Schwarzschild metric (II)

Are you ready?
We were supposed to end up with the demonstration, but it actually is pretty damn long, so in this post I'm going to just obtain the equations as functions of A, B, C and D. And I'll let the final steps of the demonstration for the next post. So two points already pendant:

1. Why metric tensor has only diagonal components? That is, why there are no cross-terms?
2. Applying $g_{\mu\nu}$ into the $R_{\mu\nu}$ equation.

Warning! Math ahead!

Edit: some calculation errors across the document.

Notation warning!

We have not talked about Einstein summation convention I think. And a lot of important stuff. Anyway, it doesn't matter. What I'm trying to say is normally a greek index in a tensor notation (under this context) means that the index is iterated between 0 and 3. You will see along this demonstration I happily change numerical indices with the name of the corresponding coordinate ($t, r, \theta, \varphi$). Obviously, these letters will be equivalent to its numerical indices ($0, 1, 2, 3$), and are not iterated. Any other greek index (as $\alpha$), is worth an iteration.

The off-diagonal terms question

This one has been hard to find, mainly because there are a lot of demonstrations over there and none of them explain (IMO) the question quite clearly. I think I have finally a good answer, and it has to do with symmetry and the initial conditions we have impossed in the problem. The general form of the metric takes this aspect.
$$
ds^2=\sum_{\mu\nu} g_{\mu\nu}dx^{\mu}dx^{\nu} = g_{00}dx^0dx^0+g_{01}dx^0dx^1+ ... + g_{32}dx^3dx^2+g_{33}dx^3dx^3
$$
And we have just "assumed" that the metric is diagonal, that is, all cross terms $g_{\mu\nu}= 0 ~\text{if } \mu \ne \nu$. This fact symplifies the problem greatly, because allows us to only find 3 or 4 functions/constants in order to find the final solution.  But simplifying is a consequence, not a cause. What are the causes for this? One by one, in the next list.

1. We are pretty sure that the cross-terms between $\theta$ and $\varphi$ are zero because we have an isotropic geometry. That means, we don't have a "preferred" direction in space regarding latitude and longitude, because our choosen coordinate system allows symmetry (ans so do the gravitational source). All that assures $g_{\theta r}=g_{r \theta} = g_{\varphi r}=g_{r \varphi}=g_{\theta \varphi}=g_{\varphi\theta} = 0$. You should ask yourself, what would happen if instead of a sphere we would have choosen another shape as source? Well, among other changes, we would probably have crossed terms because of no isotropy. 
2. There is not a preferred direction in time, so $g_{t \varphi}=g_{\varphi t}=g_{t \theta}=g_{\theta t} = 0 $. We can state this because of the static initial condition.
3. Also, we have akready said that we wanted a static and unchanged gravitational field. Saying a gravitational field is static is a very restrictive condition, much more restrictive than saying it is stationary, but for now, it sufices with saying that $g_{tr}=g_{rt} = 0$

There is another very big consequence more than simplyfiying to 3 or 4 constants. Just look at the fact that:
\begin{equation}\label{eq.Christoffel}
\Gamma^{\mu}_{\nu\sigma}=\frac{1}{2}g^{\mu\lambda}\left \{g_{\lambda\nu,\sigma}+ g_{\lambda\sigma,\nu}+g_{\nu\sigma,\lambda}\right \}
\end{equation}
The calculation of Christoffel Symbols is going to be a little easier than it could be. Explained later. If you want a hint, have a good look at the $\lambda$ index in the last equation. 

Applying $g_{\mu \nu}$ to $R_{\mu \nu}$

Now comes the pain! No, not really. It's easier than it seems. We have:
\begin{equation}\label{eq.Vacumm}
R_{\mu \mu} - \frac{1}{2}Rg_{\mu \mu} = 0 \text{ with } \mu=0...3
\end{equation}
That is actually four equations. We are going to need $R_{00}$, $R_{11}$, $R_{22}$ and $R_{33}$, and of course, $R=g^{\mu \mu}R_{\mu \mu}$. With this in mind...
\begin{equation}\label{eq.Ricci1}
R_{\mu\nu}= R^{\beta}_{\mu\nu\beta}= \Gamma^{\beta}_{\mu\beta,\nu}- \Gamma^{\beta}_{\mu\nu,\beta}+\Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\alpha\nu}-\Gamma^{\alpha}_{\mu\nu}\Gamma^{\beta}_{\alpha\beta}
\end{equation}
That means, for instance that:
\begin{equation}\label{eq.Ricci2}
R_{tt}= R^{\beta}_{tt,\beta}= \Gamma^{\beta}_{t\beta,t}- \Gamma^{\beta}_{tt,\beta}+\Gamma^{\alpha}_{t\beta}\Gamma^{\beta}_{\alpha t}-\Gamma^{\alpha}_{tt}\Gamma^{\beta}_{\alpha\beta}
\end{equation}
Taking into account equation (\ref{eq.Christoffel}) and...
\begin{equation}
\begin{array}
gg_{tt}=g_{00} = A\\
g_{rr}=g_{11} = B\\
g_{\theta\theta}=g_{22} = Cr^2\\
g_{\varphi\varphi}=g_{33} = Dr^2sin^2\theta
\end{array}
\end{equation}
It is expected for you to believe that I have made the calculations by hand. Not cheating, although I had the solutions to make comprobations. First, a couple of points

1. Christoffel symbols formulae operate with covariant and contravariant forms for $g$. In this diagonal, easy case, turns out $g_{\mu \nu}=\frac{1}{g^{\mu\nu}}$. It is a consequence of not having cross terms. In case of cross terms we would have to calculate a 4x4 matrix inverse. Not cool.
2. A, B, C and D are actually constants or/and functions of $r$. Actually isotropy condition only assures they can be functions of $r$ and/or $t$, but for the sake of brevity I'm not going to demonstrate why they are not functions of $t$. A good explanation can be found here. Demonstration is slightly different, but the point remains: if you continue assuming that functions are $f(r,t)$, in the end it turns out, they are only $f(r)$ 

With that in mind, it turns out that:

1. $\Gamma_{\nu \sigma}^\mu$ when $\mu \ne \nu$ are all zero!. This happens because of the contravariant metric terms: if $\mu \ne \nu$ they are as null as the covariant.
2.  Do you remember equation (\ref{eq.Christoffel})? Well, it turns out that nasty $\lambda$ term is not as scary as you may think. Again, no cross-term in the metric are responsible. If we had cross-terms in the metric, we should have to iterate $\lambda$ from $\lambda = 0...3$ and we would have four bracket expressions instead of one to calculate.
3. Only derivatives in $r$ are not null (and derivatives in $\theta$ for $g_{\varphi \varphi}$. This symplifies to only 13 Christoffel symbols $\ne 0$ (from the 64 possible).
4. Just a notation clarification: $\dot A$ will mean $\frac{\partial A}{\partial r}$ from now on.

With this in mind, the 13 second kind Christoffel symbols are...

\begin{equation} \label{eq.Christoffels}
 \begin{array}
  \Gamma\Gamma_{tr}^t=\Gamma_{rt}^t=\frac{\dot A}{2A}\\

 \Gamma_{tt}^r=\frac{\dot A}{2B}~~~~~~\Gamma_{tt}^t=\frac{\dot B}{B} ~~~~~~ ~~~~~~\Gamma_{\theta\theta}^t=-\frac{1}{2B}[2Cr+\dot C r^2]\\
\Gamma_{\varphi\varphi}^r=\frac{1}{2B}\left ( 2Drsin^2\theta+\dot D r^2sin^2\theta \right )\\
\Gamma_{t\theta}^\theta=\Gamma_{\theta t}^\theta=\frac{1}{r}~~~~~~ \Gamma_{\varphi\varphi}^\theta = -\frac{D}{C}sin\theta cos\theta\\
\Gamma_{r \varphi}^\varphi = \Gamma_{\varphi r}^\varphi = \frac{1}{r} +\frac{\dot D}{2D} ~~ ~~~~\Gamma_{\theta \varphi}^\varphi = \Gamma_{\varphi \theta}^\varphi = \frac{cos \theta}{sin \theta}\\

 \end{array}
\end{equation}
Half of the nasty work done!
Let's attack the other half, shall we? If you remember equations (\ref{eq.Ricci1}) and (\ref{eq.Ricci2}), you will remember Ricci tensor elements $R_{\mu\nu}$, are obtained from terms already calculated in the group of equations (\ref{eq.Christoffels}). I'm going to summarize the calculations. First an example in why I'm summarizing. $R_{tt}$, for instance, would be defined as...
\begin{equation}\label{eq.Ricci3}
\begin{array}
RR_{tt} = \Gamma_{tt,t}^t - \Gamma_{tt,t}^t+\Gamma_{tt}^\alpha\Gamma_{\alpha t}^t- \Gamma_{00}^\alpha\Gamma_{\alpha t}^t +\\
+ \Gamma_{tr,t}^r - \Gamma_{tt,r}^r+\Gamma_{tr}^\alpha\Gamma_{\alpha t}^r- \Gamma_{tt}^\alpha\Gamma_{\alpha r}^r +\\
+\Gamma_{t\theta,t}^\theta - \Gamma_{tt,\theta}^\theta+\Gamma_{t\theta}^\alpha\Gamma_{\alpha t}^\theta- \Gamma_{tt}^\alpha\Gamma_{\alpha \theta}^\theta +\\
+\Gamma_{t\varphi,t}^\varphi - \Gamma_{tt,\varphi}^\varphi+\Gamma_{t\varphi}^\alpha\Gamma_{\alpha t}^\varphi- \Gamma_{tt}^\alpha\Gamma_{\alpha \varphi}^\varphi \\
\end{array}
\end{equation}
As you may suspect, $\alpha$ is actually an index from 0 to 3 (that is, from $t$ to $\varphi$), so every (non-vanishing) pair up there is actually a summation on $\alpha$. So let's summarize, please :).

1. Look at the first line in equation (\ref{eq.Ricci3}). Now look at this item on the list. Now, back to the first line in the equation. Can you see it? Yessss... It cancels. And it actually happens in every $R_{\mu\mu}$. It will be the second line in $R_{rr}$ and the third in $R_{\theta\theta}$ and the fourth in $R_{\varphi \varphi}$
2. Luckily for us, partial derivatives play in our favour. If you look closely to the equation above, you will find there are terms with partial derivatives over $t, \theta $ and $\varphi$. As you may suspect, most of them go to zero. 

So, it turns out that...
\begin{equation}\label{eq.Rtt}
\begin{array}
RR_{tt}=-\Gamma_{tt,r}^r+\Gamma_{tr}^t \Gamma_{tt}^r- \Gamma_{tt}^r \Gamma_{rr}^r-\Gamma_{tt}^r \Gamma_{r\theta}^\theta- \Gamma_{tt}^r \Gamma_{r\varphi}^\varphi =\\
=-\frac{1}{4B^2}(2\ddot AB-2\dot B A-\dot A \dot B)+\frac{\dot A^2}{4AB}+\frac{\dot A}{B}\frac{1}{2r}-\frac{\dot A}{B}\left (\frac{1}{r}+\frac{\dot D}{2D} \right)
\end{array}
\end{equation}

If you think it's a long term, wait until we get to the Ricci scalar.

\begin{equation}
\begin{array}
RR_{rr}=-\Gamma_{\theta\theta}^r \Gamma_{rt}^t-\Gamma_{\theta\theta,r}^r+ \Gamma_{\theta r}^\theta \Gamma_{\theta\theta}^r-\Gamma_{\theta\theta}^r \Gamma_{rr}^r+\Gamma_{\theta\varphi,\theta}^\varphi +\Gamma_{\theta\varphi}^\varphi \Gamma_{\varphi\theta}^\varphi-\Gamma_{\theta\theta}^r \Gamma_{r\varphi}^\varphi =\\
=\frac{2\ddot A A -\dot A^2}{4A^2}-\frac{\dot B}{Br}-\frac{1}{r^2}+\frac{2\ddot D D-2\dot D^2}{4D^2}+\left( \frac{1}{r}+\frac{\dot D}{2D}\right)^2-\frac{\dot D\dot B}{4B}
\end{array}
\end{equation}

\begin{equation}
\begin{array}
RR_{\theta\theta}=-\Gamma_{tt,r}^r+\Gamma_{tr}^t \Gamma_{tt}^r- \Gamma_{tt}^r \Gamma_{tt}^r-\Gamma_{tt}^r \Gamma_{r\theta}^\theta- \Gamma_{tt}^r \Gamma_{r\varphi}^\varphi =\\
=\frac{1}{2B}\left( \dot C r^2+2Cr \right) \left [ \frac{\dot A}{2B}+\frac{\dot B}{2B}+\frac{\dot C}{2D}\right]+\\
+\frac{1}{4B^2}\left[ (\ddot C r^2+2C+4\dot Cr)2B-2\dot B(\dot C r^2+2Cr )\right]-1
\end{array}
\end{equation}

\begin{equation}
\begin{array}
RR_{\varphi\varphi}=-\Gamma_{tt,r}^r+\Gamma_{tr}^t \Gamma_{tt}^r- \Gamma_{tt}^r \Gamma_{tt}^r-\Gamma_{tt}^r \Gamma_{r\theta}^\theta- \Gamma_{tt}^r \Gamma_{r\varphi}^\varphi =\\
= \frac{F}{2B}\left [ \frac{\dot A}{2A}-\frac{\dot B}{2B}+\frac{\dot D}{2D}\right]-\frac{Dsin^2\theta}{C}-\frac{2\dot F B - 2\dot B F}{4B^2}+\frac{\dot D C - \dot C D}{C^2}sin\theta cos \theta\\
\text{with }~~~ F=sin^2\theta(2Dr+\dot D r^2)
\end{array}
\end{equation}

And the lovely Ricci scalar...
\begin{equation}\label{eq.Rscalar}
\begin{array}
RR=g^{\mu\nu}R_{\mu\nu}=g^{tt}R_{tt}+g^{rr}R_{rr}+g^{\theta\theta}R_{\theta\theta}+g^{\varphi\varphi}R_{\varphi\varphi}=\\
=-\frac{1}{A}\left (\frac{1}{4B^2}(2\ddot AB-2\dot B A-\dot A \dot B)+\frac{\dot A^2}{4AB}+\frac{\dot A}{B}\frac{1}{2r}-\frac{\dot A}{B}\left (\frac{1}{r}+\frac{\dot D}{2D} \right)\right)+\\
+\frac{1}{B}\left(\frac{2\ddot A A -\dot A^2}{4A^2}-\frac{\dot B}{Br}-\frac{1}{r^2}+\frac{2\ddot D D-2\dot D^2}{4D^2}+\left( \frac{1}{r}+\frac{\dot D}{2D}\right)^2-\frac{\dot D\dot B}{4B}\right)+\\
+\frac{1}{Cr^2}\left(\frac{1}{2B}\left( \dot C r^2+2Cr \right) \left [ \frac{\dot A}{2B}+\frac{\dot B}{2B}+\frac{\dot C}{2D}\right]+\\
+\frac{1}{4B^2}\left[ (\ddot C r^2+2C+4\dot Cr)2B-2\dot B(\dot C r^2+2Cr )\right]-1\right)+\\
+\frac{1}{Dr^2sin^2\theta}\left(\frac{F}{2B}\left [ \frac{\dot A}{2A}-\frac{\dot B}{2B}+\frac{\dot D}{2D}\right]-\frac{Dsin^2\theta}{C}-\frac{2\dot F B - 2\dot B F}{4B^2}+\frac{\dot D C - \dot C D}{C^2}sin\theta cos \theta \right)
\end{array}
\end{equation}

So finally, if we combine equation (\ref{eq.Vacumm}) with equations (\ref{eq.Rtt}) to (\ref{eq.Rscalar}), we will have the four differential equations we need to end this post. Until I discover some mistake in the formaulae or some gramatic abomination in my English :)

Sadly, I have to finish this post at this point. As you may have notice, equations are extremely long, even if I use cheap tricks like that $F$ definition. The reason for this is the use of C and D, functions that I had to simplify in the first place. The first thing I will do in the next post will be exactly this, in order to obtain the final equations. You are probably wondering about the reasons for C and D. I mean, in all the demonstrations I have found so far, the problem is only solved for A and B functions of $r$. There are not C and D, because of symmetry reasons (they are the functions for $\theta$ and $\varphi$) or just renamed as 1 because other reasons I could not grasp. Before being able to say "nahh, they are just $C=D=1$", I want to go deeper in the justification. A bad decision anyway: equations are much more complicated this way and it's very likely I had some mistake in one place or another...

Oh man. That was hard... Next time I'm gonna use an algebra program or something. See you in Schwarzschild metric (III)...