This Schwarzchild Schwarzschild topic is getting out of hand!
Just two more posts. In this one we are going to find the final form of the metric. Next one is about interpretations, conclusions and such, which turns out is a lot more faster than deriving the metric itself...
So, here we go again... Math ahead!
Consequences of $C=D=1$
Math, here we go. $C = 1 \implies \dot C = 0$ and $D = 1 \implies \dot D = 0$. And, Ricci terms get a lot easier.
\begin{equation}
R_{tt} = -\frac{\ddot A}{2B}+\frac{\dot A \dot B}{4V^2}+\frac{\dot A^2}{4AB}-\frac{\dot A}{rB}
\end{equation}
\begin{equation}
R_{rr} = \frac{\ddot A}{2A}-\frac{\dot A^2}{4A^2}-\frac{\dot A\dot B}{4AB}-\frac{\dot B}{Br}
\end{equation}
\begin{equation}
R_{\theta\theta} = \frac{r\dot A}{2AB}+\frac{1}{V}-\frac{r\dot B}{2B^2}-1
\end{equation}
It turns out angular terms are almost the same thing. This is going to make one of the four equations a little bit useless.
R_{tt} = -\frac{\ddot A}{2B}+\frac{\dot A \dot B}{4V^2}+\frac{\dot A^2}{4AB}-\frac{\dot A}{rB}
\end{equation}
\begin{equation}
R_{rr} = \frac{\ddot A}{2A}-\frac{\dot A^2}{4A^2}-\frac{\dot A\dot B}{4AB}-\frac{\dot B}{Br}
\end{equation}
\begin{equation}
R_{\theta\theta} = \frac{r\dot A}{2AB}+\frac{1}{V}-\frac{r\dot B}{2B^2}-1
\end{equation}
It turns out angular terms are almost the same thing. This is going to make one of the four equations a little bit useless.
\begin{equation}
R_{\varphi\varphi}= R_{\theta\theta}sin^2\theta
\end{equation}
And the monster, of course...
R_{\varphi\varphi}= R_{\theta\theta}sin^2\theta
\end{equation}
And the monster, of course...
\begin{equation}
R = -\frac{\ddot A}{AB}+\frac{\dot A \dot B}{2AB^2}+\frac{\dot A^2}{2A^2B}-\frac{2\dot A}{ABr}+\frac{2\dot B}{B^2}+\frac{2}{r^2}(1-\frac{1}{B})
\end{equation}
Now, we have just to put this terms in the Einstein equation for vacuum, and we'll have our equation system.
The differential equations
\begin{equation}\label{eq.Rtt}
0=R_{tt}-\frac{A}{2}R \rightarrow 0=\frac{\dot B}{rB^2}+\frac{1}{r^2}(1-\frac{1}{B})
\end{equation}
\begin{equation}\label{eq.Rrr}
0=R_{rr}-\frac{B}{2}R \rightarrow 0=-\frac{\dot A}{rAB}+\frac{1}{r^2}(1-\frac{1}{B})
\end{equation}
\begin{equation}\label{eq.Rtheta}
0=R_{\theta\theta}-\frac{B}{2}R \rightarrow 0=-\frac{\dot A}{A}+\frac{\dot B}{B}-\frac{r\ddot A}{A}+\frac{r\dot A \dot B}{2AB}+ \frac{r \dot A^2}{2A^2}
\end{equation}
Next result is not surprinsing if we recall there is an angular symmetry.
\begin{equation}\label{eq.Rvarphi}
R_{\varphi\varphi}- \frac{1}{2}rsin\theta R = R_{\theta\theta}+\frac{r}{2}R
\end{equation}
So we lose an equation, but it's ok, because we only have 2 unknown functions.
With all of the above in mind, our system of differential equations comprises equations from (\ref{eq.Rtt}) to (\ref{eq.Rtheta})
0=R_{tt}-\frac{A}{2}R \rightarrow 0=\frac{\dot B}{rB^2}+\frac{1}{r^2}(1-\frac{1}{B})
\end{equation}
\begin{equation}\label{eq.Rrr}
0=R_{rr}-\frac{B}{2}R \rightarrow 0=-\frac{\dot A}{rAB}+\frac{1}{r^2}(1-\frac{1}{B})
\end{equation}
\begin{equation}\label{eq.Rtheta}
0=R_{\theta\theta}-\frac{B}{2}R \rightarrow 0=-\frac{\dot A}{A}+\frac{\dot B}{B}-\frac{r\ddot A}{A}+\frac{r\dot A \dot B}{2AB}+ \frac{r \dot A^2}{2A^2}
\end{equation}
Next result is not surprinsing if we recall there is an angular symmetry.
\begin{equation}\label{eq.Rvarphi}
R_{\varphi\varphi}- \frac{1}{2}rsin\theta R = R_{\theta\theta}+\frac{r}{2}R
\end{equation}
So we lose an equation, but it's ok, because we only have 2 unknown functions.
With all of the above in mind, our system of differential equations comprises equations from (\ref{eq.Rtt}) to (\ref{eq.Rtheta})
The first equation, (\ref{eq.Rtt}) is the key in solving this thing. It only depends on a function, which in differential equations, trust me, is a good thing.
Solving the system
The key for finding $A$ and $B$ is solving the equation in $B$ first. You can write it in Wolfram Alpha like a little bitch, or you can solve it by hand. It's not difficult. And we ain't little bitches, are we, bitches?
\begin{equation}
0=\frac{\dot B}{rB^2}+\frac{1}{r^2}(1-\frac{1}{B}) \rightarrow \frac{dB}{B^2-B}=-\frac{dr}{r}
\end{equation}
We are extra lucky! We can integrate $B$ and $r$ separately. The "hard" part is in the left. We have to find 2 fractions in order to make the integral easy. It turns out that:
$$
\frac{1}{B^2-B} = -\frac{1}{B}+\frac{1}{B-1}
$$
So...
$$
-\int {\frac{dB}{B}} + \int{\frac{dB}{B-1}} = -\int{\frac{dr}{r}} \rightarrow -ln(B)+k_1+ln(B-1)+k_2 = -ln(r) +k_3
$$
\begin{equation}
0=\frac{\dot B}{rB^2}+\frac{1}{r^2}(1-\frac{1}{B}) \rightarrow \frac{dB}{B^2-B}=-\frac{dr}{r}
\end{equation}
We are extra lucky! We can integrate $B$ and $r$ separately. The "hard" part is in the left. We have to find 2 fractions in order to make the integral easy. It turns out that:
$$
\frac{1}{B^2-B} = -\frac{1}{B}+\frac{1}{B-1}
$$
So...
$$
-\int {\frac{dB}{B}} + \int{\frac{dB}{B-1}} = -\int{\frac{dr}{r}} \rightarrow -ln(B)+k_1+ln(B-1)+k_2 = -ln(r) +k_3
$$
First thing first, those $k$'s there are constants and we can assemble them in one only constant. Second thing, this expression is asking for applying logaritmic sum and substraction properties. After that, exponentials enter into play.
$$
ln\left (\frac{B-1}{B}\right )=-ln(r)+k_4 \rightarrow \frac{B-1}{B}= \frac{1}{r}e^{k_4} \rightarrow \frac{B-1}{B}= \frac{k}{r}
$$
Finally...
\begin{equation}
\left ( 1- \frac{k}{r} \right ) B = 1 \rightarrow B=\frac{1}{1-\frac{k}{r}}
\end{equation}
$$
0= \frac{\dot A }{A}\left (1-\frac{k}{r} \right) - \frac{1}{r}\left ( 1- \left (1-\frac{k}{r}\right )\right) ~~~~\rightarrow~~~~ \\ \frac{dA}{A}=\frac{kdr}{r^2-kr}
$$
We have actually almost the same equation than the previous one. Just two differences. First, that sign change in left term, is going to invert the solution. Second, we actually have two different integration constants. The first one $k$, belongs to $A$ and $B$ solutions while $l$ belongs only to $A$.
\begin{equation}
A= l \left( 1- \frac{k}{r} \right)
\end{equation}
Yes! We have found $A$ too! Well, not entirely. Let's find the constants!. Here comes another dirty trick: we are going to get them through Newtonian limit.
It is not too dirty, actually. And it allows to use the metric for something else than having a fancy expression.
Well, the objective is to achieve through the metric and the geodesic equation, a term for $\frac{d^2r}{d\tau^2}$. We aim to do that because we know that for a particle of mass $m$ falling into a mass $M$, in the Newtonian limit...
\begin{equation}
F=ma \rightarrow a=\frac{F}{m} \implies \frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}
\end{equation}
First we drop out all metric terms but $ds$ and $dt$ in the metric. We assume $dx^i$ are actually zero except for the time term. Also we have to notice that $ds=\frac{d\tau}c$, with $c$ the speed of light, because proper distance is proper time multiplied by speed of light.
$$
ds^2=A(r)dt^2 \implies \frac{d^2t}{d\tau^2}=\frac{c^2}{ l \left( 1- \frac{k}{r} \right)}
$$
Now, we have to find the geodesic equation for $r$. We know all terms but time just vanish in the geodesic equation (Newtonian limit $F=ma$ establishes this because it only takes into acount time variation in $r$), so...
\begin{equation}\label{eq.geo}
\frac{d^2x^\lambda}{d\tau^2}+\Gamma^{\lambda}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0 \rightarrow \frac{d^2r}{d\tau^2}=-\Gamma^{r}_{tt}\left (\frac{dt}{d\tau}\right)^2
\end{equation}
If you remember the last post, $\Gamma_{tt}^{r}=\frac{\dot A}{2B}$ and thus...
$$
\frac{\dot A}{2B}=\frac{l}{2}\frac{k}{r^2}\left(1-\frac{k}{r}\right)
$$
If we combine this last equation with $\frac{d^2t}{d\tau^2}$ into equation (\ref{eq.geo})...
\begin{equation}
\frac{GM}{r^2}=c^2\frac{k}{r^2} \implies k=\frac{GM}{c^2}
\end{equation}
We find $k$!! What is the value for $l$? Let's find out. Let's have a look at the metric with all terms out but time. And check for dimensions.
$$
ds^2=l \left(1-\frac{GM}{c^2r}\right)dt^2
$$
Let's do $ds$ dimensionless. It is a distance, but it is a dimensionless distance. Now, the term $\frac{GM}{c^2r}$ is an inverse squared distance. If we multiply it by a squared velocity, we will have an inverse squared time, which makes the time interval dimensionless because $dt^2$. We need a squared velocity... So... $c^2$? Does it make any sense to put any other velocity? Demostrations I have seen overthere, use $c=1$ and forget about everything else, which make sense too. But I guess I'm going to let $c^2$ and see what happens. So $l=c^2$. Good job! We got it!
$$
ln\left (\frac{B-1}{B}\right )=-ln(r)+k_4 \rightarrow \frac{B-1}{B}= \frac{1}{r}e^{k_4} \rightarrow \frac{B-1}{B}= \frac{k}{r}
$$
Finally...
\begin{equation}
\left ( 1- \frac{k}{r} \right ) B = 1 \rightarrow B=\frac{1}{1-\frac{k}{r}}
\end{equation}
$A(r)$ and the integration constants.
We have found the value for $B$. Now, we can find $A$ substituing $B$ in the second equation in the system . Allow me to go to the point a little.$$
0= \frac{\dot A }{A}\left (1-\frac{k}{r} \right) - \frac{1}{r}\left ( 1- \left (1-\frac{k}{r}\right )\right) ~~~~\rightarrow~~~~ \\ \frac{dA}{A}=\frac{kdr}{r^2-kr}
$$
We have actually almost the same equation than the previous one. Just two differences. First, that sign change in left term, is going to invert the solution. Second, we actually have two different integration constants. The first one $k$, belongs to $A$ and $B$ solutions while $l$ belongs only to $A$.
\begin{equation}
A= l \left( 1- \frac{k}{r} \right)
\end{equation}
Yes! We have found $A$ too! Well, not entirely. Let's find the constants!. Here comes another dirty trick: we are going to get them through Newtonian limit.
It is not too dirty, actually. And it allows to use the metric for something else than having a fancy expression.
Well, the objective is to achieve through the metric and the geodesic equation, a term for $\frac{d^2r}{d\tau^2}$. We aim to do that because we know that for a particle of mass $m$ falling into a mass $M$, in the Newtonian limit...
\begin{equation}
F=ma \rightarrow a=\frac{F}{m} \implies \frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}
\end{equation}
First we drop out all metric terms but $ds$ and $dt$ in the metric. We assume $dx^i$ are actually zero except for the time term. Also we have to notice that $ds=\frac{d\tau}c$, with $c$ the speed of light, because proper distance is proper time multiplied by speed of light.
$$
ds^2=A(r)dt^2 \implies \frac{d^2t}{d\tau^2}=\frac{c^2}{ l \left( 1- \frac{k}{r} \right)}
$$
Now, we have to find the geodesic equation for $r$. We know all terms but time just vanish in the geodesic equation (Newtonian limit $F=ma$ establishes this because it only takes into acount time variation in $r$), so...
\begin{equation}\label{eq.geo}
\frac{d^2x^\lambda}{d\tau^2}+\Gamma^{\lambda}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0 \rightarrow \frac{d^2r}{d\tau^2}=-\Gamma^{r}_{tt}\left (\frac{dt}{d\tau}\right)^2
\end{equation}
If you remember the last post, $\Gamma_{tt}^{r}=\frac{\dot A}{2B}$ and thus...
$$
\frac{\dot A}{2B}=\frac{l}{2}\frac{k}{r^2}\left(1-\frac{k}{r}\right)
$$
If we combine this last equation with $\frac{d^2t}{d\tau^2}$ into equation (\ref{eq.geo})...
\begin{equation}
\frac{GM}{r^2}=c^2\frac{k}{r^2} \implies k=\frac{GM}{c^2}
\end{equation}
We find $k$!! What is the value for $l$? Let's find out. Let's have a look at the metric with all terms out but time. And check for dimensions.
$$
ds^2=l \left(1-\frac{GM}{c^2r}\right)dt^2
$$
Let's do $ds$ dimensionless. It is a distance, but it is a dimensionless distance. Now, the term $\frac{GM}{c^2r}$ is an inverse squared distance. If we multiply it by a squared velocity, we will have an inverse squared time, which makes the time interval dimensionless because $dt^2$. We need a squared velocity... So... $c^2$? Does it make any sense to put any other velocity? Demostrations I have seen overthere, use $c=1$ and forget about everything else, which make sense too. But I guess I'm going to let $c^2$ and see what happens. So $l=c^2$. Good job! We got it!
The metric
And finally!!!!
\begin{equation}
ds^2= \left ( 1- \frac{2GM}{c^2r}\right )c^2dt^2- \frac{dr^2}{\left( 1- \frac{2GM}{c^2r}\right ) }-r^2d\theta^2-r^2sin^2\theta d\varphi^2
\end{equation}
F*ck this, I'm tired. I need to check everything out before the conclusions, so conclusions in the next post. There are a couple of loose ends, like the dimension problem and corrections in the equations of all previous posts (I have made a lot of mistakes), so take everything here (as usual) with a grain of salt.
No comments:
Post a Comment