
 
\begin{equation}
 
\begin{equation}
Line 779: Line 778:
 
\end{equation}
 
\end{equation}
 
Here $P^\mu$ is the total momentum and $\mu \equiv E_{c.m.}$ is the energy in the center of
 
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
+
mass frame introduced in (\ref{cm})
 
\begin{align}
 
\begin{align}
 
P^{\mu }&=m_{1}u_{1}^{\mu }+m_{2}u_{2}^{\mu };\\
 
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}
 
\mu ^{2}&=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma (1,2).\label{mu}
 
\end{align}
 
\end{align}
%\begin{equation}
+
 
%\gamma (1,2)=-u_{1\mu }u_{2}^{\mu }.
+
%\end{equation}%
+
 
Then
 
Then
 
\begin{align}
 
\begin{align}
Line 800: Line 797:
 
\end{align}
 
\end{align}
  
It follows from (\ref{e1cm}), (\ref{e2cm}) that%
+
It follows from (\ref{e1cm}), (\ref{e2cm}) that
 
\begin{equation}
 
\begin{equation}
 
( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu ,
 
( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu ,
Line 838: Line 835:
 
=\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu }
 
=\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu }
 
=\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu },
 
=\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu },
\end{equation}%
+
\end{equation}
whence%
+
whence
 
\begin{equation}
 
\begin{equation}
 
\alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha
 
\alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha
Line 867: Line 864:
 
\begin{equation}
 
\begin{equation}
 
E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel},  \label{e2}
 
E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel},  \label{e2}
\end{equation}%
+
\end{equation}
 
where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$.
 
where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$.
  
Line 873: Line 870:
 
\begin{equation}
 
\begin{equation}
 
(E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2),
 
(E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2),
\end{equation}%
+
\end{equation}
 
where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get
 
where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get
 
\begin{align}
 
\begin{align}
Line 893: Line 890:
 
\begin{equation}
 
\begin{equation}
 
E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu .
 
E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu .
\end{equation}%
+
\end{equation}
 
Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to
 
Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to
 
\begin{equation}
 
\begin{equation}
 
EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2},
 
EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2},
\end{equation}%
+
\end{equation}
 
as it should be.
 
as it should be.
 
   </p></div>
 
   </p></div>
Line 925: Line 922:
 
\begin{equation}
 
\begin{equation}
 
U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }.
 
U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }.
\end{equation}%
+
\end{equation}
 
There is no Lorentz factor or relative velocity for two photons.
 
There is no Lorentz factor or relative velocity for two photons.
  
Line 943: Line 940:
 
=-k_{1\,\mu}U_{c.m.}^{\mu }
 
=-k_{1\,\mu}U_{c.m.}^{\mu }
 
=-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2},
 
=-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2},
\end{equation}%
+
\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,
 
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.
 
so in the CM frame the energies of both photons should be equal.
Line 973: Line 970:
 
V^{i}=\frac{U^{i}}{U^{0}},\qquad  
 
V^{i}=\frac{U^{i}}{U^{0}},\qquad  
 
\alpha =\frac{1}{U^{0}},\label{obs}
 
\alpha =\frac{1}{U^{0}},\label{obs}
\end{equation}%
+
\end{equation}
 
the same as for the case of massive particle. The difference is only that for photons we
 
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 $\frac{m}{\sqrt{1-w^{2}}}$.
 
cannot express $\nu _{loc}$ in terms of mass and velocity as $\frac{m}{\sqrt{1-w^{2}}}$.
  
ZAMO frame%
+
ZAMO frame
 
\begin{equation}
 
\begin{equation}
 
\nu _{\infty }-\omega L=N\nu _{loc}.
 
\nu _{\infty }-\omega L=N\nu _{loc}.

Revision as of 11:40, 1 May 2013

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 $dl$?


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 $w$.


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.


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.


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.


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 $t=const$.


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 $g_{0i}=g_{t\phi }\delta _{i}^{\phi }$.


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 $U^\mu$.


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).


Problem 11

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


Problem 12

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


Problem 13

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


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.


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.


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?


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.


Problem 18

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


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.


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 $\Omega _{\pm }$.


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.


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 $\alpha_2$.


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 $\mu$.


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.